CUET (UG) Question Bank Chapter-wise and Topic-wise General Test | For 2024 Exam 9789359582245, 9359582247

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For 2024 Exam

BEST SELLER

CHAPTER-WISE & TOPIC-WISE

CHAPTER-WISE & TOPIC-WISE

QUESTION BANK

QUESTION BANK Includes SOLVED PAPERS (2021 - 2023)

GENERAL TEST

GENERAL TEST

Section III (Compulsory) Strictly as per the Latest Exam Pattern issued by NTA

The ONLY book you need to Ace CUET (UG)

SECTION III (Compulsory)

1

2

3

4

5

100% Exam Readiness

Concept Clarity

Fill Learning Gaps

Extensive Practice

Valuable Exam Insights

with 2023 CUET(UG) Exam Papers

with Revision Notes & Chapter

with Smart Mind Maps & Concept

With 800+ Practice Questions

With Tips & Tricks to ace CUET (UG)

Fully Solved

Analysis

Videos

of Previous Years

in 1st Attempt

(1)

1st EDITION

YEAR 2024

ISBN

“9789359582245”

SYLLABUS COVERED

CUET (UG) General Test

PUBLISHED BY

C OPYRIG HT

RESERVED BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

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DI SC L A IM ER

This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

Kindle Edition

Contents l Preface l  Oswaal Books Expert Tips to Crack CUET (UG) in the First Attempt Examination Structure for CUET (UG) & Syllabus l 

4-4 5-5 6-8

l  CUET (UG) Question Paper-2023 - 21st May 2023 Shift-1 l  CUET (UG) Question Paper-2023 - 30th May 2023 Shift-1 Ø Topic-1 : Mirror Images

General Awareness •  Chapter-1 – Indian History

25 - 41

Ø Topic-1 : Ancient History Ø Topic-2 : Medieval History Ø Topic-3 : Modern History

28 31 34

•  Chapter-2 – World And Indian Geography

63 - 74 75 - 85 86 - 104

Ø Topic-1 : Physics Ø Topic-2 : Chemistry Ø Topic-3 : Biology

•  Chapter-2 – Simplification

•  Chapter-3 – Algebraic Expressions

•  Chapter-6 – Computer Science & Technology 105 - 108 •  Chapter-7 – General Knowledge 109 - 134

•  Chapter-2 – Direction and Distance •  Chapter-3 – Coding and Decoding •  Chapter-4 – Alphabet Series •  Chapter-5 – Number Series •  Chapter-6 – Ranking •  Chapter-7 – Blood Relation •  Chapter-8 – Mathematical Operations •  Chapter-9 – Venn-Diagram •  Chapter-10 – Seating Arrangement •  Chapter-11 – Syllogism •  Chapter-12 – Statement and Conclusion •  Chapter-13 – Figure Completion •  Chapter-14 – Figure Formation •  Chapter-15 – Figure Counting •  Chapter-16 – Non Verbal Reasoning •  Chapter-17 – Word Formation •  Chapter-18 – Clock and Calendar Ø Topic-1 : Clock Ø Topic-2 : Calendar

•  Chapter-19 – Dice and Cubes •  Chapter-20 – Mirror and Water Image Ø Topic-1 : Mirror Images

245 - 251 246

Ø Topic-1 : Average Ø Topic-2 : Problems on Ages

253 255

Ø Topic-1 : Percentage Ø Topic-2 : Profit and Loss Ø Topic-3 : Discount

259 261 263

Ø Topic-1 : Simple Interest Ø Topic-2 : Compound Interest

271 273

248 252 - 257

• Chapter-6 – Ratio and Proportion, Alligation & Partnership 265 - 269 •  Chapter-7 – Simple Interest & Compound Interest 270 - 276

136 138 139

•  Chapter-8 – Time & Work and Pipes & Cisterns 277 - 282 •  Chapter-9 – Speed, Time and Distance 283 - 289

141 - 144 145 - 148 149 - 151 152 - 154 155 - 159 160 - 162 163 - 165 166 - 171 172 - 175

Ø Topic-1 : Circular Arrangement Ø Topic-2 : Linear Arrangement/Row Arrangement

238 - 244

•  Chapter-5 – Percentage, Profit-Loss and Discount258 - 264

135 - 140

Ø Topic-1 : Alphabet Analogy Ø Topic-2 : Number Analogy Ø Topic-3 : Word Analogy

233 234

Ø Topic-1 : Algebraic Identities Ø Topic-2 : Fundamental Operations and Equations

•  Chapter-4 – Average and Problems on Ages

110 111 111 115 119 121 123 125

Logical Reasoning •  Chapter-1 – Analogy

228 230

Ø Topic-1 : BODMAS Rule, Square Root and Cube Root 239 Ø Topic-2 : Fraction and Decimal Fractions, Surds and Indices241

89 94 97

Ø Topic-1 : Art and Culture Ø Topic-2 : Science and Technology Ø Topic-3 : Awards and Honours Ø Topic-4 : Books and Authors Ø Topic-5 : Decades and Dates Ø Topic-6 : Important Organisations Ø Topic-7 : Sports Ø Topic-8 : Miscellaneous

227 - 237

Ø Topic-1 : Progressions and Series Ø Topic-2 : Basic Concepts of Number System Ø Topic-3 : Division Algorithm, Arithmetic and Geometric Progressions Ø Topic-4 : Problems on Numbers, H.C.F. and L.C.M.

44 52

•  Chapter-3 – Indian Polity And Constitution •  Chapter-4 – Indian Economy •  Chapter-5 – General Science

225

Quantitative Aptitude •  Chapter-1 – Number System

42 - 62

Ø Topic-1 : World Geography Ø Topic-2 : Indian Geography



1 - 13 14 - 24

Ø Topic-1 : Speed, Time and Distance & Boat and Stream284 Ø Topic-2 : Problems on Trains 287

•  Chapter-10 – Mensuration

290 - 298

Ø Topic-1 : 2D Mensuration Ø Topic-2 : 3D Mensuration

291 295

•  Chapter-11 – Geometry

299 - 313

Ø Topic-1 : Basic Concepts and Coordinate Geometry Ø Topic-2 : Triangle Ø Topic-3 : Circle

173 174

•  Chapter-12 – DATA Interpretation

176 - 180 181 - 184 185 - 188 189 - 193 194 - 198 199 - 204 205 - 207 208 - 213

314 - 323

Ø Topic-1 : Tables Ø Topic-2 : Graphs Ø Topic-3 : Pie Chart

•  Chapter-13 – Trigonometry

315 317 320

324 - 334

Ø Topic-1 : Degree and Radian Measure and Trigonometric Ratios Ø Topic-2 : Standard Identities of Trigonometry and Height and Distance

210 212

300 304 308

326 330

• Chapter-14 – Permutations and Combinations & Probability 335 - 338 •  Chapter-15 – Set Theory 339 - 341 •  Chapter-16 – Measures of Central Tendency 342 - 344

214 - 219 220 - 226 222

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PREFACE

The National Testing Agency took over the responsibility for conducting CUET in 2021. The Common University Entrance Test (CUET) had been introduced as the assessment test for admission to UG programs across all Central Universities. This has been done to ensure that students across the country, whether from urban areas or rural/ remote areas have access to the same opportunities/environment for assessment. As an all-India test conducted by the National Testing Agency (NTA), it provides a single-window opportunity for admissions, and streamlines the admissions process, and standardizes the evaluation process. A full Computer Based Test (CBT), the question paper is divided into four sections: Sections IA & IB contain language-specific questions, Section II comprises domain questions, and Section III contains questions on general topics. All questions are MCQ based. The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET (UG) scores are mandatory required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG) provided by NTA. Almost 1.92 million candidates registered for CUET (UG) in 2023. Candidates have been quite anxious about appearing for CUET (UG), however, with the right preparation strategy and resources, you can secure a good rank in CUET (UG). This book is a step in that direction, helping candidates refine their examination strategy, and bridging the gap between their present preparation level and a rank-securing preparation level.

How will this book benefits students?

This book has been designed to help candidates successfully crack CUET (UG) General Test. Prepared by the Editorial Board of Oswaal Books, it is designed towards helping candidates up-leveling their preparation level through blended learning. • • • • •

100% Exam Ready With 2023 CUET(UG) Exam Papers – Fully Solved with Explanations Concept Clarity: With Revision Notes & Chapter Analysis with updated pattern Extensive Practice With 800 + Practice Questions of Previous Years (2021-2023) Fill Learning Gaps with Smart Mind Maps & Concept Videos Valuable Exam Insights With Tips & Tricks to ace CUET (UG) in 1st Attempt

The features of the book make it a must-have for anyone preparing for CUET (UG) General Test. We hope it will help students to supplement their CUET (UG) preparation strategy and secure a high rank. We wish the readers all the best in their CUET (UG) preparation!

All the best!



Team Oswaal Books

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Oswaal Expert Tips toinCrack Oswaal Books ExpertBooks Tips to Crack CUET (UG) the First CUET (UG) in the First Attempt

empt

Excited about your UG but unsure if you will get admission to your preferred university? In a major announcement by the chairman of the University Grants Commission, the N onal Te ng Agency will be conduc g the Common Univers es Entrance Test (CUET (UG) ) for undergraduate programs in Central Univers es for the upcoming academic session. However, the UGC Chairperson also stated that CUET (UG) will not just be limited to admissions to Central Univers es. Many prominent private univers es have indicated tha hey would also like to adopt a common entrance exam for undergraduate admissions and take admissions on the basis of CUET (UG) scores. This makes CUET (UG) a very important examina n in itself and hence it becomes mandatory to be aware o he s & tricks that could help you ace the exam on the first empt.

The first step is to understand ern of the ex on. The ons, CUET includes three on 1 includes que on based on languages, sec n 2 includes 27 domain-based subjects and sec n 3 includes General Test. The syllabus of the upcoming Common University Entrance Test, CUET is completely based on the syllabus of class 12 th . No ques n will be asked from class 11th syllabus.

While preparing for the exam, i s important to iden y the importan opics and prac e important ques ons from those topics. Pra ce important que ons through Oswaal Ques n Bank and Sample n Papers, Li ng Ques topics also helps in iden ying the weak areas that need special e. The aspirants can effort and start preparing to focus on the areas tha hey consider to be tough, followed by the ones that are their strengths.

Make a habit of preparing notes f ro m t h e b e g i n n i n g o f t h e prepar on. It will not only help in c but making the study syst also make the revision o he syllabus easy even when you e to revise. might have limited

Collec g and preparing from the appropriate study material cannot be ignored as irrelevant. The books chosen by the aspirants to study from should be on the lines of the current syllabus and the ones that could help you with revision before the ex on.

Make sure to revise as much as possible. The revision will help the aspirants in keeping the concepts fresh in their minds un the day of the ons. They may refer to a few good final ex pr ce ques ons and concise revision notes to achieve their desired results.

Devote a sufficient amount o me to all the ons of the ex ons. This requires a wellmade plan and an honest adherence to the said plan. Priori e the most importan opics or the topics that the aspirants are not familiar with to be able to master them in e.

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1712

With this said, an important ques n tha s gaining ground amongst students who will be appearing for this exam is if they should take coaching to get themselves ready for the exams. The answer is a simple no, the exam will simply not require any coaching as it is completely based on the Class 12th syllabus which will be quite fresh in students' minds as they will be just out of school. All they need is a good revision and prac e of ques ons from Oswaal Ques n Bank and Sample Ques n Papers for CUET (UG) prepar ons.

Examination Structure for CUET (UG) CUET (UG) – is consisted of the following 4 Sections:    

Section IA Section IB Section II Section III

– – – –

13 Languages 20 Languages 27 Domain specific Subjects General Test

Choosing options from each Section is not mandatory. Choices should match the requirements of the desired University. Broad features of CUET (UG) - are as follows: Section

Subjects/ Tests

Section IA Languages

There are 13* different languages. Any of these languages may be chosen.

Section IB Languages Section II Domain

Section III General Test

*

Questions to be Attempted 40 questions to be attempted out of 50 in each language

Question Type

Duration 45 Minutes for each language

There are 20** Languages. Any other language apart from those offered in Section I A may be chosen.

Language to be tested through Reading Comprehension (based on different types of passages–Factual, Literary and Narrative, [Literary Aptitude and Vocabulary]

There are 27*** Domains specific sub- 40 Questions to be jects being offered under this Section. attempted out of 50 A candidate may choose a maximum of Six (06) Domains as desired by the applicable University/Universities.

• Input text can be used for MCQ Based Questions • MCQs based on NCERT Class XII syllabus only

45 Minutes for each Domain Specific Subjects.

For any such undergraduate programme/ programmes being offered by Universities where a General Test is being used for admission.

• Input text can be used for MCQ Based Questions • General Knowledge, Current Affairs, General Mental Ability, Numerical Ability, Quantitative Reasoning (Simple application of basic mathematical concepts arithmetic/ algebra geometry/ ensuration/s tat taught till Grade 8), Logical and Analytical Reasoning

60 Minutes

50 Questions to be attempted out of 60

Languages (13): Tamil, Telugu, Kannada, Malayalam, Marathi, Gujarati, Odiya, Bengali, Assamese, Punjabi, English, Hindi and Urdu

** Languages (20): French, Spanish, German, Nepali, Persian, Italian, Arabic, Sindhi, Kashmiri, Konkani, Bodo, Dogri, Maithili, Manipuri, Santhali, Tibetan, Japanese, Russian, Chinese. *** Domain Specific Subjects (27): 1.Accountancy/ Book Keeping 2.Biology/ Biological Studies/ Biotechnology/Biochemistry 3.Business Studies 4.Chemistry 5.Computer Science/ Informatics Practices 6.Economics/ Business Economics 7.Engineering Graphics 8.Entrepreneurship 9.Geography/Geology 10.History 11.Home Science 12.Knowledge Tradition and Practices of India 13.Legal Studies 14.Environmental Science 15.Mathematics 16.Physical Education/ NCC /Yoga 17.Physics 18.Political Science 19.Psychology 20.Sociology 21.Teaching Aptitude 22.Agriculture 23. Mass Media/ Mass Communication 24.Anthropology 25.Fine Arts/ Visual Arts (Sculpture/ Painting)/Commercial Arts, 26. Performing Arts – (i) Dance (Kathak/ Bharatnatyam/ Oddisi/ Kathakali/Kuchipudi/ Manipuri (ii) Drama- Theatre (iii) Music General (Hindustani/ Carnatic/ Rabindra Sangeet/ Percussion/ Non-Percussion), 27. Sanskrit • A Candidate can choose a maximum of any 3 languages from Section IA and Section IB taken together.However, the third language chosen needs to be in lieu of 6th domain specific Subject chosen by the candidate - as applicable (so the maximum number of tests to be taken remains 9 only i.e. 2 Languages+6 Domain Specific Subjects+1 General Test OR 3 Languages+5 Domain Specific Subjects+1 General Test: flexibility being provided to help a candidate apply for many Universities depending on their eligibility conditions). • Section II offers 27 Subjects, out of which a candidate may choose a maximum of 6 Subjects. • Section III comprises General Test.

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Contd... • For choosing Languages from Section IA and IB and domain specific Subjects from Section II and General Test under Section III, the Candidate must refer to the requirements of his/her intended University. Mode of the Test

Computer Based Test-CBT

Test Pattern

Objective type with Multiple Choice Questions

Medium

13 languages (Tamil, Telugu, Kannada, Malayalam, Marathi, Gujarati, Odiya, Bengali, Assamese, Punjabi, English, Hindi and Urdu) Section IA & IB: Language to be tested through Reading Comprehension (based on different types of passages–Factual, Literary and Narrative [Literary Aptitude & Vocabulary]

Syllabus

Section II : As per NCERT model syllabus as applicable to Class XII only Section III : General Knowledge, Current Affairs, General Mental Ability, Numerical Ability, Quantitative Reasoning (Simple application of basic mathematical concepts arithmetic/algebra geometry/mensuration/stat taught till Grade 8), Logical and Analytical Reasoning

Level of questions for CUET (UG) : All questions in various testing areas will be benchmarked at the level of Class XII only. Students having studied Class XII Board syllabus would be able to do well in CUET (UG). Number of attempts: If any University permits students of previous years of class XII to take admission in the current year also, such students would also be eligible to appear in CUET (UG). Choice of Languages and Subjects: Generally the languages/subjects chosen should be the ones that a student has opted in his latest Class XII Board examination. However, if any University permits any flexibility in this regards, the same can be exercised under CUET (UG) also. Candidates must carefully refer to the eligibility requirements of various Central Universities in this regard. Moreover, if the subject to be studied in the Undergraduate course is not available in the list of 27 Domain Specific Subject being offered, the Candidate may choose the Subject closest to his/her choice for e.g. For Biochemistry the candidate may choose Biology.

GENERAL AWARENESS

SYLLABUS

1. Indian History → Prehistoric Period, The Harappan Civilisation, The Vedic Period, Jainism and Buddhism, Magadha Empire, The Mauryan Empire, The Sangam Period, The Gupta Period, Age of Harshavarthana, Rashtrakutas, Gangas, Pallavs, The Cholas 2. World and Indian Geography → 2.1. (World Geography): Universe, Solar System, Planets, Sun, Earth, Structure of Earth’s Interior, Plate Tectonic Theory, Volcanism, Rocks, Landforms, Atmosphere, Jet Stream, Oceans, Continents, Important Facts → 2.2. (Indian Geography): India at a Glance, Physical Features, Islands, Drainage System of India, Climate, Agriculture, Soils of India, Mineral Resources, Transport 3. Indian Polity and Constitution → Constitution, Preamble, Sources of the Constitution, Parts of the Constitution, Amendability of the Preamble, Schedules of the Constitution, Union and its Territories, Citizenship, Rights, Directive Principles of State Policy, Union Government, State Government, Local Government, Elections, Planning Commission, NDC, Finance Commission, Amendments of the Constitution 4. Indian Economy → Economy, Economic Growth, Indian Economy, National Income of India, Planning of India, India’s New Economic Policy, Flagship Programmes of Government of India, Agriculture, Indian Financial System, Indian Currency System, Macro Economic, Problems in India. 5. General Science → 5.1. (Physics): Units, Motion, Friction, Gravitation, General Properties of Matter, Heat and Thermodynamics, Thermal Expansion, Waves, Light, Electricity and Magnetism → 5.2. (Chemistry): Physical and Chemical Changes, Matter; Atom, Molecule and of Element, Corrosion, Renewable and Non-renewable Natural Resources, Coal, acids, Bases and Salts, Some important Compounds in Everyday life, Polymers → 5.3. (Biology): Living World, The Cell, Nucleic Acids, Human Systems, Respiratory System, Vitamins, Major Enzymes of Digestion, Blood (Lymphatic System), Central Nervous System, Some Human Diseases Caused by Viruses and Bacteria / Fungi, Animal/Human Diseases Caused by Fungi, Important Vaccines Discoverer, Ebola Virus, Ecology, Pollution, Biotechnology, Some Important Branches of Biology, Some Important Discoveries, Some Important Antibiotics, Environment and Ecology 6. Computer Science and Technology → Computer, Components of Computer, Memory, Hardware, Software, Networking, Security Threats, Internet, Some commonly used terms, Super Computer. 7. General Knowledge → Art & Culture, Important Inventions & Discoveries Science & Tech, Awards & Honours, Books & Authors, Decodes & Dates, Important Organizations, Sports, Miscellaneous.

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Contd... QUANTITATIVE APTITUDE

Number System → Introduction to Numbers; Various types of Numbers; Even numbers; Odd numbers, Prime numbers; Rational numbers; Irrational number; Tests of divisibility; Standard results; Division algorithm; Arithmetic Progression; Geometric Progression; Unit place value in the Product of Numbers; Problems on Numbers & H.C.F and L.C.M. 2. Operations → BODMAS Rule; Squares; Cubes; Square Roots; Cube Roots; Fractions; Types of fractions; Decimals; Surds and Indices; Properties of Surds and Indices. 3. Algebraic Expressions → Constants; Variables; Algebraic expressions; Equations; Linear equation; Quadratic equations; Factors; Solutions. → 3.1. Average and problems on Ages → Average; Properties of Averages; Problems on Ages → 3.2. Percentage, Profit-Loss and Discount → Percentage meaning; Fraction representation, Consumption & Expenditure, Reducing & Exceeding prices, Voters in an Election; Percentage increase or increasing and decreasing problems ; Profit, Loss; Buy-sell problems ; Discounts, Successive discount problems; Tax Miscellaneous Questions. → 3.3. Ratio and Proportion, Alligation & partnership → Ratio; Proportion; Properties of Ratio and Proportion; Alligation; Partnership; Active Partner; Sleeping Partner; Types of Partners; Shares and Partners; Coins and Rupees; Income and Expenditure → 3.4. Simple Interest & Compound Interest → Interest; Principal; Rate of Interest; Time; Compound Interest; Difference between Compound and Simple Interest; Miscellaneous questions on Simple Interest and Compound Interest. → 3.5. Time & Work and Pipes & Cisterns → Time and Work; Work and Wages; Efficiency of worker; worker leaves or joins; Man−day work done; Pipes and Cisterns; Inlet; Outlet; Quantity and Time; Miscellaneous Question. → 3.6. Speed, Time and Distance → Speed; Time; Conversion of Units; Average Speed; Speed of Stream; Upstream; Downstream; Problems on Trains in same direction; Problem on train cross each other in opposite direction; Relative Speed. 4. Mensuration → Area and Perimeter of triangles, Rhombus, Quadrilaterals, Circles; Volume of 3D figures; Surface area; Total surface area; Cylinder, Sphere, Hemisphere, Cube, Cuboid, Cone, Frustum of a cone; Standard Results; Equilateral triangle; Parallelogram; Rhombus, Trapezium; Regular Hexagon etc. 5. Geometry → Lines; Angles; Bisectors; Polygons; Coordinate geometry; Types of triangles; Congruency; Similar triangles; Circles; Chord; Tangents; Secant; Cyclic quadrilateral; Circumcircle; Incircle. 6. DATA Interpretation → Data Interpretation; Tables; Bar Graphs; Line Graphs; Pie Charts; Mix Graph. 1.

LOGICAL REASONING

1. Analogy → Alphabet Analogy; Number Analogy; Word Analogy; Analogy based on prime number and mixed operation based Analogy. → 1.1. Alphabet Series → Finding the next or missing alphabet in a series. → 1.2. Number Series: N2 series, N2 + 1 series, N2 – 1 series, Cube series, Prime number series, Mixed Operation series etc. 2. Direction and Distance : Concept of direction, Different types of directions, Orientation, Turns, Concept of Shadow, Finding distance, Finding shortest distance, Different types of questions- Based on turns and rotations, Based on shadow, Based on finding shortest distance, Based on coded form, Based on seating arrangements, Based on clock timing indication. Coding and Decoding : Letter coding, Alphabetical coding, Direct letter coding, Number coding, Deciphering message coding/Message coding, Substitution coding, Mixed coding and Alpha-Numeric coding. 3. Ranking : Age-Based Comparison, Height Based Comparison, Integrated Height and Age-Based comparison, Linear seating arrangement comparison. No. of elements between two-element, Position interchanges in a queue, Puzzle based comparison. 4. Blood Relation : Dialogue/Conversation Based, Puzzles, Coding-Decoding, Linear seating arrangement base blood relations, Circular seating arrangement based blood relations, Mixed blood relation. 5. Mathematical Operations : Interchanging operators, Symbols, Balancing the equation and Trick based mathematical operations. Venn Diagram : Relation Based Venn Diagram, Analysis Based Venn Diagram. 6. Seating Arrangement → Facing towards the center, facing outwards/opposite/outside the center, facing inside and outside the center Rectangular or square arrangement. Triangular arrangement (facing the centre, facing opposite to the centre). 7. Syllogism → Positive conclusion, Negative conclusion, Either-OR/Complementary pair, No conclusion. 8. Statement and Conclusion → One statement multiple conclusions, Two or more than two statements, Multiple conclusions. Statement and Assumptions, Statement and Course of Action, Cause and Effect, Statement and Arguments, Inference. → 8.1. Figure Completion → Based on movement, Based on Rotation 45°/90°/135°/180°, Symmetry of image/ Water and Mirror Concept, Miscellaneous Concept. → 8.2. Figure Formation → Assembling main figure from split figures, Fragmentation of the main figure into pieces. Making up a figure from given components, Making up a three dimensional figure by paper folding. → 8.3. Figure Counting → Based on triangles, Based on squares, Based on Rhombus, Based on rectangles, Based on the circle, Based on combined figures, Counting of straight lines. 9. Non-Verbal Series → Figure series, Figure series based on rotation, Based on addition of symbols, Based on deletion of symbols, Based on replacement, Based on replacement and arrangements, Similar Figure present in questions figures, Simultaneous operations of rotation, addition and replacement. 

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CUET (UG) Question Paper - 2023 National Testing Agency 21st MAY 2023 – SHIFT 1

Section - III (General Test) Examination Duration: 60 Minutes

Maximum Marks - 250

General Instructions:

Marking scheme of the test: (a) There are 60 questions asked in the section- III. Attempt only 50 questions. (b) Correct answer or the most appropriate answer will be given five marks (+5). (c) Any incorrect option marked will be given minus one mark (–1). (d) Unanswered/Marked for review will be given no mark (0).

1. If the word ‘LEADER’ is coded as 20-13-9-12-13-26. How would you write “LIGHT” (A) 20-16-15-17-22 (B) 20-17-15-16-28 (C) 20-15-16-18-23 (D) 20-16-17-15-27 Ans. Option (B) is the correct. Explanation: Logic: Table of Alphabetical series (point to remember) Alphabets

A

B

C

D

E

F

G

H

I

J

K

L

M

Positional value

1

2

3

4

5

6

7

8

9

10

11

12

13

Alphabets

Z

Y

X

W

V

U

T

S

R

Q

P

O

N

Positional value

26

25

24

23

22

21

20

19

18

17

16

15

14

L ⇒ 12 + 8 = 20 E ⇒ 5 + 8 = 13 A⇒1+8=9 D ⇒ 4 + 8 = 12 E ⇒ 5 + 8 = 13 R ⇒ 18 + 8 = 26 Therefore, LIGHT can be coded as L ⇒ 12 + 8 = 20 I ⇒ 9 + 8 = 17 G ⇒ 7 + 8 = 15 H ⇒ 8 + 8 = 16 T ⇒ 20 + 8 = 28 So “LIGHT” will be coded as = 20 - 17 - 15 - 16 - 28. 2. In a class boys stand in a single line. One of the boys is seventeenth in order from both the ends. How many boys are in the class? (A) 34

(B) 33

(C) 32

(D) 27

Ans. Option (B) is correct. Explanation: Given: One of the boys is seventeenth in order from both ends. Hence, Total boys in the class = 16 + 1 + 16 = 33

3. If the 2nd half of the letters of the word INTERMEDIATE are reversed and placed before 1st half of the letters, which letter will be 2nd to the right of 10th letter from the right? (A) A (B) D (C) E (D) I Ans. Option (B) is correct. Explanation: Word: INTERMEDIATE Reverse 2nd half and placed before 1st half ETAIDEINTERM 10 - 2 = 8th from right, which is D.

2

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

4. Arrange the following in meaningful sequence: (A) Key (B) Door (C) Lock (D) Room Choose the most appropriate answer from the options given below: (A) A, C, B, D (B) D, C, A, B (C) B, A, D, C (D) C, B, D, A Ans. Option (A) is correct. Explanation: (A) Key

(C) Lock

(B) Door

(D) Room

5. Which of the following is a metal (A) Carbon (B) Mercury (C) Sulphur (D) Iodine Ans. Option (B) is the correct. Explanation: Mercury is a metal. It is a chemical element represented by the symbol Hg with atomic number 80 and an atomic mass 200.59. It is a d-block element which is present in period 6 and group 12 of the periodic table. It is the only metal liquid at room temperature, which is why it is used in thermometers to detect body temperature. Carbon is a non-metal which is represented by the symbol C. It has an atomic number of 6 and an atomic mass of 12. It is a p-block element belonging to period 2 and group 14 of the periodic table. Sulphur is a non-metal which is represented by the symbol S. It has an atomic number of 16 and an atomic mass of 32. It is a p-block element belonging to period 3 and group 16 of the periodic table. It is yellow-coloured and crystalline in nature. Iodine is a non-metal which is represented by the symbol I. It is a halogen with an atomic number 53 and atomic mass 126.9. It is a p-block element and belongs to the periodic table’s group 17 and period 5. 6. Match List I with List II LIST I Movement A. Quit India Movement

LIST II Purpose I. To achieve self Government in India of demand for large political representation

B. Civil II. Compelled British’s to Disobedience leave India Movement C. Non Cooperation Movement

III. Refusal to obey certain laws orders or Commands of the Government

D. Home rule league Movement

IV. Indians resigning their title boycotting foreign goods and Government institutions refused to pay taxes

Choose the most appropriate answer from the options given below: (A) A-I, B-III, C-II, D-IV (B) A-II, B-III, C-I, D-IV (C) A-III, B-II, C-IV, D-I (D) A-II, B-III, C-IV, D-I Ans. Option (D) is the correct. Explanation: Quit India Movement was started by Mahatma Gandhi in which he demanded the end of British rule in India. This was launched at the Bombay session of All India Congress Committee on 8 August 1942. Gandhiji’s famous slogan of ‘Do or Die’ was the driving force of this movement. The phrase ‘Quit India’ was coined by Yusuf Meherally. However, this movement was suppressed by British forces. The Civil Disobedience Movement was launched under the leadership of Mahatma Gandhi in 1930 as part of the larger Indian National Congress’s campaign for freedom. He urged Indians to peacefully protest by refusing to cooperate with British authorities. Indians boycotted British institutions, refused to pay taxes, and participated in acts of non-cooperation. There were mass protests, demonstrations, and acts of civil disobedience taking place throughout the country. The Salt March or Dandi March was one of the important events of Civil Disobedience Movement. Non-Cooperation Movement was a movement in which Indians united against British colonial rule with the suspension of cooperation with British-run institutions such as schools, colleges, courts, and legislative councils. Indians boycotted British goods and promoted the use of Indianmade products. However, this movement was suspended in 1922 due to a violence incidence at Chauri-Chaura, Uttar Pradesh. Home rule league movement was started by Annie Besan and it aimed to achieve selfgovernment in India within the British Raj. She formed the All India Home Rule League in September 1916. She was the first woman president of the Indian Congress. 7. Choose the one which is different from the rest three. (A) 431 (B) 162 (C) 831 (D) 232 Ans. Option (C) is correct. Explanation: Logic: 431  4  3  1  12 162  1  6  2  12 831  8  3  1  24  12 232  2  3  2  12 Hence, we can see that option (C) is not following the same pattern as others. 8. X got 98 marks in his exam which is 56 % of the total marks. What is the maximum marks of the exam? (A) 150 (B) 175 (C) 200 (D) 225

3

CUET (UG) Solved Paper 2023 (21st May Shift -1)

Ans. Option (B) is correct. Explanation: Let the total marks be x. According to question, 56% of x = 98 56 x = 98 100 98  100 x  175 56 Hence,the maximum marks of the exam is 175. 9. Which of the following speeds is the least? (A) 50 meter / second (B) 50 meter/minute (C) 70 km / hour (D) 5 km / minute Ans. Option (B) is correct. Explanation: Logic: Solving by options, (A) 50 m/sec

50  m   60  s  70 km 70  1000 700  m  (C)   hour 3600 36  s  5 km 5  1000 500  m  (D)   min 60 6  s  (B)

50 m/minute =

Hence, option (B) is correct answer. 10. When seen through a mirror, a clock shows 3:30. What is the correct time? (A) 2: 30 (B) 8: 30 (C) 5: 30 (D) 4: 30 Ans. Option (B) is correct. Explanation: The correct time is: (11 : 60 - 03 : 30) = 8 : 30 11. A is 3 years younger than C but one year older than D. D is one year older than B but 4 years younger than C. C is 15 years old. What is the age of B in years (A) 13 (B) 12 (C) 11 (D) 10 Ans. Option (D) is correct. Explanation: Let the present age of C = x Present age of D = y Age of A = x - 3 Age of C = 15 years A is 3 years younger than C Hence, A’ s age = 15 - 3 = 12 years Age of D = C - 4 = 15 - 4 = 11 years Age of D = Age of B + 1 Age of B = 11 - 1 = 10 years

12. What is the smallest square number which is divisible by 4,6 and 32? (A) 100 (B) 196 (C) 96 (D) 576 Ans. Option (D) is correct. Explanation: LCM of 4, 6, 32 =2×2×3×8 = 96 Now find the multiple of 96. 96 is not a perfect square number. Hence, 576 is correct answer. 13. The only Indian who received noble prize in literature is. (A) Bankim Chandra Chatterjee (B) Toradutt (C) R.K. Narayan (D) Rabindra Nath Tagore Ans. Option (D) is the correct. Explanation: Rabindra Nath Tagore is the only Indian to receive the noble prize in Literature in the year 1913 for his novel Gitanjali. He was the first Indian and Asian to be awarded the prize. Rabindra Nath Tagore was a Bengali writer, poet, composer, philosopher, painter, and social reformer. He is the writer of India’s national anthem-Jana Gana Mana and Bangladesh’s national anthem- Amar Shonar Bangla. Some of his best known works are Ghare-Baire , Gore etc. The Nobel Prize in Literature is a literary honour given by the Swedish Academy and is one of the five Nobel Prizes established by the will of Alfred Nobel in 1895. The Nobel laureate is honoured with a gold medal, reward money and a diploma with citation. The first Nobel Prize in literature was given in 1901 to the French poet and essayist Sully Prudhomme. 14. Who won the ‘Noble Prize’ for in the field of “physiology or medicine”? (A) C.V. Raman (B) Jagdish Chandra Bose (C) Homi Jehangir Bhabha (D) Har Gobind Khorana Ans. Option (D) is the correct. Explanation: Har Gobind Khorana won the Noble Prize in the field of physiology or medicine in the year 1968. He shared the prize along with Marshall W Nirenberg and Robert W. Holley. The three of them were awarded for their interpretation of the genetic code and its function in protein synthesis. Har Gobind Khorana was an Indian American biochemist who has won many national and international awards for his research. The Nobel Prize in Medicine is awarded by the Nobel Assembly at the Karolinska Institute for outstanding discoveries in physiology or medicine every year. It is one of the five Nobel

4

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

Prizes that are a part of the will of Alfred Nobel in 1895. The Nobel laureate is honoured with a medal, reward money and a diploma with citation. The first Nobel Prize in literature was given in 1901 to the German physiologist, Emil von Behring, for his work on serum therapy and the development of a vaccine against diphtheria.

Ans. Option (C) is correct. Explanation:

15. The relationship between the values of a country’s imports and its exports is called. (A) Balance of Trade (B) Balance of Payment (C) Balance of currency (D) Bill of exchange Ans. Option (A) is the correct. Explanation: The relationship between the value of country’s imports and exports is called Balance of Trade(BOT). It is simply the difference between country’s imports and exports of goods is also known as net exports or commercial balance. If the import is more than the export, it is known as trade deficit and if the exports are more than imports, it is known as trade surplus. Balance of Payment is basically the summary of country’s economic transactions with the world during an accounting year. This includes all the inflow and outflow of funds that have taken from the country. The BOP statement is indicative of country’s surplus or deficit of funds. The Bill of exchange is defined under the Negotiable Instruments Act 1881. It is a written instrument that directs a person to pay the designated sum of money to the bearer of the instrument. It basically ensures a timely payment. It involves three parties- the drawee is the party that pays the sum, the payee receives that sum, and the drawer is the one that obliges the drawee to pay the payee. 16. Find out which of the answer figures (1), (2), (3) and (4) completed the figure matrix?

A B

A B

A B

B

B

B

A B

A B

?

(A)



(B)

A

A

B

B

(C)



(D)

A

A

B

B

A B

A B

A B

B

B

B

A B

A B

?

Logic: Follow the pattern and the symmetry. Hence, A B 17. Match List I with List II LIST I River

LIST II City

A.

Mahanadi

I.

Ludhiana

B.

Godavari

II.

Cuttack

C.

Sutlej

III

Lucknow

D.

Gomti

IV.

Nasik

Choose the most appropriate answer from the options given below: (A) A-II, B-IV, C-III, D-I (B) A-II, B-IV, C-I, D-III (C) A-IV, B-II, C-I, D-III (D) A-III, B-I, C-II, D-IV Ans. Option (B) is the correct. Explanation: Mahanadi River flows through Odisha and Chattisgarh and ends in Bay of Bengal. It is a Peninsular river in East Central India. The city of Cuttack is situated on the banks of the Mahanadi river. Godavari is the second longest River of India after Ganga. It passes from the states of Maharashtra, Andhra Pradesh, Telangana, Chhattisgarh, Odisha and empties into Bay of Bengal. Nasik is situated on the banks of Godavari river. Sutlej is the longest of the five Rivers that flows through Punjab in India and Pakistan. It is the easternmost tributary of Indus River and is also known as Satadru. Ludhiana stands on the old bank of Sutlej River. Gomti River is a tributary of the River Ganges and originates in Pilibhit, Uttar Pradesh. It flows through Lucknow and supply water to the city. 18. The point (-2, 3) lies in which quadrant? (A) I (C) III Ans. Option (B) is correct.

(B) II (D) IV

5

CUET (UG) Solved Paper 2023 (21st May Shift -1)

21. Which of the following players didn’t receive Medal in Tokyo Olympics 2020?

Explanation:

(A) PV Sindhu

3

(2, 3)

(B) Neeraj Chopra x axis

2

(D) Ravi Kumar Dahiya Ans. Option (C) is correct.

(2,3)3)lies lies quadrant. (-2, inin II II quadrant. 19. If selling price of 80 articles is equal to the cost price of 100 articles, then find the gain percentage. (A) 30 % (B) 25% (C) 40 % (D) 50 % Ans. Option (B) is correct. Explanation: Selling Price of 80 articles = C.P of 100 articles SP 100 5 = = CP 80 4 Gain% 

(C) Abhinav Bindra

SP  CP  100 CP

 SP    1   100 CP   5     1   100 4  54   100 4 = 25% 20. A railway half-ticket costs half the full ticket. However the reservation charge for all the tickets is constant. One full reserved ticket for a journey is ` 525. If the cost of one full and one half reserved ticket for the same journey is ` 850, then what is the reservation charge per ticket? (A) ` 120 (B) ` 150 (C) ` 125 (D) ` 115 Ans. Option (C) is correct. Explanation: Let the reservation charge be x Cost of full ticket = y Price of one full reserved ticket = 525 x + y = 525 y Cost of 1 full and 1 half ticket  x  y  x  2 3y 2x   850 2 3y 2  525  y    850 2 2(1050 - 2y) + 3y = 850 × 2 y = 2100 - 1700 = 400 x = 525 - y x = 525 - 400 = `125

Explanation: Abhinav Bindra didn’t receive medal in Tokyo Olympics 2020. He is a retired shooter and the first Indian to win a gold medal at the Olympics. He had won the gold medal in shooting at the 2008 Summer Olympics. PV Sindhu won the Bronze medal in women’s singles badminton at Tokyo Olympics 2020. She is the first Indian woman and second athlete to win two individual Olympic medals. Ravi Kumar Dahiya won the Silver medal in men’s 57 kg freestyle wrestling at Tokyo Olympics 2020, freestyle wrestling. Neeraj Chopra became India’s second individual Olympic champion, after Abhinav Bindra, with his men’s javelin throw gold at Tokyo 2020. It was India’s first track-and-field medal at any Olympic Games. 22. Which book is Written by Dr. S. Radha Krishnan? (A) The world’s largest democracy (B) India divided (C) Indian Philosophy (D) India Priceless heritage Ans. Option (C) is the correct. Explanation: Dr. Sarvapalli Radhakrishnan wrote the book, ‘Indian Philosophy’. He was the second President (1962 to 1967) of India. The History of the World’s Largest Democracy is a non-fiction book which is written by Indian historian Ramachandra Guha. The famous book ‘India divided’ was written by Dr. Rajendra Prasad. The book “India Priceless heritage” was written by N.A. Palkhivala, was an Indian lawyer and a noted jurist. x x x x 23. If the median of , x , , and (where x > 0 ) is 8, 4 2 5 3 then the value of x will be (A) 24 (B) 32 (C) 8 (D) 16 Ans. Option (A) is the correct. Explanation: Arrange in increasing order x x x x , , , ,x 5 4 3 2 Since number of terms = 5 ⇒ odd Hence, Median 

n1 th observation 2

6

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

8

51 th 2

= 8 3= rdobservation  x = 8 × 3 = 24

20  25 100

 20  25 

= 50% Single equivalent % increase if 50%, 30% 50  30  50  30  100 = 95%

x 3

24. The minimum number of colours to required paint all sides of a cube that no two adjacent faces may have the same colour is. (A) 5 (B) 4 (C) 3 (D) 6 Ans. Option (C) is correct.

28. Find out which of the figures out of given option can be formed from the pieces given in fig (X)

Explanation: Opposite face has the same colour and there are six faces in a cube so we need 3 different colours to do the need full. 25. What sum of money will amount to ` 520 in 5 years and to ` 568 in 7 years on simple interest? (A) ` 400 (B) ` 120 (C) ` 510 (D) ` 220 Ans. Option (A) is correct. Explanation: A = P + SI P×R×5 ...(1) 520 = P + 100 PRT 568  P  100 2 PR 568  520  100 48 PR = 2 100 Now,put in equation 1 520 = P + 24 × 5 P = 520 - 120 = 400

Fig X (A)



(B)

(C)



(D)

Ans. Option (B) is correct. Explanation:

26. Here are some words translated from an artificial language Holo polo means base ball Moto prot means my India Prot shot means India won Which word could be mean “All India Radio” (A) Holo polo prot (B) Kud prot nid (C) Prot polo nid (D) Polo nid prot Ans. Option (C) is correct.

Fig X

Logic: Follow the pattern and the symmetry. Hence,

Explanation: Code for India is Prot. Hence, code for All Radio should be different from the codes given in the question. Hence, All India Radio-Kud Prot nid 27. Find a single equivalent increase if the number is successively increased by 20 %, 25 % and 30%? (A) 75 % (B) 85 % (C) 95 % (D) 35 % Ans. Option (C) is correct. Explanation: Single equivalent % increase if 20%, 25% ab

ab 100

is correct answer. 29. Match List I with List II LIST I Physical Quantity A. B. C. D.

Electric charge Force Power Energy

LIST II Units I. II. III. IV.

Newton Coulamb Joule Watt

7

CUET (UG) Solved Paper 2023 (21st May Shift -1)

Choose the most appropriate answer from the options given below: (A) A-I, B-II, C-IV, D-III (B) A-II, B-I, C-IV, D-III (C) A-II, B-IV, C-I, D-III (D) A-III, B-II, C-IV, D-I Ans. Option (B) is correct. Explanation: The SI unit of electric charge is Coulomb and is represented by the symbol ‘C’. Coulomb is named after French physicist Charles-Augustin de Coulomb. 1C is approximately equal to the charge of 6.242 × 1018 protons or electrons. The SI unit of Force is Newton and is represented by N. It is defined in the Second Law of Motion and is named after Issac Newton. 1N is defined as the force required to accelerate 1 kilogram of mass at the rate of one meter per second in the direction of applied force. The SI unit of Power is Watt and is named after James Watt. 1W is the power produced when 1 Joule of work is done for 1 second. The SI unit of Energy is Joule and is named after the English physicist James Prescott Joule. 1 Joule is the amount of work done when a force of 1 newton displaces a mass through a distance of 1 metre in the direction of the force applied. 30. In a row of 40 children, A is 13th from the left end and B is ninth from the right end. How many children are there between A and C if C is forth to the left of B. (A) 13 (B) 14 (C) 15 (D) 16 Ans. Option (B) is the correct. Explanation: Left

12

A

14

13 from left th

C

B 8

The entry-level rank for the Indian navy is SubLieutenant. The Person then gets promoted to Lieutenant and then to Lieutenant Commander. 32. Statement I : Constitution is the frame work for the governance of a country which delegates power and authority to the executive, legislative and judiciary. Statement II : It serves a country in maintaining good relationships with her neighboring countries. In the light of the above statements, choose the most appropriate answer from the options below. (A) Both Statement I and Statement II are correct (B) Both Statement I and Statement II are incorrect (C) Statement I is correct but Statement II is incorrect (D) Statement I is incorrect but Statement II is correct Ans. Option (C) is the correct. Explanation: Constitution provides the framework for governance and delegates the power and authority to the judiciary, executive, and legislative of the country. It outlines the rights and duties of a citizen in order to build a prosperous nation. 33. How many terms are there in the A.P. 3, 7, 11, ... 407? (A) 100 (B) 101 (C) 99 (D) 102 Ans. Option (D) is correct. Explanation: 3, 7, 11…………………407 = an 407 = ,a 3 d  a2  a1  7  3  4 an  a   n  1  d

Right

13 9 from right th

th

407  3   n  1 4

40 Hence,14 is correct answer.

31. Major is related to Leiutenant in the same way as ‘Squadron Leader is Related to? (A) Pilot officer (B) Flying Attendant (C) Group captain (D) Flying officer Ans. Option (D) is the correct. Explanation: Major is related to Lieutenant in the same way as Squadron leader is related to Flying Officer. A lieutenant is an entry-level rank for the commissioned officer in the Indian army. The person then gets promoted to Captain and then to Major. Similarly, the entry-level rank for the Indian air force is Flying officer. The person then gets promoted to Flight Lieutenant and then to Squadron Leader.

407  3   n  1 4 404  n1 4 101 + 1 = n n = 102 34. What is the probability that any non-leap year will have 53 Sundays? (A)

1 53

1 7 Ans. Option (C) is correct. (C)

(B)

2 53

(D)

2 7

Explanation: As a result, there are 52 Sundays in a non-leap year. But one leftover day apart from those 52 weeks can be either a Monday, Tuesday,

8

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

Wednesday, Thursday, Friday, Saturday, or a Sunday. Therefore, the probability of getting 53 Sundays in a non-leap year is 1/7. 35. Match List I with List II LIST I A.

Blue Revolution

LIST II I.

Increase in crop yield and Agricultural Products

B.

White Revolution

II.

C.

Yellow Revolution

III. Increase of Fish Production

37. A man walks 2 km towards East and then he turns to South and walks 6 km. Again he turns to East and walks 4 km, after this he turns to North and walk 14 km. How far is he from his starting point? (A) 10 km (B) 15 km (C) 20 km (D) 25 km Ans. Option (A) is correct. Explanation: C

Increase in Oil-Seeds Production

Green IV. Increase in the field of Revolution milk production Choose the most appropriate answer from the options given below: (A) A-IV, B-III, C-II, D-I (B) A-III, B-IV, C-II, D-I (C) A-I, B-III, C-II, D-IV (D) A-III, B-I, C-II, D-IV Ans. Option (B) is the correct.

8 km 4 km

Start A

2 km

D.

Explanation: Blue Revolution is the improvisation, domestication, production and cultivation of fish, aquatic plants and animals in order to get economic benefits from the same. It was also known as Neel Kranti Mission and was launched by Dr. Hiralal Chaudhari and Dr. Arun Krishnsan in 1985. White Revolution focused on increasing the milk production of the country. It is also known as Operation Flood and was launched by Dr Verghese Kurien, who is also known as the father of white revolution. Yellow Revolution was launched to increase the production of edible oil in the country, especially from mustard and sesame seeds in 1986. Sam Pitroda is known as the father of the Yellow Revolution in India. Green Revolution was launched to increase crop yield and agricultural products in India. It led to an increase in the production of food grains especially wheat and rice. MS Swaminathan is known as the father of Green Revolution in India. 36. A bag contains 5 black, 3 white and 2 red balls. Three balls are drawn in suecession. What is the probability that the first ball is red, the second ball is black and the third ball is white? 3 1 (A) (B) 10 24 1 1 (C) (D) 2 10 Ans. Option (A) is correct. Explanation: E: event that first ball is red, second ball is black and 3rd ball is white. 2 5 3 1 P  E     10 9 8 24

End

6

B 6 km

4 km In Δ ABC,

 AC 

2

 8  6 2

2

= 64 + 36 = 100

= AC

= 100 10 km

38. Find the perimeter of a rhombus whose one diagonal is 16 cm long and area is 240 cm2. (A) 68 cm (C) 24 cm Ans. Option (A) is correct.

(B) 30 cm (D) 36 cm

Explanation: Areaof  R hom bus  240 

1  d1  d2 2

1  16  d2 2

d2 = 30 cm In rhombus diagonal bisect each other  2 2 sideof    8   15    64  225   r hom bus  289  17



Perimeter of Rhombus = 4 × 17 = 68 cm 39. From the figure, what is the value of x? A 50 x B (A) 50° (C) 60° Ans. Option (D) is correct.

120 C

D

(B) 120° (D) 70°

9

CUET (UG) Solved Paper 2023 (21st May Shift -1)

Ans. Option (C) is the correct.

Explanation: A 50˚

x

120˚

B C D Exterior angle of a triangle = Sum of opposite interior angles 120  50  x x  120  50  70

40. Consider the Diagram. 500 Candidates appeared in an Examination comprising test in English, Hindi and Maths. The Diagram gives number of students who failed in different tests. What is the % percentage of student who failed at least two subjects? English

Hindi 30

75

10

Explanation: The Major constituent of natural gas is Methane. Natural gas is a fossil fuel energy source. It is composed mainly of methane (CH4), smaller amounts of natural gas liquids (NGLs, which are also hydrocarbon gas liquids), and nonhydrocarbon gases, such as carbon dioxide and water vapor. It is formed from the remains of plants and animals that have been buried for millions of years under the surface of earth. These mix with sand, calcium carbonate, silt and change into natural gas as a result of high temperature and pressure. 42. Statements. I. Some cars are black II. Some Lions are cars Conclusion I. Some blacks are Lions II. No black is Lion (A) Only I follow (B) Only II follow (C) Either I or II follows (D) None follows Ans. Option (C) is correct. Explanation:

5 12

Lion

12

Black

Since there is no definite relation between lion and Black. Hence, either some blacks are lions or no black is lion

50

43. Match List I with List II LIST-I Books

Math (A) 6.8% (B) 7.8 % (C) 1.0 % (D) 0.078 % Ans. Option (B) is correct. Explanation: English

Hindi 30

Car

75

10 5 12

12

LIST-II Authors

A. India wins freedom

I.

B.

EL Mahatma Gandhi

The guide

R K Naravan

C. India from midnight III Abdul Kalam to Millenium Azad D. Conquest of self

IV. Shashi Tharoor

Choose the most appropriate answer from the options given below: (A) A-I, B-III, C-IV, D-II (B) A-III, B-I, C-IV, D-II (C) A-III, B-I, C-II, D-IV

50 Math

Number of students who failed in at least 39 × 100 = 7.8% subjects = 500 41. Major Constituent of natural gas is. (A) Propane (B) Butane (C) Methane (D) Carbon

(D) A-I, B-IV, C-II, D-III Ans. Option (B) is the correct. Books

Authors

India Wins Freedom

Maulana Azad

The Guide

R. K. Narayan.

India from Midnight Shashi Tharoor to Millennium Conquest of self

Mahatama Gandhi

10

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

Ans. Option (C) is correct.

44. Match List I with List II LIST-I Deficiency

LIST-II Diseases

A. Insulin

I.

Kwashiorka

B. Protein

II.

Scurvy

C. Thyroxin

III. Diabetes

D. Vitamin C

IV

(C) A-II, B-I, C-IV, D-III (D) A-IV, B-III, C-II, D-I Ans. Option (A) is the correct. Diseases

Doctors

1 5 6 2

Professor

3

4 Married People 1 number indicates doctors who are not married.

Goitre

Choose the correct answer from the options given below: (A) A-III, B-I, C-IV, D-II (B) A-I, B-III, C-IV, D-II

Deficiency

Explanation:

47. The average of 12 numbers is 15 and the average of the first two numbers is 14. What is the average of the remaining numbers? (A) 15 (B) 15.2 (C) 14 (D) 14.2 Ans. Option (B) is correct. Explanation:

Insulin

Diabetes

Protein

Kwashiorkor

Average =

Thyroxin

Goitre

Vitamin C

Scurvy

Sum of first 2 numbers = 14 × 2 = 28 28  remaining number sum 15  12 Remaining number sum = 15 × 12 - 28 = 152 152 152 Average remaining    15.2 12  2 10

45. Find the angle of elevation of the Sun, when the 1 times the length of the shadow of a tree is 3 height of the tree. (A) 30° (B) 45° (C) 60° (D) 90° Ans. Option (C) is correct.

48. Match List-I with List-II

Explanation: Sun Tree

sum of all numbers  Total number of all numbers

A. B.

LIST-I Scientists Har Gobind Khorana C.V Raman

I. II.

x  Tree shadow 1 x 3  1  tan   x /  x  3  ⇒ tan    3 ⇒ tan   tan60 ⇒   60

46. In figure out which Number indicate doctors who are not married (A) 2 (B) 4 (C) 1 (D) 6

C. D.

Jagdish Chandra III. Bose Aryabhata IV

LIST-II Discoveries Discovery of Zero Genetic composition of cell Scattering of light Measurement of plant growth

Choose the most appropriate answer from the options given below: (A) A-II, B-III, C-I, D-IV (B) A-III, B-II, C-IV, D-I (C) A-II, B-III, C-IV, D-I (D) A-I, B-III, C-II, D-IV Ans. Option (C) is the correct. Scientists

Discoveries

Har Gobind Khorana

Genetic composition of cell

CV Raman

Scattering of Light

Jagdish Chandra Bose

Measurement of Plant growth

Aryabhatta

Discovery of Zero

11

CUET (UG) Solved Paper 2023 (21st May Shift -1)

49. Bhoodan-Gram Dan Movement was initiated by. (A) Mahatma Gandhi

Cities

(B) Vinoba Bhave (C) Shri Ram Chandra Reddy (D) Sardar Patel Ans. Option (B) is the correct. Explanation: Bhoodan Gram Dan Movement or Land Gift Movement was initiated by Acharya Vinoba Bhave. It was a voluntary land reform movement initiated at Pochampally village, in the then Andhra Pradesh. It was launched in 1951 and as a part of this movement, wealthy land owners voluntarily gave some percentage of their land to landless people. These landless labourers could use this and for growing crops or settling down on the same. They were not allowed to resell it or use it for any commercial activity. 50. The ratio of ages of 2 boys is 3: 7. After 2 years, the ratio of their ages will become 5: 9. The ratio of their ages after 10 years will be (A) 15 : 16 (B) 5 : 17 (C) 17 : 18 (D) 13 : 17 Ans. Option (D) is correct. Explanation: Let the age of first boy = 3x Let the age of second boy = 7x After 2 years 3x  2 5  7x  2 9 27x + 18 = 35x + 10 8x = 8 x=1 Ratio of their ages after 10 years 

Ans. Option (A) is the correct.

3x  10 3  1  10 13    13 : 17 7 x  10 7  1  10 17

51. Match List I with List II Cities with their nicknames LIST I Cities

LIST II Nickname

A.

Nagpur

I.

Diamond City

B.

Sural

II.

Pink City

C.

Jaisalmer

III. Orange City

D.

Jaipur

IV.

Golden City

Choose the most appropriate answer from the options given below: (A) A-III, B-I, C-IV, D-II (B) A-I, B-III, C-IV, D-II (C) A-III, B-I, C-II, D-IV (D) A-II, B-I, C-III, D-IV

Nickname

Nagpur

Orange City

Surat

Diamond City

Jaisalmer

Golden city

Jaipur

Pink City

52. Which of the following is not a ‘state’? (A) Nagaland (B) Manipur (C) Laddakh (D) Meghalaya Ans. Option (C) is the correct. Explanation: Ladakh is not a State. It is a Union Territory and became one on 31 October 2019 as a part of the Indian Government’s Jammu and Kashmir Reorganisation Act 2019. The erstwhile Jammu and Kashmir was stripped of its special status and reconstituted into two Union Territories – Ladakh without legislature and Jammu and Kashmir with the legislature. The first Lt. Governor of Ladakh Union Territory was RK Mathur. This is for the first time in the history of India that a state is being converted into two Union Territories. India has 28 states and 9 union territories. 31 October is observed as the ‘National Unity Day’ to mark the birth anniversary of the country’s first home minister Sardar Vallabhbhai Patel. 53. Find the angle traced by hour hand of a correct clock between 7 pm o’ clock and 2 am o’ clock. (A) 200° (B) 210° (C) 310° (D) 290° Ans. Option (B) is correct. Explanation: Angle made by hour hand in 1 hour = 30° 7 pm to 2 am = 7 hours. Angle traced = 30° × 7 = 210° 54. Find the angle of elevation of the Sun, when the length of the shadow of a tree is 3 times the height of the tree. (A) 30° (B) 45° (C) 60° (D) 90°

12

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

Ans. Option (B) is correct.

Ans. Option (A) is correct. Explanation: Sun Tree x  Tree shadow  3 x tan   x / ⇒ tanθ =



3x



1

3 ⇒ tan   tan30

⇒   30 55. Which of the following is not the satellite launched by India? (A) Explorer (B) APPLE (C) Bhaskar (D) INSAT Ans. Option (A) is the correct. Explanation: Explorer 1 was the first successful satellite launched by the United States on 31 January 1958. It was launched as a response to Soviet Union’s Sputnik 1. It is credited for the discovery of radiation belts around Earth held in place by the planet’s magnetic field. These were later named as Van Allen belts in honor of their discoverer. APPLE stands for Ariane Passenger Payload Experiment. It was India’s first communication satellite launched by ISRO on 19 June 1981. The satellite was carried on a non-metallic bullock cart to avoid reflections off the metal plates, affecting the satellite’s antenna. Bhaskara I was the first experimental remote sensing satellite built by ISRO and named after Indian mathematician Bhaskara. The satellite was launched to collect data on hydrology, oceanography, forestry and telemetry. It was launched from Volgograd Launch Station, Russia. INSAT is Indian National Satellite System. It is a space-based satellite system that provides television, communication, and meteorological services to India and its neighbouring countries. 56. The area of a circle is numerically equal to its circumference. Find the diameter of the circle. (A) 2 unit (B) 4 unit (C) 1 unit (D) 5 unit

Explanation: Area of circle = πr2 Circumference of circle = 2πr According to question πr2 = 2πr r=2 Hence, diameter = 2r = 2 × 2 = 4 units 57. Find the next term in the alpha-numeric series D4T, F9R, H20P,  J43N. (A) L 90 M (B) N 90 N (C) L 90 L (D) J 90 L Ans. Option (C) is correct. Explanation: D4T, F9R, H20P,  J43N. Logic: D  2  F, F  2  H , H  2  J , J  2  L T  2  R , R  2  P, P  2  N , N  2  L 4 × 2 + 1 → 9 , 9 × 2 + 2 → 20 , 20 × 2 + 3 → 43, 43 × 2 + 4 → 90 Hence, L90L is the correct answer. 58. Match List I with List II LIST I LIST II Diet deficiency Disease A. Deficiency of I. Anaemia Vitamin B B. Deficiency of II. Beri-Beri Vitamin A C. Deficiency of Iron III. Goitre D. Deficiency of IV Nisht blindness Iodine Choose the most appropriate answer from the options given below: (A) A-IV, B-II, C-III, D-I (B) A-II, B-IV, C-I, D-III (C) A-I, B-III, C-IV, D-II (D) A-III, B-I, C-II, D-IV Ans. Option (B) is the correct. Explanation: Diet Deficiency

Disease

Deficiency of Vitamin B

Beri Beri

Deficiency of Vitamin A

Night Blindness

Deficiency of Iron

Anaemia

Deficiency of Iodine

Goitre

59. A and B can do a work in 9 days and 12 days respectively. If they work on alternate days starting with A, then in how many days will the work be completed? (A) 36 days (B) 10 days 1 (D) 13 days (C) 10 days 4

CUET (UG) Solved Paper 2023 (21st May Shift -1)

Ans. Option (C) is correct. Explanation: A 9

B 12

days 4

3 (efficiency)

(Total work =LCM(9, 12)) 36 So, if they work on alternate days for two days, they will complete a total of 4 + 3 = 7 units of work. Therefore, if they work on alternate days for 10 days, they will complete 7 × 5 = 35 units of work. The remaining work will be 36 - 35 = 1 unit. Time taken by A to finish the remaining 1 work = 4 1 1 So, total time 10   10 days 4 4

13 60. If today is Saturday then what will be the day on 363rd day? (A) Sunday (B) Monday (C) Thursday (D) Friday Ans. Option (D) is the correct. Explanation: Considering that there are 7 days in a week. Since 363 is not an exact multiple of 7, we need to find the remainder when dividing 363 by 7. 363 ÷ 7 = 51 with a remainder of 6. This means that after 51 weeks, there will be 6 days left. Adding these 6 days to Saturday, we get: Saturday + 6 days = Friday So, the day on the 363rd day will be Friday.



CUET (UG) Question Paper - 2023 National Testing Agency 30th MAY 2023 – SHIFT 1

Section - III (General Test) Examination Duration: 60 Minutes

Maximum Marks - 250

General Instructions:

Marking scheme of the test: (a) There are 60 questions asked in the section- III. Attempt only 50 questions. (b) Correct answer or the most appropriate answer will be given five marks (+5). (c) Any incorrect option marked will be given minus one mark (–1). (d) Unanswered/Marked for review will be given no mark (0).

1. Fins the value of 6 + 6 + 6 + 6 + ……… . (A) 3 (B) 2 (C) 6 (D) 8 Ans. Option (A) is the correct.

Ans. Option (D) is the correct. Explanation:

Explanation: Let x =

6 + 6 + 6 + 6 + ……… squaring both sides, x 2 = 6 + 6 + 6 + 6 + ……… x2 = 6 + x x2 - x - 6 = 0 x2 - 3x + 2x - 6 = 0 x(x - 3) + 2(x - 3) = 0 x - 3 = 0, x + 2 = 0 x = 3, x = -2 x cannot be negative, Hence, x = 3 is correct answer. 2. Which among the following will continue the same series as established by the given figure.

? (A)



(B)

(C)



(D)

? Follow the pattern and the symmetry. Hence,

is correct. 3. Match List - I with List - II. List - I (Programme)

List - II (Year of Beginning)

(A) Vande Matram Scheme

(I)

2000

(B) Swajal Dhara Yojana

(II)

2005

(C) Bharat Nirman Yojana

(III) 2002

(D) Janshree Bima Yojana

(IV) 2004

Choose the most appropriate answer from the options given below: (A) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (B) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (C) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) (D) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) Ans. Option (A) is the correct. Explanation: Vande Matram Scheme was launched on 9th Feb 2004 with the objective to improve access to safe motherhood services and to reduce maternal and neonatal deaths significantly. Under the scheme, the gynaecologists having

15

CUET (UG) Solved Paper 2023 (30th May Shift -1)

practice would devote the 9th of every month for free check-up of pregnant women who would be provided with medicines and contraceptives by the government. Swajal Dhara Yojana was launched in 2002. It aimed at achieving self-sufficiency in drinking water in rural India by 2004. It was launched by the Department of Drinking Water Supply under the then Ministry of Rural Development. The funds were provided by the Central government to set 1 lakh hand pumps in rural areas and revival of 1 lakh traditional sources of water and providing drinking water to 1 lakh schools in rural areas. Bharat Nirman Yojana was launched on 16 December 2005 under the Ministry of Rural Development. It focused on the development of six areas- Road, Water Supply, Electricity, Irrigation, Telecommunication and housing in rural areas of the country. Janshree Bima Yojana was a joint initiative of the Central Government and Life Insurance Corporation (LIC) of India and was launched in 2000. Under this life insurance coverage was offered to both urban and rural people, who are below and marginally above the set poverty line. 4. Which country has launched the world’s first commercial moon lander? (A) Japan (B) Russia (C) India (D) U.S.A. Ans. Option (A) is the correct. Explanation: A Japanese startup is pace has launched its own private lander to the Moon aboard a SpaceX rocket from Cape Canaveral, Florida. Its mission is named HAKUTO-R, which means a white rabbit residing near the moon. The carrier deployed a NASA satellite into moon’s orbit in order to search for traces of water. It also stationed two robotic covers, one from JAXA space agency and the four-wheeled Rashid explorer of UAE. JAXA is the national space agency of Japan. 5. Arrange the following group of letters arranged in the dictionary from first to last. (A) AMDNOR (B) RANDOM (C) ADMNOR (D) ANMDOR (E) ANDMOR Choose the most appropriate answer from the options given below : (A) (C), (A), (E), (B), (D) (B) (C), (A), (D), (E), (B) (C) (C), (A), (E), (D), (B) (D) (C), (B), (A), (E), (D) Ans. Option (B) is the correct. Explanation: ADMNOR AMDNOR ANDMOR ANMDOR RANDOM

6. Statement : European economy is dependent mainly on forests. Conclusions : (I) Europe wants only maintenance of forests to improve economic conditions. (II) Trees should be preserved to improve the Europe economy. Which conclusion will follow on the basis of given statement? (A) Only Conclusion I follow (B) Only Conclusion II follow (C) Both Conclusion I and II follows (D) Either Conclusion I and II follows Ans. Option (B) is the correct. Explanation: Conclusion I is rejected because of the use of the ‘only’ word as this word says it has only one way. Hence, only conclusion II is the answer. 7. Life span of ‘RBC’ is ___________. (A) 48 Hours (B) 24 Hours (C) 12 Hours (D) 120 days Ans. Option (D) is the correct. Explanation: RBC stands for Red Blood Corpuscles or Erythrocytes. These are the carriers of fresh oxygen in the body as they contain haemoglobin. Haemoglobin is the iron-containing oxygen-transporting pigment present in the RBCs. RBCs are made in bone marrows and have a typical lifespan of 120 days. 8. A sector as shown in the figure is assembled to form a cone. What is the base radius (in cm) of the cone 22   so formed?  π =  7  

9 cm 210

(A) 9 (B) 5.25 (C) 6.75 (D) 4.5 Ans. Option (B) is the correct. Explanation: θ × 2 πr 360° Circumference of cone = length of arc of circle 210 22 2 πR = ×2× ×9 360 7 Lengthofarc =

R=

7 × 9 = 5.25 cm 12

16

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

9. In which city of Punjab, railway coaches are manufactured? (A) Ludhiana (B) Patiala (C) Kapurthala (D) Bhatinda Ans. Option (C) is the correct.

Ans. Option (A) is the correct. Explanation:

Explanation: Railway coaches are manufactured in Kapurthala at the Rail Coach Factory. It is the coach manufacturing unit of Indian railways and is located on the Jalandhar-Firozpur railway line. RCF was established in 1985. 10. Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Milk production in India is low as compared to many countries of the world. Reason (R) : The animal rearers in India are poor In the light of the above statements, choose the most appropriate answer from the options given below: (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (C) (A) is correct but (R) is not correct (D) (A) is not correct but (R) is correct Ans. Option (C) is the correct. Explanation: Here, assertion (A) is true because milk production is low in India as compared to other countries in the world because of the lack of good breed and improper feed. However, reason (R) is false because it isn’t the fact that the animal rearers in India are poor. And even if it is true, it isn’t the correct reason which is mentioned. Hence, it will be considered as false.

C

B

13. Statements: (I) No tree is desktop. (II) All desktops are computers. Conclusion : (I) Some desktops are tree. (II) Some trees are computer. Which conclusion will follow on the basis of given statement? (A) Only I follow (B) Only II follow (C) Both I and II follows (D) Neither I nor II follows Ans. Option (A) is the correct. Explanation: Desktop

Trees

Computer

2 a 2 − 3b 2 2 = , then a : b = a2 + b 2 41 (A) 5: 4 (B) 25: 16 (C) 16: 25 (D) 4: 5 Ans. Option (A) is the correct.

From the Venn diagram (1) No desktop is tree (2) Can’t say definitely some trees are computers or not. They can be. Hence, neither conclusion I nor II follows. 14. Which number is in the square, ellipse and triangle?

Explanation: 2 a 2 − 3b 2 2 = a2 + b2 41 41 (2a2 − 3b2) = 2 (a2 + b2) 82a2−123b2 = 2a2+ 2b2 80a2 = 125b2 a 2 125 25 = = b2 80 16

a 5 b =4     a:b=5:4

A+

D+ B is wife of D.

11. If

2

E+

2

12. In a family of 6 members, D is the only son of A, E is the grandfather of D. B is the daughter-in-law of C. F is the uncle of D. C is wife of A. What is the relationship of B with D? (A) Wife (B) Sister (C) Niece (D) Aunt

4 6 1

10

5

9

2

7

11

3 (A) 7 (B) 6 (C) 5 (D) 1 Ans. Option (A) is the correct. Explanation: 4 6 1

5

10 9

2

7 3

11

7 number is in the square, ellipse and triangle.

17

CUET (UG) Solved Paper 2023 (30th May Shift -1)

15. A dice is thrown twice. Find the probability of getting an odd number in the second throw and a multiple of 3 in the first throw. 1 1 (B) (A) 3 2 1 1 (C) (D) 9 6 Ans. Option (C) is the correct. Explanation: Favourable outcomes (3, 1) (3, 3) (3, 5) (6, 1) (6, 3) (6, 5) Total outcomes = 6 × 6 = 36 P (getting an odd number in second throw and a multiple of 3 in first throw. 6 1 = = 36 6 16. Find the missing term in the given series? 1 4, 6, 9, 13 ,_________? 2 3 1 (B) 20 (A) 22 4 4 1 (C) 19 (D) 17 2 Ans. Option (B) is the correct. Explanation: 4, 6, 9, 13 ½ ……… Logic: (4 × 3) ÷ 2 = 6 (6 × 3) ÷ 2 = 9 27 1 (9 × 3) ÷ 2 = = 13 2 2 81 1  27  Next × 3 ÷ 2 = = 20  2  4 4 17. Study the following arrangement carefully and answer the question given below: R D A K 7 B I 3 M J E N 67 U Z V 1 W 3 H 4 F Y 8 P 6 How many such numbers are there in the above sequence each of which is immediately preceded by a consonant and immediately followed by a vowel? (A) Zero (B) One (C) Two (D) Three Ans. Option (A) is the correct. Explanation: RDAK7B13MJEN67UZV1W3H4FY8P6 Logic: Look for the below mentioned pattern Consonant ← Number → Vowel. Hence, there is no such number. 18. A rectangular room can be partitioned into two equal square rooms by a 7 meter long partition. What is the floor area of the rectangular room in m 2? (A) 49 (B) 98 (C) 147 (D) 196 Ans. Option (B) is the correct.

Explanation:

7

7

7

7

7

7 7 Length of rectangular room = 7 + 7 = 14 m Breadth of rectangular room = 7 m Area of Floor = L × B = 14 × 7 = 98 m2 19. If E = 5, PEN=35 then PAGE =? (A) 27 (B) 29 (C) 28 (D) 36 Ans. Option (B) is the correct Explanation: E = 5, PEN 35 then PAGE =? Logic: Table of Alphabetical series (point to remember) Alphabets Positional value Alphabets Positional value

A B C D E F

G H I

J

1 2 3 4

7

8

9

10 11 12 13

Z Y X W V U T

S

R Q P O N

5

6

K L M

26 25 24 23 22 21 20 19 18 17 16 15 14

E=5 PEN = 16 + 5 + 14 = 35 PAGE = 16 + 1 + 7 + 5 = 29

20. If sin 2θ = cos 40°, then the smallest positive value of θ is : (A) 25° (B) 20° (C) 40° (D) 30° Ans. Option (A) is the correct. Explanation: sin2θ = cos 40° sin2θ = sin (90°− 40°) On comparing both sides 2θ = 50° θ = 25° 21. The 10th term of the A.P. 1, 5, 9, 13,... , is : (A) 36 (B) 33 (C) 37 (D) 41 Ans. Option (C) is the correct. Explanation: The 10th term of the A.P. 1, 5, 9, 13, ..., is: First term of A.P. ⇒ a = 1 d = a2 - a2 = 5 - 1 = 4 an = a + (n - 1) d a10 =1 + (10 - 1) (4) = 37

18

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

22. Find the odd pair out?

It is known as Green Fuel and is an environment-friendly alternative automotive fuel that plays an important role in reducing vehicular greenhouse gas emissions and environmental pollution significantly.

(A) 11 : 125 (B) 33 : 1093 (C) 35 : 1229 (D) 17 : 295 Ans. Option (D) is the correct. Explanation: (11)2 + 4 = 121 + 4 = 125 (33)2 + 4 = 1089 + 4 = 1093 (35)2 + 4 = 1225 + 4 = 1229 (17)2 + 4 = 289 + 4 = 293 Hence, option 4 is the odd pair out. 23. Atmospheric Pressure is measured ______________. (A) Pedometer (B) Thermometer (C) Odometer (D) Barometer Ans. Option (D) is the correct.

by

Explanation: Atmospheric Pressure is measured by Barometer, which consists of a long glass tube open at one end and closed at the other. The tube is filled with mercury and the barometer works by balancing mercury. Atmospheric Pressure is measured in Pascal. A pedometer is used to count the number of steps. It is usually a portable and electronic or electromechanical device that detects the motion of a person’s hands or hips. A thermometer is a device used to measure temperature or a temperature gradient. Odometer is an instrument used to measure the distance travelled by a vehicle. It may be electronic, mechanical or electromechanical. 24. Which Indian Meteorological satellite has renamed as Kalpana-I in 2003? (A) INSAT-3A (B) METSAT (C) INSAT-3C (D) GSAT-2 Ans. Option (B) is the correct. Explanation: METSAT was renamed as Kalpana - 1 on 5th February 2003. It was named after the Indian-born American Astronaut Dr. Kalpana Chawla, who died on 1 February 2003 in the US Space Shuttle Columbia disaster. METSAT was the first satellite in the series of exclusive meteorological satellites built by ISRO. METSAT was the first exclusive meteorological satellite built by ISRO. 25. “CNG” Stands for (A) Common Natural Gas (B) Compressed Natural Gas (C) Common Effective Natural Gas (D) Compressed Neutral Gas Ans. Option (B) is the correct. Explanation: CNG stands for Compressed Natural Gas. Basically, CNG is natural gas compressed under a pressure of 200 to 250 kg/cm2. With this compression, it occupies less than 1% of its volume at atmospheric pressure. CNG primarily consists of methane.

26. In which of the following Union Territories of India is the “National Meat and Polutry Processing Board” located? (A) Chandigarh (B) Puducherry (C) Delhi (D) Dadra and Nagar Haveli Ans. Option (C) is the correct. Explanation: National Meat and Poultry Processing Board (NMPPB) India is located in New Delhi. It was an autonomous body and was launched on 19th Feb 2009. It served as a single window service provider for producers/manufacturers and exporters of meat and meat products, for promoting & regulating the meat industry. However, the board has been winded up by the government in 2023. 27. Find the missing figure from the given options. ++

++

++

(A)

(B)

(A)

?

+

+

+

(C)

(D)

(B)

+

+

+ (C)

+

+

(D)

+ Ans. Option (D) is the correct. Explanation: ++

++

++

?

+

(A) (B) (D) Follow the logic and the (C) pattern. Hence, +

+

28. A certain weight (in kg) is divided into two parts such that 5 times the first part added to 11 times the second part makes 7 times the whole weight. The ratio of the first part to the second part is : (A) 1: 2 (B) 5: 11 (C) 2: 1 (D) 2: 3 Ans. Option (C) is the correct.

19

CUET (UG) Solved Paper 2023 (30th May Shift -1)

Explanation: Let the first part be x. Hence, second part be y. According to the question, 5x + 11y = 7 (x + y) 5x + 11y = 7x + 7y 11y − 7y = 7x − 5x 4y = 2x x 4 2 = = = 2 :1 y 2 1

States

Punjab, Rajasthan, and Gujarat Pakistan Himachal Pradesh, Uttarakhand, Sikkim, Arunachal Pradesh

West Bengal, Sikkim, Arunachal Bhutan Pradesh and Assam D

65 45

x

C (A) 65

B

(B) 70

(C) 75 (D) 45 Ans. Option (B) is the correct Explanation: A

D 65 O

45

x

B C ∠COA + ∠OAC + ∠ACO = 180° (by angle sum property) ∠COA + 90° + 45° = 180° ∠COA + 180° - 135° = 45° Now ∠BOD = ∠COA = vertically opposite angle In ∆BOD ∠BOD + x + 65° − 45° = 70° x = 180 − 65° − 45° = 70 30. Match List - I with List - II. List - I (States)

List - II (Boarding Countries)

(A) Punjab, Rajasthan, Gujarat

(I)

(B) Himachal Pradesh, Uttarakhand, Sikkim, Arunachal Pradesh (C) West Bengal, Mizoram, Meghalay, Tripura, Assam

(II) Pakistan

(D) West Bengal, Sikkim, Arunachal Pradesh, Assam

China

West Bengal, Mizoram, Megha- Bangladesh laya, Tripura, and Assam

29. From the figure, find x. A

O

Bordering Countries

Bhutan

(III) China

(IV) Bangladesh

Choose the most appropriate answer from the options given below : (A) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (B) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) (C) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (D) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) Ans. Option (D) is the correct.

31. Match List - I with List - II. List - I (Writer)

List - II (Book)

(A) (B) (C) (D)

(I) (II) (III) (IV)

Plato Leo Tolstoy Pt. Vishnu Sharma Sri Aurobindo Ghosh

Panchatantra War and Peace The Life Divine Republic

Choose the most appropriate answer from the options given below : (A) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (B) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (C) (A)-(IV), (B)-(I), (C)-(II), (D)-(III) (D) (A)-( II), (B)-(I), (C)-(III), (D)-(IV) Ans. Option (A) is the correct. Writer Plato Leo Tolstoy Pt. Vishnu Sharma

Book Republic War and Peace Panchatantra

Aurobindo Ghosh

The Life Divine

32. Keoladeo Bird Sanctuary is located in (A) Rajasthan (B) Delhi (C) Haryana (D) Punjab Ans. Option (A) is the correct. Explanation: Keoladeo National Park or Keoladeo Ghana National Park is situated in Bharatpur, Rajasthan. It is famous as a bird sanctuary that hosts thousands of birds during the winter. It is a UNESCO world heritage site and was declared the same in 1985. It is home to 370 species of birds and animals. It was earlier known as Bharatpur Bird Sanctuary. 33. The average weight of A, B and C is 60 kg. If the average weight of B and C is 50 kg and that of A and B is 52 kg, then what is the weight of B? (A) 26 kg (B) 24 kg (C) 30 kg (D) 32 kg Ans. Option (B) is the correct. Explanation: Total weight of A, B and C = 60 × 3 = 180 kg Weight of B and C = 50 × 2 = 100 kg Weight of A and B = 52 × 2 = 104 kg So, the average weight of B =[(weight of B and C)+(Weight of A and B)] - (Total weight of A, B and C) = [100 + 104] - 180 = 24 kg

20

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

34. Degreewise break-up of expenditure of a family in a month is shown in the pie chart. Total amount spent in a month is ` 45800. Investment Commuting 54 79.2 Entertainment 36 Shopping 68.4

Medicines 39.6 Groceries 82.8

What is the amount spent by the family on commuting? (A) ` 10076 (B) ` 10354 (C) ` 6870 (D) ` 8702 Ans. Option (A) is the correct. Explanation:

Investment 54

Commuting 79.2

Entertainment 36 Shopping 68.4

Medicines 39.6 Groceries 82.8

Amount spent on commuting

79.2 × 45800 360 = ` 10076 =

35. Who said “INC (Indian National Congress) should distinguish between begging and claiming the Rights”. (A) Aurobindo Ghosh (B) Bal Gangadhar Tilak (C) Bipin Chandra Pal (D) Lala Lajpat Rai Ans. Option (B) is the correct. Explanation: INC should distinguish between begging and claiming the rights was said by Bal Gangadhar Tilak. He was an Indian nationalist, independence activist, and teacher. He was one of the strongest advocates of Swaraj and was the first leader of the Indian independence movement. He was also conferred with the title of ‘Lokmanya.’ His most famous slogan was ‘Swaraj is my Birth right and I shall have it.’ He is also known as ‘Father of Indian Unrest.’ 36. An equation of the type y = kx (k ≠ 0) represents a line: (A) passing through the origin (B) intersecting the coordinate axes at two different points (C) parallel to x-axis (D) parallel to y-axis

Ans. Option (A) is the correct. Explanation: An equation of the type y = kx (k = 0) represents a line: Logic: Y = kx When x = 0, y = 0 Hence, the line is passing through the origin. 37. How many times Mithali Raj played in ICC Women’s Cricket World Cup? (A) 4 (B) 3 (C) 5 (D) 6 Ans. Option (D) is the correct. Explanation: Mithali Raj played in the iCC’s women’s world cup for six times in 2000, 2005, 2009, 2013, 2017 and 2022. She is the first woman cricketer to hold this record. She is a former Indian cricketer and captain of the Indian cricket team from 2004 to 2022. She is hailed as one of the greatest women cricketers of all times. She is the only female cricketer to surpass the 7,000run mark in Women’s One Day International matches. She is the first player to score seven consecutive 50s in ODIs. She has been awarded with Arjuna Award, Padma Shri, and Major Dhyan Chand Khel Ratna Award in 2021. 38. Which amongst the following is known as “Island of Pearls”? (A) Australia (B) Zanjibar (C) Bahrain (D) Panama Ans. Option (C) is the correct. Explanation: Bahrain is known as the island of Pearls because it is a hub of natural pearls. It is an island country situated between Qatar and the northeastern coast of Saudi Arabia. 39. Study the following arrangement carefully and answer the question given below. M D E K 7 B I 4 M J N E 76 Z U V W I 3 H 4 F How many such numbers are there in the above sequence each of which is immediately preceded by a consonant and immediately followed by a vowels? (A) Zero (B) One (C) Two (D) Three Ans. Option (A) is the correct. Explanation: MDEK7B14 MJNE76ZUVW13H4F Logic: Look for the below mentioned pattern Consonant ← Number → Vowel. Hence, there is no such number. 40. Mohan is 14th from the left end in a row of 40 boys. What is his position from the right end? (A) 21th (B) 24 th st (C) 27 (D) 25 th Ans. Option (C) is the correct.

21

CUET (UG) Solved Paper 2023 (30th May Shift -1)

Explanation: Position of Mohit from right end = 40 + 1 - 14 = 27 41. On compound interest, ` 2,000 amounts to ` 2,226.05 in 2 years. What is the rate of interest per annum? (A) 5% (B) 5.5% (C) 6% (D) 6.5% Ans. Option (B) is the correct. Explanation: P = ` 2000 A = ` 2226.05 T = ` 2 years Using,

R   A = P1 +   100 

T

R   2226.05 = 2000  1 +   100  R  222605  = 1 +  200000  100 

2

2

44521  R  =  1 + 40000  100  211 R −1= 100 200 11 × 100 = R 200 R = 5.5% per annum 42. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. The diameter of the base 22   of the cylinder is  Use π =  : 7   (A) 2 cm (B) 4 cm (C) 5 cm (D) 0.5 cm Ans. Option (A) is the correct. Explanation: Curved surface area of right circular cylinder = 2πrh 22 88 = 2 × × r × 14 7 88 × 7 =1 r= 2 × 22 × 14  Diameter = 2r = 2 × 1 = 2 cm 43. A wholeseller sells a jacket to a retailer at a profit of 5% and the retailer sells it to a customer at a profit of 10%. If the customer pays ` 4,158 for the jacket, then what was the cost price of the jacket for the wholesaler? (A) ` 3,500 (B) ` 3,400 (C) ` 3,300 (D) ` 3,600 Ans. Option (D) is the correct. Explanation:  105   110  SellingPriceof Jacket = CP ×  ×   100   100  4158 × 100 × 100 CP = = `3600 105 × 110

44. Which Union Territory of India is the smallest as per its area? (area wise smallest UT) (A) Lakshadweep (B) Puducherry (C) Chandigarh (D) Delhi Ans. Option (A) is the correct. Eplanation: Lakshadweep is the smallest Union Territory with an area of 32 square kilometres. It’s made up of 12 atolls, 3 reefs, and 5 submerged banks out of which ten Islands are inhabited. The capital of Lakshadweep is Kavaratti. The Union Territory was formed in 1956 and it was named Lakshadweep in 1973. The current administrator of Lakshadweep islands is Praful Khoda Patel. 45. A clock is set right at 5 a.m. The clock loss 16 min in 24 h. What time will be right time when the clock indicate 10 p.m. on the 4 th day? (A) 12:30 p.m. (B) 12:00 p.m. (C) 11:00 p.m. (D) 11: 15 p.m. Ans. Option (C) is the correct. Explanation: The clock loses 16 minutes in 24 hours ⇒ 60 minutes in 90 hours. So, 89 hours of this clock = 90 hours in an accurate clock. Time from 5 a.m. to 10 p.m. of 4th day = 72 + 12 + 5 = 89 hours. Which means 90 hours have passed originally. So, the correct time now is 11 p.m. 46. Match List - I with List - II. List – I (Animals)

List - II (Natural Habitat)

(A)

Elephants

(I)

Sunderban National park

(B)

Camel

(II)

Gir Forest National park

(C)

Indian Lion

(III)

Thar Desert

(D)

Tiger

(IV)

The Hot Wet forest of Assam and Kerala

Choose the most appropriate answer from the options given below : (A) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (B) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (C) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (D) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) Ans. Option (A) is the correct. Animals

Natural Habitat

Elephants

Hot Wet forest of Assam and Kerala

Camel

Thar Desert

Indian Lion

Gir Forest National Park

Tiger

Sunderban National Park

47. The ascending order of the numbers 0.8, 0.88, 0.808, 0.08 is (A) 0.88, 0.808, 0.8, 0.08 (B) 0.808, 0.8, 0.08, 0.88 (C) 0.08, 0.8, 0.88, 0.808 (D) 0.08, 0.8, 0.808, 0.88 Ans. Option (B) is the correct.

22

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

Explanation: Ascending order means ⇒ Increasing order 0.08, 0.8, 0.88, 0.808 48. Find the wrong number of the series. 6400, 3200, 1600, 600, 400, 200, 100, 50 (A) 200 (B) 1600 (C) 100 (D) 600 Ans. Option (D) is the correct. Explanation: 6400, 3200, 1600, 600, 400, 200, 100, 50 Logic: 6400 ÷ 2 = 3200 3200 ÷ 2 = 1600 1600 ÷ 2 = 800 800 ÷ 2 = 400 400 ÷ 2 = 200 200 ÷ 2 = 100 100 ÷ 2 = 50 Hence, in the given series there should be 800 in place of 600. 49. Nine boys can complete a work in 360 days, 18 men can complete the same work in 72 days and 12 women can complete it in 162 days. In how many days, can 4 men, 12 women and 10 boys together, complete the work? (A) 79 (B) 81 (C) 83 (D) 85 Ans. Option (B) is the correct. Explanation: 9 boys can do in 360 days ⇒ 1 boy = 9 × 360 = 3240 18 men can do in 72 days ⇒ 1 man = 18 × 72 ⇒ 1296 12 women in 162 days ⇒ 1 woman = 12 × 162 ⇒ 1944 Hence, 5 boys = 2 men = 3 women ⇒ 2 men = 5 boys 10 boys = 4 men, 12 women 8 men Now, 10 boys + 12 women + 4 men 4 men + 8 men + 4 men = 16 men (18 × 72 ) 16 men complete in = 16 ⇒ 81 days Hence, number of days in which work can be completed together is 81 days. 50. Crescograph, a device for measuring growth in plants was invented by which Indian Scientist? (A) lagdish Chander Bose (B) C. V. Raman (C) Homi J. Bhabha (D) Vikram Sarabhai Ans. Option (A) is the correct.

Explanation: Crescograph, a device for measuring growth in plants was invented by Sir Jagadish Chandra Bose. CV Raman is known for the discovery of the Raman effect for which he was awarded with Nobel Prize in Physics in the year 1930. Homi Jehangir Bhabha is known as the father of the Indian nuclear programme. He was the founding director and professor of Tata Institute of Fundamental Research (TIFR) and Bhabha Atomic Research Centre. He was the first chairman of the Indian Atomic Energy Commission and secretary of the Department of Atomic Energy. Vikram Sarabhai was an Indian physicist and astronomer. He is regarded as the Father of the Indian Space Program. He was the first chairman of the Indian Space Research Organisation (ISRO). 51. Find the missing term in 1, 4, 10, 22, 46, ___________? (A) 100 (B) 98 (C) 96 (D) 94 Ans. Option (D) is the correct. Explanation: 1, 4, 10, 22, 46,…. Logic: 1×2+2=4 4 × 2 + 2 = 10 10 × 2 + 2 = 22 22 × 2 + 2 = 46 46 × 2 + 2 = 92 52. Today is Monday. After 60 days it will be? (A) Thursday (B) Tuesday (C) Friday (D) Wednesday Ans. Option (C) is the correct. Explanation: Today is Monday

60 = 4 odd days 7 Monday +4 days =Friday Day after 60 days =

53. The sum of the fourth proportional of 4, 5, 16 and the mean proportional of 4, 16, is (A) 25 (B) 28 (C) 30 (D) 32 Ans. Option (B) is the correct. Explanation: Fourth proportion a c = b d Let the 4th proportional be x 4 16 = 5 x 16 × 5 x= = 20 4 Mean proportional of 4, 16 is sum = 20 + 8 = 28

4 × 16 = 8

23

CUET (UG) Solved Paper 2023 (30th May Shift -1)

54. _______ is called the father of Indian Archaeology. (A) James Princip (B) John Marshell (C) Alexender Cunninghum (D) Mortimer Wheeler Ans. Option (C) is the correct. Explanation: Alexander Cunningham is called the father of Indian Archaeology. He was the first Director General of the Archaeological Survey of India (ASI). James Prinsep was an English scholar, orientalist and antiquary. He was the founding editor of the Journal of the Asiatic Society of Bengal. John Marshall was the Director-General of the Archaeological Survey of India from 1902 to 1928 under whom the cities of Harappa and Mohenjo Daro were excavated. Mortimer Wheeler was the Director General of the Archaeological Survey of India from 1944 to 1948. 55. Which of the following article of Constitution of India is related to the Finance Commission? (A) Article 21A (B) Article 280 (C) Article 324 (D) Article 124 Ans. Option (B) is the correct. Explanation: Article 280 of the Indian Constitution is related to Finance Commission. It states that President has the power to constitute a Finance Commission that consists of a Chairman and four other members to be appointed by the President for five years. The finance commission makes recommendations on tax distribution between state and union governments. Article 21A of the Indian constitution states that government will provide free and compulsory education to children between six to fourteen years of age as a fundamental right. It was inserted by the 86th Constitution Act, of 2002. Article 324 of the Indian constitution is related to the Election Commission which will supervise, direct, and control the national and state elections, election to the office of the president of India and the office of vice-president of India. Article 124 of the Indian constitution is related to the establishment and Constitution of Supreme Court. 56. A, B and C started a business by investing ` 40,000, ` 50,000 and ` 60,000 respectively. After 8 months B withdrew ` 10,000. After one more month C withdrew ` 40,000. Find the ratio in which profit earned at the end of the year will be distributed. (A) 4: 5: 6 (B) 2: 2: 1 (C) 12: 14: 15 (D) 10: 12: 13 Ans. Option (C) is the correct. Explanation: 40,000 : 50,000 : 60,000 40,000 × 12 : 50,000 × 8+40000 × 4: 60000 × 9 + 20000 × 3

10000(48) : 10000 (40+16) : 10000 (54+6) 48 : 56 : 60 12 : 14 : 15 57. Complete the figure matrix :

? (A)

(B)

(c)

(D)

Ans. Option (B) is the correct. Explanation:

? Logic: Follow the pattern and the symmetry. Hence,

58. Two men P and Q start together from place A at 3 km/h and 3.75 km/h respectively for place B. Q reaches half an hour before P. What is the distance that each one has covered? (A) 7 km (B) 7.2 km (C) 7.5 km (D) 7.8 km Ans. Option (C) is the correct. Explanation: Distance = Speed × Time Distance travelled by P = 3t   1  Distance travelled by Q = 3.75  t −      2  1 According to the question, 3t = 3.75  t −   2

24

OSWAAL CUET (UG) Chapter-wise & Topic-wise Question Bank GENERAL TEST

3.75 2 0.75t × 2 = 3.75 3.75 t= = 2.5 hours 0 . 75 × 2      3.75t − 3t =

Distance covered = 3 × 2.5 = 7.5 Km 59. Who won the T20 Cricket World Cup-2022 held in Australia? (A) Australia (B) England (C) India (D) Pakistan Ans. Option (B) is the correct. Explanation: England won the T20 Cricket World Cup 2022 held in Australia. The final match was played between Pakistan and England at Melbourne Cricket Ground. Sam Curran of England was named the Player of the match and also the Player of the tournament.

60. Starting from point P, Sunil walked 30 m towards South. He turned to his left and walked 40 m. Then, he again turned to his left and walked 30 m. Turning again to his left he walks 15 m and stops there at point N. How far is Sunil from starting point P? (A) 30 m (B) 15 m (C) 25 m (D) 10 m Ans. Option (C) is the correct. Explanation: Start

P N 15 m

30 m

End

30 m W

N E

R L

L S 40 m R ­ Distance between starting and end point = 40 − 15 = 25 m. 

GENERAL AWARENESS : INDIAN HISTORY

25

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Kingdom

26

GENERAL AWARENESS : INDIAN HISTORY

27

Chapter

1

Indian History

Chapter Analysis Concept Name Ancient History, Medieval History and Modern History

Topic-1 Ancient History  Revision Notes • The development of archaeology and the study of archaeological remains help us understand the lives of prehistoric people since there are no available written records. • Archaeological remains­—stone tools, pottery, artefacts and metal implements were used by the people of prehistoric age. • The important sites of the Old Stone Age (2 mn to 10000 BCE) in India were Soan valley and Patwar plateau in the Shivalik Hills on North India, Bhimbetka in Madhya Pradesh, Adamgarh Hills in Narmada Valley, Kurnool in Andhra Pradesh and Attirampakkam near Chennai. • The food was obtained by hunting and gathering edible plants and tubers.

Mesolithic or Middle Stone Age (10000 BCE–6000 BCE)

• The famous sites of middle stone age Langhanj were in Gujarat, Adamgarh in Madhya Pradesh, some places of Rajasthan, Uttar Pradesh and Bihar. • In this age horticulture and primitive cultivation started.

Neolithic Age (6000 BCE–4000 BCE)

• The important sites of this age were—Kashmir valley, Chirand in Bihar, Belan Valley in UP, several places in the Ocean, South India—Maski, Brahmgiri, Halles, Kodikal, Paiyapalli, Guntur. • The wheels were used to make pottery. • The mud houses were built instead of grass huts.

Metal Age of Chalcolithic (Copper-Stone 4000 BC -2000 BC) Period

• When metal came into use, the age is called the Metal Age or Chalcolithic Age. major sites of this period are Ahar (Rajasthan), Diamabad (Maharashtra) and Navdatoli (Madhya Pradesh). • This was the first time when instead of stone Copper was adopted to made tools.

Iron Age (1200 BCE–600 BCE)

•  The important sites of this age were—Hallur, Maski, Nagarjunakonda, Adichanallur. • In India Iron age is belonged to the Vedic Culture. Weapons, tools now started to made with Iron.

The Harappan Civilisation

• It was first called the Indus Valley Civilisation, but later it was renamed Indus Civilisation. • The earliest excavations in the Indus valley were done at Harappa in West Punjab and Mohenjodaro in Sindh (Both now in Pakistan). • The excavations at Mehargarh 150 miles to the northwest of Mohenjodaro reveal the existence of pre Harappan culture. • The sites of Amri and Kot Diji remain the evidence for the early Harappan stage. • The total span of this culture should be between 2300 and 1750 BCE. • The important sites of this civilisation are—Kot Diji in Sindh, Kalibangan in Rajasthan, Ropar in Punjab, Banawali in Haryana, Lothal, Surkotoda, Dholavira in Gujarat.

2021

2022

2023

Additional Questions

3

1

4

7

• ‘The Great Bath’ was found at Mohenjodaro. It was the most important public place measuring 39 feet in length, 23 feet in breadth and 8 feet in deep. It must have served as a ritual bathing site. The floor of the bath was made of burnt bricks. • Foreign trade was mainly conducted with Mesopotamia, Afghanistan and Iran. • Pashupati (proto-Siva)—the chief male deity represented in seals as a sitting yogic position with three faces and two horns. He is surrounded by four animals (elephant, tiger, rhino and buffalo). • People of this period used to worship ‘Pashupati’. • It is believed that the decline of Harappan culture was due to natural calamities. • The destruction of forts is mentioned in Rig Veda.

Vedic Period

• The cities of Harappan culture had declined by 1500 BCE. Around this period, Indo-Aryans, who spoke Sanskrit, entered North-West India from the Indo-Iranian region. • Arctic region, Germany, Central Asia and Southern Russia are believed to be the original home of Aryans. • The word ‘Veda’ is derived from the root ‘vid’, which means ‘superior knowledge’. There are four Vedas: • Rig Veda: It is earliest of all the four Vedas , consisting of 1028 hymns sung in praise of various gods. • Yajur Veda: It consists of various details of rules to be observed at the time of sacrifice. • Sama Veda: It is set to tune to chant during the sacrifice. It is called the book of chants and the origins of Indian music are traced in it. • Atharva Veda: It consists of details of rituals. • Other sacred words are Brahmanas, Upanishads, Aranakyas, Epics Ramayana and Mahabharata.

Rig Vedic Age and Early Vedic Period (1500–1000 BCE)

• The Rig Veda refers to Sapta Sindhu or land of seven rivers— Jhelum, Chenab, Ravi, Beas, Sutlej, Indus and Saraswati. • The women poets of this period were Apala, Viswara, Ghosa and Lopamudra • The important Rig Vedic gods were—Prithvi (earth), Agri (fire), Vayu (wind), Varuna (rain) and Indra (thunder).

Later Vedic Period (1000–600 BCE) • Later Vedic literature mentioned several tribal groups and kingdoms. • Now society was divided into four Varnas (Brahamin, Vaishya, Kshatriya & Shudras) • The later Vedic texts also refer to the three divisions in India— Aryavarta, Madhyadesa, Dakshinapatha. • It saw the rise of Buddhism and Jainism. Upavedas and Associated vedas Upavedas Ayurveda (Medicine) Gandharva veda (Music) Dhanurveda (Archery) Arthaveda (Science of wealth)

Vedas Rig veda Sama veda Yajur veda Atharva veda

29

GENERAL AWARENESS : INDIAN HISTORY

Rigvedic Names of Indian Rivers River

Rigvedic Name

• Dhana Nanda kept the empire together and had a strong army. 16 Mahajanapads

Capitals

Indus

Sindhu

Anga

Champa

Jhelum

Vitasta

Magadha

Girivraja/Rajgriha

Chenab

Askini

Kasi/Kashi

Kasi

Beas

Vipasa

Vasta

Kausambhi

Ravi

Purushani

Kosala

Shravasti (northern) Kushavathi (southern)

Sutlej

Sutudari

Surasena

Mathura

Panchala

Ahichchatra and Kampilya

Life of Vardhamana Mahavira (539—467 BCE)

Kuru

Indraprastha

• Vardhamana Mahavira, the 24 Tirthankara of Jain tradition, born at Kundagrama to Siddharta and Trisala, embraced asceticism at 30, wandering for 12 years. • In the 13th year, he attained Kevala Jnana, becoming Mahavira and Jina. His followers, Jains, practice Jainism. Preaching for 30 years, he died at 72 in Pava near Rajgriha. Mahavira organized the Sangha to spread teachings. • Jainism found patronage from rulers like Chandragupta Maurya, Kharvela of Kalinga, and dynasties in South India. • Jainism has two sects: Shvetambaras (white-clad) led by Sthalbahu and Digambaras (skyclad or naked) led by Bhadrabahu. • The fi rst Jain council in Pataliputra (3rd century BCE) was led by Sthulbadhra. The second council in Vallabhi (512 CE), presided over by Devarddhigani kshamashramana, led to the fi nal compilation of 12 Angas and 12 Upangas.

Matsya

Viratanagara

Chedi

Sathivati

Gandhara

Taxila

Kamboja

Poonch/Punch

Jainism

th

Buddhism

Life of Mahatma Buddha (567—487 BCE)

• Siddhartha Gautam also known as Lord Buddha was born in 567 BCE in Lumbini Garden, founded Buddhism. Witnessing old age, disease, death, and asceticism prompted him to seek truth. • He left home at 29, attained enlightenment at 35 under a bodhi tree in Bodhgaya, and delivered the fi rst sermon in Sarnath. • He died at 80 in Kushinagar. • Buddha had two types of disciples: monks (bhikshu) and lay worshippers (upasakas), with notable monks like Sariputta and Moggallanna. • The fi rst council was at Rajagriha, the second at Vaishali, and the third at Patliputra under Ashoka. • The fourth, led by Kanishka, occurred in Kashmir and included Ashvaghosha. The Tripitakas, Buddhist texts in Pali, comprise Sutta pitakas, Vinaya pitakas, and Abhidhamma pitakas.

Magadha Empire

• The Buddhist literature Anguttara Nikaya gives a list of 16 great kingdoms called Sixteen Mahajanapadas. • Finally in the 6th century BCE, only four kingdoms survived— Vatsa, Avanti, Kosala and Magadha. • Vatsa—The capital was Kaushambi, near modern Allahabad (Prayagraj). • Avanti—The capital was Ujjain, the ruler was Pradyota. • Kosala—The capital was Shravasti (Ayodhya), the ruler was king Prasenjit • Magadha—The capital was Rajgriha, the rulers were Bimbisara, Ajatsatru. • Bimbisara won a battle against Brahmdutta of Anga and expanded his kingdom. He also built friendly relationships with Avanti. Ajatshatru fought for 16 years against Kosala and Vaishali, making him more powerful. • Udayin followed Ajatshatru as the ruler, and Shishunaga brought Avanti under the Magadhan empire. • Mahapadma Nanda ruled strongly for ten years and had eight sons, with Dhana Nanda being the last one to rule.

Asmaka or Assaka Potali/Podama Vajji

Vaishali

Malla

Kusinagara

The Mauryan Empire

• Kautilya’s Arthashastra was written by Kautilya, a contemporary of Chandragupta Maurya, most important literary source of the history of the Maurya. • Kautilya was also known as Indian Machiavelli. • The manuscript of Arthshastra was fi rst discovered by R. Sharma Shastri in 1904. • Megasthenes was the Greek ambassador in the court of Chandragupta Maurya. He wrote Indica. • Chandragupta Maurya (322—298 BCE) was the founder of the Mauryan empire. • Bindusara was called by the Greeks ‘Amitraghatha’ meaning slayers of enemies. • He supported the Ajivikas, a religious sect Bindusara appointed his son Ashoka as the governor of Ujjain. • The most important event of Ashoka’s reign was victorious war with Kalinga in 261 BCE. • As an effect of the Kalinga war, Ashoka embraced Buddhism under the infl uence of Buddhist monk, Upagupta. • The stupa of Sanchi is the only remaining of the Mauryan period. • The caves presented to Ajivikas by Ashoka and his son Dasaratha remain important heritage of the Mauryas. • The caves at Barabar hills near Bodhgaya are wonderful pieces of Mauryan Architecture. • The Mauryan rule was replaced by Shunga dynasty. • The founder of the Shunga dynasty was Pushyamitra Shunga. • Pushyamitra followed Brahmanism. • Devabhuti was last Shunga ruler who was murdered by minister Vasudeva Kanva, the founder of the Kanva dynasty. • Satavahanas is also known as the Andhras established their independent rule after the decline of the Mauryas in the Deccan. • Simuka was the founder of the Satavahanas dynasty, succeeded by Krishna, who extended the kingdom up to Nasik. • Gautamiputra Satkarni (106—130 BCE) was the greatest ruler of Satavahanas dynasty. • The Satavahanas built Chaityas and Viharas. • Vashisthiputra Pulumavi repaired the old Amaravati stupa. • The Kushanas were a branch of the Yuchi tribe. • The founder of the Kushanas dynasty was Kujula Kadphises or Kadphises. • His son Wima Kadphises or Kadphises 2nd conquered the whole of Northwestern India as far as Mathura. • The most important ruler of the Kushana dynasty was Kanishka (78—120 CE). • He was the founder of the Saka era which started from 78 CE.

30 • His capital was Purushapura, Mathura was another important city in his empire. • He followed Buddhism. • Last important ruler of this dynasty was Vasudeva.

The Sangam Period

• The Sangam Age constitutes an important chapter in the history of South India. According to Tamil legends, there existed three Sangams popularly called Muchangam. • The fi rst Sangam held at Madurai was attended by gods and legendary stages but no literary work of this Sangam was available. • The second Sangam was held at Kapadapuram but all the literary works had perished except Tolkappiyam. • The third sangam at Madurai was founded by Moda Thirumaram. • The corpus includes Tolkappiyam Ettuhogai, Pattuppattu, Panthinenkil-Kanakku, and the two epics—Silappathikaram and Manimegalai. • The Ashokan inscriptions mentioned the Cholas, Cheras and Pandyas on the South of the Mauryan Empire. • The excavations at Arikkamedu, Poompuhar, Kodumanal and other places reveal the overseas commercial activities of the Tamils. • The Tamil country was ruled by three dynasties namely Cheras, Cholas, and Pandyas. • The Capital of Cheras was Vajji and important seaports were Tondi and Musiri. • Perum Sorru Udhiyan Cheralatan, Imayavaramban Nedum Cheralathan and Cheran Senguttuvan were famous rulers of this dynasty. • The Capital of Cholas was fi rst located in Vraiyur and then shifted to Puhar. • The famous king of Cholas was Karaikala. • The capital of Pandyas was Madurai. • Maduraikkanji, written by Mangudi Maruthanar describes the socio-economic condition of the Pandyas. • The last famous king was Uggira Peruvaludhi. • The reason of decline was due to invasion of Kalabhras. • The primary deity was Murugan or Seyon. • Mayon (Vishnu), Vendan (Indiran), Varunan and Korravai were also worshipped.

The Gupta Period

• The eminent rulers of the Gupta empire were: • Chandragupta I (320—335 or 340 CE) • Ramagupta (late fourth century) • Chandragupta II (nearly 380—413 CE or 415 CE) • Kumargupta (near about 415—455 CE) • Skandagupta (455—467 CE) • Purugupta (467—469 CE) • Buddhagupta (477 to nearly 500 CE) • Devichandraguptam and Mudhrarakshasam written by Visakhadutta provide information regarding the rise of Guptas. • The Chinese traveler Fa-Hien, who visited India during the region of Chandragupta II, has left a valuable account of the social, economic, and religious conditions of the Gupta empire. • The founder of the Gupta dynasty was Sri Gupta. • Chandragupta I was the fi rst to be called Maharajadhiraja (the great king of kings). • Samudragupta was hailed as ‘Indian Napoleon’ because of his military achievements. • Military achievements—The Allahabad pillar inscription speaks of his magnanimity to his foes, his polished intellect, his poetic skill, and his profi ciency in music—his image depicting him with Veena is found in the coins issued by him. • Rudrasimha III, the last ruler of Saka Satrap was defeated by Chandragupta II. • Chandragupta II also defeated the confederacy of enemy chiefs in Vanga.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• Fa-Hien, a china pilgrim, who visited India during the reign of Chandragupta II. • Fa-Hien provided valuable information on the religious, social, and economic condition of the Gupta empire. • Kumargupta was the son and successor of Chandragupta II. • He laid the foundation of Nalanda University which emerged as an institution of international reputation. • Gupta power disappeared due to the Hunas invasions and later by the rise of Yasodharman in Malwa. • Most of the Gupta kings were Vaishnavas. They performed Ashvamedha sacrifi ces. • Both the Nagara and Dravidian styles of art evolved during this period. • Religious literature like Puranas were composed during this period.

Field of Mathematics, Astronomy and Astrology

• Aryabhatta was a great mathematician and astronomer who wrote the book Aryabhatiyam in 499 CE. It deals with mathematics and astronomy. Aryabhatta was the fi rst to declare that the earth was spherical and that it rotates on its axis. • Varahamihira Composed Pancha Siddhanta, the fi ve astronomical systems. • Brihadsamhita is a great work by him in Sanskrit literature. His Brihadjataka is a standard work on astrology. • Field of Medicine. • Vagbhata was the last of the great medical trio of ancient India. • He was the author of Ashtangasamgraha. • Charaka and Susruta were two great scholars that lived before the Gupta age.

Age of Harshavardhana (606—647 CE)

• Harshacharitra was written by Banabhatta (the court poet of Harsha). • Travel accounts of Hiuen Tsang, the Chinese traveller who visited India in the 7th century CE. • The dramas written by Harsha were Ratnavali, Nagananda, and Priyadarshika. • The Madhuben plate inscription and the Sonpat inscription are helpful to know the ecology of Harsha. • The Banskhera inscription contains the signature of Harsha, a larger kingdom in North India. • The founder of the family of Harsha was Pushyabhuti. • Prabhakarvardhana was the fi rst important king of the Pushyabhuti dynasty. His capital was Thaneswar, North of Delhi. • Harsha lead a campaign against the ruler of Sindh. • Hiuen Tsang converted him to Mahayana Buddhism. • A religious assembly was organized by Harsha at Kannauj to honor the Chinese pilgrim Hiuen Tsang that went on continuously for 23 days. • A brick temple of Lakshmana at Sirpur belongs to the period of Harsha. • Nalanda University was patronized by successors of Kumar Gupta and later by Harsha.

Rashtrakutas

• The dynasty was founded by Dantidurg; Krishna I built the Kailasha temple at Ellora. Amonghavarsha of Rashtrakuta dynasty wrote the fi rst Kannada Poetry Kavirajmarga. Rashtrakutas are credited for building cave shrine Elephants, dedicated to Shiva.

Gangas

• The dynasty ruled Orrissa, Narsimhadeva constructed the Sun temple at Konark; Anantvarman built the Jagannath Temple at Puri; and Kesaris who used to rule before Gangas built the Lingaraja Temple at Bhubaneshwar.

Pallavas

• The dynasty was founded by Simhavishnu and its capital was Kanchi. Narsimhavarmas founded the town of Mamallapuram (Mahabalipuram) and built rock-cut rathas and even pagodas.

31

GENERAL AWARENESS : INDIAN HISTORY

• Palas, with their capital at Monghyr is known for Dharmapala, their second king, who founded the Vikramashila University and revived the Nalanda University. • The greatest ruler of Pratiharas and bhoja (also known as Mihir, Adivaraha) • Khajuraho temple was built during the reign of Chandellas of Bundelkhand. • Chalukya of varapi-founded by Jayasimha was contemporary to Harshavardhan. • Rajputs were divided into four clans: Pratiharas (S Rajasthan), Chauhans (E Rajasthan), Chalukya/Solankis (Kathiawar), Parmaras (Malwa).

The Cholas

• The founder was Vijayalaya and the capital was Tanjore. • Aditya I wiped out the Pallavas and weakened the Pandyas. • Purantaka I captured Madurai but was defeated by the Rashtrakuta ruler Krishna III at the Battle of Takkolam. • Rajaraja I (AD 985—1014) led a naval expedition against the Shailendra empire (Malaya Peninsula) and conquered Northern Sri Lanka; constructed Rajarajeshwari (or Brihadeshvara) Shiva temple at Tanjore. • Rajendra I (AD 1014—1044) annexed the whole of Sri Lanka; took the title of Gangaikonda and founded Gangaikonda Cholapuram. • The dancing fi gure of Shiva (Nataraja) belongs to the Chola period. Local self-government existed.

S. No.

Kingdom

Capital

Founder

6.

Chandela of Jejakabhukti

Khajuraho, Mahoba, Kalinjar

Nannuk Chandela

7.

Kalchuri/Haihaya of Chedi

Tripuri

Kokkala I

8.

Gadhawal/Rathore of Kannauj

Kannauj

Chandradeva

9.

Tomar of Haryana and Delhi

Dhillika

Anangpal Singh Tomar

10.

Sisodiya of Mewar

Chittor

Bappa Rawal Hammir I

Arab Conquests of Sind (712 CE)

 Revision Notes

• The religion Islam was born at Mecca in Arabia. • The Founder was Prophet Mohammad. • He migrated to Medina in 622 CE, which was the starting point of the Muslim calendar and the Muslim era called Hijra. • After eight years, he returned to Mecca with his followers. He died in 632 CE. • The followers of Muhammad set up an empire called the Caliphate. • The Umayyad and the Abbasida were called the Caliphs. • The Arab conquest occurred in 712 CE by Muhammad-binQasim. • He was the commander of the Umayyad Kingdom. • Qasim extended the conquest further into Multan. • Qasim organised the administration of Sind. • The expansion of Muslims in India was obstructed by powerful Pratihara Kingdom in Western India.

Early Medieval Indian Period

Turkish Invasions

Topic-2

Medieval History

• After the death of Harshavardhana, there was a lack of political unity in North India and the country was split up into a number of kingdoms. • The important kingdoms in North India were Kashmir, Gandhara, Sindh, Gujarat, Kannauj, Ajmer, Malwa, Bengal and Assam.

Rajput Kingdoms

• The dominance of Rajputs began from the 7th and 8th centuries and lasted till the Muslim conquest in the 12th century. There are various theories about the origin of Rajputs—such as, they are considered as descendants of foreign invaders and of the Kshatriyas. Modern scholars agree that nearly all Rajput clans originated from peasant or pastoral communities. • Their chief occupation was trade and agriculture prospered. • They engaged in various wars to establish their supremacy over other rulers.

Rajput Kingdoms S. No.

Kingdom

Capital

Founder

1.

Chauhan/Chahaman of Delhi-Ajmer

Delhi

Vasudeva

2.

Pratihara/Parihar of Kannauj

Avanti, Kannauj

Nagabhatt I

3.

Pawar/Parmar of Malwa

Ujjain, Dhar

Seeak II ‘Sri Harsha’

4.

Chalukya/Solanki of Kathiyawar

Anihalvada

Mularaja I

5.

Rashtrakuta of Malkhand

Malkhand/ Manyakheta

Dantidurg (Danti Varman II)

• Mahmud Ghaznavi, the ruler of Ghazni have made 17 raids into India. The initial raids were directed against the Hindu rulers. • In AD 1001, the Hindu ruler Jaipal was once again defeated by Mahmud. His son Anandpala was defeated in the Battle of Waihind (AD 1008–09). • Ghaznavi plundered Somnath temple in AD 1025. • In AD 1026, Mahmud defeated the Jats. He died in AD 1030. • He patronised three persons, contemporary to him, Firdausi, Alberuni and Utbi. • Alberuni wrote Kitab-ul-Hind. • In 1173, Muhammad Ghori ascended the throne at Ghazni. • He moved towards Punjab and Ganga valley, but was defeated by Prithviraj, the ruler of Delhi in the fi rst Battle of Tarain in 1191. • In the second battle of Tarain in 1192, Prithviraj was defeated by Muhammad Ghori. • Prithviraj was allowed to rule over Ajmer for sometimes, but was executed on a charge of conspiracy later. • Prithviraj Raso was written by the Chand Bardai, the court poet of Prithviraj. It depicts the life story of Prithviraj and his love for Sanyogita. • The defeat laid the foundation of Muslim rule in India. • Later, in 1194 Jaichand of Kannauj was also defeated at Battle of Chandawar.

Delhi Sultanate

• The Muslim invasions into India had ultimately resulted in establishment of Delhi Sultanate which existed from A.D. 1206 to 1526. Five different dynasties—the Slave, the Khilji, the Tughlaq, the Sayyids and the Lodhis—ruled under the Delhi Sultanate. • The Slave dynasty was also called Mamluk (Slave) dynasty. This dynasty ruled Delhi from AD 1206 to 1290. • Three dynasties were established during this period— (i) Qutbi dynasty (1206–1210) founded by Qutbuddin Aibak. (ii) First Ilbari dynasty (1210–1266) founded by Iltutmish. (iii) Second Ilbari dynasty (1266–1290) founded by Balban.

32 • Qutbuddin Aibak was a slave of Muhammad Ghori, who made him the governor of his Indian possessions. • Aibak made Lahore his capital after the death of Ghori. • Muslim writers call Aibak ‘lakh Baksh’ or giver of lakhs because he gave liberal donations to charities. • Aibak patronized the great scholar Hasan Nizami. • Iltutmish belonged to the Ilbari tribe and hence his dynasty was named as Ilbari dynasty. • The Mongol Policy of Iltutmish saved India from the wrath of Chengiz Khan. • Minhaji-us-Siraj, Taj-ud-Din, Nizam-ul-Mulk, Muhammad Janaidi Malik, Qutb-ud-din Hasan and Fakhr-ul-Mulk-Isami were his contemporary scholars who added grandeur to his court. • Apart from completing the construction of Qutub Minar at Delhi (started by Qutbuddin Aibak), the tallest stone tower in India (238 ft.), he built a magnifi cient mosque at Ajmer. • Iltutmish nominated his daughter Raziya as his successor (First female ruler of India). • Ghiyasuddin Balban, who was known as Ulugh Khan, served as Naib or regent to Sultan Nasiruddin Mahmud (younger son of Iltutmish). • He established a separate military department Diwan-i-Arz and reorganized the army. • The advent of Khilji Dynasty (1290–1320) marked the Zenith of Muslim imperialism in India. The founder of the Khilji dynasty was Jalaluddin Khilji. • Alauddin Khilji treacherously murdered his father-in-law Jalaluddin Khilji and usurped the throne of Delhi. • In 1301, Alauddin marched against Ranthambore and after a three month’s siege it fell. The Rajput women committed jauhar or selfimmolation. • Alauddin Khalji died in 1316. • He patronized poets like Amir Khusrau and Amir Hasan. • He built famous gateway known as Alai Darwaza and constructed a new capital at Siri. • The founder of the Tughlaq dynasty was Ghiyasuddin Tughlaq. Ghiyasuddin laid the foundation of Tughlaqabad near Delhi. • Muhammad Bin Tughlaq was the only Delhi Sultan who had received a comprehensive literary, religious and philosophical education. • In 1327 he made extensive preparations for the transfer of the royal household, and the Ulema and Sufi s from Delhi to Devagiri, which was renamed as Daulatabad. • Muhammad Bin Tughlaq issued copper coins at par with the value of the silver tanka coins. • Tughlaq increased the land revenue on the farmers of GangaYamuna Doab. • The rebellion of Ahsan Shah resulted in the establishment of the Madurai Sultanate. In 1336 the Vijayanagar kingdom was founded. In 1347 Bahmani kingdom was established. • In Gujarat Taghi rose in revolt against the Sultan, who spent nearly three years chasing him. Muhammad Bin Tughlaq’s health deteriorated and he died in 1351. • Firoz Shah Tughlaq appointed Khan-i-Jahan Maqbul, a Telugu Brahmin convert as Wazir (Prime Minister). • About 30 new towns were built during his reign. The famous amoung them was Firozabad near Red Fort in Delhi, now called Firoz Shah Kotla. • Firoz patronized scholars like Barani and Afi f. • Before his departure from India in 1399, Timur appointed Khizr Khan as governor of Multan. He captured Delhi and founded the Sayyid dynasty in 1414. • Muhammad Shah died in 1445 and was succeeded by his son Alam Shah (1445–51), the weakest of Sayyid princes. • The Lodis (1451–1526), who succeeded Sayyids, were Afghans. Bahlol Lodi was the fi rst Afghan ruler while his predecessors were all Turks. He died in 1489 and succeeded by his son, Sikandar Lodi. • Sikandar Lodi (1489–1517) was the greatest of the three Lodi sovereigns.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• He destroyed many Hindu temples and imposed many restrictions on Hindus. • Sikandar Lodi was succeeded by his elder son, Ibrahim Lodi, who was arrogant ruler. • Babur marched against Delhi. He defeated and killed Ibrahim Lodi in the fi rst battle of Panipat (1526). • The Afgan kingdom lasted for only seventy-fi ve years. • The Turks introduced arches, dome, lofty towers or minarets and decorations using Arabic script. • The Quwwat-ul-Islam mosque near Qutub minar in Delhi was built by using the materials obtained from destroying many Hindu and Jain temples. • The Kotla fort at Delhi was the creation of Firoz Tughlaq. The Lodi garden in Delhi was the example of architecture of the Lodis. • Amir Khusrau introduced many new ragas such as ghora and sanam. He evolved a new style of light music known as Qawwali by blending the Hindu and Iranian Music, he is credited for the invention of Tabla and Sitar.

Bhakti Movement

• Sufi sm had its origin in Persia and spread into India in the eleventh century. • Some sufi saints: 1. Shaikh Ismail of Lahore (fi rst sufi saint) 2. Khwaja Moinuddin Chishti (Ajmer) • Bahauddin Zakariya came under the infl uence of another famous mystic, Shihabuddin Suhrawardi. • The sufi s believed service to humanity was tantamount to service to god. • The sufi s lay stress on inner purity. • In the ninth century Shankaracharya gave a new orientation to Hinduism. • Shankaracharya was born in Kaladi in Kerala. • His doctrine of Advaita or Monism was too abstract to appeal to common man. • There was a reaction against the Advaita concept of Nirguna Brahman (God without attributes) with the emergence of the idea of Saguna Brahman (God with attributes). • In the twelfth century, Ramanuja, who was born at Sriperumbudur near modern Chennai, preached Visishtadvaita. • In the thirteenth century, Madhava from Kannada region propagated Dvaita or dualism or Jivatma and Paramatma. • Nimbarka and Vallabhacharya were preachers of Vaishnavaite Bhakti in modern Telangana region. • Surdas was disciple of Vallabhacharya and he popularised Krishna cult in North India. • Mirabai was great devotee of Krishna and she became popular in Rajasthan, for her Bhajans. • Tulsidas was worshipper of Rama and composed the famous Ramcharitmanas, the Hindi version of Ramayan. • In the 14th and 15th centuries, Ramananda, Kabir and Nanak remained great apostles of Bhakti cult. • Ramananda was born at Allahabad, originally a follower of Ramanuja. He later founded his own sect and preached his principles in Hindi at Banaras and Agra. • Kabir was born near Banaras. Kabir’s object was to reconcile Hindus and Muslims and establish harmony between the two religions. • His followers are called Kabirpanthis. • Guru Nanak was founder of the Sikh religion. He was born in Talwandi near Lahore. • Chaitanya was well-known saint and reformer of Bengal who popularised the Krishna cult. • Jnanadeva was the founder of Bhakti movement in Maharashtra in the thirteenth century. He wrote a commentary on Bhagvad Gita called Dnyaneshwari. • Namadeva preached the gospel of love. He opposed idol worship and priestly domination. • Ekanatha opposed caste distinction and was sympathetic towards the lower class. • Tukaram, a contemporary of King Shivaji, was responsible for creating a background for Maratha nationalism.

33

GENERAL AWARENESS : INDIAN HISTORY

Vijayanagar and Bahamani Kingdoms Vijayanagar Empire

• The Four dynasties—Sangama, Saluva, Tuluva and Aravidu— ruled Vijayanagar from AD 1336 to 1672. • Krishnadevaraya’s Amuktamalyada, Gangadevi’s Madhuravijayam and Allasani Peddanna’s Manucharitam are some of the indegenous literature of this period. • The greatest ruler of Sangama dynasty was Deva Raya II. After his death, the Sangama dynasty became weak. • The next dynasty, Saluva dynasty was founded by Saluva Narasimha, who reigned only for a brief period (1486–1509). • The Tuluva dynasty was founded by Vira Narasimha. The greatest of the Vijayanagar rulers, Krishna Devaraya belonged to the Tuluva dynasty. • Eight eminent scholars known as Ashtadiggajas were at his royal court. Allasani Peddanna was the greatest and he was called Andhrakavita Pitamaga. His important work includes Manucharitam and Harikathasaram. • Pingali Suranna and Tenali Ramakrishna were other important scholars. • Krishna Deva Raya authored a Telugu work, Amuktamalyada and sanskrit works, Jambavati Kalyanam and Ushaparinayam. • He also built the famous Vittalaswami and Hazara Ramaswamy temples at Vijayanagar. • He also built a new city Nagalapuram in memory of his queen Nagaladevi. Besides, he built a large number of Rajagopuram. • The most important temples of the Vijayanagar style were found in the Hampi ruins. Vittalaswamy and Hazara Ramaswamy temples are the best examples of this style. • The Varadharaja and Ekambaranatha temples at Kanchipuram stand as example for the magnifi cence of the Vijayanagar style of temple architecture. • The founder of the Bahmani Kingdom was Alauddin Bahman Shah, also known as Hasan Gangu in 1347. Its capital was Gulbarga. • Ahmed Wali Shah shifted the capital from Gulbarga to Bidar. The power of the Bahmani kingdom reached its peak under the rule of Muhammad Shah III. • The success of Muhammad Shah was due to the advice and services of his minister Mahmud Gawan. • The Bahmani kingdom reached its peak under the guidance of Mahmud Gawan. He was a Persian merchant. He possessed a great knowledge of mathematics. • He made endowments to build a college at Bidar which was built in Persian style of architecture. He was also a military genius. • His conquests include Konkan, Goa and Krishnadasa—Godavari delta. • By the year 1527, the Bahmani kingdom had disintegrated into fi ve independent sultanates. They were Ahmednagar, Bijapur, Berar, Golkonda and Bidar, also known as Deccan Sultanates.

The Mughal Empire Babur (1526–1530)

• Babur was the founder of the Mughal Empire in India. Babur succeeded his father Umar Shaikh Mirza as the ruler of Farghana. • On the eve of Babur’s invasion of India, there were fi ve prominent Muslim rulers—the Sultans of Delhi, Gujarat, Malwa, Bengal and Deccan—and two prominent Hindu rulers—Rana Sangha of Mewar and Vijayanagar Empire. • Babur proclaimed himself as “Emperor of Hindustan”. • Rana Sanga of Mewar marched against Babur and in Battle of Khanwa (near Agra) held in 1527, Babur won a decisive victory over him. Babur assumed the title of Ghazi. • He was also a great scholar in Arabic and Persian languages. • He wrote his memoir, Tuzuk-i-Baburi in Turkish language.

Humayun (1530–1540)

• Humayun was the eldest son of Babur. • Humayun marched against Sher Shah and in the Battle of Chausa, held in 1539, Sher Shah destroyed the Mughal army and Humayun escaped from there.

• Humayun reached Agra to negotiate with his elder brother. But as they were not cooperative, Humayun was forced to fi ght with Sher Shah alone in the Battle of Bilgram in 1540. This battle was also known as the Battle of Kannauj. Humayun was thoroughly defeated by Sher Shah.

Sur Interregnum (1540–1555)

• The founder of the Sur dynasty was Sher Shah. • His empire consisted of the whole of North India except Assam, Nepal, Kashmir and Gujarat. • He built a new city on the banks of river Yamuna near Delhi. He also built a mausoleum at Sasaram, which is considered as one of the masterpieces of Indian architecture. • Malik Muhammad Jayasi wrote the famous Hindi work Padmavat during his reign. • In 1555, Humayun defeated Afghans and recovered the Mughal throne.

Akbar (1556–1605)

• Akbar was one of the greatest monarch of India. • In the second Battle of Panipat in 1556, Hemu (commander-inchief of Afghanistan) was almost on the point of victory. But an arrow pierced his eye and he became unconscious. His army fl ed and fortune favoured Akbar. The Mughal victory was decisive. • He conquered Northern India from Agra to Gujarat and then from Agra to Bengal. He strengthened the North West frontier. • Buland Darwaza was constructed by Akbar at Fatehpur Sikri after victory over Gujarat in AD 1572. • He was married to Harkha Bai, daughter of Rajput ruler Bharmal. • The fi rst English man, who visited Akbar’s court was Ralph Fitch in AD 1585. • He abolished Jaziyah (AD 1564). He believed in Sulh-i-kul (peace to all). • He built Ibadat Khana (Hall of Prayer) at Fatehpur Sikri; issued ‘Degree of Infallibility’ (AD 1579). He formulated religious order Din-i-Ilahi (AD 1582). Birbal was the fi rst to embrace it. • Land revenue system was called Todar Mal Bandobast or Zabti system which included measurement of land, classifi cation of land and fi xation of rent. He introduced Mansabdari system (holder of rank) to organise nobility and army. • The Navratnas of Akbar’s court were Todarmal, Abul Fazal, Faizi, Birbal, Tansen, Abdur Rahim Khan-i-Khana, Mullah-do-Pyaza, Raja Man Singh and Fakir Aziao-Din.

Jahangir (AD 1605–1627)

• He executed the fi fth Sikh guru, Guru Arjan Dev. • His greatest failure was loss of Kandahar to Persia in AD 1622. • He married Mehr-un-Nisa in AD 1611 and conferred the title of Nurjahan on her. • He established Zanjir-i-Adal at Agra fort for the seekers of royal justice. • Englishmen who visited his court were Captain Hawkins (1608) and Sir Thomas Roe (1615). • The famous painters in his court were Abdul Hassan, Ustad Mansur and Bishandas.

Shahjahan (AD 1628–1658)

• Shahjahan annexed Ahmadnagar while Bijapur and Golconda accepted his overlordship. • He secured Kandahar in AD 1639. • Two Frenchmen, Bernier and Tavernier and an Italian adventurer Manucci visited his court. • He built Moti Masjid and Taj Mahal at Agra, Jama Masjid and Red Fort at Delhi. • His reign is considered as the Golden Age of the Mughal Architecture.

Aurangzeb (Alamgir) (AD 1658–1707)

• Aurangzeb became victorious after the brutal war of succession among his brothers Dara, Shuja and Murad.

34

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• Rebellions during his rule were Jat peasantry at Mathura, Satnami peasantry in Punjab and Bundelas in Bundelkhand. • The annexation of Marwar in AD 1658 led to a serious rift between Rajput and Mughals after the death of Raja Jaswant Singh. • He executed ninth Sikh guru, Guru Tegh Bahadur in AD 1675. • Mughal conquest reached territorial climax during his reign. • He was called Darvesh or a Zinda Pir. He forbade Sati. • He conquered Bijapur (AD 1686) and Golconda (AD 1687) and reimposed Jaziya in AD 1679. • He built Biwi ka Makbara on the tomb of his queen Raba-udDurani at Aurangabad; Moti Masjid within Red Fort, Delhi; and the Jami or Badshahi Mosque at Lahore. Causes behind the fall of Mughal Empire • Weak and incompetent successors • Wars of successions. • Aurangzeb’s Deccan, Religious and Rajput policies. • Jagirdari crisis • Growth of Marathas and other regional powers. • Foreign invasions of Nadir Shah (1739) and Abdali.

Later Mughals

• Bahadur Shah I (1707–1712)–His original name was Muazzam. He was titled as Shah Alam I. • Jahandar Shah (1712–1713)–He ascended the throne with the help of Zulfi kar Khan. He abolished Jaziya. • Farrukhsiyar (1713–1719)–He lacked the ability and knowledge to rule independently. His reign was the emergence of the Sayyid Brothers. • Muhammad Shah (1719–1748)–Nadir Shah invaded India and took away Peacock throne and Kohinoor diamond. • Ahmad Shah (1748–1754)–Ahmad Shah Abdali (General of Nadir Shah) marched towards Delhi and the Mughals ceded Punjab and Multan. • Alamgir (1754–1759)–Ahmed Shah occupied Delhi. Later, Delhi was plundered by Marathas. • Shah Alam II (1759–1806)–He could not enter Delhi for two years. • Akbar II (1806–1837) was pensioner of East India Company. He gave the title ‘Raja’ to Ram Mohan Roy. • Bahadur Shah II (1837–1857)–He was the Last Mughal Emperor who was made premier during the 1857 revolt.

Literature of Mughal Period Author

Work

Babur

Tuzuk-i-Baburi

Abul Fazal

Ain-i-Akbari, Akbarnama

Jahangir

Tuzuk-i-Jahangir

Hamid Lahori

Padshahnama

Darashikoh

Majma-ul-Bahrain

Mirza Md Qasim

Alamgirnama

Marathas (AD 1674–1818) Shivaji (AD 1627–1680)

• He was born at Shivner to Shahji Bhonsle and Jijabai. His religious teacher was Samarth Ramdas and guardian was Dadaji Kondadev. • The Treaty of Purandar (AD 1665) was signed between Shivaji and Mughals. • His Coronation was held at Raigarh (AD 1674) and he assumed the title of Haindava Dharmoddharak (protector of Hinduism). • Ashtapradhan (eight ministers) helped him in administration. These were Peshwas, Sar-i-Naubat (military), Mazumdar or Amatya (accounts), Waqenavis (intelligence), Surunavis (correspondence), Dabir or Sumanta (ceremonies), Nyaydhish (Justice) and Panditrao (Religious & Charity works).

• The successors of Shivaji were Sambhaji, Rajaram and Shahu. • Shahu fought Battle of Khed in AD 1708.

Peshwas (AD 1713–1818)

• Balaji Vishwanath was fi rst Peshwa, who concluded an agreement with the Sayyid Brothers (the king makers in the history) by which Mughal Emperor Farukhsiyyar recognised Shahu as the king of Swarajya. • Baji Rao was considered as the ‘greatest exponent of guerilla tactics after Shivaji’. Maratha power reached its zenith and system of confederacy began. He defeated Siddis of Janjira; conquered of Bassein and Salsette from Portuguese. • Balaji Baji Rao was known as Nana Sahib. The third battle of Panipat in AD 1761 between Marathas and Ahmed Shah Abdali gave a big jolt to Maratha empire.

Sikh Gurus

• Guru Nanak Ji (1469–39) founded Sikh religion. • Guru Angad (1539–52) invented Gurumukhi. • Guru Amardas (1552–74) struggled against Sati system, and Purdah system and established 22 Gadiyans to propagate religion. • Guru Ramdas (1574–81) founded Swarna Mandir (Golden Temple) and composed Adi Granth, which later expanded into Guru Granth Sahib. Guru Arjan Das (1581-1606) Compiled the fi rst offi cial edition of Sikh Scripture, Adi Granth. • Guru Hargobind Singh (1606–44) established Akal Takht, and fortifi ed Amritsar. • Guru Har Rai (1644–61) provided medical care to Dara Shikoh. • Guru Har Krishan (1661–64) • Guru Tegh Bahadur (1664–75) • Guru Gobind Singh (1675–1708) was the last Guru who founded the Khalsa. After him sikh guruship ended.

Topic-3

Modern History

 Revision Notes Advent of the Europeans Portuguese

• In 1498, Vasco-da-Gama reached the port of Calicut during the reign of King Zamorin, who was the Hindu ruler of Calicut. • They had settlements in Daman, Salsette, Chaul, and Bombay (west coast) Santhome (near Madras), and at Hooghly. • The second Governor of India, Alfonso de Albuquerque arrived in 1509 and captured Goa in AD 1510. • The fi rst governor of India was Francisco de Almeida.

Dutch

• In AD 1602, the Dutch East India Company was formed. • In the Battle of Bedara, held in AD 1759, Dutch were defeated by the English. As per the agreement, the Dutch gained control over Indonesia and the British over India, Sri Lanka, and Malaya. • They set up their fi rst factory of Masulipatnam in 1605. Their other factories were Pulicat, Chinsura, Patna, Balasore, Nagapattinam, Cochin, Surat, Karaikal, and Kasimbazar.

English

• In 1599, the English East India Company was formed under a charter granted by Queen Elizabeth in 1600. Jahangir granted a Farman to Captain William Hawkings permitting the English to erect a factory at Surat (1613). • In 1615, Sir Thomas Roe succeeded in getting an imperial Farman to trade and establish a factory in all parts of the Mughal Empire by ruler Jahangir. • In 1690, Job Charnock established a factory at Sutanuti. In 1698, following the acquisition of Zamindari of three villages of Sutanuti, Kalikata, and Govindpur, the city of Calcutta was founded. Fort William was set up in 1700. • In 1717, John Surman obtained a Farman from Farrukhsiyar, which gave large concessions to the company. This Farman has been called the Magna Carta of the company.

35

GENERAL AWARENESS : INDIAN HISTORY

• In the battle of Plassey held in 1757, English defeated Sirajuddaula, the nawab of Bengal. • The battle of Buxar was held in 1764 in which captain Munro, defeated joint forces of Mir Qasim (Bengal, Shujauddaula (Awadh) and Shah Alam II (Mughal).

• The Third Maratha war (1817–18) and dissolution of Maratha confederation and creation of Bombay Presidency took place under his reign. • The Pindari war and establishment of Ryotwari system was done by Thomas Munro (1820).

Danes

Lord Amherst (AD 1823–28)

• In 1616, the Danish Eash India Company was formed. • The Danish colony ‘Tranquebar’ was established on the Southern Coromandel coast of India. • Serampur (Bengal) and Tranquebar (Tamil Nadu) sold their settlements to the English in 1845.

• The First Burmese war (1824–26), Treaty of Yandaboo (1826), and capture of Bharatpur (1826) took place under his reign.

Governor–Generals of India

Lord William Bentick (AD 1828–35)

French

• Colbert formed the French East India Company in 1664 under State Patronage. The First French factory was established at Surat by Francois Caron in 1668. A factory at Masulipatnam was set up in 1669. • The french were defeated by the English in the Battle of Wandiwash (1760).

• The Charter Act of 1833 was passed. He was made the fi rst Governor-General of India. Before him, the designation given was Governor-General of Bengal. • He carried out Social reforms like the prohibition of the Sati system (1829) and the elimination of thugs (1830). • Macaulay’s recommendations English was made the medium of higher education. He also suppressed female infanticide and child sacrifi ce.

Governor-Generals of Bengal

Lord Metcalfe (AD 1835–36)

Warren Hastings (AD 1772–85)

• He brought the dual government to an end by the Regulating Act, 1773. • The Act of 1781 made a clear demarcation between the jurisdiction of the Governor General-in-Council and Supreme Court at Calcutta. • Pitt’s India Act (1784), Rohilla War (1774), the fi rst Maratha War (1778–1782), the Treaty of Salbai with Marathas (1782) and the second Mysore war (1780–84) happened under his rule. The Asiatic Society of Bengal (1784) was founded in Calcutta by Sir William Jones. • Charles Wilkins, in 1785, translated the Bhagavad Gita into English.

Lord Cornwallis (AD 1786–93)

• The third Mysore War (1790–92) and Treaty of Seringapatnam (1792) happened during his rule. • He introduced permanent settlement in Bengal and Bihar (1793). • He is called the Father of Civil Services in India. He introduced judicial reforms by separating revenue administration from judicial administration and established a system of circles (thanas, headed by a Daroga (an Indian)). • The translation of Abhigyan Shakuntalam in English was done by William Jones in 1789.

Sir John Shore (AD 1793–98)

• He played an important role in the introduction of Permanent settlement. • The Battle of Kharda between the Nizams and the Marathas took place in 1795.

Lord Wellesley (AD 1798–1805)

• He introduced the subsidiary alliance in 1798. His fi rst alliance was with the Nizam of Hyderabad followed by Mysore, Tanjore, Awadhi, the Peshwa, the Bhonsle, and the Scindia. • The Treaty of Bassein (1802) and the second Maratha war took place under his reign.

• He was known as the ‘liberator of the press’ in India.

Lord Auckland (AD 1836–42)

• The First Afghan war (1838–42), a disaster for English happened during his rule.

Lord Ellenborough (AD 1842–44)

• He brought an end to the Afghan war. • The War with Gwalior (1843) and Annexation of Sind by Charles Napier (1843) happened during his rule.

Lord Hardinge (AD 1844–48)

• The First Anglo-Sikh war (1845–46) and Treaty of Lahore (1846) happened during his rule. • He gave preference to English educated persons in employment.

Lord Dalhouise (AD 1848–56)

• Introduction of Doctrine of Lapse and annexation of Satara (1848), Jaipur and Sambalpur (1849), Bhagat (1850), Udaipur (1852), Jhansi (1853), Nagpur (1854), and Awadh (annexed in 1856 on account of Maladministration) happened during his rule. • He laid down the fi rst railway line between Bombay and Thane (1853). Telegraph line between Calcutta and Agra and Postal reforms (fi rst issue of the Indian stamp in Karachi in 1854) with the Post Offi ce Act were introduced during his rule. • Widow Remarriage Act, 1856 (the main force being Ishwar Chand Vidyasagar) was brought under Lord Dalhousie. • He started the public works department, Grand Trunk Road work, and harbor of Karachi, Bombay, and Calcutta were developed. • Charter Act, 1853–Selection to civil service through competitive examination. • He started engineering college at Roorkee made Shimla, the summer capital of India.

Viceroys of India

Lord Canning (AD 1856–62)

Lord Minto 1 (AD 1807–13)

• He concluded the Treaty of Amritsar with Maharaja Ranjit Singh (1809). • The Charter Act of 1813 was passed under his reign.

• He was the last Governor-General and fi rst Viceroy. He withdrew Doctrine of Lapse. • Revolt of 1857 and the mutiny took place. Indian Penal Code 1860 was passed. • The government of India Act, 1858, which ended the rule of the East India Company, was passed. • The Universities of Calcutta, Bombay, and Madras were established in 1857.

Lord Hastings (AD 1813–23)

Lord Elgin (AD 1862)

George Barlow (1805–07)

• Vellore Mutiny took place on July 10, 1806.

• The Anglo Nepal war (1814–16) and Treaty of Sagauli (1816) took place under his reign.

• Wahabi Movement was suppressed in the time period of Lord Elgin.

36

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Lord John Lawrence (AD 1864–69)

Lord Minto (AD 1905–10)

Lord Mayo (AD 1869–72)

Lord Hardinge (AD 1910–16)

• He established the High Courts at Calcutta, Bombay, and Madras in 1865. • Telegraphic communication was opened with Europe. • He created the Indian Forest Department. • He organised the statistical survey of India and for the fi rst time in Indian history and census was held in 1817. • He started the process the process of fi nancial decentralization in India and established the Department of Agriculture and Commerce. • He established the Rajkot college at Kothiawar and Mayo College at Ajmer for Indian Princess. • He was the only Viceroy to be murdered in offi ce by a Pathan convict in the Andamans in 1872.

Lord Northbrooke (AD 1872–76)

• Kuka Rebellion in Punjab and Famine in Bihar happened during his tenure.

Lord Lytton (AD 1876–80)

• He was known as the ‘Viceroy of Reverse Character’. • Royal Titles Act of 1876 and the assumption of the title of ‘Empress of India’ by Queen Victoria, the Delhi Durbar in January 1877 were passed during his tenure. • Vernacular Press Act (also called the ‘Gagging Act’ to restrain the circulation of printed matter) and Arms Act (made it mandatory for Indians to acquire a license in arms) of 1878 were passed during his tenure.

Lord Ripon (AD 1880–84)

• The fi rst factory Act of 1881 prohibited child labor under the age of Local Self-Government was introduced in 1882. • Lord Ripon is regarded as ‘the founding father of local selfgovernance’ in India. • An Education commission was appointed under Sir William Hunter in 1882 to improve primary and secondary education. • The Ilbert Bill Controversy (1883) enabled Indian district magistrates to try European criminals.

Lord Dufferin (AD 1884–88)

• The Third Burmese War took place from AD 1885-86. • The Indian National Congress was established in 1885.

Lord Lansdowne (AD 1888–94)

• The Factory Act granted weekly holidays and stipulated working hours for women and children. • The Civil services were divided into Imperial, Provincial and subordinate services. • The Indian Councils Act of 1892 increased the size of Legislative councils in India. • The Durand Commission defi ned the Durand Line between British India and Afghanistan, in 1893.

Lord Elgin II (AD 1894–99)

• The Southern uprisings of 1899 were seen. The great famine of 1896–1897 and the Lyall Commission on Famine was established.

Lord Curzon (AD 1899–1905) • In 1902, a commission was appointed under Sir Thomas Raleigh to suggest reforms regards universities, the Indian Universities Act of 1904 was passed based on its recommendations was passed. • Ancient Monuments Preservation Act of 1904 was passed. Thus, the Archaeological Survey of India was established. • Agricultural Research Institute was established at Pusa in Delhi, He partitioned Bengal in 1905.

• Swadeshi Movement took place from (1905–08); the foundation of Muslim League took place in 1906. Surat Sessions and split in the Congress took place in (1907). Morley Minto Reforms were introduced in (1909). • The capital was shifted from Calcutta to Delhi (1911). • The Delhi Darbar and Partition of Bengal were cancelled in 1911. • The Hindu Mahasabha was founded in 1915 by Pandit Madan Mohan Malaviya.

Lord Chelmsford (AD 1916–21)

• In 1915, Gandhi returned to India and founded the Sabarmati Ashram (1916). Champaran Satyagraha (1917), Satyagraha at Ahmedabad (1918), Kheda Satyagraha (1918). • August declaration (1917) by Montague, and then-Secretary of State, and Montford reforms or Government of India Act of 1919. • Rowlatt Act (March 1919) and the Jallianwala Bagh Massacre took place on (13th April 1919). • The Khilafat Committee was formed and Khilafat Movement started (1919–20). • Non-cooperation movement took pace from 1920-22. • Women’s University was founded at Poona in (1916).

Lord Reading (AD 1921–26)

• Repeal of Rowlatt Act, Chauri-Chaura incident. RSS was founded in 1925. The formation of Swaraj Party happened on January 1, 1923. • Moplah Rebellion (1921) took place. The Kakori train robbery took place of August 1, 1925. • Communal Riots of 1923–25 happened in Multan, Amritsar, Delhi, etc.

Lord Irwin (AD 1926–31)

• In 1927, Simon Commission visited India. Congress passed the Indian Resolution in 1929. • The Dandi March (12th March 1930). Civil Disobedience Movement (1930). • The fi rst Round Table Conference was held in 1930 in England. Gandhi-Irwin Pact was signed. • The Lahore Session of Congress and Poorna Swaraj Declaration were held in (1929).

Lord Willington (AD 1931–36)

• The Second Round Table conference was held in 1931 and the third was held in 1932. • The Government of India Act (1935) was passed. • The Communal Awards (16th August 1932) assigned a separate electorate. Gandhiji went on an epic fast to protest against this division.

Lord Linlithgow (AD 1936–43)

• Congress Minister’s resignation celebrated as ‘Deliverance Day’ by the Muslim League (1939), the Lahore Resolution (23rd March 1940) of the Muslim League demanding a separate state for the Muslims. (It was at this session that Jinnah propounded his two-Nation Theory.) Outbreak of World War II in 1939. Cripps Mission came to India in 1942. The Quit India Movement was started on 8th August 1942.

Lord Wavell (AD 1943–47)

• The cabinet Mission plan was announced on 16th May 1946. • The First meeting of the Constituent Assembly was held on 9th December 1946. • Arranged the Shimla Conference on 25th June 1945 with the failure of talks between the Indian National Congress and Muslim League. • The Election to the Constituent Assembly were held and an interim government was appointed under Nehru.

37

GENERAL AWARENESS : INDIAN HISTORY

Lord Mountbatten (March to August 1947)

• He was the Last Viceroy of British India and the fi rst Governor– General of free India. • The partition of India was decided by the 3rd June Plan or Mountbatten Plan. • He retired in June 1948 and was succeeded by C. Rajagopalchari, the fi rst and last Indian Government – General of Free India. • The Indian Independence Act was passed by the British Parliament on 4th July, by which India became independent on 15th August 1947.

The Revolt of 1857

• The revolt of 1857 started at Meerut on 10th May 1857. • Political causes–The policy of Doctrine of Lapse, Subsidiary Alliance, Annexation of Awadh, Disrespect shown to Mughal Emperor. • Economic causes–Heavy taxation, evictions, Discriminatory Tariff Policy against Indian products, and destruction of traditional handicrafts that hit pleasants, artisans, and small zamindars. • Military discrimination–As Indian soldiers were paid low salaries, they could not rise above the rank of subedar and were racially insulted. • Grievances of Sepoys–The introduction of the Enfi eld rifl e, and its cartridge of which were greased with animal fat, provided the spark. • A rebellion broke out among Sepoys of Meerut on 10th May 1857 which later spread to other parts of the country.

Causes of Failure

• The Nizam of Hyderabad, the Raja of Jodhpur, Scindia of Gwalior, the Holkar of Indore, the rulers of Patiala, Sindh and Kashmir, and the Rana of Nepal provided active support to the British. • Comparative lack of effi cient leadership. Impact of Revolt • The control of Indian administration was passed on the British crown by the Government of India Act, 1858. • Reorganisation of the army. • After the revolt, the British pursued the policy of Divide and rule.

Chief National Activities

The Indian National Congress

• It was formed in 1885 by AO Hume, a retired civil servant. • The fi rst session was held in Bombay under the Presidentship of WC Bannerjee in 1885, attended by 72 delegates from all over India. • The fi rst two decades of INC are described in history as those of moderate demands and a sense of confi dence in British justice and generosity. • Moderate leaders were Dada Bhai Naoroji, Badruddin Tayabji, Gopal Krishna Gokhale, Surendranath Banerjee and Anand Mohan Bose.

Partition of Bengal (1905)

• The Partition of Bengal was announced by Lord Curzon on 16th October 1905 through a royal proclamation, reducing the old province of Bengal in size by creating East Bengal and Assam out of the rest of Bengal.

Swadeshi Movement (1905)

• Swadeshi movement had its origin in the anti-partition movement of Bengal. Lal, Bal, Pal, and Aurobindo Ghosh played an important role. INC took the Swadeshi call fi rst at the Banaras Session, 1905, presided over by GK Gokhale.

Muslim League (1906)

• Muslim League was set up in 1906 by Aga Khan, Nawab Salimullah of Dhaka and Nawab Mohsin ul Mulk • The league supported the Partition of Bengal and opposed the Swadeshi Movement, demanding special safeguards to its community and a separate electorate for Muslims.

• This led to communal differences between the Hindus and Muslims. • The demand for Swaraj was put forth in the Calcutta Session in December 1906. • The INC, under the leadership of Dadabhai Naoroji, adopted ‘swaraj’ (self-government) as the goal of the Indian people.

Surat Session (1907)

• The INC split into two groups: the Extremists and the Moderates, due to the debate on the nature of Swadeshi Movement. • Extremists were led by Lal, Bal, Pal, while the moderates by GK Gokhale.

Morley-Minto Reforms (1909)

• The reforms envisaged a separate electorate for Muslims, besides other constitutional measures. • Lord Minto came to be known as the ‘Father of Communal Electorate’.

Ghadar Party (1913)

• It was formed by Lala Hardayal, Taraknath Das and Sohansingh Bakhna. It was headquartered in San Francisco. • The name was taken from a weekly paper, Ghadar, which had been started on 1st November 1913 to commemorate the 1857 revolt.

Home Rule Movement (1916)

• It was started by BG Tilak (April 1916) at Poona and Annie Besant and S. Subramania Iyer at Adyar, near Madras (September 1916) • The objective was self-government of India in the British Empire. • During this movement, Tilak raised the Slogan ‘Swaraj is my Birth Right and I shall have it!’

Lucknow Pact (1916)

• This was a pact between INC and Muslim League following a war between Britain and Turkey leading to Anti-British feelings among Muslims. Both organizations jointly demanded dominion status for the country congress accepted a separate electorate for Muslims.

August Declaration (1917)

• After the Lucknow Pact, the British policy was announced which aimed at increasing associations of Indians in every branch of the administration for the progressive realization of responsible government in India as an integral part of the British Empire. This came to be called the August Declaration. • The Montague-Chelmsford reforms or the Act of 1919 were based on this declaration.

Rowlatt Act (18th March 1919)

• This gave unbridled powers to the government to arrest and imprison suspects without trial. This law enabled the government to suspend the right of Habeas Corpus, which had been the foundation of civil liberties in Britain. • Rowlatt Satyagraha was started against the act. This was the fi rst countrywide agitation by Gandhiji.

Jallianwala Bagh Massarce (13th April 1919)

• People were agitated over the arrest of Dr. Saifuddin Kitchlew and Dr. Satya Pal on 10th April 1919. • General Dyer fi red at people who assembled in the Jallianwala Bagh, Amritsar. Michael O’Dwyer was Lt. Governor of Punjab at that time. The Hunter Commission was appointed to enquire into it. • Rabindranath Tagore returned his knighthood in protest. • It was Sardar Udham Singh killed Michael O’Dwyer in Caxton Hall, London on March 13, 1940.

Khilafat Movement (1920)

• This movement was started by the Ali brothers, Mohd. Ali and Shaukat Ali. It was jointly led by the khilafat leaders and the Congress.

38 • The reason to led this movement was, the Muslims were alighted by the treatment with Turkey by the British in the treaty that followed the First World War.

Non-cooperation Movement (1920)

• In 1920 Congress adopted the new programme of non-violent Non-cooperation under the leadership of Gandhiji. • The Non-cooperation movement aimed to readdress the wrongs done to Punjab and Turkey. Another aim was to reattain Swaraj. • Many Indians resigned their job, foreign clothes were burnt, strikes all over the country. • By the beginning of 1922 about 30,000 persons were in Jail. • Early in February 1922 Gandhiji decided to launch a no-tax campaign in the Bardoli district in Gujarat. • However in Chauri-Chaura people turned violent and set fi re to a police station causing the death of 22 policemen. • When the news reached Gandhiji, he decided to call off the Noncooperation movement.

Chauri-Chaura Incident (1922)

• The Congress session at Allahabad in December 1921, decided to launch a Civil Disobedience Programme, Gandhiji was appointed as its leader. • Before it could be launched, a multitude of people at ChauriChaura (near Gorakhpur) clashed with the police and burnt 22 policemen on 5th February 1922. This compelled Gandhiji to withdraw the Non-cooperation Movement on 12th February 1922.

Civil Disobedience Movement

• There was mass participation by women country-wide. • The Garhwal soldiers refused to fi re on the people at Peshwar. • The Civil Disobedience Movement began with Dandi March. • In May 1934 the entire Civil Disobedience Movement was completely called off.

Dandi March (1930)

• It was also called the Salt Satyagraha. • Gandhiji started his march from Sabarmati Ashram on 12th March 1930 for the small village Dandi to break the Salt Law.

Swaraj Party (1923)

• The founder of the Swaraj party were Motilal Nehru, CR Das, and NC Kulkarni. They demanded that the nationalist should end the boycott of the Legislative councils enter them and expose them. • CR Das was the President of the Swaraj Party.

Simon Commission (1927)

• In November 1927 the British Government appointed the Simon Commission to look into the working of the Government of India Act of 1919 and to suggest changes. • The commission consisted of Englishmen without a single Indian representative. • The commission arrived in India in February 1928 and was met with country-wide protests. • Majority of members of the Central Legislative Assembly boycotted the Commission. • Anti-Simon Committees were formed all over the country to organize demonstrations and hartals wherever the commission went. • Peaceful demonstrators were beaten by the Police at many places. Lala Lajpat Rai was assaulted and soon after died.

The Nehru Report (1928)

• After boycotting the Simon Commission, all political parties constituted a committee under the Chairmanship of Motilal Nehru to evolve and determine the principles for the constitution of India.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Lahore Session (1929)

• Under the Presidentship, of JL Nehru, the INC declared Poorna Swaraj (complete independence) as its ultimate goal, on 19th December, at the Lahore session. • The tri-colored fl ag adopted on 31st December 1929, was unfurled and on 26th January 1930 was fi xed as the First Independence Day, to be celebrated every year. • Later 26th January was chosen as India’s Republic Day.

First Round Table Conference (1931)

• The First Round Table Conference was held on 12th November 1930 in London to discuss Simon Commission. • It was the fi rst conference arranged between the British and Indians as equals. • Hindu Mahasabha and Muslim League participated in it. • The conference failed due to the absence of the Indian National Congress.

Second Round Table Conference (1931)

• The Second Round Table was held in London from 7th September to 1st December 1931. • Gandhi and Indian National Congress participated in it. • Civil unrest had spread throughout India again, and upon return to India, Gandhi was arrested along with other Congress leaders.

Third Round Table Conference (1932)

• It proved fruitless as most of the national leaders were in the prison.

Gandhi-Irwin Pact (1931)

• In 1931 Gandhiji and some other leaders were released from Jail. • In March an agreement known as Gandhi-Irwin Pact was signed under which the Civil Disobedience Movement was called off. • The government promised to release all the political prisoners except those charged with acts of violence. • The Congress agreed to participate in the Second Round Table Conference which had been called to consider a scheme for a new constitution for India.

The Communal Award (16th August 1932)

• It was announced by Ramsay McDonald. It showed the Divide and Rule policy of the British. • It envisaged communal representation of depressed classes, Sikhs and Muslims. • Gandhiji opposed it and started fast unto death in Yerwada jail Pune (Maharashtra).

Poona Pact/Gandhi-Ambedkar Pact (25th September 1932) • The idea of a separate electorate for the depressed classes was abandoned, but seats reserved for them in the provincial Legislative were increased. • Thus Poona Pact agreed upon a joint electorate for upper and lower classes.

Demand for Pakistan

• In 1930, it was suggested by Iqbal that the North-West provinces and Kashmir should be made Muslim states within the federation. • The term ‘Pakistan’ was given by Chaudhary Rehmat Ali in 1933. • Muslim league fi rst passed the proposal of separate Pakistan in its Lahore session in 1940. It was known as Jinnah’s Two-Nation Theory. • It was drafted by Sikandar Hayat Khan, moved by Fazul Haq, and the second by Khaliquzzamah. • In December 1943, the Karachi session of the Muslim League adopted the Slogan Divide and Quit.

August Offer (8th August 1940)

• It offered—(i) Dominion status in the unspecifi ed future, (ii) A post-war body to enact the constitution. (iii) To expand to

39

GENERAL AWARENESS : INDIAN HISTORY

Governor General’s Executive Council to give full weightage to the minority opinion. • INC rejected this but was accepted by the Muslim league.

Important Facts Indus Valley Sites Site

The Cripps Mission (1942)

River

• In March 1942, Sir Stafford Cripps came to India to hold talks with the Indian Leaders. • However, the talks broke down as the British were not willing to promise independence even after the II world war was over and rejected the Congress proposal for the formation of a national government during the war. • After the failure of talks with Cripps, the Congress prepared to launch the third mass movement against British rule.

Harappa

Ravi

Mohenjodaro

Indus

Chanhudaro

Indus

Lothal

Bhogava

Kalibangam

Ghaggar

Dholavira

Lumi

Indian National Army (INA)

Banavali

Saraswati

• Subash Chandra Bose escaped to Berlin in 1941 and set up the Indian League there. • In July 1943, he joined the INA in Singapore. Rash Behari Bose handed over the leadership to him. • INA had three fi ghting brigades, named after Gandhi, Azad, and Nehru. • Rani of Jhansi Brigade was an exclusive woman force - INA had its headquarters at Rangoon and Singapore.

Causes behind the fall of Mughal Empire

The Cabinet Mission Plan (1946)

Some Important Religious Institution and Founder

• • • • • •

Weak and incompetent successors Wars of succession Aurangzeb’s Deccan, religious and Rajput policies Jagirdari crisis Growth of Marathas and other regional powers. Foreign invasions of Nadir Shah (1739) and Abdali

• The Members of Cabinet Mission Plan were Pethick Lawrence, Stafford Cripps and AV Alexander, Lord Wavell was the viceroy of India at that time. • The main proposals were— • Rejection of demand for a full-fl edged Pakistan. • Loose Union under a center with the Center’s control over the defense and foreign affairs. • Provinces were to have full autonomy and residual powers. • Provincial legislatures would elect a Constituent Assembly. • The Muslim league accepted it on 6th June 1946. Congress also partially accepted this plan.

Brahmo Samaj (1828)

Raja Ram Mohan Roy

Young Bengal Movement (1826–31)

H. L. Vivian Derozio

Tattvabodhisini Sabha (1839)

Debendranath Tagore

Wahabi Movement Rohilakhand

Syed Ahmed of Rai

(1820)

Bareilly

Prarthana Samaj Bombay (1867)

Atmaram Pandurang

Arya Samaj (1875) Bombay

Dayanand Saraswati

Aligarh Movement (1875)

Syed Ahmed Khan

Ram Krishna Mission (1897) Belur

Vivekananda

Formation of Constituent Assembly (December 1946)

Servants of Indian Society (1905)

Gopal Krishna Gokhale

• The Constituent Assembly was formed on 9th December 1946. • Dr. Rajendra Prasad was elected as its President.

Mountbatten Plan (also called 3rd June plan) (3rd June 1947)

Some Important Movements Movement

Leaders

Pagal Panthi (1825–1835)

Karam Shah, Tipu Shah

The plan formulated by Lord Mountbatten outlined that: • India was to be further divided into India and Pakistan. • There would be a separate constitutional Assembly for Pakistan to frame its constitution. • The Princely states would enjoy the liberty to either join India or Pakistan or could even remain independent. • Bengal and Punjab will be partitioned and a referendum in NWFP and Sylhet district of Assam would be held. • A separate state of Pakistan would be created. Boundary Commission was to be headed by Radcliffe.

Moplah Rebellion (1921)

Sayyid Ali, Sayyid Fazi

Indigo Revot (1860)

Digambar Biswas, Bishnu Biswas, Harish Chandra Mukherjee

Poona Sarvajanik Sabha (1870)

By MG Ranade

Eka Movement (1921)

Madari Pasi

Champaran Satyagraha (1917)

Gandhiji, Dr. Rajendra Prasad

Kheda Satyagraha (1918)

Gandhiji

Partition and Independence

All India Kisan Sabha (1918)

Indira Narain Dwivedi, Madan Mohan Malviya

Awadh Kisan Sabha (1920)

JL Nehru, Baba Ram Chandra

• India Independence Act, 1947 was implemented on 15th August 1947. • It abolished the sovereignty of British Parliament. Dominions of India and Pakistan were created. • Each dominion was to have a Governor-General. Pakistan was to comprise Sindh, British Baluchistan, NWFP, West Bengal, and East Bengal. • The fi rst Home Minister, Sardar Vallabhai Patel integrated all the states by 15th August 1947. • Kashmir, Hyderabad, Junagarh, Goa (with Portuguese), and Pondicherry (with French) later acceded to Indian Federation.

Andhra Ryots Association (1928) NG Ranga All India Kisan Sabha (1936)

Swami Sahajananda

Bijolia Movement (1905, 1913, 1916, 1927)

Sitaram Das, Vijay Pathik Singh

Telangana Movement (1946)

Communists

40

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions 1. Who amongst the following was shocked by Chauri-Chaura incident which took place in Gorakhpur and withdrew Non-Cooperation movement on 5th Feb 1922? (2023) (1) Mohandas Karamchand Gandhi (2) Sardar Vallabhbhai Patel (3) Motilal Nehru (4) Lala Lajpat Rai 2. One of the Harappan site known as Kalibangan is located in which state of India? (2023) (1) Rajasthan (2) Gujarat (3) Madhya Pradesh (4) Punjab 3. Which of the following country shares the shortest border with India? (2023) (1) Bhutan (2) Myanmar (3) Nepal (4) Pakistan 4. Who discovered ‘Harappa Civilization’ in 1921? (2023) (1) B. D. Banerjee (2) R. D. Banerjee (3) John Marshal (4) Dayaram Sahni 5. Which among the following are considered as Vedangas? (2022) (a) Kalp (b) Ayurveda (c) Vyakaran (d) Manu Smriti (e) Jyotish Choose the correct answer from the options given below: (1) (a), (b) and (d) only (2) (b), (d) and (e) only (3) (a), (c) and (e) only (4) (a), (c) and (d) only 6. The British Cabinet Mission which came to India in March 1946 did not have as its member: (2021) (1) Lord Pethick-Lawrence (2) Sir Stafford Cripps (3) A.V. Alexander (4) Campbell H. Johnson 7. What was the main agenda of the Swaraj Party? (2021) A. Council entry B. Politics of obstruction from within C. Constructive rural word (1) A only (2) C only (3) A and B only (4) A, B and C 8. Who is regarded as India’s fi rst labour leader much before the beginning of Trade Union Movement? (2021) (1) Sorabjee Shapoorjee Bengalee (2) Narayan Meghaji Lokhande (3) Lala Lajpat Rai (4) B.P. Wadia 9. The First Health Minister of Independent India was _____. (1) Maulana Abdul Kalam Azad (2) Sardar Vallabhbhai Patel (3) Vijayalakshmi Pandit (4) Rajkumari Amrit Kaur 10. In which year was the Non-Cooperation Movement launched? (1) 1877 (2) 1920 (3) 1856 (4) 1919 11. Chauri Chaura incident took place in? (1) 1902 (2) 1912 (3) 1922 (4) 1932 12. From whom did the French East India Company acquire Pondicherry? (1) From the Portuguese (2) The ruler of Golconda (3) Sultan of Bijapur (4) Snatching 13. In which of the following years sea travel has been made mandatory for Indian soldiers? (1) 1854 AD (2) 1856 AD (3) 1857 AD (4) 1858 AD 14. Where did the leader of the Individual Satyagraha movement, Acharya Vinoba Bhave started this movement? (1) Nashik (2) Poona (3) Pochampali (4) Nagpur 15. Which emperor’s army did Maharana Pratap fi ght in the Battle of Haldighati? (1) Mahmud Shah (2) Humayun (3) Ibrahim Lodi (4) Akbar

1. (1) 9. (4)

Answer Key 2. (1) 3. (1) 4. (4) 5. (3) 6. (4) 7. (3) 8. (2) 10. (2) 11. (3) 12. (3) 13. (2) 14. (3) 15. (4)

Answers with Explanations 1. Option (1) is correct. Mohandas Karamchand Gandhi was shocked by the Chauri Chaura incident and withdrew from the Non-Cooperation movement. After the Chauri Chara incident, Mahatma Gandhi called off the Non-Cooperation movement. Chauri Chaura Incident took place on 4th February 1922 at Chauri Chaura in the Gorakhpur district of Uttar Pradesh. The police fi red on the protestors participating in the Non-Cooperation movement and in retaliation the protesters had set the police station on fi re. This led to the halt of the Non-Cooperation movement on a national level. 2. Option (1) is correct. Kalibangan is an ancient site of Indus valley civilization and is situated in Hanumangarh, Rajasthan. The site had remains of both pre-Harappan and post-Harappan remains which included copper and produced pottery, and use of baked bricks etc. It is also the site of the earliest recorded earthquake in 2600 BC. The seven most important sites of Indus Valley Civilization at-Mohenjo Daro, Harappa, Kalibangan, Lothal, Chanhudaru, Dholavira, and Banawali. 3. Option (1) is correct. Of the given options, India shares the shortest border with Bhutan. Although, India shares the shortest border with Afghanistan of 106 km, the longest border with Bangladesh of 4096.7 km. Bordering Countries

States

Length of the border

Bhutan

West Bengal, Sikkim, Aruna- 699 km chal Pradesh and Assam

Myanmar

Arunachal Pradesh, Manipur, 1643 km Mizoram, and Nagaland

Nepal

Sikkim, West Bengal, Uttar 1751 km Pradesh and Uttrakhand

Pakistan

Jammu & Kashmir, Punjab, 3323 km Rajasthan, and Gujarat 4. Option (4) is correct. Harappa was discovered by archeologist Dayaram Sahni in 1921. Harappa civilisation is also known as the Indus Valley Civilization. It is one of the oldest civilisations of the world and is believed to have existed from approximately 3300 BCE to 1300 BCE. Some of the famous cities of Harappa civilisation are Harappa, Lothal, Dholavira, Mohenjodaro, and Kalibangan and these were found near the Indus River in the Sindh (Sind) region. The civilisation is known for developing the fi rst accurate system of standardised measures and weights. 5. Option (3) is correct. Vedangas are six auxiliary disciplines associated with the study and understanding of the Vedas. These are Shiksha, Chhandas, Vyakarana, Nirukta, Kalpa and Jyotisha. The Vedangas are the last treatises of the Vedic Literature. 6. Option (4) is correct. Cabinet Mission was sent in February 1946 to India by the Attlee Government (British Prime Minister). It had three British cabinet members – Pethick Lawrence, Stafford Cripps, & and A.V. Alexander. Its aim was to discuss the transfer of power from British to Indian leadership.

GENERAL AWARENESS : INDIAN HISTORY

Another person, Lord Wavell was not a member of the Cabinet Mission but was involved in it. Its main objectives were to formulate a constitution-making body, to establish an Executive Council, and to make an agreement with the Indian leaders for framing of a constitution for India. 7. Option (3) is correct. The Swaraj Party or the Congress-Khilafat Swarajya Party was formed by C R Das and Motilal Nehru on 1st January 1923. Mahatma Gandhi withdrew the Non-cooperation Movement (because of the Chauri Chaura incident) in 1922. It created a lot of disagreements among leaders of the Congress Party. Some wanted to continue non-cooperation, others wanted to end the legislature boycott and contest elections. The former were called no-changers and such leaders included Rajendra Prasad, Sardar Vallabhai Patel, C Rajagopalachari, etc. The others who wanted to enter the legislative council and obstruct the British government from within were called the pro-changers. These leaders included C R Das, Motilal Nehru, Srinivasa Iyengar, etc. 8. Option (2) is correct. Narayan Meghaji Lokhande was the fi rst leader to organize labour movement in India. He is considered the father of trade union movement in India. The trade union movement in India started with the forming of Bombay Mill Hands Association in 1890 by N.K. Lokhands. The fi rst clearly registered trade-union is considered to be the Madras Labour Union founded by B.P. Wadia in 1918. Lala Lajpat Rai was an freedom fi ghter, and politician; and was popularly known as Punjab Kesari. In 1878, Sorabjee Shapoorji Bengalee drafted a bill for providing better working conditions to the labourers and tried to pass in the Bombay Legislative Council. 9. Option (4) is correct. Rajkumari Amrit Kaur existed an Indian activist and politician. She was selected the foremost Health Minister of India in 1947 and stayed in seat until 1957.

41 10. Option (2) is correct. The Non-cooperation Movement founded by Mahatma Gandhi on August 1, 1920, lived the fi rst gathering activity organized national during India’s struggle for liberation. Nagpur session in December 1920 described the non-Cooperation agenda. If Non-cooperation existed virtually dragged release, articulated Gandhiji, India would defeat Swaraj within a year. 11. Option (3) is correct. The Chauri Chaura incident accepted position on 4th February 1922 in Chauri Chaura in the Gorakhpur community of the British Union Province of India. 12. Option (3) is correct. The French East India Company acquire Pondicherry from the Sultan of Bijapur. French came to India in 1664 and initiated their trading movements. At the identical duration, they were again discovering further forms to Develop in India. Pondicherry was made up of four territories like Yanam, Male, Karaikal, and Pondicherry. 13. Option (2) is correct. In 1856, The East India Company expired a recent rule in which ocean journey retains been created compulsory for Indian fi ghters. In those period, numerous individuals in the government acknowledged that if they struck the sea they would fail their belief and caste. 14. Option (3) is correct. Behind the defeat of August’s request, Gandhi selected to create a unreasonable satyagraha on an particular cause by a periodic designated someone in every locality. Vinoba Bhave was the fi rst to offer the satyagraha and Nehru, the second. In Pochampali the leader of the individual Satyagraha movement, Acharya Vinoba Bhave started this movement. 15. Option (4) is correct. The Battle of Haldighati was fought in the year 1576 AD. The Battle was fought between Rana Pratap Singh, the Hindu Rajput ruler of Mewar in Rajasthan and Raja Man Singh of Amber, the general of the Mughal Emperor Akbar.

Deforms

Rocks

Atmosphere

Jetstreams

Oceans

Solar System Sun

Theories of Plates of Earth

Earthquakes

Earth

Earth at Glance

Longitude

Latitude

First Level

Second Level

Trace the Mind Map

Third Level



Classification of Planets

Meteors

Structure of Earth’s InteriorLayers of Earth

Asteroids

World Geography

Volcanism

Continents

Universe-facts about universe

42 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Railways Soils in India

Agriculture

Climate

Physical Features

Indian Geography

Mineral Resources

Air Transportation

Water Transport

Transport

Road Transport

India at a Glance Imp.facts

Himalayan River System

Second Level

Trace the Mind Map

Peninsular River System

First Level

Drainage

Islands

Coastal Plains

Great Plains

Peninsular Plateau

Himalaya

Third Level



GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

43

World and Indian Geography

Chapter

2

Chapter Analysis Concept Name World & Indian Geography

Topic-1 World Geography  Revision Notes Universe

Scan to know

Sirius The Brightest star outside solar system is Sirius. It is also called Dog star. Proxima Centauri The closest star to our solar system is Proxima Centauri (4.2 light years away) followed by Alpha Centauri (4.3 light years away) and Barnard’s star (5.9 light years away).

Description It is the distance that light can travel in one year at the speed of 3 × 108 m/s. It is equal to almost 9.46 × 1012 km.

Astronomical It is the average distance between the Sun and unit the Earth. 1 AU = 150 million km. Parsec

2023

Additional Questions

1

4

8

Biggest Planet Biggest Satellite

Jupiter Ganymede

Blue Planet Green Planet Brightest Planet Brightest star outside solar system Closest Star to solar system Coldest Planet Evening Star Farthest Planet from Sun Planet with maximum number of satellites

(Jupiter) Earth Uranus Venus Sirius (Dog star) Proxima Centauri Neptune Venus Neptune Saturn (Overtook

Fastest revolution in solar system Hottest Planet Densest Planet Fastest rotation in solar system Morning Star Nearest Planet to Earth Nearest Planet of Sun Red Planet Slowest Revolution in solar system Slowest Rotation in solar system Smallest Planet Smallest Satellite Earth’s Twin Only Satellite with an atmosphere like Earth

Jupiter) Mercury Venus Earth Jupiter Venus Venus Mercury Mars Neptune Venus Mercury Deimos (Mars) Venus Titan

Solar System

• The solar system consists of the Sun, eight planets, satellites and thousands of heavenly bodies such as comets, asteroids and meteors. • The centre of the solar system is the Sun. All heavenly bodies revolves around it. • Sun is the nearest star to the Earth.

Asteroids

• Asteroids are also called ‘Planetoids’ and ‘Minor Planets’. • These are the small planetary bodies that revolve around the Sun and are found between the orbits of Mars and Jupiter.

Meteors and Meteorites

Units of Distance Light year

2022

2

Important Facts about Universe

•  Cosmology is the study of universe. more about this topic Universe is all of space and time and their contents including planets, stars, galaxies and all other forms of matter and energy. • A galaxy is a vast system of billions of stars, dust and light gases bounded by their own World gravity. There are 100 billion galaxies in the Geography universe, with each galaxy having, 100 billion stars, on an average. • Our galaxy is Milky Way (or the Akash Ganga). • The nearest galaxy to the Milky Way is Andromeda. • Big Bang was an explosion of concentrated matter in the universe. It occurred around 15 billion years ago. It led to formation of galaxies, stars and other heavenly bodies. • It is believed that universe should be filled with radiation called the ‘Cosmic Microwave Background’. •  Cosmic Background Explorer (COBE) and the Wilkinson Microwave Anistropy Probe (WMAP) are two missions launched by NASA to study these radiations. • Stars are heavenly bodies. These are made up of hot burning gases. They shine by emitting their own lights. • Some stars have very high gravitational power as they have three times greater mass than that of sun so that even light cannot escape from its gravity and hence are called Black Holes. • Comets are made up of frozen gases. They move around the sun in elongated elliptical orbit with the tail, always pointing away from the sun. •  The sky is divided into units and these units enable the astronomers to identify the position of stars. These units are called constellations. There are 88 known constellations. • The heavenly bodies that revolve around the planets are called satellites. The Moon is the natural satellite of Earth.

Unit

2021

It is the distance from the Earth to a star that has parallax of I arc second. The length is about 3.262 light years.

• Meteors are fragments of rocks, coming towards the Earth. • These are also known as ‘shooting stars’. • These are formed due to collision among the asteroids. • Meteorites are the meteors that do not burnt completely in Earth’s atmosphere and they land on the Earth. • Meteorites are composed of nickel – iron alloy (10% nickel and 90% iron) and silicate minerals, in various proportions.

Classification of Planets

• Inner Planets – Mercury, Venus, Earth and Mars. • Outer Planets – Jupiter, Saturn, Uranus and Neptune.

45

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

Inner Planets

Outer Planets

They are called as Terrestrial They are called as Jovian or or Rocky planets Gaseous Planets They are nearer to the sun

They are far away from the sun.

• Dwarf Planet – According to International Astronomical Union (IAU), it is a celestial body in direct orbit of the sun, that is massive enough that its shape is controlled by gravitational forces, but has not cleared its neighbourhood. For example – Pluto, Ceres, Makemake, Haumea, Eris.

Components of Solar System

• The Sun, eight planets (excluding Pluto) and their respective satellites. • Five Dwarf planets namely Ceres, Pluto, Eris, Makemake and Haumea. • Interstellar debris like asteroids, meteoroids, comets. • The electrically charged gases are known as Plasma. • Interplanetary dust particles known as cosmic dust which pervade between planets in the solar system. • Small Solar system bodies except the planets, dwarf planets and satellites.

The Sun

• The Sun accounts for more than 99% mass of the solar system. Due to this, the sun exerts immense gravitational pull on the planets that keep them rotating around it in defi nite elliptical orbit. • Photosphere is the glowing surface of the sun that we see, above the photosphere is the red coloured Chromosphere and beyond it is the magnifi cent Corona, which is most easily visible during eclipses. • The major source of energy of the solar system is the Sun. The energy is provided by nuclear fusion reaction, that converts hydrogen into helium in the core of the sun. • Superimposed on sun’s white light are hundred of dark lines called Fraunhofer lines. Each line indicates the elements which exist as gases in the Sun’s atmosphere. Facts about the Sun • The centre of the solar system is the sun. • It has a surface temperature of about 6000°C. • It is one of the stars in the Milky Way galaxy. • It takes 224 million years to complete one circle (revolution) around the center of Milky Way. This is called Cosmic Year. • The Sun is 1300000 times bigger than the Earth in terms of volume. • The sunlight takes 8 minutes 16.6 second to reach to the Earth.

• Average distance from the Earth—149600000 km. • Diameter—1391980 km. • Temperature of the core—15000000˚C • Rotational speed—25.38 days (with respect to equator) 33 days (with respect to poles).

Earth

• Earth is almost spherical, fl attened a little at the poles with slight bulge at the centre (equator), hence it is an oblate spheroid. • Nearest position of the Earth to the Sun is ‘Perihelion’. • Farthest position of the Earth to the Sun is ‘Aphelion’. • The interior of Earth is composed of three layers the Crust, the Mantle and the Core. • The interior of Earth is explained by Eduard Suess on the basis of chemical composition as SiAl, SiMa and NiFe. • The upper part of the crust is SiAl (Silicon-Aluminium) • The lower part of the crust is SiMa (Silicon-Magnesium). • The outer part of the core is NiFe (Nickel-Iron) • The spinning of Earth on its imaginary axis from West to East in one day is called Rotation of the Earth. It results in causation of day and night, tides. • The motion of Earth in elliptical orbit around the Sun in one year is called Revolution of the Earth. It results in the change of seasons.

Latitudes

• Latitudes are the imaginary lines drawn on the Earth’s surface, parallel to the equator. Equator (0°) is the biggest latitude that divides Earth in two equal hemispheres (North and South). Tropic of Cancer

23.5½° N

Tropic of Capricon

23.5½° S

Arctic Circle

66.5½° N

Antarctic Circle

66.5½° S

• The distance between each degree of latitude equals to 111 km. • Equator is the most important latitude.

Longitudes (Meridians)

• A series of semicircle that run from pole to pole passing through the equator are the Meridians. • Prime Meridian passes through Greenwich near London, divides the Earth in Eastern and Western hemisphere. Its value is 0°. • Longitude has very important function i.e., it determines local time in relation to Greenwich Mean Time (GMT). • 1° change of longitude corresponds to 4 minutes difference in time.

A Comparative Study of the Planets of the Solar System Planet

Special Characteristics

Mercury

It is the smallest and innermost planet. It has no atmosphere. It has a cratered surface, much like the moon.

Venus

It is also called as the veiled planet as it is surrounded by thick clouds. It is known as Evening and Morning star. It is the brightest planet of the Solar System because of almost 70% albedo. It contains 90-95% CO2. The night and day temperature are almost the same. The Earth is neither too hot nor too cold. It is called as the Blue Planet, as 71% of the surface is covered with water. It is known as Red Planet. It has a thin atmosphere comprising of Nitrogen and Argon.

Earth Mars

Rotation and Revolution time Rotation: 58.65 days. Revolution: 88 days (Fastest revolution in the solar system). It has the slowest rotational speed. It has almost equal rotation and revolution. Rotation: (clockwise) 243 days and Revolution: 224.7 days. Rotation: 24 hours. Revolution: 365 days and six hours. Rotation: 24.6 hours (almost equal to earth) Revolution : 687 days.

Important Physical Satellite System Properties It has the maximum diurnal No Satellite. range of temperature. Rotates from East to West No Satellite. unlike the other planets. It is the hottest planet.

It is the densest of all and is unique for the presence of higher forms of life. It is marked by dormant volcanoes. Olympus Mons is the highest mountain, which is three times of mount Everest.

Moon is the only natural satellite. Two satellites: Phobos and Deimos

46 Jupiter

Saturn

Uranus

Neptune

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

It is the largest planet. It has a mass 2.5 times greater than the combined mass of all the remaining plants, satellites and asteroids put together. It contains Hydrogen, Helium, Methane and Ammonia. A great red spot is detected on it. It is the second largest planet and it is surrounded by a set of eight rings, which are made of primordial dust and ice particles. It is unique as its axis of rotation is inclined at 98° to its orbital plane.

Fastest rotational velocity It is too massive to solidify as a (9.8 hours) planet, but not massive enough Revolution: 11.8 years. to develop nuclear fusion and became a star. It gives off more energy than it receives from the Sun, because of the heat inside. Rotation: 10.3 hours Revolution: 20.5 years.

Unlike others, which spins on their axis, Uranus rolls, apparently from North to South. Rotation: 17.2 hours Revolution: 84 years. It is the farthest planet from the Sun. Rotation: 16 hours and It has a dynamic atmosphere, which Revolution 165 years. contains an Earth sized spot called the Great Dark Spot that is similar to the Jupiter’s Great Red Spot.

Statistics Data of the Earth

It has the least density of all the It has 82 Satellites planets. the largest being 30 percent less denser than water. Titan. Surrounded by a system of It has 27 satellites. faints rings. The prominent are: Miranda, Ariel, etc. It has 5 faint rings. It appears vivid blue in colour.

5.975 5 × 2024

Volume

1083 × 1012 km3

Total Surface Area

510 million sq km

Rotation Speed

23 hr, 56 min and 4.100 s

• Selenology is the study of moon. • It is known as fossil planet. • It is the natural Satellite of the Earth.

Revolution Speed

365 days, 5 h and 45.51 s

Moon at a Glance

Longest day

21st June, (Summer Solstice) Sun is vertically overhead at Tropic of Cancer

Shortest day

22nd December, (Winter Solstice) Sun is vertically overhead at Tropic of Capricorn

Escape velocity

112 km/s

Mean surface temperature 14°C

Rotation of Earth

• The Earth has an imaginary axis, on which it spins from west to East in 23 hrs 56 min and 4.09 second. • The maximum speed of rotation is at equator (1667 km/hr). It decreases towards poles, where it is zero. The implications of rotation of Earth are: 1. Causation of day and night 2. Rise and fall of tides everyday 3. Change in direction of winds and ocean currents. 4. A difference of one hour between two meridians, which are 15° apart.

Revolution of Earth

• The motion of Earth in elliptical orbit around the Sun is revolution. • It takes 365 days, 5 hours, 48 minutes and 45.51 seconds. • It leads to one extra day in every fourth year.

The Revolution results in –

1. Change in season 2. Variation of length of days and nights at different times of year. 3. Shifting of the wind belts.

Equinox

• Under this situation, when days and nights are equal, the Sun is vertically overhead at the equator. It happens on two days of

It has 14 satellites. The prominent are Triton and Nereid.

the year i.e., 21st March (Vernal Equinox) and 23rd September (Autumnal Equinox).

Mass

Dates when days and March 21 (Vernal Equinox) 23rd September (Autumnal Equinox) nights are equal

It has 67 satellites. Some of the prominent satellites are: Europa, Callisto, and Gannymede

Moon

Distance from earth – 384400 km Diameter – 3475 km Mass (with respect to Earth) – 1:8.1 Ratio of gravitational pull of Moon and Earth – 1:16 Highest Mountain – 18046 ft (Hugyens Mountain) Time taken by moonlight to reach Earth – 1.3s Rotational speed – 3680 kmph Speed of Revolution around Earth – 3680 kmph Revolution period around Earth – 27 days, 7 h, 43 min and 11.47 s Rotation period – 27 days, 7 h, 43 min and 11.47 s Atmosphere – Absent Part of moon not visible from Earth – 41% Maximum distance from Earth (Apogee) – 406000 km Minimum distance from Earth (Perigee) – 356400 km Circumference – 11000 km International Date Line (IDL) • It is the longitude where the date changes by exactly one day when it is crossed. • 180° East and 180° West meridians are the same line, which is called the International Date Line. • Crossing Date line from West to East — addition of 1 day Crossing Date line from East to West — subtraction of 1 day • Recently, Samoa Island decided to shift itself on west side of IDL. Indian Standard Time (IST) • The Earth takes approximately 24 hours to complete one rotation i.e., it takes 24 hours to complete 360° of its rotation. • Indian Standard Time is calculated on the basis of 82.5°E longitude which passes through Uttar Pradesh, Madhya Pradesh, Odisha, Chhattisgarh and Andhra Pradesh. • IST is 5 hr 30 min ahead of GMT.

Eclipses

When the light of the Sun or the Moon is blocked by another body, the Sun or the Moon is said to be in eclipse. • Solar Eclipse–It is caused, when the Moon revolving around the Earth comes in between the Earth and the Sun, thus making a part or whole of the Sun invisible from a particular part of the Earth. Thus, the eclipse can be partial or complete.

47

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

• Lunar Eclipse–When the Earth comes between the Moon and the Sun, the shadow cast by the Earth on the Moon results in a lunar eclipse.

Structure of Earth’s Interior

• The interior of Earth is classifi ed in three parts namely–he Crust, the Mantle and the Core.

The Crust: The Outer Layer

• The Crust is the outer most and thinnest layer of Earth. The denisty of this layer is the least. The thickness varies from 8 km to 40 km. • The upper layer of crust is called SiAl (Silica And Aluminium), as this layer is rich in lighter minerals. The average density of this layer is 2.7 gm/cm3. • Due to the presence of minerals like Silica And Magnesium in the lower part of Earth’s crust, it is known as SiMa (Silica and Magnesium).

Composition of Earth’s Crust Element Oxygen Silicon Aluminium Iron Calcium Sodium Potassium Magnesium

Percent 46 28 8 6 3.6 3.8 2.6 1.5

The Mantle: the Middle Layer

• It is the intermediate layer of the Earth in terms of both its location and density. It is the about 2900 km in thickness, composed of minerals in a semi solid state. • It is further divided in two layers: upper mantle and lower mantle. The upper part of the mantle is called the asthenosphere, which is about 250 km thick. • The average density of this layer is about 5.68 gm/cm3.

The Core: the Inner Layer

• The core is the innermost layer of Earth. It occupies the centre of Earth. It is about 3500 km in radius. • This layer is also known as NiFe (Nickel and Iron). • The temperature of the core is between 2200°C and 2750°C. • Density of this part of the Earth is 13.6 gm/cm3.

Continental Drift Theory

• This theory was given by Alfred Wegener, in 1915. This theory explains the origin and evolution of the continents and the oceans. • According to this theory, about 250 million years ago, there was only one huge continental landmass. This was named as Pangaea. It was surrounded by huge mass of water body named as Panthalassa. • The present state of continents and oceans is due to the break up of Pangaea. • The Northern drift cuts Pangaea from East to West creating Laurasia in the North and Gondwana land in south. • Tethys, a shallow sea, was situated between the Laurasia and Gondwanaland.

Plate Tectonic Theory

• It is a scientifi c theory that describes the large scale motions of Earth’s lithosphere. • There are seven major plates and more than fi fteen small plates which comprises whole Earth’s lithosphere. • The major plates are – 1. North American Plate 2. South American Plate 3. African Plate 4. Eurasian Plate 5. Pacifi c Plate 6. Antarctic Plate 7. Indo-Australian Plate

• Depending upon the type of movement, plate margins are of three types – 1. Divergent Plates (Constructive margins) 2. Convergent Plates (Destructive margins) 3. Parallel Plates (Conservative margin or Transform Boundary plate).

Weathering

The process by which rocks are chemically or physically disintegrated into fragments.

Earthquakes

• Any sudden disturbance below the Earth’s surface may produce vibrations or shaking in Earth’s crust and some of these vibrations, when reach the surface, are known as earthquakes. • The magnitude of an earthquake is measured by Richter Scale. • The intensity of earthquake waves is recorded by Seismograph. • Intensity of shaking is measured on the modifi ed Mercalli Scale. • Focus is the point beneath the Earth where earthquake originates. • Epicentre is the point just above the focus on the Earth’s surface.

Volcanism

• Sudden eruption of hot magma, rock boulders, ash from inside the earth.

Types of Volcanoes

• Active–The ones which erupt erupt frequently, e.g., Mauna Loa (Hawaii), Etna (Italy), Vesuvius (Italy), Stromboli (Mediterranean Sea). • Dormant–The ones which have not erupted for quite sometime, e.g., Fujiyama (Japan), Krakatoa (Indonesia), Barren Island (India). • Extinct–The ones which have not erupted for several centuries, e.g., Arthur’s Seat, Edinburgh, Scotland. • Ring of Fire–These are hundreds of active volcanoes found on the land near the edges of the Pacifi c Ocean.

Important Volcanic Mountains Name Vesuvius St. Helena Krakatoa Mauna Loa Cotopaxi Fujiyama Popocaipetl Ojas del Salado

Country Italy USA Indonesia USA (Hawaii) Ecuador Japan Mexico Chile – Argentina

Tsunami

Large ocean wave that is caused by sudden motion on the ocean fl oor. The motion could be due to an earthquake, volcanic eruption or underwater landslide.

Rocks

Rocks are made up of individual substance, called minerals, found mostly in solid state. Rocks are classifi ed into three major types • Igneous rocks are formed by the solidifi cation of the molten magma, e.g., Mica, Granite, etc. • Sedimentary rocks are formed due to accumulation of rock particles and organic matter in layers, under tremendous pressure, e.g., Gravel, Peat, Gypsum, etc. • Metamorphic rocks were originally igneous or sedimentary but later changed due to pressure, heat or action of water, e.g., Gneiss, Marble, Quartzite, etc. Type of Rock Original Rock Metamorphic Rock Igneous Granite Gneiss Igneous Basalt Green-stone Sedimentary Limestone Marble Sedimentary Coal Graphite Sedimentary Sandstone Quartzite Sedimentary Shale/Clay Slate, Mica, Schist

48

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Landforms

Famous Plateaus of the World

There are three major landform mountain, plateaus and plains.

Mountains

An uplifted portion of the Earth’s surface is called a hill or a mountain.

Mountains are classified into following four types

• Fold Mountains These are formed by folding of crustal rocks by compressive forces, e.g., Himalayas (Asia), Alps (Europe). • Block Mountains When great blocks of the Earth’s crust are raised or lowered during the last stage of mountain building, block mountains are formed, e.g., Vosges in France, Black Forest mountains in Germany. • Volcanic Mountains These are formed by the matter thrown out from the volcanoes, and are also known as mountains of accumulation, e.g., Mt Mauna Loa in Hawaii, Mt Popa in Myanmar. • Residual or Dissected Mountains They are known as relict mountains or mountains of circum-denudation. They owe their present form to erosion by different agencies, e.g., Nilgiris, Girnar and Rajmahal.

Major Mountain Ranges Range Andes Himalayas, Karakoram

Location South America South Central Asia

Length (km) 7200 5000

and Hindukush Rockies Great Dividing Range Atlas Western Ghats Caucasus Alaska Alps

North America East Australia North-West Africa Western India Europe USA Europe

4800 3600 1930 1610 1200 1130 1050

Major Mountain Peaks Mountain Peak

Plateau

Situation

Tibetan Plateau

Between Himalayas and Kunlun Mountains

Deccan Plateau

Southern India

Arabian Plateau

South-West Asia

Plateau of Brazil

Central-Eastern South America

Plateau of Mexico

Mexico

Plateau of Columbia

USA

Plateau of Madagascar Madagascar Plateau of Alaska

North-West North America

Plateau of Bolivia

Andes Mountains

Great Basin Plateau

South of Columbia Plateau, USA

Colorado Plateau

South of Great Basin Plateau, USA

Atmosphere

• The vast expanse of air, which envelops the earth all around is called the atmosphere. It extends to thousands of kilometres. • Signifi cance of Atmosphere– (i) Atmosphere acts as a fi lter as it absorbs the various harmful ultraviolet radiations of sun. (ii) It supports life forms in biosphere. (iii) Source of various life supporting gases like Oxygen, CO2 etc.

Structure of Earth’s Atmosphere Layer

Height (km) 0–18 km

Contains 75% of the gases in the atmosphere. As height increases, temperature decreases (about 6.5° C/km ascent).

Stratosphere

18–50 km

This layer contains the Ozone layer. The temperature remains fairly constant in the lower part but increases slowly with increase in height due to presence of Ozone gas. At upper layer temperature is almost 0°C.

Mesosphere

50–80 km

This is the coldest region of the atmosphere. The temperature drops to about –100°C.

Ionosphere

80–600 km

Radio waves are bounced off the ions and refl ect waves back to the Earth. This generally helps radio communication.

Location

Mt Everest (Highest in the world)

Nepal-Tibet

K2 (Godwin Austen)

India (PoK)

Dhaulagiri

Nepal

Annapurna

Nepal

Gurla Mandhata

Tibet

Tirich Mir

Pakistan

Aconcagua

Argentina

Cotopaxi

Ecuador

Kilimanjaro

Tanzania

Plains

A relatively low-lying and fl at land surface with least difference between its highest and lowest points is called a plain.

Classification of Plains Structural Plain

Erosional Plain

Depositional Plain

These are formed due to the upliftment of a part of the sea fl oor.

These are formed when the elevated tract of land is worn down to a plain by the process of erosion.

These are formed by fi lling up of sediments into depressions along the foothills, lakes and seas.

Plateaus

Plateaus are fl at, table like, upland areas with rough top surface and steep side walls.

Feature

Troposphere

Magnetosphere Above 600 km Upper part of Exosphere is called Magnetosphere. The temperature keeps on rising constantly at high rate. Thermosphere

85–600 km

Extent of Atmosphere

The upper part of Thermosphere contains only the lighter gases like helium and hydrogen.

• The altitude of atmosphere is not known, but it is estimated that it extends till 10,000 km above Earth’s surface. • The vertical distribution is not uniform. • Almost 98% of the atmospheric mass is limited to an altitude of 30 km from the surface of Earth.

49

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

Composition of Atmosphere

• The atmosphere is composed of gases, water vapour and particles. • The percentage composition of various gases in atmosphere upto 50 km is given below: Gases

Percentage composition

Significance

Nitrogen 78.08%

Acts as dilutent and is generally chemically inactive.

Oxygen

21%

Inhaled by biotic components for survival

Argon

0.93%

Inert Gas

Carbon Dioxide

0.3%

Being a green house gas, it maintains the temperature of the lower atmosphere

Neon

0.0018%

Inert Gas

Helium

0.005%

Inert Gas

• It also regulate temperature preventing the Earth from becoming too hot or too cold. • The major constituents of air in the atmosphere are Nitrogen (78%), Oxygen (21%), Argon (0.93%) and Carbon dioxide (0.03%). • Water vapour, dust particles, smoke, salts and other impurities are present in air in varying quantities.

Greenhouse Effect and Global Warming

• A Green House Gas (sometimes abbreviated GHG) is a gas in the atmosphere that absorbs and emits radiation within the thermal infrared range. This process is the fundamental cause of the greenhouse effect. • The primary greenhouse gases in the Earth’s atmosphere are water vapour, carbon dioxide, methane, nitrous oxide and ozone. • In the solar system, the atmosphere of Venus, Mars and Titan also contain gases that causes greenhouse effects. • Global warming is the increase of Earth’s average surface temperature due to the effect of greenhouse gases, such as carbon dioxide emissions from burning fossil fuels or from deforestation. This is a type of greenhouse effect.

Ozone Layer

• The Ozone layer is located within the stratosphere. • It is about 24 km above the Earth’s surface. • The layer consists of ozone gas molecules that are formed as the sunlight reacts with oxygen. • This layer is also known as ‘The Blanket of Globe’, as it protects life on Earth by fi ltering the Sun’s dangerous UV rays. • Due to increased pollution on the Earth, chemicals such as chlorofl uorocarbons (CFCs) are destroying this protective ozone layer, which could lead to increased health risks and damage agricultural and aquatic ecosystem. • Montreal Protocol on substances, that deplete the ozone layer is an international treaty designed to protect the ozone layer from chlorofl uorocarbons (CFCs) effective from 1889.

Pressure System of Earth

• The pressure exerted by the atmosphere due to its weight, above a unit area of the Earth’s surface is called atmospheric pressure. It is measured by Mercury Barometer. • Major pressure belts of the Earth are Equatorial Low, Sub-Tropical High, Sub-Polar Low and Polar High.

Winds

Due to horizontal differences in air pressure, air fl ows from the areas of high pressure to the areas of low pressure. The horizontal movement of the air is called wind. The types of winds are given below:

• Planetary Winds (i) Trade Winds: These are permanent east to west prevailing winds that fl ow in the Earth’s equatorial region. (ii) Westerlies: These are permanent winds that blow in the middle lattitude. These are also called roaring forties, the Furious Fifties and Shrieking or Screaming Sixties. (iii) Polar winds: These begin near the North and South poles. (iv) Periodic winds: These fl ow at regular intervals. They originate due to the difference in temperature and pressure. (v) Local winds: These occurs due to the movement of air between high and low pressure systems withion small areas. (vi) Cyclones: It is a large mass of air that rorates around a strong centre of low atmospheric pressure. (vii) Anticyclones: It is a large mass of air slowly rotates around a centre of high atmospheric pressure in clockwise direction in the Northern hemisphere and in anticlockwise direction in the Southern hemisphere. (viii) Hurricanes: These are tropical cyclone over tropical or subtropical waters.

List of Local Winds Name

Nature of Wind

Chinook

Hot, dry wind in Rockies, also called ‘Snow Eater’.

Fohn

Hot, dry wind in the Alps.

Khamsin

Hot, dry wind in Egypt.

Sirocco

Hot, moist wind from Sahara to Mediterranean Sea. It is also known as Blood Rain.

Solano

Hot, moist wind from Sahara towards Iberian Peninsula.

Harmattan Hot, dry wind blowing outwards from the interior of Western Africa. Also called Guinea Doctor. Bora

Cold, dry wind blowing outwards from Hungary to the North of Italy (near Adriatic Sea).

Mistral

Very cold wind, which blows from the Alps over France.

Punas

Cold, dry wind blowing down towards the Western side of Andes.

Blizzard

Very cold winds in Tundra region.

Purga

Cold wind in Russian Tundra.

Levanter

Cold wind in Spain.

Norwester Hot wind in New Zealand. Santa Ana Hot wind in South California in USA.

Jet Stream

• The strong and rapidly moving circumpolar Westerly air circulation in a narrow belt of a few hundred kilometers width in the upper limit of troposphere is called Jet stream.

Humidity

• The content of water vapour present in the air at a particular time and space is referred as Humidity. • Hygrometer is an instrument used to measure Humidity.

Clouds

• Clouds are the mass of small water droplets or tiny ice crystals. • Clouds are classifi ed in four categories – 1. High Clouds (6-20 km) 2. Middle Clouds (2.5-6 km) 3. Low Clouds (below 2.5 km) 4. Clouds with Great Vertical Extent (upto 9000 m)

Rainfall

• On the basis of its origin, rainfall may be classifi ed into three categories –

50

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

1. Convectional Rainfall – Occurs due to thermal convection current. 2. Frontal or Cyclonic Rainfall – Occurs due to upward movement of air. 3. Orographic Rainfall – Occurs due to ascent of air forced by a mountain barrier.

Weather and Climate

• Weather refers to the atmospheric state of a particular area for a particular time. • The average weather conditions over a large area and for a long time are called the climate of a place.

• Yangtze river or Chang Jiang is the longest river of Asia. • Laos is the only land locked country in South-East Asian peninsula. • Amur river forms the boundary between Russia and China. • Caspian sea is the world’s largest lake. • Abhors, Afridis, Khirgiz, Kurds, Nicobaris and Mundals are main tribes of Asia. • Asia’s wealth is in its agricultural resources which includes fertile soil and water for irrigation.

Africa

• There are four oceans. In order of their size they are: Pacifi c Ocean, Atlantic Ocean, Indian Ocean and Arctic Ocean. • The average depth of oceans on Earth is about 4 km.

• • • • •

Coral Reefs

•

Oceans

• Due to accumulation and compaction of skeletons of lime secreting organisms known as Polyps, the Coral Reefs are formed. • They are mainly found in the tropical oceans and seas, because they require high mean annual temperature ranging between 20°C to 25°C. • Corals do not live in deeper water due to lack of suffi cient sunlight and oxygen. • They are formed mainly along Western coast.

Ocean Currents

• Ocean current is a continuous, directed movement of ocean water generated by the forces acting upon it, such as breaking waves, wind, coriolis effect, temperature and salinity difference and tides caused by the gravitational pull of the Moon and the Sun. • Ocean currents are of two types – Warm Currents and Cold Currents.

Tides

• The periodic rise and fall of the sea level as a result of the gravitational forces between the Earth, the Moon and the Sun is called a tide. • The time interval between two tides is 24 hours 50 minutes.

Continents

• The word continent means continuous or connected land. • There are seven main divisions of land, or seven continents on Earth.

Area of the Continents Continent Asia Africa North America South America Antarctica Europe Australia

Asia • • • •

Percent of Earth Area 29.8 20.3 16.2 11.9 9.5 6.7 5.7

There are 48 countries in Asian continent. The latitudinal extent of Asia is 10° S and 80° N. The longitudinal extend of Asia is 25° E and 170° W. The oceans and seas that lies in Asian continent are – Arctic Ocean, Pacifi c Ocean, Indian Ocean, Red Sea, Gulf of Aden, Persian Gulf, Gulf of Oman, Arabian Sea, Bay of Bengal, China Sea, Yellow Sea, Okhotsk Sea and Bering Sea. • The straits of Asia are Strait of Malacca, Bering Strait, Hormuz Strait, Dardenelles Strait. • The islands are – Kurile, Sakhalin, Honshu, Shikoku, Hokkaido, Taiwan, Borneo, Sumatra, Java, Celebes, New Guinea, Philippines, Sri Lanka, Bahrain, Cyprus.

• • • • • •

It is the second largest continent of the Earth. It includes 54 countries. The latitudinal Extent of Africa is 35°S and 37°N. The longitudinal Extent of Africa is 16°W and 51°E. The oceans and seas are Indian Ocean, Red Sea, Atlantic Ocean, Madagascar Sea, Gulf of Guinea, Mediterranean Sea. The straits are Strait of Bab-el – Mandeb, Strait of Gibraltar, Mozambique Channel. The islands are Madagascar, Cape Verde Islands, the Comoros, Zanzibar, Mauritius, Seychelles. The entire continent is a plateau. Lake Victoria is the largest lake of Africa. Nasser lake is man-made lake. Nile River is longest river of the world. Africa is rich in mineral deposits and these minerals are mostly found on the plateau region.

North America • • • • •

North America is the third largest continent of the Earth. It includes 23 countries and more than 12 small islands. Latitudinal Extent of this continent is 7°N and 84°N. Longitudinal Extent is 20°W and 180°W. The major deserts of North America are Chihuahuan, Colorado, Mojave, Sonoran. • The oceans and seas are Atlantic Ocean, Caribbean Sea, Gulf of California, Gulf of Alaska, Bering Sea, Hudson Bay. • The islands are Greenland, Baffi n, Victoria, New Foundland, Cuba, Jamaica, Haiti, Bermuda, Hawaii. • Canada has the longest coastline in the world. • Lake superior is the largest fresh water lake in the world. • Anglo – America is rich in iron ore.

South America • • • • • • • •

South America is the fourth largest continent of the Earth. There are 13 countries in South America. The latitudinal Extent is 12°N and 55°S. The longitudinal Extent is 35°W and 81°W. The strait are Strait of Magellan, Drake Passage. The deserts are Atacama, Patagonia. It has the world’s highest waterfall, i.e., Angel Fall. South America is rich in minerals such as petroleum, aluminum, bauxite, copper, gold, lead, nickel, nitrates, diamond, coal, etc.

Europe • • • • •

It is the sixth largest continent of the Earth. It includes 50 countries and is part of peninsula of Eurasia. The latitudinal extent is 35°N and 75°N. The longitudinal extent is 25°W and 65°E. The oceans and seas are– Atlantic ocean, Arctic ocean, Mediterranean Sea, Caspian Sea, Black Sea, White Sea, North Sea, Norwegian Sea, Baltic Sea, Gulf of Bothnia, Gulf of Finland, Bay of Biscay, Aegean Sea, Adriatic Sea. • The straits are Dardenelles Strait, English Channel, Straits of Gibraltar. • The lakes are Lake Ladoga, Onega, Peipus, Vanern, Vattern. • The islands are British Isles, Iceland, Sardinia, Sicily, Crete.

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GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

Australia • • • • • • •

It is an island continent. The latitudinal Extent is 12°S and 38°S. The longitudinal Extent is 114°E and 154°E. The island is Tasmania Mt Kosciusko (2228 m) is the highest peak of Australia. Great Barrier Reef is world’s largest coral reef. Coal, iron ore, bauxite, uranium, gold, petroleum are major mineral resource of Australia.

Antarctica • • • • •

It is the seventh largest and least populous continent of Earth. It is known as the Continent for Science. Mt. Erebus is the only active volcano on Antarctica. Mt. Vinson (4892 m) is the highest peak of Antarctica. It is the only continent which is completely frozen, therefore, known as White Continent. • About 98% of Antarctica is covered by ice that averages 1.9 m of thickness. • Scientifi c studies have shown Antarctica to be rich in gold, platinum, coal, nickel, hydrocarbon and iron ore.

World Geography: Important Facts Important Lakes of the World Lake Caspian Sea Superior Victoria Huron Michigan Tanganyika Baikal Great Bear Great Slave

Location Asia Canada and USA Africa Canada and USA USA Africa Russia Canada Canada

Major Rivers of the World River Nile Amazon Yangtze Mississippi Missouri Yenisei Huang Ho Ob Congo Amur Lena Mekong Niger

Origin Victoria lake Andes (Peru) Tibetan Kiang Plateau Itaska Lake (USA) Tannu-Ola Mountains Kunlun Mountains Altai Mountains, Russia Lualaba and Luapula rivers North East China Baikal Mountains Tibetan Highlands Guinea

Waterfalls Waterfall Angel Falls Tugela Falls Mongo Yosemite Catarata Yumbilla

Location Venezuela South Africa Norway United States Peru

Important Canals of the World Panama Suez Erie Kiel

Pacifi c Ocean with Caribbean Sea Mediterranean Sea to Red Sea Atlantic Ocean to Great Lakes North Sea to Baltic Sea

Famous Grasslands of the World Grassland

Country

Steppe

Eurasia

Pustaz

Hungary

Prairie

USA

Pampas

Argentina and Uruguay (South America)

Veld

South Africa

Downs

Australia

Canterbury

New Zealand

Great Deserts of the World Name Country/Region Sahara (Libyan, Nubian) North Africa Australian (Gibson, Simpson), Great Australia Victorian, Great Sandy Arabian (Rub al Khali, An-Nafud) Dasht-e-Lut (Barren Desert) Dasht-e-Kavir (Salt Desert) Desierto de Sechura Atacama Patagonia Kalahari Namib

Arabia Iran Iran Peru North Chile Argentina Botswana Namibia

Minerals of the World Mineral

Leading Producer

Gold

China

Bauxite

Australia

Copper

Chile

Platinum

South Africa

Chromium

South Africa

Vanadium

China

Antimony

China

Tungsten

China

Phosphate

China

Manganese

China

Diamond

Russia (Botswana, in terms of value)

Iron ore

China

Petroleum

USA

Agriculture Agricultural Produce Coffee Rubber Tea Oil Palm Cocoa Coconut Date Palm Cotton Wheat Maize Fruits and Vegetables Wool Rice Cloves

Leading Producer Brazil Thailand China Indonesia Ghana Indonesia Egypt China China USA China Australia China Zanzibar

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Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Highest and Lowest Points of the Continent Continent

Highest (m)

Lowest (m)

Asia

Mt Everest (8848)

Dead Sea (−427)

Africa

Mt Kilimanjaro (5895) Lake Assal (−155)

North America Denali (6198)

Death Valley (−86)

South America Mt Aconcagua (6960) Lagun del Carbon (−105) Antarctica

Vinson Massif (4892) Deep lake, Vestfold hills (−50)

Europe Australia

Mt El’ brus (5642)

Caspian Sea (−28)

Puncak Jaya (4884)

Lake Eyre (−18)

Important International Boundary Lines Name

In Between

Radcliffe Line (1947)

India and Pakistan (Indo-Pak)

Name

In Between

McMahon Line (1914) India and China (Indo-China) Durand Line (1893)

Pakistan and Afghanistan

Hindenburg Line

Germany and Poland

Maginot Line

France and Germany

Oder Neisse Line

Germany and Poland

Siegfried Line

Fortifi cation between Germany and France

38th Parallel Line

North and South Korea

th

49 Parallel Line

USA and Canada

24th Parallel Line

Pakistan claims that it is the boundary between India and Pakistan in Rann of Kachcha

17th Parallel Line

North Vietnam and South Vietnam

Important Industrial Cities City

Industry

City

Industry

Anshan (China)

Iron and Steel

Los Angeles (USA)

Petroleum

Baku (Azerbaijan)

Petroleum

Lyon (France)

Silk Textiles

Belfast (Ireland)

Ship-building

Magnitogorsk (Russia)

Iron and Steel

Birmingham (UK)

Iron and Steel

Manchester (UK)

Cotton Textile

Chicago (USA)

Meat Packing

Milan (Italy)

Silk Textile

Detroit (USA)

Automobile

Multan (Pakistan)

Pottery

Havana (Cuba)

Cigars

Munich (Germany)

Lenses

Hollywood (USA)

Films

Nagoya (Japan)

Automobiles

Johannesburg (South Africa)

Gold Mining

Philadelphia (USA)

Locomotives

Kansas City (USA)

Meat Packing

Pittsburg (USA)

Iron and Steel

Kawasaki (Japan)

Iron and Steel

Plymouth (USA)

Ship-building

Kimberley (South Africa)

Diamond Mining

Rourkela (India)

Iron and Steel

Krivoi Rog (Ukraine)

Iron and Steel

Sheffi eld (UK)

Cutlery

Leeds (UK)

Woollen Textiles

Vladivostok (Russia)

Ship-building

Leningard (Russia)

Ship-building

Wellington (New Zealand)

Dairy Products

Deepest Point of Oceans

Topic-2

Name

Ocean

Trench

Challenger Deep

Pacifi c

Mariana

Brownson Deep

Atlantic

Puerto Rico

Factorian Deep

Southern

South Sandwich

Unnamed Deep

Indian

Java

Molloy Hole

Arctic

Unnamed Trench

Cyclones of the World Regional Name

Region

Typhoons

China Sea

Tropical Cyclones

Indian Ocean

Hurricanes

Caribbean Sea

Tornadoes

USA

Willy Willies

Northern Australia

Indian Geography

 Revision Notes India at a Glance

• India is the seventh largest country area wise. The Area of India is 32,87,263 km2, which is 2.4% of world’s area. • India is the second most populous country of the world. It has 17.44% of population of the world, about 1.21 billion. • India is a subcontinent and it is situated in Northern and Eastern Hemisphere. • India shares its boundary with seven countries namely – Bangladesh (4096 km), China (3488 km), Pakistan (3310 km), Nepal (1752 km), Myanmar (1643 km), Bhutan (578 km) and Afghanistan (106 km). • The Tropic of Cancer (23½°N latitude), in India, passes through eight states – Gujarat, Rajasthan, Madhya Pradesh, Chhattisgarh, Jharkhand, West Bengal, Tripura and Mizoram. • The islands are Andaman and Nicobar Islands in the Bay of Bengal; Lakshadweep, Amindivi and Minicoy in the Arabian Sea. • India lies midway between the Far East and Middle East. The trans-Indian Ocean routes connecting the industrially developed countries of Europe in the West and the under developed countries of East Asia pass close by Indian subcontinent.

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GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

It is surrounded by Arabian Sea in the South-West and Bay of Bengal in the South-East.

Important Facts

Himadri: Greater Himalayas

• It is the continuous range consisting of the loftiest peaks with an average height of 6000 meters. • It contains all the prominent Himalayan peaks.

Latitudinal extent

8°4′ North to 37°6′ North

Longitudinal extent

68°7′ East to 97°25′ East

North-South extent

3214 km

East-West extent

2933 km

Land Frontiers

15200 km

Total Coastline

7516.6 km

Number of States

28

Union Territories

Eight (After bifurcation of J & K in Jammu and Kashmir and Ladakh and merger of Dadar and Nagar Haveli with Daman and Diu)

Important Passes in Greater Himalayas Pass

Land Neighbours

Pakistan, Afghanistan, China, Nepal, Bhutan, Bangladesh and Myanmar

Longest Coastline

Gujarat

Active volcano

Barren Island in Andaman and Nicobar Islands

Southern-most tip

Kanyakumari

Northern-most point

Indira Col

Western-most point

Sir Creek in Kutch District

Eastern-most point

Kibithu (Arunachal Pradesh)

Location

Connectivity

Karakoram

Ladakh

India to China

Burzi La

PoK

Kashmir Valley to Gilgit

Zoji La

Ladakh

Srinagar to Leh

Shipki La

Himachal Pradesh

Shimla to Gartok (Tibet)

Jelep La

Sikkim

Sikkim to Lhasa (Tibet)

Yang Yap

Arunachal Pradesh

Entry of Brahmaputra river

Himachal: Middle Himalayas or Lesser Himalayas

• The lesser Himalayas have a gradual elevation towards the Dhauladhar and the Pir Panjal ranges. • The rise is more rapid in Shimla hills, to the south of which is high peak of Choor Chandni (3647 m).

Important Passes in Middle Himalayas Pass

Location

Connectivity

Pir Panjal Jammu and Kashmir Jammu-Srinagar road passes from this pass.

• The 82°30'E longitude is taken as the standard Meridian of India, as it passes through the middle of India (from Naini, near Allahabad). • India is separated from Sri Lanka from Gulf of Mannar and Palk Strait, on the South-Eastern side.

Banihal

Jammu and Kashmir Jammu-Srinagar NH-44 passes from this pass. Jawahar tunnel is located here.

Rohtang

Himachal Pradesh

Kullu valley with Lahul and Spiti valley in Himachal Pradesh

Shiwalik: Outer Himalayas

Indian States & UTs Situated on the Border Country

Border

Pakistan (4)

Gujarat, Rajasthan, Punjab, Jammu and Kashmir

Afghanistan (1)

Ladakh

China (5)

Ladakh, Uttarakhand, Himachal Sikkim, Arunachal Pradesh

Nepal (5)

Uttar Pradesh, Uttarakhand, Bihar, West Bengal, Sikkim

Bhutan (4)

(b) Himachal (Lesser Himalayas) (c) Shiwalik (Outer Himalayas)

Pradesh,

Sikkim, West Bengal, Assam, Arunachal Pradesh

• These are the southernmost Himalayan mountain range. • Their average altitude varies from 600 to 1500 metres. • They are known as Jammu Hills in Jammu and Dalfa, Miri, Abor, Mishmi Hills in Arunachal Pradesh.

Peaks of India Peak (Name)

Height (in m)

State/UTs

Mt K2

8611

Jammu and Kashmir

Kanchenjunga

8598

Sikkim

Bangladesh (5) West Bengal, Assam, Meghalaya, Tripura, Mizoram

Nanda Devi

7817

Uttarakhand

Kamet

7756

Uttarakhand

Physical Feature

Saltoro Kangri

7742

Ladakh

The physiographic divisions of India are – (a) The Himalayan Range of Mountain (b) The Peninsular Plateau (c) The Great Plains of India (d) The Coastal Plains (e) The Islands of India

The Himalayas

• Himalayas are young fold mountains of tertiary period, which were folded over Tethys sea due to inter-continental collision. • They are one of youngest fold mountain ranges in the world and comprise mainly sedimentary rocks. • The Himalayas stretch across the north-eastern portion of India. • It covers approximately 1,500 mi (2,400 km) and passes through India, Pakistan, Afghanistan, China, Bhutan and Nepal. • They consists of three parallel ranges – (a) Himadri (Greater Himalayas)

Kangto

7090

Arunachal Pradesh

Reo Purgyil

6816

Himachal Pradesh

Saramati

3841

Nagaland

Sandakphu

3636

West Bengal

Khayang

3114

Manipur

Anaimudi

2695

Kerala

Dodda Betta

2636

Tamil Nadu

The Great Plains

• The Great plains of northern India, also known as Ganga–Sutlej Plains, are transitional belt between the Himalayas and Peninsular India. • The Great plains cover an area of 7,74,000 km2 (3,00,000 square miles) having west–east length of 2400 km and north-south width of 144 km.

54

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Bhangar and Khadar Bhangar

Khadar

• These are low plains. • The deposit of fresh alluvium Formed of older alluvium every year brought by the Himalayas rivers makes the belt of Northern plains. • This belt ends in Khadar

• This belt ends in Terai.

South to North Doabs Doab • Bist Doab • Bari Doab • Rechna Doab • Chaj Doab • Sind Sagar Doab

Region • Between Beas and Sutlej • Between Beas and Ravi • Between Ravi and Chenab • Between Chenab and Jhelum • Between Jhelum and Indus

The Peninsular Plateau (Lava Plateau)

• The Peninsular Plateau is a tableland composed of old crystalline igneous and metamorphic rocks. • It was formed due to the breaking and drifting of the Gondwanaland and thus, making it a part of oldest land mass. • The plateau has broad and shallow valleys and rounded hills. • Some plateaus of Peninsular India are –

The Central Highland

• Then Central Highland Plateau lie to the North of the Narmada river covering a major area of Malwa plateau.

The Deccan Plateau

• This Plateau is a triangular land lying to the south of the river Narmada.

Meghalaya Plateau

• Garo-Raj-Mahal gap separates this plateau from main block of the peninsular plateau. From East to West, the plateau is covered by Garo, Khasi, Jaintia and Mikir hills.

The Bundelkhand Upland

• It is located to the south of Yamuna river between Central Indian plateau and the Vindhayan range. It is composed of granite and gneiss.

• The Maikal range forms a connecting link between Vindhya and Satpura.

Eastern and Western Ghats Eastern Ghat

Western Ghat

These are Located East to These are Located West to Deccan Plateau Deccan Plateau They are parallel to Eastern They are parallel to Western Coast, i.e., Coromandel, Coast, i.e., Konkan, Kannad, Northern Circar, etc. Malabar, etc. Mahanadi, Cauveri, Narmada, Tapi, Sabarmati and Godavari, Krishna, etc. rivers Mahi, etc. rivers are drawn in are drawn in this land. this land. Jindhagada with an altitude Anaimudi with an altitude of of 1690 m is the highest peak. 2695 m is the highest peak.

The Coastal Plains

On the basis of location and active geomorphological processes, it can be divided into – (a) Eastern Coastal Plains (b) Western Coastal Plains

The Coastal Plains Eastern Coast

Western Coast

Smooth outline

Dissected outline

Occurrence of deltas

Occurrence of estuaries

Broad

Narrow

Long rivers

Small rivers

Islands

• India’s islands can be broadly divided into two groups: 1. Andaman and Nicobar group 2. Lakshadweep group. • Archipelago is the group of islands.

Andaman and Nicobar Group

Telangana Plateau

• It is located in Western Andhra Pradesh, comprising the NorthEastern part of the Deccan Plateau and has an area of about 148,000 sq. km.

• It is located in Bay of Bengal. • There are 325 islands in Andaman group and 247 islands in Nicobar group. • Andaman is separated from Nicobar by Ten degree channel. • Duncan passage lies between South Andaman and Little Andaman groups.

Karnataka Plateau (Mysore Plateau)

Lakshadweep Group

• It is located in Karnataka. It has an area of 73000 sq. miles.

Chota Nagpur Plateau

• It covers mostly Jharkhand, Chhattisgarh and Purulia region of West Bengal.

Hill Ranges of Peninsular India Aravali Range

• These are the world’s oldest fold mountains running in North-East to South-East direction. • It is an example of relict mountain.

• It is located in Arabian Sea. • Minicoy is second largest and southernmost island and Andrott island is largest island of this group. • Nine Degree channel separates the rest of Lakshadweep from Minicoy. • Eight Degree channel separates Lakshadweep group from Maldives.

Drainage in India

Satpura Range

• Satpura is a series of seven mountains running in East-West direction, south of Vindhya and in between the Narmada and Tapi. • It comprises of Rajpipla hills, Mahadeo hills and Maikal range.

• Water drains in two directions of the main water divide line of India. Ninety percent of the rivers drain into Bay of Bengal and the rest drains into Arabian Sea. • In India, the rivers can be divided into two main groups. 1. Himalayan rivers. 2. Peninsular rivers.

Vindhyan Range

The Himalayan River System

• This acts as a water divide between Ganga river system with the South Indian river system.

• The Himalayan river system is divided into three major river systems.

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GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

The Indus System

• Indus, also known as Sindhu, originates from a glacier near Bokhar Chu in the Tibetan region near Mansarovar lake. • Its most important tributaries are Jhelum, Chenab, Ravi, Beas and Sutlej.

Climatic Regions of India Type

Area

Tropical Rain Forests

Western Ghats, West High temperature Coastal Plains, Parts throughout the year, of Assam heavy seasonal rainfall, annual rainfall 200 cm annually (May to November)

Tropical Savanna Climate

Most of Peninsular Dry winters, annual region (except rainfall varies from 76 leeward side of cm to 150 cm. Western Ghats)

Tropical SemiArid-Steppe Climate

Rainshadow belt running Southward from Central Maharashtra to Tamil Nadu.

The Ganga System

• Ganga rises from the Gangotri glacier near Gaumukh (3900 m) in the Uttarakashi district of Uttarakhand. • The left bank of tributaries of Ganga are Ramganga, Gomti, Kali or Sarda, Gandak, Kosi, Ghaghara. • The right bank tributaries of Ganga are Yamuna and Son.

The Brahmaputra System

• It is one of the largest river of the world. • Brahmaputra forms large number of riverine islands. Majuli is the largest riverine island in the world.

The Peninsular River System

Characteristic

Low rainfall varies from 38 cm to 80 cm and temperature from 20° to 30°C.

Peninsular rivers can be divided into two groups: (i) East fl owing Rivers (or Delta forming rivers) • East fl owing rivers drain into Bay of Bengal. • Except some small rivers, East fl owing rivers do not form estuaries. (ii) West fl owing Rivers (or Estuaries forming rivers) • West fl owing rivers form Estuaries. • West fl owing rivers drain into Arabian Sea.

Tropical and Punjab, Haryana and Temperature varies S u b - t r o p i c a l Kachchh region. from 12°-35°C. Steppes Tropical Desert

Western parts of Scanty rainfall (mostly Barmer, Jaisalmer in form of cloud burst), and Bikaner districts high temperature. of Rajasthan and parts of Kachchh

Important Rivers of India

Humid Subtropical climate with dry winters

South of Himalayas

Mild winters extremely summers.

Mountain Climate

Mountainous region (above 6000 m or more)

Rainfall varies from 63.5 cm to 254 cm. (Mostly during SouthWest Monsoon)

Name Ganges Sutlej Indus Ravi Beas Jhelum Yamuna Chambal

Originates from Gangotri Glacier Manasarovar, Rakas Lakes Near Manasarovar Lake Kullu Hills near Rohtang Pass Near Rohtang Pass Verinag in Kashmir Yamunotri Singar Chouri Peak, Vindhyan escarpment Ghaghara Matsatung Glacier Kosi Near Gosai Dham Peak Betwa Vindhyanchal Son Amarkantak Brahmaputra Near Mansarovar Lake Narmada Amarkantak Tapti Betul District in Madhya Pradesh Mahanadi Raipur District in Chhattisgarh Luni Aravallis Ghaggar Himalayas Sabarmati Mewar hill, Aravallis Krishna Western Ghats Godavari Nasik district in Maharashtra Cauveri Brahmagiri Range of Western Ghats Tungabhadra Western Ghats

Falls into Bay of Bengal Chenab Arabian Sea Chenab Sutlej Chenab Ganga Yamuna Ganga Ganga Yamuna Ganga Bay of Bengal Gulf of Khambat Gulf of Khambat Bay of Bengal Rann of Kachchh Near Fatehabad Gulf of Khambat Bay of Bengal Bay of Bengal Bay of Bengal Krishna

Climate

Monsoon

A type of wind system, in which there is almost complete reversal of prevailing wind direction. Types 1. South West Monsoon (June to September) 2. North East Monsoon (September to December)

and hot

Season of India • • • •

Winter Season: Mid December to Mid March Summer Season: Mid March to May Rainy Season: June to September. Season of Retreating Monsoon: October to Mid December.

Agriculture

India is essentially an agricultural land. Two-thirds of its population still lives on agriculture. It includes farming, animal rearing and fi shing.

Agricultural Seasons in India

There are three major crop seasons in India.

Kharif

These are sown in June/July, harvested in September/October, e.g., rice, jowar, bajra, ragi, maize, cotton and jute.

Rabi

These are sown in October/December, harvested in April/May, e.g., wheat, barley, peas, rapeseed, mustard, sesame.

Zaid

These are sown in February/March harvested in May/June, e.g., urad, moong, melons, etc.

Green Revolution

It is the phrase generally used to describe the spectacular increase that took place during 1960s and is continuing in the production of foodgrains in India. The components of Green Revolution are High Yield Variety Seeds, Irrigation, Use of Fertilisers, Use of Insecticides and Pesticides, Command Area Development Programme, Consolidation of Holdings, etc.

56

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Natural Vegetation of India Name

Climatic Requirement

Feature

Important Species

Found in Area

Tropical Wet Evergreen

Rainfall > 250 cm Temperature 25-27°C, Humidity 80% or more

Dense forest, tall trees

Mesa, Dhup, White Cedar, Jamun, Bamboo, Agar and Hopes

North-East India, Western slopes of Western Ghats, Andaman and Nicobar Islands

Tropical Semi-Evergreen

Rain > 200-250 cm. Tempearture 24-27°C, Humidity 80%

Evergreen mixed with deciduous, Height 24-36 cm

Semul, Rosewood, Indian Chestnut, Kusum, Mesua

Lower slopes of Eastern Himalayas. Odisha Coast

Tropical Dry Evergreen

Areas receive rain from North-East Monsoon, Temperature 28°C, Humidity 74%

Presence of canopy, low height, about 9-12 m

Khimi, Jamun, Tamarind, Neem, Cane

Coromandel Coast of Tamil Nadu

Tropical Moist Deciduous

Moderate rainfall of 150200 cm Temperature 2627°C, Humidity 60-80%

Trees shed their leaves in the dry season

Sal, Teak, Sandalwood, Western, Ghats, Eastern Ebony, Mahua, Shisham coastal plains, Eastern Plateau

Tropical Dry Deciduous

Rainfall < 150 cm, Dry season

Undergrowth is shrubby and grassy, trees shed their leaves in the dry season

Sal, Teak, Khair, Palash, Uttar Pradesh, Tamil Nadu, Tendu, Laurel Western Ghats, Rajasthan and West Bengal

Tropical Thorny

Rainfall 50-70 cm, Temperature 25-27°C, Humidity < 47%

Trees are stunted (6-9 m), trees have long roots, sharp spines and glossy leaves to conserve water

Babul, Acacia, Khair, Khejri

South-Western Punjab, Western Haryana and Uttar Pradesh, Western Madhya Pradesh, Kachchh and Saurashtra, Rajasthan

Tidal/Littoral Mangrove

Rainfall > 200 cm, high water salinity and areas are fl ooded regularly

Trees are evergreen, breathing roots called pneumatophores

Keora, Amur, Sundari, Agar, Bhendi, Nipa

Delta regions of Ganga, Mahanadi, Godavari and Krishna

Soils in India Types

States where Found (Occurrence)

Composition

Crops Grown

Alluvial

Punjab, Haryana, Uttar Pradesh, Bihar Rich in potash and lime but defi cient in Large variety of rabi and kharif and Jharkhand nitrogen and phosphorus. crops such as wheat, rice, sugarcane, cotton and jute etc.

Black soil (Regur soil)

Deccan Plateau, Valleys of Krishna and Rich in iron, lime, aluminium, Cotton sugarcane, Godavari, Andhra Pradesh, Madhya magnesium, calcium, but lacks in tobacco, wheat and rice. Pradesh and Tamil Nadu. nitrogen, phosphorus and humus.

Red

Eastern parts of Deccan Plateau, Tamil Rich in iron and potash, but defi cient in Wheat, rice, cotton sugarcane Nadu, Goa, Odisha and Meghalaya. lime, nitrogen, phosphorus and humus. and pulses.

Laterite

Summits of Eastern and Western Rich in iron but poor in silica, lime Tea, coffee, rubber, cashew and Ghats, Assam hills, Andhra Pradesh, phosphorus, potash and humus. millets. Karnataka, West Bengal and Odisha.

Desert

West and North-West India, Rajasthan, Rich in soluble salts, but defi cient in Generally unsuitable for North Gujarat and Southern Punjab. organic matter. cultivation, but with irrigation useful for cultivation of droughtresistant lime; millets, barley, cotton, maize and pulse.

Mountain

Hills of Jammu and Kashmir, Rich in iron and humus, but defi cient With fertilisers, tea, fruits and Uttarakhand and Assam hills. in lime. medicinal plants can be grown.

jowar,

Saline (Reh, Drier Parts of Bihar, Jharkhand, Uttar Many salts such as sodium, magnesium Unfi t for agriculture. Kallar, Usar, Pradesh, Haryana, Punjab, Rajasthan and calcium. Thur, Rukar) and Maharashtra. and Alkaline Peaty Marshy

and Kerala, coastal regions of Odisha, Tamil Contain large amount of soluble salts Useful for Nadu and Sundarbans of West Bengal. and organic matter, but lack in potash cultivation. and phosphates.

Mineral Resources Types of Minerals

Metallic Iron ore, copper, aluminium, tin, lead, gold and silver. Non-metallic Coal, mica, manganese, petroleum and sulphur.

rice

and

jute

Radioactive Uranium and thorium Gondwana rocks (Chhotanagpur Plateau) are the richest mineral deposits in India.

57

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

Mineral Resources of India Mineral Coal

States West Bengal, Jharkhand, Odisha, Madhya Pradesh and Chhattisgarh Copper Madhya Pradesh, Rajasthan, Jharkhand, Karnataka Gold Karnataka, Andhra Pradesh Iron Karnataka, Chhattisgarh and Jharkhand Bauxite Odisha, Jharkhand, Gujarat and Madhya Pradesh Mica Jharkhand, Andhra Pradesh and Rajasthan Petroleum Assam, Gujarat, Mumbai High Basin (South of Mumbai High) Uranium Jharkhand, Rajasthan, Andhra Pradesh and Karnataka Thorium Kerala Coast, Rocks of Aravalli in Rajasthan Silver, Zinc Rajasthan, Andhra Pradesh, Karnataka (Kolar mines) and Lead Diamond Panna (Madhya Pradesh), Banda (Uttar Pradesh)

Transport Railways

• Indian Railway system is the second largest in Asia and the fourth largest in the world. • The longest railway platform in India is now Gorakhpur with a stretch of around 1.3 km. Railway Zone

Headquarters

Central

Mumbai (CST)

Eastern

Kolkata

Northern

New Delhi

North-Eastern

Gorakhpur

North-East Frontier

Maligaon-Guwahati

Southern

Chennai

Central

Secunderabad

South-Eastern

Kolkata

Western

Mumbai Churchgate

East Coast

Bhubaneshwar

East Central

Hajipur

North Central

Prayagraj

North-Western

Jaipur

South-Western

Hubli

West Central

Jabalpur

South-East Central

Bilaspur

Kolkata Metro

Kolkata

South Coast Railway

Visakhapatnam

• The fi rst train ran in India between Bombay and Thane, a stretch of 34 km on 16th April, 1853. • The second train ran between Howrah and Hooghly in 1854. • The fi rst electric train in India was Deccan Queen. It was introduced in 1929 between Bombay and Poona. • The longest train route is of ‘Vivek Express’ from Dibrugarh in Assam to Kanyakumari in Tamil Nadu. It covers a distance of 4273 km (2655 miles). • The fi rst Metro train was introduced in Kolkata (West Bengal) on 24th October, 1984. The two stations connected were Dumdum and Belgachhia. • In 1990, Konkan Railway has been started between Goa, Maharashtra and Karnataka. • Delhi metro rail was started in 2002 on 25th December between Shahdra and Tees Hazari. • Rapid metro train has been started in Gurgaon (Haryana) on 14th November 2013.

• Vande Bharat Express also known as Train 18, is an Indian semihigh speed electric (India’s fastest train) train made by Integral Coach Factory, Chennai, under Make in India Programme. The fi rst Vande Bharat ran between New Delhi and Varanasi, Uttar Pradesh. • Delhi-Meerut Regional Rapid Transit System (RRTS) is an 82.15 km long, under-construction, semi-high speed rail corridor connecting Delhi, Ghaziabad and Meerut. • The Lucknow-New Delhi Tejas Express is the fi rst Indian train operated by private operators, IRCTC, a subsidiary of Indian Railway. • The Indian Railways operate in three different gauges, i.e., Broad Gauge (distance between rails is 1.676 m), Metre Gauge (distance between rails is 1.00 mm) and Narrow Gauge (distance between rails is 0.762 or 0.610 m).

Air Transport

• In 1935, the ‘Tata Air Lines’ started its operation between Mumbai and Thiruvananthapuram and in 1937 between Mumbai and Delhi. • In 1953, all the private airline companies were nationalised and Indian Airlines and Air India came into existence. • Vayudoot Limited started in 1981 as a private air carrier and later on it merged with Indian Airlines. • International Airports Authority of India and National Airports Authority were merged in 1995 to form Airports Authority of India. • The Authority manages the Civil Aviation Training College at Allahabad and National Institute of Aviation Management and Research at Delhi.

Major International Airports in India International Airports Rajiv Gandhi International Airport Calicut International Airport Chhatrapati Shivaji International Airport Kempe Gowda International Airport Goa Airport in Vasco da Gama City Netaji Subhash Chandra Bose International Airport Thiruvananthapuram International Airport

City Hyderabad Calicut Mumbai Bengaluru Goa Kolkata Thiruvananthapuram Lokpriya Gopinath Bordoloi International Airport Guwahati Sardar Vallabhbhai Patel International Airport Ahmedabad Indira Gandhi International Airport Delhi Chennai International Airport Chennai Sri Guru Ram Dass Jee International Airport Amritsar Pakyong Airport (First green fi eld airport in Sikkim Northeast region)

Road Transport

• India has one of the largest road networks in the world (48 lakh km approx). It consists of National highways, State highways; major/other district roads and rural roads. • NH 44 (3745 km) is the longest highway of India (Srinagar to Kanyakumari). • NH 548 is the shortest National Highways with the length of 5 km. • The North-South and East-West Corridor (NS-EW) is the largest ongoing expressway project in India. It is the second phase of the National Highways Development Project (NHDP) and involves building 7300 km of six lane expressway connecting Srinagar, Kanyakumari, Porbandar and Silchar. • Maharashtra has the maximum length of surfaced roads in India. • Eastern Peripheral Expressway or Kundli-Ghaziabad-Palwal Expressway is a 6-lane expressway passing through the states of Haryana and Uttar Pradesh. • India’s longest greenfi eld six lane expressway, named as AgraLucknow expressway was inaugurated in Uttar Pradesh.

58

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Bharat Mala It is a major highway, economic corridor and expressway development scheme of Government in India. Launched in 2015, it is the biggest road Construction Plan in the country (approximately 83,677 km). Government of India has decided to construct a greenfi eld major port at Vadhavan in Gujarat under Sagarmala Project.

Important National Highways (New Numbering) NH NH 1 NH 4 NH 7 NH 10 NH 21 NH 32 NH 40 NH 44

Connects Uri-Barmula-Srinagar-Kargil-Leh Mayabandar-Port Blair-Chiriyatapu Fazilka-Patiala-Rudraprayag-Mana Siliguri-Gangtok Jaipur-Agra-Bareilly Chennai-Puducherry-Nagapatinam Kurnool-Chittoor-Ranipet Srinagar-Ludhiana-Agra-Sagar-HyderabadKanyakumari

Water Transport

As per the National Waterways Act, 2016, 111 Waterways have been declared as National Waterways including the six existing NWs given below: NW1 Allahabad to Haldia on Ganga river 1620 km NW2 Sadia to Dhubri on Brahmaputra river 891 km NW3 Kollam to Kottapuram (along Champakara 168 km and Udyogmandal Canal) NW4 Kakinada to Marak-kanam along Godavari 1095 km and Krishna river NW5 Manglagarhi to Paradeep and Talcher to 623 km Dhamara along Mahanadi and Brahmani NW6 Lakhipur to Bhanga on Barak river 121 km

13 Major-Ports in India Western Coast

Eastern Coast

Kandla (Child of Partition) Paradip (exports raw iron to Gujarat Japan) Odisha Mumbai (busiest and biggest) Vishakhapatnam (deepest port) Maharashtra Andhra Pradesh

JL Nehru (fastest growing) Chennai (oldest and artifi cial) Maharashtra Tamil Nadu Marmagao (naval base also) Ennore (most modern in private Goa hands) Tamil Nadu Mangalore (exports Kudremukh Tuticorin (Southernmost) Tamil Nadu iron-ore) Karnataka Cochin (natural harbour) Kerala

Port Blair (strategically important) Andaman and Nicobar Islands Enayam Port (Tamil Nadu)

Note: Kandla port was renamed as Pt. Deen Dayal Upadhyay Port in 2017.

Important Indian Towns on Rivers Town Jamshedpur Delhi Kanpur Surat Ferozpur Prayagraj Varanasi Haridwar Badrinath Ludhiana Srinagar Ayodhya Ahmedabad Patna Kota Jabalpur Panji Ujjain Guwahati Kolkata Cuttack Hyderabad Nasik Lucknow

River Subarnarekha Yamuna Ganga Tapti Sutlej At the confl uence of the Ganga and Yamuna Ganga Ganga Alaknanda Sutlej Jhelum Saryu Sabarmati Ganga Chambal Narmada Mandavi Kshipra Brahmaputra Hooghly Mahanadi Musi Godavari Gomti

Important River Projects and their Beneficiary States Important

River

Purpose

Beneficiary States

Bhakra Nangal Project

Sutlej

Power and irrigation

Punjab, Himachal Pradesh, Haryana and Rajasthan

Damodar Valley

Damodar

Power, irrigation and fl ood control

Jharkhand and West Bengal, shared by Madhya Pradesh

Hirakud

Mahanadi

Power and irrigation

Odisha

Tungabhadra Project

Tungabhadra

Power and irrigation

Andhra Pradesh and Karnataka

Nagarjunasagar Project

Krishna

Power and irrigation

Andhra Pradesh and Telangana

Gandak River Project

Gandak

Power and irrigation

Bihar, Uttar Pradesh, Nepal (joint venture of India and Nepal)

Kosi Project

Kosi

Flood control; Power and irrigation

Bihar

Farakka Project

Ganga, Bhagirathi

Power, irrigation, avoid accumulation of silt to improve navigation

West Bengal

Beas Project

Beas

Irrigation and power

Rajasthan, Haryana, Punjab and Himachal Pradesh

Indira Gandhi Canal Project (Rajasthan Canal Project)

Sutlej, Beas and Ravi

Irrigation

Rajasthan, Punjab and Haryana

59

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

Important

River

Purpose

Beneficiary States

Chambal Project

Chambal

Power and irrigation

Madhya Pradesh and Rajasthan

Kakrapara Project

Tapti

Irrigation

Gujarat

Ukai Project

Tapti

Power and irrigation

Gujarat

Tawa Project

Tawa (Narmada)

Irrigation

Madhya Pradesh

Poochampad Project

Godavari

Irrigation

Telangana

Malaprabha Project

Malaprabha

Irrigation

Karnataka

Durgapur Barrage

Damodar

Irrigation and navigation

West Bengal and Jharkhand

Mahanadi Delta Project

Mahanadi

Irrigation

Odisha

Iddukki Project

Periyar

Hydroelectricity

Kerala

Koyna Project

Koyna

Hydroelectricity

Maharashtra

Power and irrigation

Uttar Pradesh and Uttarakhand

Multipurpose power and irrigation

Uttar Pradesh and Madhya Pradesh

Ramganga Project

Multipurpose Ramganga near Kalagarh

Matatila Project

Betwa

Tehri Dam Project

Bhilangana, Bhagirathi Hydroelectricity

Uttarakhand

Rihand Scheme

Rihand

Hydroelectricity

Uttar Pradesh

Kundah Project

Kundah/Bhavani

Hydroelectricity and irrigation

Tamil Nadu

Important Lakes of India Name of Lake

State/UTs

Important Fact

Chilika Lake

Odisha

It is largest brackish water lake of India.

Kolleru Lake

Andhra Pradesh

It is a freshwater lake.

Loktak Lake

Manipur

It is freshwater lake having inland drainage.

Lonar Lake

Maharashtra

It is a meteorite crater lake in Buldhana area of Maharashtra. The water is highly charged with Sodium carbonates and Sodium chloride.

Pangong Lake

Jammu and Kashmir

It is a salty lake.

Pulicat Lake

Tamil Nadu & Andhra Pradesh border

It is a saline and lagoon lake.

Sambhar Lake

Rajasthan

It is a shallow lake which is saline, located near Jaipur.

Tso Moriri Lake

Jammu & Kashmir

It is a salty lake.

Vembanad Lake

Kerala

It is a lagoon lake and largest lake by surface area.

Wular & Dal Lakes

Jammu and Kashmir

Wular lake was created due to tectonic activities and is largest fresh water lake of India.

Some of the Important Waterfalls of India Waterfall

Height (km)

River

State

Kunchikal

455

Varahi

Karnataka

Jog/Gersoppa

260

Sharavati

Karnataka

Rakim Kund

168

Gaighai

Bihar

Chachai

127

Bihad

Madhya Pradesh

Kevti

98

Mahana

Madhya Pradesh

Sivasamudram

90

Cauveri

Karnataka

Chief Crops and Producing States Type

Name

Major Producers

Cereals

Wheat Rice Gram Barley Bajra

Uttar Pradesh, Punjab and Madhya Pradesh West Bengal and Uttar Pradesh Madhya Pradesh, Maharashtra and Rajasthan Maharashtra, Uttar Pradesh and Rajasthan Rajasthan, Maharashtra and Gujarat

Cash Crops

Sugarcane Poppy

Uttar Pradesh and Maharashtra Uttar Pradesh and Himachal Pradesh

60

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Oil Seeds

Coconut Linseed Groundnut Rapeseed and Mustard Sesame Sunfl ower

Kerala and Tamil Nadu Rajasthan, Madhya Pradesh and Haryana Gujarat, Andhra Pradesh and Tamil Nadu Rajasthan, Madhya Pradesh and Haryana Gujarat, West Bengal and Karnataka Karnataka, Andhra Pradesh and Maharashtra

Fibre Crops

Cotton Jute Silk Hemp

Maharashtra and Gujarat West Bengal and Bihar Karnataka and Kerala Madhya Pradesh and Uttar Pradesh

Plantations

Coffee Rubber Tea Tobacco

Karnataka and Kerala Kerala and Karnataka Assam and Kerala Gujarat, Maharashtra and Madhya Pradesh

Spices

Pepper Cashewnuts Ginger Turmeric

Kerala, Karnataka and Tamil Nadu Kerala, Tamil Nadu and Andhra Pradesh Kerala and Uttar Pradesh Andhra Pradesh and Odisha

Forests of India Forest Type

Distribution

Climatic Conditions

Characteristics

Species

Tropical Evergreen Forests

• Rainy slopes of Western Ghats. • NE India except Arunachal Pradesh. • Eastern part of West Bengal and Odisha. • Andaman and Nicobar Islands.

• Rainfall > 200 cm • Relative Humidity > 70% • Average temperature is about 24°C. • Hot and humid climate.

• Height of trees is 40 to 60 m. • Leaves are dark green and broad.

Mahogany, Mahua, Bamboo, Cones, Ironwood, Kadam, Irul, Jamun, Hopea, Rubber tree, Toon, Telsur, etc.

Tropical Moist Deciduous Forests

• Eastern parts of Sahyadris (Western Ghats). • North Eastern part of Peninsula. • Middle and lower Ganga valley. • Foothills of Himalayas in Bhabar and Tarai region. • These cover about 20% India’s forest area.

• 100 to 200 cm rainfall per annum. • Moderate temperature.

• 30 to 40 m high trees. • Due to defi ciency of water, they shed their leaves in spring (onset of summer).

Sal, Teak, Arjun, Mulberry, Kusum, Sandalwood, Siris, Haldi, Khair, Mango, Banyan tree, etc.

Tropical Dry Deciduous Forests

• Large parts of Maharashtra and Andhra Pradesh. • Parts of Punjab, Haryana and Eastern parts of Rajasthan. • Northern and Western parts of Madhya Pradesh. • Tamil Nadu. • Southern parts of Uttar Pradesh.

• 50 to 100 cm rainfall. • Moderate humidity.

• 6 to 15 m high. • Roots are thick and long.

Teak, Sal, Bamboo, Mango, Acacia, Neem, Shisham, etc.

Dry Forests or Arid Forests

• Rajasthan and adjoining areas of Haryana, Gujarat and Punjab. • Rainshadow area of peninsular India.

• Low rainfall (less than 50 cm per annum). • Relative humidity is less.

• Thorny vegetation. • Roots are very long. • Leaves are small.

Cactus, Thorny bushes, Kikar, Babool, Date palm, Acacia, Khair, Euphorbias, etc.

Objective Type Questions 1. Match List - I with List - II. List - I (Boundary Lines) (1) 38th Parallel Line (I) (2) Durand Line (II) (3) MacMohan Line (III)

(2023) List - II (Countries Sharing borders) Between U.S.A and Canada Between India and China Between North and South Korea

(4)

49th Parallel Line (IV)

Between India and Afghanistan

Choose the most appropriate answer from the options given below: (1) (1)-(II), (2)-(III), (3)-(IV), (4)-(I) (2) (1)-(III), (2)-(IV), (3)-(II), (4)-(I) (3) (1)-(IV), (2)-(III), (3)-(II), (4)-(I) (4) (1)-(III), (2)-(IV), (3)-(I), (4)-(II)

61

GENERAL AWARENESS : WORLD AND INDIAN GEOGRAPHY

2. The central rice research station is situated in (2023) (1) Raipur (2) Chennai (3) Bangalore (4) Cuttak 3. Which of the following country shares the shortest border with India? (2023) (1) Bhutan (2) Myanmar (3) Nepal (4) Pakistan 4. The only continent through which Tropic of Cancer, Equator and Tropic of Capricorn passes? (2023) (1) Asia (2) Australia (3) Antarctica (4) Africa 5. _______ of the total surface area of the earth is covered with water? (2022) (1) 72.5% (2) 66.7% (3) 70.8% (4) 71.6% 6. The mating and nesting around of 50% of the world’s olive ridley turtles, the smallest and abundant sea turtles, is located in: (2021) (1) Goa coast (2) Odisha coast (4) Gulf of Kutch (3) Malabar coast 7. Which of the following is the most toxic air pollutant? (2021) (1) Arsenic (2) Asbestos (3) Benzene (4) Potassium chloride 8. What is meant by the term “mid-night sun”? (1) Twilight (2) Rising Sun (3) Very bright moon (4) Sun shining in the polar circle for long time 9. As one moves from the polar region towards equator the diversity of plant and animal, species: (1) increases (2) decreases (3) no changes (4) No option is correct. 10. Who had used the term “Ecology” fi rst time? (1) Charles Darwin (2) Robert Whittaker (3) Ernst Haeckel (4) Arthur Tansley 11. Algae belongs to which of the following levels of the ecosystem? (1) Decomposers (2) Destructors (3) Producers (4) Consumers 12. To which habitat is the turtle Petri adapted? (1) Desert (2) Forest (3) Sea (4) Marshes 13. With what bio-region is the term “Steppe” associated? (1) Grasslands (2) Tropical forests (3) Savanna (4) Coniferous forests 14. ‘Terra rossa’ is a Latin word which means: (1) hot area (2) red terrain (3) lateritic region (4) region near the poles 15. According to Census 2011, which state has the highest density of population? (1) West Bengal (2) Bihar (3) Uttar Pradesh (4) Maharashtra 1. (2) 9. (1)

Answer Key 2. (4) 3. (1) 4. (4) 5. (3) 6. (2) 7. (3) 8. (4) 10. (3) 11. (3) 12. (3) 13. (1) 14. (2) 15. (2)

Answers with Explanations 1. Option (2) is correct. Boundary Lines 38th Parallel Line Durand Line MacMohan Line 49th Parallel Line

Countries sharing boundaries Between North Korea and South Korea India and Afghanistan India and China USA and Canada

2. Option (4) is correct. The Central Rice Research Institute is situated in Cuttack, Odisha. It was established on 23rd April 1946 at Bidyadharpur,

Cuttack, Odisha with an experimental farm land of 60 hectares provided by the Government of Odisha. Its founder Director was Dr. K Ramiah, who was an eminent rice breeder. In 1966, the administrative control of the Institute was transferred to the Indian Council of Agricultural Research (ICAR). CRRI has two research stations- rice research on rainfed upland ecologies at Hazaribag, Jharkhand and rice research on fl ood prone rainfed lowland ecologies at Gerua, Assam. 3. Option (1) is correct. Of the given options, India shares the shortest border with Bhutan. Although, India shares the shortest border with Afghanistan of 106 km, the longest border with Bangladesh of 4096.7 km. Bordering Countries Bhutan Myanmar Nepal Pakistan

Length of the border West Bengal, Sikkim, Aruna- 699 km chal Pradesh and Assam Arunachal Pradesh, Manipur, 1643 km Mizoram, and Nagaland Sikkim, West Bengal, Uttar 1751 km Pradesh and Uttrakhand Jammu & Kashmir, Punjab, 3323 km Rajasthan, and Gujarat States

4. Option (4) is correct. Africa is the only continent through which the Equator, Tropic of Cancer and Tropic of Capricorn passes. Africa is the second largest continent after Asia with an area of 30.3 million kilometre square. Equator is the imaginary 0 degree latitude that divides the Earth into Northern and Southern Hemispheres. It is the longest latitude on the Earth. The Tropic of Cancer is 23 degrees 27 minutes latitude that lies in the Northern Hemisphere. It is the northernmost point where the Sun can be directly overhead at noon during the summer solstice in the Northern Hemisphere. The Tropic of Capricorn is 23 degrees 27 minutes latitude that lies in the Southern Hemisphere. It is the southernmost point where the Sun can be directly overhead at noon during the summer solstice in the Southern Hemisphere. 5. Option (3) is correct. About 70.8 % of the Earth’s surface is water-covered. This is in various forms such as oceans, rivers, lakes, glaciers and ice sheets. The oceans hold about 96.5 percent of all Earth’s water. 6. Option (2) is correct. The coast of Orissa in India is the largest mass nesting site for the Olive-ridley, followed by the coasts of Mexico and Costa Rica. The Olive Ridley turtles are the smallest and most abundant of all sea turtles found in the world. These turtles got their name from their olive-colored carapace, and they are carnivores in nature. They are found in warm waters of the Pacifi c, Atlantic and Indian oceans. The Gahirmatha Marine Sanctuary located in Odisha is known as the world’s largest breeding colony (rookery) of sea turtles. Protection Status: 1. Wildlife Protection Act, 1972: Scheduled 1 2. IUCN Red List: Vulnerable 3. CITES: Appendix I 7. Option (3) is correct. Hazardous air pollutants, also known as toxic air pollutants, are those pollutants that are known or suspected to cause cancer or other serious health effects. Benzene is the most toxic air pollutant among the given options. Examples of toxic air pollutants include: 1. benzene, found in gasoline;

62 2. perchloroethylene, emitted from some dry cleaning facilities; and 3. methylene chloride, used as a solvent and paint stripper by a number of industries. Other air toxics include dioxin, asbestos, toluene, and metals such as cadmium, mercury, chromium, and lead compounds. 8. Option (4) is correct. The term midnight sun is above the horizon at midnight in the arctic or antarctic summer. 9. Option (1) is correct. As one drives from the polar part towards the equator, the diversity of plant and animal species rises because the warmer temperature about the equator delivers a tremendous quantity of resources, better solar energy, and warmer temperatures demanded by ectothermic species. 10. Option (3) is correct. Ecology is described as the extension of analysis concerning the connection of an organism with one another and with the physical surroundings. 11. Option (3) is correct. Most varieties of algae are categorized as producers within an ecosystem because they can make their food.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

12. Option (3) is correct. Petric island is residence to the standard smudged turtle and Snapping. They are adjusted to living in the sea, swampy waters, and sandy river bases. 13. Option (1) is correct. Steppe is connected with Grasslands. The Steppe is a dry, cold grassland developed in all continents besides Australia and Antarctica. 14. Option (2) is correct. ‘Terra Rossa’ is a Latin term that represents Red Landscape. When weathering limestone rock carries, the Clay retained in the stones is left-back, and this red clay soil is named ‘Terra Rossa’. This kind of red soil is likewise made by weathering of dolomite rock. 15. Option (2) is correct. Bihar has the most elevated inhabitants viscosity of 1106 individuals per sq. km. with a resident’s thickness of 1028 individuals per sq. km. Uttar Pradesh’s is 829 persons per sq. km, Kerala’s is 860 persons per sq. km

Constitution

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

63

Chapter

3

Indian Polity and Constitution

Chapter Analysis Concept Name Indian Polity and Constitution

 Revision Notes Framing of the Indian Constitution • The idea to have a Constitution was first Scan to know more about given by MN Roy (A pioneer of Communist this topic Movement in India). • The Constitution was framed by the Constituent Assembly of India, set-up on 16th May 1946, in accordance with the Cabinet Mission Plan, under the chairmanship of Indian Sachidanand Sinha, initially. Dr. Rajendra Constitution Prasad and HC Mukherjee were elected as the President and Vice-President respectively on 11th December 1946. BN Rau was appointed as the Constitutional Advisor. • The total membership of the Constituent Assembly was 389, of these 292 were representatives of British States; 93 were representatives of Princely States and 4 were from the Chief Commissioner’s Provinces of Delhi, Ajmer-Merwara, Coorg and British Baluchistan. • The Chairman of the Drafting Committee was Dr. BR Ambedkar, also known as the Father of the Indian Constitution.

Making of the Constitution • The Constituent Assembly was framed in 1946, under the scheme formulated by Cabinet Mission Plan. • There were total 389 members, out of these 296 were elected to represent the British India and 93 seats were allotted to the Princely states. • 292 members, out of 296 members, were to be elected by the provincial legislatures while four members were to represent the chief commissioner’s provinces of Delhi, Ajmer – Merwara, Coorg and British Baluchistan. • The first meeting of Constituent Assembly was held on 9th December, 1946, and reassembled on 14th August, 1947, as the sovereign Constituent Assembly for the dominion of India. • It took 2 years 11 months and 18 days to finalise the Constitution. • On March 12, 1949, Objective Resolution was adopted. It was moved in the first session of the Constituent Assembly (on 13th December, 1946), by Pandit Jawaharlal Nehru. • Dr. Sachidanand Sinha became the provincial President of the Constituent Assembly. • Dr. Rajendra Prasad and HC Mukherjee were elected as the permanent President and Vice President of the Assembly respectively. • BN Rau was appointed as the constitutional advisor of the Assembly. • The Drafting Committee was appointed on 29th August, 1947 • Dr. BR Ambedkar who was the chairman of the Drafting Committee, submitted a Draft Constitution of India to the President of the Assembly on 21st February, 1948. • Other members of drafting committee were N Gopal Swamy Ayyangar, Alladi Krishna Swami Ayyar, KM Munshi, Mohammad Saad-ul-lah, N Madhava Rau and TT Krishnamachari.

2021

2022

2023

Additional Questions

3

1

3

8

Interim Government (1946) Members

Portfolios Held

Jawaharlal Nehru

External Affairs & Commonwealth Relations

Sardar Vallabhbhai Patel

Home, Information & Broadcasting

Dr Rajendra Prasad

Food & Agriculture

Dr John Mathai

Industries & Supplies

Jagjivan Ram

Labour

Sardar Baldev Singh

Defence

CH Bhabha

Works, Mines & Power

Liaquat Ali Khan

Finance

Abdur Rab Nishtar

Posts & Air

Asaf Ali

Railways & Transport

C Rajagopalachari

Education & Arts

II Chundrigar

Commerce

Ghaznafar Ali Khan

Health

Joginder Nath Mandal

Law

Note: Interim government was formed from the newly elected Constituent Assembly.

Enactment of the Constitution • The Constituent Assembly took 2 years, 11 months and 18 days to complete the Constitution. • Some of the provisions related to Citizenship, Elections, Provisional Parliament etc. were given immediate effect. • The Constitution was adopted on 26th November 1949, contained a Preamble, 395 Articles divided into 22 Parts and 8 Schedules. Presently, it has 448 Articles divided into 25 Parts and 12 Schedules. • The enforcement of Constitution was delayed till 26th January because, in 1929, on this day Indian National Congress demanded Poorna Swaraj in Lahore Session, Chaired by JL Nehru. • The Constitution came into force on 26th January, 1950, known as Republic Day of India. The Constituent Assembly adopted our National Flag on 22nd July, 1947. It was designed by Pingali Venkayya.

Preamble

• It is the preface or the introduction of the Constitution. It is an integral part of the Constitution. The interpretation of the Constitution is based on the spirit of the Preamble. • The Objective Resolution was drafted and moved by Pandit Jawaharlal Nehru and adopted by the Constitution Assembly, consequently became the Preamble. • The idea of the Preamble was borrowed from the Constitution of USA. • The words, Socialist, Secular and Integrity were added by the 42nd Constitutional Amendment Act in 1976.

65

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

The Preamble

WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN, SOCIALIST, SECULAR, DEMOCRATIC, REPUBLIC and to secure to all its citizens: JUSTICE, Social, Economic and Political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and of opportunity; and to promote among them all ; FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT, AND GIVE TO OURSELVES THIS CONSTITUTION.

Sources of the Indian Constitution UK

USA

• • • • • • • •

Rule of Law Cabinet System Prerogative Writs Parliamentary Government Bicameral Parliament CAG Offi ce Single Citizenship Law making procedures

• Written Constitution • Vice-President as the Ex-offi cio Chairman of Upper House • Fundamental Rights • Supreme Court • Independence of Judiciary and Judicial Review • Preamble

Parts of the Constitution Part-I (Article 1-4):

Deals with territory of India, formation of new states, alterations of boundary, names of existing states.

Part-II (Article 5-11):

Deals with various provisions of citizenship.

Part-III (Article 12-35):

Deals with Fundamental Rights of Indian citizens. (Article 31-Dealing with the right to property was deleted by 44th amendment.)

Part-IV (Article 36-51):

Deals with Directive Principles of State Policy.

Part-V (Article 52-151):

Deals with Government at the Union level. Duties and functions of Prime Minister, Ministers, President, Attorney General, Parliament – Lok Sabha and Rajya Sabha, Comptroller and Auditor General.

Part-VI (Article 152-237):

Deals with Government at state level. Duties and Functions of Chief Minister and his Ministers, Governor, State Legislature, High Court, Advocate general of the state.

Part-VII (Article 238):

Deals with States, was dated in 1956, by the 7th Amendment.

Part-VIII (Article 239- Deals with Union Territories. 241): Part-IX (Article 243- Deals with Panchayats. Added by 243-0): 73rd Amendment Act in 1992.

• Fundamental Duties

Part-IX-A (Article Deals with the Municipalities. Added 243P-243ZG): by 74th Amendment Act in 1992.

Australia

• Concurrent List • Joint sitting of Parliament

Japan

• Procedure established by law

Part-IX-B (Article The Cooperative Societies (243243ZH-243ZT): ZH to 243-ZT) (added by 97th Amendment Act, 2011).

Germany

• Suspension of Fundamental Rights during the Emergency

Erstwhile USSR

Canada

Ireland

• Scheme of federation with a strong Centre • Distribution of powers between the Centre and the States and placing Residuary Powers with the Centre • Concept of Directive Principles of State Policy • Method of election of the President

South Africa • Procedure for amendment of the constitution and election of member of Rajya Sabha France

• Republic and the ideals of Liberty, Equality and Fraternity in the Preamble

Main Features • Bulkiest Written Constitution in the World • Combination of Rigidity and Flexibility • Parliamentary System of Government • Federal System with a Unitary bias • Fundamental Rights and Duties • Directive Principles of the State Policy • Integrated and Independent Judiciary • Single Citizenship • Emergency Powers • Universal Adult Franchise

Part-X (Article 244- Relations between the centre and the 244A): states. Part-XI (Article 245- Deals with distribution of revenue 263): between Union and States. Part-XII (Article 264- Deals with distribution of 300A): revenue between Union and States, Appointment of Financial Commission (Article 280), contracts liabilities etc. Part-XIII (Article 301- Relates to Trade, Commerce and 307): Intercourse within the Territory of India. Part-XIV (Article 308- Deals with UPSC and State Public 323): Service Commission. Part-XIV A (Article Deals with Tribunals (Added by 42nd Amendment). 323A-323B): Part-XV (Article 324- Deals with Elections (Also Election 329A): Commission). Part-XVI (Article 330- Deals with special provisions for 342): Scheduled Castes and Scheduled Tribes and Anglo-Indian Representation. Part-XVII (Article 343- Offi cial language. 351): Part-XVIII (Article 352- Deals with Emergency Provisions. 360):

66

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Part-XIX (Article 361- Exemption of criminal proceeding 367): for their Offi cial Acts of President and Governors and Miscellaneous. Part-XX (Article 368):

Deals with Amendment of the Constitution

Part-XXI (Article 369- Temporary, Transitional and Special 392): Provision (Article 369 gives temporary powers to the Parliament to make laws for state list).

Fazal Ali Commission

• The States Reorganisation Commission (SRC) was constituted by the Central Government of India in August 1953 to recommend the reorganisation of state boundaries. • In October 1955, after two years of study, the Commission, comprising Justice Fazal Ali, K. M. Panikkar and H. N. Kunzru, submitted it’s report.

Union Territories

Name of States and UTs and their Territorial Extent.

• Jammu & Kashmir, Ladakh, National Capital Territory of Delhi, Puducherry, Andaman & Nicobar are headed by the Lieutenant Governor. Daman and Diu, Dadar and Nagar Haveli have a common administrator. • From 26th January 2020 both Daman and Diu and Dadar and Nagar Haveli have been merged into single UT named Dadar and Nagar Haveli and Daman and Diu. • Lakshadweep, Chandigarh is also governed by an Administrator. • J&K, Delhi and Puducherry have legislative assemblies. • There are total 8 Union Territories – Delhi, Puducherry, Daman and Diu & Dadar and Nagar Haveli, Chandigarh, Lakshadweep, Andaman and Nicobar Islands, Jammu and Kashmir and Ladakh. • By the 69th Constitutional Amendment Act 1991, Delhi was given the status of National Capital Territory of India.

2nd Schedule

Salaries and Emoluments of President, Governor etc.

3rd Schedule

Forms of Oath and Affi rmations of Members of Legislatures, Ministers and Judges.

Citizenship

4th Schedule

Allocation of seats in the Rajya Sabha.

5th Schedule

Administration and control of Scheduled Areas and Scheduled Tribes.

6th Schedule

Administration of Tribal Areas in the State of Assam, Meghalaya, Tripura and Mizoram.

7th Schedule

Distribution of Powers between the Union and the State Government.

7th Schedule

Languages (22 scheduled languages)

9th Schedule

Validation of Certain Acts and Regulations (added by 1st Amendment, 1951).

Part-XXII (Article 393- Concerns the short title, 395): commencement, authoritative text in Hindi and repeals.

Amendability of the Preamble

• The Preamble has been amended only once. That is, by 42nd constitutional amendment act, 1976 when three new terms were added – Socialist, Secular and Integrity.

Schedules of the Constitution Schedule st

1 Schedule

Subject

10th Schedule Anti-Defection Law Amendment, 1985). 11th Schedule

(added

by

52nd

Powers, authority and responsibilities of Panchayats. It has 29 matters (added by 73rd Amendment, 1992).

12th Schedule Powers, authority and responsibilities of Municipalities. It has 18 matters (added by 74th Amendment, 1992).

Union and Its Territory

• Articles 1-4 under Part-I of the Constitution deals with the Union and its Territories. • Article 1, describes India as a Union of States. • Territories of India are classifi ed into three categories – 1. Territories of the States 2. Union Territories 3. Territories that may be acquired by the Government of India at any time. • At present there are 28 states and 8 union Territories, after unifi cation of Daman and Diu and Dadar and Nagar Haveli.

Evolution of States and Union Territories Dhar Commission

• On June 17, 1948, Rajendra Prasad, the president of the Constituent Assembly, set up the Linguistic Provinces Commission (aka Dhar Commission) to recommend whether the states should be reorganised on linguistic basis or not.

• Indian Constitution deals with the citizenship from articles 5-11 under Part II. Articles 5 to 8 deal with the citizenship of (a) Persons domiciled in India (b) Persons migrated from Pakistan (c) Persons Migrated to Pakistan but later returned (d) Persons of Indian origin residing outside India.

Acquisition of Citizenship • A citizen of a country enjoys full membership of the country. • Indian Constitution provides a single and uniform citizenship for the entire country. • The Citizenship Act of 1955, prescribes fi ve ways of acquiring citizenship viz birth, decent, registration, naturalisation and incorporation of territory.

1. By Birth

• Every person born in India, on or after 26th January, 1950 but before 1st July, 1987 shall be a citizen by birth, irrespective of the nationality of his parents. • Every person born in India between 1st July, 1987 and 2nd December, 2004, shall be citizen of India, given either of his/ her parents is a citizen of India. • Every person born in India on or after 3rd December, 2004, shall be citizen of India, if at least one parent is citizen of India and the other is not an illegal migrant at the time of birth. • Enemy aliens and the children of foreign diplomats posted in India cannot acquire Indian citizenship by birth.

2. By Descent

• A person born outside India on or after 26th January, 1950, is a citizen of India by descent if his/her father was a citizen of India by birth. • A person born outside India on or after 10th December, 1992 is a citizen of India by descent if his/her either parent was a citizen of India by birth.

3. By Registration

• The Central Government may, on an application, register as a citizen of India any person if he belongs to any of the following categories-

67

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

(a) A person of Indian origin, residing in India for 7 years. (b) A person of Indian origin, who is ordinarily resident in any country or place outside undivided India. (c) Persons, who are married to citizens of India and resident of India for 7 years. (d) Minor children of persons, who are citizen of India.

4. By Naturalisation

• It can be acquired by a foreigner, who has resided in India for 12 years.

5. By Incorporation of Territory

• If any new territory become a part of India, the Government of India specifi es the person of that territory to be citizens of India. • For example, Pondicherry in 1962.

Loss of Citizenship • The Citizenship Act, 1955 prescribes three ways of losing Citizenship—

1. By Renunciation

If a person gives up his Indian Citizenship voluntarily.

2. By Termination

W hen an Indian citizen voluntarily acquires the citizenship of another country, his citizenship can be terminated by Government of India by law.

• Freedom from attendance of religious instructions or religious worship in certain educational institutions (Article 28).

Cultural and Educational Rights

• Protection of interest of minorities (Article 29). • Right of minorities to establish and administer educational institutions (Article 30). Freedom of Press is implicit in the Article 19. Article 20 and 21 cannot be suspended even during National Emergency (Article 352). Right to Property under Article 19 (1) (f) was repealed by the 44th Amendment Act, 1978, and was made a legal right under Article 300A.

Rights to Constitutional Remedies

• Right to move to the Supreme Court (Article 32) and the High Courts (Article 226) in case of violation of the Fundamental Rights. BR Ambedkar called Article 32 as the Heart and Soul of the Constitution. • 5 Writs of Habeas Corpus, Mandamus, Prohibition, Certiorari and Quo-Warranto can be issued under this.

Types of Writs Writ

Meaning

Habeas Corpus

D eprivation of citizenship by the Government of India on the basis of acquisition of citizenship by fraud, helping an enemy during a war or being disloyal to the constitution.

You may have To release a person who has been the body detained unlawfully whether in prison or in private custody.

Mandamus

We command

Fundamental Rights

To secure the performance of public duties by lower court, tribunal or public authority.

Certiorari

To be certifi ed

To quash the order already passed by an inferior court, tribunal or quasi judicial authority.

Prohibition

The act of stopping something

To prohibit an interior court from continuing the proceedings in a particular case where it has no jurisdiction.

Quo Warranto

What is your To restrain a person from holding authority a public offi ce for which he is not entitled.

3. By Deprivation

Right to Equality (Article 14–18)

• Equality before law and equal protection of the laws (Article 14). • Prohibition of discrimination on grounds only of religion, race, caste, sex, place of birth only (Article 15). • Equality of opportunity in matters of public employment (Article 16). • Abolition of untouchability (Article 17). • Abolition of titles (Article 18).

Right to Freedom (Article 19–22)

• Protection of certain rights regarding speech and expression, assembly, association, movement, residence, and profession (Article 19). • Protection in respect of conviction for offences (Article 20). • Protection of life and personal liberty (Article 21). • Protection against arrest and detention in certain cases (Article 22).

Right to Education

• Article 21A states that the state shall provide free and compulsory education to all children of the age of 6–14 years.

Rights against Exploitation (Article 23–24)

• Prohibition of human traffi cking and forced labour (Article 23). • Prohibition of employment of children in any factories, etc (Article 24).

Rights to Freedom of Religion (Article 25-28)

• Freedom of conscience and right to profess, practice and propagate one’s religious beliefs. (Article 25) • Freedom to manage religious affairs (Article 26). • Freedom from taxation for promotion of any particular religion (Article 27).

Intended Purpose

Directive Principles of State Policy

• The Directive Principles of State Policy are enumerated in Part IV of the constitution from Articles 36 to 51. • Dr. BR. Ambedkar described these principles as ‘novel features’ of the Constitution. • The Directive Principles of State Policy along with Fundamental Rights contain the philosophy and the soul of the Constitution, some of the important DPSP’s are (i) State to secure a social order for the promotion of welfare of the people (Article 38). (ii) Organisation of village Panchayats (Article 40). (iii) To secure right to work, education and public assistance in cases unemployment, old age etc. (Article 41). (iv) Uniform Civil Code for the citizens (Article 44). (v) Promotion of educational and economic interests of Scheduled Castes, Scheduled Tribes and other weaker sections (Article 46). (vi) Separation of Judiciary from Executive (Article 50). (vii) Promotion of international peace and security (Article 51).

Fundamental Duties

According to Article 51A, it shall be the duty of every citizen of India, to –

68 (a) Abide by the constitution and respect National Flag & National Anthem. (b) Follow the ideals of the freedom struggle. (c) Protect sovereignty and integrity of India. (d) Defend the country and render national services whenever called upon. (e) Promote spirit of common brotherhood. (f) Preserve composite culture. (g) Preserve natural environment (h) Develop scientifi c temper. (i) Safeguard Public Property. (j) Strive for excellence. (k) Provide opportunities for education to his child or as the case may be, ward between the age of six to fourteen years by a parent or guardian.

Union Government The President

• President is the executive head of the state and the fi rst citizen of India. • The 42nd Amendment of the constitution has made it obligatory on the part of the president to accept the advice of the council of Ministers. However, 44th amendment Act amended the word ‘obligatory’ and added that ‘President can send the advice for reconsideration’. • Qualifications: The person must be a citizen of India; His age should be 35 years and; should be eligible to be member of the Lok Sabha and must not hold any offi ce of profi t. • Election: A President is indirectly elected through Electoral College consisting of elected members of both the Houses of the Parliament and elected members of the Legislative Assemblies of the States and elected members of the Legislative Assemblies of Union Territories of Delhi and Puducherry. • The members of the Legislative Councils have no right to vote in Presidential election. • Tenure: The term is 5 years though there is no upper limit on the number of times a person can become the President (Article 57). He can give resignation to the Vice-President before the full-term. • President’s monthly salary is 5,00,000. • The Vice-President acts as the President, in case the President’s offi ce falls vacant. • If he is not available, then Chief Justice of India, if not then the senior most Judge of the Supreme Court shall act as the President of India. • The fi rst Chief Justice of India appointed as President was Justice M Hidayatullah (July 1969 – August 1969).

Powers

• The president appoints Prime Minister, other Ministers, Chief Justice and Judges of Supreme Court, High Courts, the Attorney General of India, the Comptroller and Auditor General, Chairman and Members of UPSC, Chief Election Commissioner and other Election Commissioners of the Election Commission, Governors, Members of Finance Commission, etc. • He can summon a joint sitting of both the Houses of Parliament, which is presided by the Speaker of Lok Sabha. • He nominates 12 members of Rajya Sabha (from amongst the persons who have special knowledge in fi eld of literature, science, art and social service). • Money Bills can be introduced in the Parliament only with his prior recommendation. • He also have the powers of Pardon, Reprieve, Remission, Respite and Commutation (Article 72). • He is supreme commander of the defence forces in India and appoints the Chiefs of the Army, the Navy and the Air Force. • He can proclaim emergency in three conditions after getting the

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

written recommendation of the Cabinet. (1) National Emergency (Article 352) – arising out of war, external aggression or armed rebellion within the country. (2) Constitutional Emergency (Article 356) – arising out of the failure of the constitutional machinery in the states. It is also known as President’s Rule or State Emergency. (3) Financial Emergency (Article 360) – arising out of threat to fi nancial stability or credit of India.

Vice-President

• Article 63 of the Constitution stipulates there shall be a VicePresident for India. • The Vice-President is indirectly elected by an electoral college, consisting of members of both the Houses of Parliament. • He can resign from his offi ce at any time by addressing the resignation to the President. • He acts as the ex-offi cio Chairman of the Rajya Sabha. In this capacity, his powers and functions are similar to those of speaker of Lok Sabha. Note: For elections of the President and the Vice-President election petitions can only be fi led with the Supreme Court.

Prime Minister

• In the scheme of Parliamentary system of government, the President is the nominal executive authority (de jure executive) and Prime Minister is the real executive authority (de facto executive). • Prime minister is the head of the government while President is the head of state. • He recommends persons who can be appointed as Ministers by the President.

Union Council of Ministers

• The total number of Ministers shall not be exceed 15% of the total strength of Lok Sabha. The Council of Ministers are collectively responsible to the Lok Sabha (Article 75).

Types of Ministers • Cabinet Ministers – They are real policy makers. The Cabinet’s consent is necessary for all important matters. • Ministers of State – They can hold either independent charge or can be attached to a Cabinet Minister.

Parliament

• The Indian Parliament consists of the President, the Lok Sabha and the Rajya Sabha (Article 79).

Rajya Sabha

• Rajya Sabha is a permanent body and it can not be subjected to dissolution. • The maximum strength is 250 out of which 238 are representatives of states and UTs and 12 are nominated by the President.

Lok Sabha

• Lok Sabha is the Lower House of the Parliament. • Its maximum strength is 550. At present, the strength of Lok Sabha is 545. • The term is fi ve years after which it automatically gets dissolved. • The Constitution provided for nomination of two Anglo-Indians to the Lok Sabha. However, this provision has not been extended after 2020.

Speaker of the Lok Sabha

• The Speaker of the Lok Sabha is the presiding offi cer and the highest authority of the Lok Sabha. • He/She conducts the business in the House and maintains decorum and order in the House.

69

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

Qualification of Members of Parliament

• He must be a citizen of India. • He must not be less than 30 years of age in case of Rajya Sabha and less than 25 years of age in case of the Lok Sabha.

Joint Session

• The President can summon a Joint session if– (a) a bill is passed by one house and rejected by another. (b) the amendments made by the other house are not acceptable to the House, where the bill originated. (c) other house do not take any action for six months on bill.

Difference of powers between two houses

• A money bill can be introduced only in the Lok Sabha on the recommendation of the President and not in the Rajya Sabha. • The fi nal power to decide whether a particular bill is a money bill or not, is vested in the Speaker of Lok Sabha.

Supreme Court

• Article 124 states the establishment and constitution of the Supreme Court. • The Supreme Court consists of 34 judges (1 CJI and 33 Judges), at present. • The Chief Justice of India and the Judges of the Supreme Court are appointed by the President under clause (2) of Article 124 of the Constitution. • The fi rst woman Judge of Supreme Court was Justice Fathima Beevi in 1987 and the second woman Judge was Sujata V Manohar in 1994. Indu Malhotra is the fi rst woman Judge of Supreme Court to be directly appointed to the Supreme Court from the Bar. • Under Article 124(3), a person to be appointed as Judge of the Supreme Court should have the following qualifi cations. (i) He should be a citizen of India. (ii) He should have been a Judge of a High Court (or High Courts in succession) for fi ve years or he should have been an advocate of a High Court (or High Courts in succession) for ten years or he should be a distinguished jurist in the opinion of the President. • The Constitution does not prescribe a minimum age for appointment as a Judge of the Supreme Court. • He holds offi ce until he attains the age of 65 years. • He can resign from his offi ce after writing to the President.

State Government Governor

• The President appoints the Governor (Article 155). • Article 156 states that the Governor holds offi ce during the pleasure of the President. • Under Article 157, the Constitution lays down the following qualifi cation for the Governor’s offi ce: (i) He must be a citizen of India. (ii) He should have completed 35 years of age. (iii) He shouldn’t be a member of either Houses of Parliament or State Legislatures. (iv) He holds offi ce for a period of fi ve years. (v) He may resign by writing a letter addressed to the President. (vi) The governor appoints the Advocate General of the State, State Election Commissioners, Chairman and Members of the State Public Service Commission, VC’s of Universities. (vii) He nominates one-sixth members of Legislative Council. His other powers are mostly similar to the President.

Chief Minister

• A Chief Minister heads a State government’s Council of Ministers and can be deputized in that role by a deputy Chief Minister. • Article 164, says that Chief Minister shall be appointed by the Governor.

• The Governor has to appoint the leader of a party or coalition that secures majority in the state Legislative Assembly elections as the Chief Minister. • A person, who is not a member of State Legislature can also be appointed as the Chief Minister but he has to become member of either House within six months otherwise he is removed.

Legislative Assembly

• The State Legislative Assembly, or Vidhan Sabha, is a legislative body in the states and union territories of India. • Each Member of a Legislative Assembly (MLA) is directly elected to serve fi ve years terms by single-member constituencies.

Legislative Council

• Its establishment is defi ned in Article 169 of the Constitution of India. • As of 2022, six out of 28 states have a State Legislative Council. These are Andhra Pradesh, Karnataka, Telangana, Maharashtra, Bihar and Uttar Pradesh. • The total strength cannot exceed one-third of the strength of the Legislative Assembly, subject to minimum of 40 members.

Strength of State Legislative Assemblies S.No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

State Andhra Pradesh Arunachal Pradesh Assam Bihar Chhattisgarh Goa Gujarat Haryana Himachal Pradesh Jharkhand Karnataka Kerala Madhya Pradesh Maharashtra Manipur Meghalaya Mizoram Nagaland Odisha West Bengal Punjab Rajasthan Sikkim Tamil Nadu Telangana Tripura Uttarakhand Uttar Pradesh

Number of Seats 175 60 726 243 90 40 182 90 68 81 224 140 230 288 60 60 40 60 147 295 117 200 32 235 119 60 70 403

Union Territories 1.

Delhi

:

70

2.

Puducherry

:

30

3.

Jammu & Kashmir

:

114 (24 in Pakistan occupied Kashmir, PoK)

70

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Strength of State Legislative Councils State Uttar Pradesh Andhra Pradesh Maharashtra Karnataka Bihar Telangana

Number of Seats 100 58 78 75 75 40

High Court

• The Constitution of India provides a High Court for each state, but the 7th Amendment Act of 1956, authorised the Parliament to establish a common High Court for two or more states or/and union territories.

• Under Article 217, the President appoints the Judges of the High Court. • A High Court’s Judge should be a citizen of India. • He should have held a judicial offi ce in the territory of India for 10 years or he should have been an advocate of a High Court (or High Courts in succession) for 10 years. • He holds offi ce until he attains the age of 62 years.

Jurisdiction and Powers of High Court

At present, a High Court enjoys the following jurisdiction and powers– • Original Jurisdiction • Writ Jurisdiction (Article 226) • Appellate Jurisdiction • Supervisory Jurisdiction • Control over subordinate courts • A court of record • Power of Judicial review

Jurisdiction and Seats of High Courts Court Name

Estd. in the Year

Territorial Jurisdiction

Seat

Mumbai

1862

Maharashtra, Dadar and Nagar Haveli, Goa, Daman and Diu

Mumbai (Bench at Nagpur, Panaji and Aurangabad)

Kolkata

1862

West Bengal, Andaman and Nicobar Islands

Kolkata (Circuit Bench at Port Blair)

Chennai

1862

Tamil Nadu and Puducherry

Chennai (Bench at Madurai)

Allahabad

1868

Uttar Pradesh

Prayagraj (Bench at Lucknow)

Karnataka

1884

Karnataka

Bengaluru (Circuit Benches at Hubli, Dharwad and Gulbarga)

Patna

1916

Bihar

Patna

Madhya Pradesh

1956

Madhya Pradesh

Jabalpur (Benches at Gwalior and Indore)

Jammu & Kashmir

1928

Jammu and Kashmir

Srinagar and Jammu

Punjab and Haryana

1875

Punjab, Haryana and Chandigarh

Chandigarh

Orissa

1948

Odisha

Cuttack

Guwahati

1948

Assam, Nagaland, Mizoram and Arunachal Pradesh

Guwahati (Bench at Kohima, Aizwal and Itanagar)

Rajasthan

1949

Rajasthan

Jodhpur (Bench at Jaipur)

Kerala

1958

Kerala and Lakshadweep

Ernakulam

Gujarat

1960

Gujarat

Ahmedabad

Delhi

1966

National Capital Territory of Delhi

New Delhi

Himachal Pradesh

1971

Himachal Pradesh

Shimla

Sikkim

1975

Sikkim

Gangtok

Chhattisgarh

2000

Chhattisgarh

Bilaspur

Uttarakhand

2000

Uttarakhand

Nainital

Jharkhand

2000

Jharakhand

Ranchi

Tripura

2013

Tripura

Agartala

Manipur

2013

Manipur

Imphal

Meghalaya

2013

Meghalaya

Shillong

Andhra Pradesh

2019

Andhra Pradesh

Amaravati

Telangana

2019

Telangana

Hyderabad

Jammu and Kashmir

• According to Article 370 of the Indian Constitution, Jammu and Kashmir have a special status. • J & K had its own Constitution, framed by a special constituent Assembly and came into being on the 26th January, 1957. • But, on August 2019, the status was revoked.

Local Government The Panchayats (Article 243-243 O)

• It was introduced by the 73rd Amendment Act, 1992 which envisaged a three tier system of local government.

These are: 1. Gram Panchayat at the village level 2. Panchayat Samiti at the block level 3. Zilla Parishad at the district level

The Municipalities (Article 243 P-243 ZG)

• It was introduced by the 74th Amendment Act, 1992 which envisages three types of urban local bodies, namely Nagar Panchayat, Municipal Council and Municipal Corporation.

71

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

Committees to Study Panchayat System Name

Established

Recommendations

Balwant Rai Mehta

1957

Establish local bodies, devolve power and authority, basic unit of decentralised government to be Block Samiti. Conceptualised Panchayati Raj Institutions (PRIs) as a 3-tier system.

K Santhanam

1963

Panchayats to have powers to levy tax on land revenue, etc. Panchayati Raj Finance Corporation to be set-up.

Ashok Mehta

1978

District to be viable administrative unit for planning, PRIs as two-tier system with Mandal Panchayat and Zila Parishad.

GVK Rao

1985

PRIs to be activated and supported. Block Development Offi ce (BDO) to be the centre of rural development.

LM Singhvi

1986

Local self-governments to be constitutionally recognised, non-involvement of political parties in PRIs.

Union and State Public Service Commissions Functions

• It conducts exams for appointment to services under the Union and the States. • It maintains continuity in administration. • The members of the UPSC and State Public Service Commissions can be removed by the President on the charges of misbehaviour, if these charges are upheld by the Supreme Court.

Elections (Article 324–329)

• The Election Commission was established on 25th January 1950 under Article 324 of the Constitution. • The fi rst Chief Election Commissioner was Sukumar Sen.

Planning Commission

• The Planning Commission was established in March, 1950 by an executive resolution of the Government of India (i.e., Union Cabinet) on the recommendation of the Advisory Planning Board constituted in 1946 under the Chairmanship of KC Neogi. • In 2015, the Planning Commission replaced by Niti Aayog.

Article 324 stipulates that the superintendence, direction and control of elections shall be vested in the Election Commission. Article 325 provides for a single electoral roll for every constituency. Also stipulates that no person shall be eligible or ineligible for inclusion in electoral rolls on the basis of race, religion, caste or sex. Article 326 stipulates that elections shall be held on the basis of adult suffrage. Every person, who is a citizen of India and is not less than 18 years of age shall be eligible for inclusion.

NITI Aayog

Political Parties

• The functions of the NDC was to review the working of national plan. The NDC was formed in 1952 to associate the states in the formulation of the plans. • The Prime Minister is the ex-offi cio chairman of NDC. • It is an extra-constitutional and extra legal body. • Since the establishment of NITI Aayog, NDC has been proposed to be abolished.

Representation of the People Act, 1951 provides for registration of the political parties with the Election Commission. There are eight (8) National Parties in India, namely BJP, Congress, BSP, NCP, CPI, CPM, Trinamool Congress and National People’s Party. A political party shall be eligible to be recognised as a National Party if (i) It secures at least 6% of the valid votes polled in any four or more states, at a general election to the House of the People or to the State Legislative Assembly; and (ii) In addition, it wins at least four seats in the House of the People from any State or States. or (iii) It wins at least 2% seats in the House of the People (i.e., 11 seats in the existing House having 543 members) and these members are elected from at least three different States. or (iv) The party gets recognition as a state party in four states. Likewise, a political party shall be entitled to be recognised as a State Party, if (i) It secures at least 6% of the valid votes polled in the State at a general election, either to the House of the People or to be Legislative Assembly of the State concerned; and (ii) In addition, it wins at least two seats in the Legislative Assembly of the State concerned. or (iii) It wins at least 3% of the total number of seats in the Legislative Assembly of the State or at least three seats in the Assembly, whichever is more.

Election Commission

• The Election Commission is an autonomous, quasi-judicial constitutional body. Its function is to conduct free and fair elections in India.

• NITI Aayog or National Institution for Transforming India Aayog is a policy think-tank of Union Government of India that replaces Planning Commission of India and aims to involve the states in economic policy-making in India. It provides strategic and technical advice to the Central and the State Governments. • Prime Minister of India heads the Aayog as its Chairperson.

National Development Council (NDC)

Finance Commission

• Article 280 of the Constitution of India provides for a Finance Commission as a quasi-judicial body. It is constituted by the President of India every fi fth year. • It consists of Chairman and 4 other members.

Functions

The Finance Commission is required to make recommendation to the President of India in the following matters: • The distribution of the net proceeds of taxes to be shared between the centre and the states and the allocation between the states, the respective shares of such proceeds. The 15th Finance Commission was appointed on November, 2017 under the Chairmanship of NK Singh.

Amendments of the Constitution (Article 368) There are two categories of amendments under Article 368 which are:1. By special majority of the Parliament that is more than fi fty percent of the total membership of each house and a majority of two-thirds of the members of each house present and voting. 2. By special majority of the Parliament and with the consent of half of the State Legislature by a simple majority. Provisions related to Federal structure can be amended through this procedure. There is a third category of the Amendment which is done by simple majority though these amendments do not come under ambit of Article 368.

72

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• It has been held by the Supreme Court in the Keshavananda Bharati Case (1973) that every provision of the Constitution is amendable under the meaning of Article 368 except the basic structure of the Constitution.

e-Governance

The word electronic in the term e-Governance implies technology driven governance. The perspective of the e-governance is the use of the technology that connects both governing and have to be governed. Generally fi ve basic models are given in e-Governance • G2C (Government to Citizens) • G2B (Government to Business) • G2E (Government to Employees) • G2G (Government to Government) • C2G (Citizens to Government) The National e-Governance Plan (NeGP) takes a holistic view of e-governance initiatives across the country, integrating them into a collective vision.

Impacts of e-Governance

e-Governance brings about two major impacts fi rstly, making the government offi ces work smart. Secondly, e-governance makes services available to the citizen at his doorstep through the internet. Some of the most successful citizen oriented e-governance projects are the Railways Reservation System, MCA 21 of the Ministry of Corporate Affairs and Bhoomi Project in Andhra Pradesh, etc.

e-districts

It is a mission mode under e-governance. Its objective under National e-Governance Policy is computerisation of services. Under it, different programmes are being conducted in following states– • Jandoot Project – Madhya Pradesh • Compact 2020 – Andhra Pradesh • Land Programme - Karnataka • Friends – Kerala

PARLIAMENTARY TERMS

Quorum It is the minimum number of members required to transact the business of the House. Article 100 of the Constitution specifi es that the Quorum of either House shall be 10% of the strength of the House. Question Hour: The fi rst hour of every sitting of Parliament is called the Question Hour. Questions usually need a 10 day notice before being answered by the concerned minister. Starred Questions: These are to be answered orally on the fl oor of the House. Supplementary questions can be asked. Unstarred Questions: These are to be answered in writing. No supplementary questions may be asked. Zero Hour: It does not formally exist in the Parliamentary procedure. The hour after Question Hour is popularly known as Zero Hour. Members raise matters which they feel are urgent. Adjournment Motion: It is the motion to adjourn the proceedings of the House, so as to take up a matter of urgent public importance. It can be moved by any member. Requires support from at least 50 members. Calling attention motion: A member may call the attention of a minister to an urgent matter and the minister may make a statement regarding it.

No confidence motion: A no confi dence motion indicates lack of confi dence of the Lok Sabha in the Council of Ministers. It can be introduced in the Lok Sabha only. If the Motion is passed, the government must resign. Constitutional Amendments First amendment Act, 1951: It led to the addition of Ninth Schedule. Seventh Amendment Act, 1956: It was necessitated on the account of reorganisation of States on a linguistic basis. Fifteenth Amendment Act, 1963: It allowed the increase of age to retirement of the Judges of High Court from 60 to 62 years. Twenty Sixth Amendment Act, 1971: It abolished the titles and special privileges of former rulers of princely states. Thirty Sixth Amendment Act, 1975: Sikkim Sikkim a State. Forty Fourth Amendment Act, 1978: Under this, the right to Property was deleted from Part III. Article 352 was amended to provide ‘Armed Rebellion’ as one of the circumstance for declaration of emergency, in place of ‘Internal Disturbance’. Seventy Third Amendment Act, 1992: With this, the institution of Panchayat Raj received constitutional guarantee, status and legitimacy. XI Schedule was added to deal with it. It also inserted Part IX, containing Articles, 243 to 243 O. Eighty Ninth Amendment Act, 2003: The Act adds Article 338 A and provides for the creation of National Commission for Scheduled Tribes. Ninety First Amendment Act, 2003: It adds the Anti-Defection Law in 10th schedule and also made a provision that the number of ministers in the Centre and State Government, cannot be more than 15% of the strength of Lok Sabha and the respective Vidhan Sabha. Ninety Third Amendment Act, 2005: It reserves seats for socially and educationally backward classes, besides the Scheduled Castes and the Scheduled Tribes, in private unaided institutions other than those run by minorities. Ninety Seventh Amendment, 2011: It deals with the amendment of Article 19(1) (i), insertion of article 43B and insertion of Part IXB. This amendment gives constitutional status to cooperatives. Ninety-Ninth Amendment Act, 2014: It deals with replacing the collegium system for the appointments of the Judges of the Supreme Court and the 24 High Courts. But Supreme Court of India declared this unconstitutional and void. One Hundredth Amendment Act, 2015: It deals with the acquiring of territories by India and transfer of certain territories to Bangladesh in pursuance of the agreements and its protocol. One Hundred First Amendment Act, 2016: It deals with Goods and Services Tax (GST) One Hundred Second Amendment Act, 2018: It provides the constitutional status to National Commission for Backward Classes. One Hundred Third Amendment Act, 2019: It provides 10% reservation to the economically backward classes of society. One Hundred Fourth Amendment Act, 2020: It extends the reservation of seats for SCs and STs in Lok Sabha and State Assemblies from seventy years to eighty years and removed the reserved seats for Anglo-Indian community in Lok Sabha and State Asse mblies. One Hundred Fifth Amendment Act, 2021: It restores power of the states to make their own OBC (other backward classes) lists. One Hundred Sixth Amendment Act, 2023: It seeks to provide 33% of seats for women in Lok Sabha and state legislative assemblies.

Objective Type Questions 1. The provision for the anti-defection law is given under which schedule of the constitution: (2023) (2) 9th Schedule (1) 10th Schedule (4) 12th Schedule (3) 8th Schedule

2. RTI (Right to Information) act came into force on? (1) October, 2005 (2) December, 2004 (3) October, 2003 (4) December, 2006

(2023)

73

GENERAL AWARENESS : INDIAN POLITY AND CONSTITUTION

3. Which of the following fundamental right has been removed from our Constitution? (2023) (1) Right against exploitation (2) Right to Constitutional remedies (3) Right to Property (4) Right to freedom of religion 4. The One hundred Sixth constitutional ammendment act provides for how much reservation for women in Lok Sabha and State Legislative assemblies? (2022) (1) 25% (2) 50% (3) 40% (4) 33% 5. The Rajya Sabha has equal powers with the Lock Sabha in: (2021) (1) the matter of creating new All India services (2) amending the Constitution (3) the removal of the government (4) making cut motions 6. The Speaker of the Lok Sabha can be removed from the offi ce by the: (2021) (1) simple majority (2) special majority (3) absolute majority (4) effective majority 7. For an Indian citizen, the duty to pay taxes is a: (2021) (1) constitutional obligation (2) moral obligation (3) legal obligation (4) fundamental duty 8. As of June 2019, which Indian state has the lowest number of districts? (1) Goa (2) Telangana (3) Sikkim (4) Arunachal Pradesh 9. As per the Table of Precedence, who among the following enjoys the highest ceremonial protocol? (1) Holders of Bharat Ratna (2) Judges of the Supreme Court of India (3) Cabinet ministers of the Union (4) Deputy Chairman of Rajya Sabha 10. Who was the founder of Bahujan Samaj Party? (1) Munshi Ram (2) Devi Lal (3) Kanshi Ram (4) Lakshman Singh 11. Name the Prime Minister who brought about a thaw in IndiaChina relations by signing the ‘Line of Actual Control’. (1) Lal Bahadur Shastri (2) P.V. Narashima Rao (3) Chandrasekhar (4) V.P. Singh 12. _________ of the Constitution of India deals with emergency due to war, external aggression or armed rebellion. (1) Article 350 (2) Article 352 (3) Article 347 (4) Article 269 13. Which article of the Constitution of India provisions of reservations of seats for scheduled castes and scheduled tribes in the House of People? (1) Article 325 (2) Article 321 (3) Article 330 (4) Article 335 14. Anti-defection law was passed in which Constitutional Amendment Act? (1) 41st Constitutional Amendment Act (2) 46th Constitutional Amendment Act (3) 48th Constitutional Amendment Act (4) 52nd Constitutional Amendment Act 15. The Judges of High Court are administered oath of offi ce by (1) The Chief Justice of High Court (2) The President of India (3) The Chief Justice fo India (4) Governor of the State 1. (1) 9. (3)

Answer Key 2. (1) 3. (3) 4. (4) 5. (2) 6. (4) 7. (3) 8. (1) 10. (3) 11. (2) 12. (2) 13. (3) 14. (4) 15. (1)

Answers with Explanations 1. Option (1) is correct. The provision for the anti-defl ection law is given under the tenth schedule of the Indian constitution. It was included by the 52nd amendment of the Indian constitution in the year 1985. This law allows the disqualifi cation of individual MPs or MLAs for a given period of time, who leave one party to join the other. The Ninth Schedule of the Indian constitution consists of the state and central government laws that cannot be challenged in the court. This schedule was inserted by the First Constitution Amendment Act, 1951. It basically consists of laws that focus on the socio-economic development of the society. The eighth schedule of the Indian constitution lists 22 offi cial languages. These are- Assamese, Bengali, Gujarati, Hindi, Kannada, Kashmiri, Konkani, Malayalam, Manipuri, Marathi, Nepali, Odia, Punjabi, Sanskrit, Sindhi, Tamil, Telugu, Urdu, Bodo, Santhali, Maithili and Dogri. There is no fi xed criteria for a language to be included in the eighth schedule of the constitution. The twelfth schedule of the Indian constitution was included by the 74th Amendment Act of 1992. It deals with the power, rights and responsibilities of the Municipality in order to help them function effectively. 2. Option (1) is correct. The RTI Act was enacted by the Indian Parliament on 15th June 2005, and it came into effect on 12th October 2005. It ensures that citizens have the right to access information held by public authorities and government departments, which in turn promote transparency and reduce corruption. However, there are some government organisations that are not in the ambit of RTI due to the nature of information they are handling. According to the Constitution of India, RTI is not included as a Fundamental Right. However, it protects and is implicit in Freedom of Expression and Speech under Article 19(1)(a) and Right to Life and Personal Liberty under Article 21 of the Constitution. 3. Option (3) is correct. Right to Property was removed from the list of fundamental rights recognised by the constitution of India. This was done through the 44th Constitutional Amendment Act of India in 1978. This was done to remove the excessive possession of land in the hands of few people. A new article 300 A was added to the constitution that made Right to Property a legal right. This allows the state to acquire property under two conditions when the acquisition should be made for a public purpose and it should provide for payment of compensation to the owner. 4. Option (4) is correct. One Hundred Sixth Amendment Act, 2023 seeks to provide 33% of seats for women in Lok Sabha and state legislative assemblies. It is also known as Women’s Reservation Bill, 2023. 5. Option (2) is correct. The Rajya Sabha (Upper House) has equal powers with the Lok Sabha (Lower House) in the matter of amending the constitution. The Rajya Sabha enjoys equal powers with Lok Sabha in other matters like the impeachment of the President, removal of the vice-president, and removal of the judges of the Supreme Court and the High Courts. Article 312 in the Constitution of India deals with the creation of new All India Services. In matter of creating All India Services, the Rajya sabha enjoys special powers. If the Rajya Sabha passes a resolution by a majority of not less than two-thirds of the members present and voting declaring

74

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

that it is necessary or expedient in the national interest to create one or more All India Services. A cut motion is a special power vested in members of the Lok Sabha to oppose a demand being discussed for specifi c allocation by the government in the Finance Bill as part of the Demand for Grants. 6. Option (4) is correct. As per Article 93 of the Indian Constitution, the Lok Sabha has a Speaker and a Deputy Speaker. In the Lok Sabha, the Speaker and the Deputy Speaker are elected from among its members by a simple majority of members present and voting in the House. As per Article 94, the Speaker can be removed by the Lok Sabha by a resolution passed by the effective majority of the house. The Speaker can also be removed on being disqualifi ed for being the Lok Sabha member under sections 7 and 8 of the Representation of the People Act, 1951. The Speaker may, at any time, resign from Offi ce by writing under his/her hand to the Deputy Speaker. 7. Option (3) is correct. For a citizen of India the duty to pay taxes is a legal obligation. As per the Direct taxation laws; income and other kinds of taxes are a legal obligation for an Indian citizen as he uses public services, receives social and political security from the state etc. The Swaran Singh Committee suggested inclusion of duty to pay taxes under fundamental duties, but it was not agreed upon. The fundamental duties were originally included in the constitution through the 42nd amendment, 1976. The Fundamental duties are more in the nature of moral obligations of citizens to help promote a spirit of patriotism and to uphold the unity of India. 8. Option (1) is correct. The lowest number of the district is in Goa. In Goa, there are only two districts. North Goa and South Goa district.

9. Option (3) is correct. The Indian Declaration of Precedence is the scale of signifi cant roles. The hierarchy of precedence in India is only to demonstrate the proper protocol. It is not suitable for the daily functioning of the Government of India. 10. Option (3) is correct. The Bahujan Samaj Party (BSP), English Majority People’s Party, is the national political party in India. It was constituted on the birth anniversary of B. R. Ambedkar, 14th April 1984, by Kanshi Ram. It symbolizes the most downward group of people in the Hindu social scenario. 11. Option (2) is correct. The Prime Minister who fetched almost a thaw in India-China connections by signing the ‘Line of Actual Control’ is P. V. Narashima Rao. 12. Option (2) is correct. National emergency under Article 352. Many Fundamental Rights of Indian residents can be broken during a national crisis. The President can claim such an emergency only founded on a documented proposal by the Cabinet guided by the Prime Minister. 13. Option (3) is correct. Article 330 in the Constitution of India 1949. 1949 provides for the reservation of seats for Scheduled castes and Tribes in the House of People. The seats shall be secured in the House of the People for (a) the Scheduled Castes (b) The Scheduled Tribes besides for the Scheduled Tribes in the liberated districts of Assam. (c) The Scheduled Tribes in the independent districts of Assam. 14. Option (4) is correct. In 1985, the Tenth Schedule of the 52nd amendment to the Constitution of India was passed by the Parliament of India to achieve this. 15. Option (1) is correct. The President of India nominated a high court judge with the discussion of the Chief Justice of India and Governor of the State. The governor of that concerned state administers the high court judge’s oath.

Agriculture

Flagship Programmes of Government of India

Indian Money Market

Indian Capital Market

Indian Financial System

Indian Currency System

Economic Growth Economy

Poverty

Macro Economic Problems in India

First Level

Second Level

Trace the Mind Map

Third Level



India’s new Economic Policies

Planning in India

National Income Aggregates

National Income of India

Unemployment

Indian Economy

Indian Economy

GENERAL AWARENESS : INDIAN ECONOMY

75

Chapter

Indian Economy

4

Chapter Analysis Concept Name Indian Economy

 Revision Notes

2023

Additional Questions

2

2

10

Gross National Product (GNP)

Scan to know more about this topic

Indian Economy

Economic Growth • Economic growth can be defined as the increase or improvement in the inflation­—adjusted market value of the goods and services produced by an economy over time.

Indian Economy • Indian Economy is the fifth largest in World (as per December 2020) by nomial GDP and third largest a purchasing power parity (PPP) basis in the year 2019. • Purchasing Power Parity (PPP) is a theory which states that exchange rates between two currencies are in equilibrium when their purchasing power is the same in each of the two countries. • India has a share of around 17.9% of the world population. • The characteristics of Indian Economy economy are: • The nature of the Indian economy is mixed, i.e., both public and private sectors coexist. • In this economy, agricultural dominance prevails in both the Gross National Product (GNP) and employment. • The following features show that Indian economy is a developing economy: (a) Low per capita income (b) Heavy population pressure (c) Prevalence of chronic unemployment and (d) Low human development indicators

Broad Sectors of Indian Economy

Primary Sector: This sector consists of agriculture, forestry, and fishing, mining, etc. Secondary Sector: In this sector manufacturing, electricity, gas, water supply and construction and construction are included. Tertiary Sector (also called service sector): The sector includes business, transport, telecommunication, banking, insurance, real estate, community, and personal services.

National Income of India

2022

1

National Income Aggregates:

Economy • The economy is the financial condition of the different sectors of a country. • The study of the economy of any country helps to helps to understand the financial condition of the population as well as the working sectors. • It also helps to compare the financial conditions of two distinct countries.

2021

• The net value of all the final goods and services produced in a country during a financial year is the Nation Income. • 1st April to 31st March is the financial year of India and it is calculated annually. • The measurement of the monetary value of assets, possessed by the citizens of a nation at a particular point in time.

• GNP refers to the money value of the total output of production of final goods and services produced by the nationals of a country during a given period of time, generally a year. Symbolically, GNP = C + G + I + (X ­– M) + (R – P) where C = Consumption of expenditure G = Government expenditure I = Investment expenditure (X – M) = Net exports (R – P) = Net Factor Income from Abroad

Gross Domestic Product (GDP)

• GDP is the total money value of all final goods and services produced within the geographical boundaries of the country during a given period of time. Symbolically, GDP = GNP where R is the income reciepts and P is the net outflow to foreign assets when R – P is 0 (Zero), GDP = GNP

Gross Value Added (GVA)

• GVA is the value of goods and services produced by an industry, sector, manufacturer, area, or region in an economy. • RBI dumped the GVA model in August 2018 and switched back to GDP to measure the economy.

Net National Product (NNP)

• NNP is the monetary value of finished goods and services produced by a country’s citizens, overseas and domestically, in a given period. • Symbolically, NNP = GNP – Depreciation. • When NNP is calculated at Factor Cost (FC) it is called National Income. This measure is calculated by deducting Indirect Taxes and adding subsidies in NNP at Market Price (MP). • NNPFC = NNPMP – Indirect Taxes + Subsidies

Net Domestic Product (NDP)

• It is calculated by subtracting the value of depreciation from GDP. • NDP = GDP – Depreciation

Personal Income (PI)

• Personal Income = National Income – Undistributed Profits of Corporations – Payment for Social Security Provision – Corporate taxes + Transfer Payments + Net interest paid by the government.

Personal Disposable Income (PDI)

• When personal direct taxes are subtracted from personal income, the obtained value is called personal disposable income, symbolically PDI = PI – Direct Taxes PDI = Consumption + Savings Planning of India Economic Planning in India

77

GENERAL AWARENESS : INDIAN ECONOMY

Planning Commission (1950) was setup under the Chairmanship of Pandit Jawaharlal Nehru (Gulzarilal Nanda was the fi rst Deputy Chairman). The basic aim of Economic Planning is to bring rapid economic growth through agriculture, industry, power, and all other sectors of the economy.

NITI Aayog

NITI Aayog or National Institution for Transforming India Aayog came into existence on 1st January 2015. It is the policy-making think-tank of government that replaces Planning Commission and aims to involve states in economic policy making. It will provide strategic and technical advice to the Central and the State Governments. The Prime Minister heads the Aayog as its

chairperson. Rajiv Kumar is the Vice-Chairperson of NITI Aayog of India. Historical Milestones Planned Economy for India

(1934)

M. Visvesvaraya

National Planning Committee

(1938)

Jawaharlal Nehru

Bombay Plan

(1944)

Gandhian Plan

(1944)

SN Agarwal

People’s Plan

(1945)

MN Roy

Sarvodaya Plan

(1950)

JP Narayan

Five Year Plans At a Glance Plan First Plan (1951–56) (Based on Harrod Domar Model) Second Plan (1956–61) (Based on PC Mahalanobis two sector model) Third Plan (1961–66) Plan Holiday (1966–69) Fourth Plan (1969–74) Fifth Plan (1974–78) Rolling Plan (1978–80) Sixth Plan (1980–85) Seventh Plan (1985–90) Two Annual Plans (1990–92) Eighth Plan (1992–97) Ninth Plan (1997–2002) Tenth Plan (2002–07) Eleventh Plan (2007–12) Twelfth Plan (2012–17)

Growth Rate Target Achieved 2.1% 3.6%

Agriculture, irrigation, electricity

4.5%

4.2%

Heavy industries

5.6%

2.8%

Foodgrains, heavy industries

5.7% 4.4%

3.3% 4.8%

Agriculture Removal of poverty

5.2% 5.0%

5.4% 6.0%

Agriculture, industries Energy, foodgrains

5.6% 6.5% 8.1% 8.0% 8%

6.6% 5.4% 7.5% 7.9% –

Human resource education Social justice Income, energy Inclusive growth Faster, sustainable, and more inclusive growth

India’s New Economic Policy (1991)

• In 1985, during the period of Prime Minister Rajiv Gandhi, new economic policy related to economic reforms was revised and implemented for the fi rst time. • In 1991, during the period of PV Narsimha Rao Government, the second wave of new economic reforms came widely known as LPG (Liberation, Privatisation, and Globalisation) reforms. • In the 1990s, macro-economic crises necessitated economic reforms in India. The crisis had the following aspects: (a) Fiscal imbalance or increasing fi scal defi cit (b) Fragile balance of payment situation (c) Infl ationary pressures in the economy. • The Balance of Payment (BOP) situation was such that India did not have enough foreign exchange reserves to pay even for three months imports. • Two distinct strands of reform measurer were prescribed by the World Bank and the IMF. These were: • Macro-economic stabilization–Demand management • Structural Adjustments supply–Side management.

Macro Economic Problems in India Poverty

• The erstwhile Planning Commission estimated poverty rate based on data collected by the National Sample Survey Organisation (NSSO). • Main Reasons for Rural Poverty were Rapid population growth, lack of capital, lack of alternate employment other than poor agriculture, illiteracy, and lack of proper implementation of Public Distribution System (PDS).

Important Sector

• Main Reasons for Urban Poverty were migration from the rural area, lack of skilled labour, lack of housing facilities, limited job opportunities in cities.

Unemployment

• Unemployment is an economic condition in which an individual actively seeking jobs remain unemployed. • It is expressed as a percentage of the total available workforce.

Types of Unemployment The types of unemployment are as follows:

Structural Unemployment

• This is a category of unemployment caused by differences between the spills passed by the unemployed population and the jobs available in the market.

Under Employment

• Underemployment is a measure of employment and labor utilization in the economy that looks at how well the labor force is being used in terms of skills, experience, and availability of work.

Open Employment

• When the labours don’t fi nd the work to do, they come under the category of open unemployment.

Disguised Unemployment

• Disguised unemployment is unemployment that does not affect aggregate economic output. • It occurs when productivity is low and too many workers are fi lling too few jobs.

78

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Frictional Unemployment

• This is the result of voluntary employment transition within an economy.

Seasonal Unemployment

• This occurs when people are unemployed at certain times of the year because they work in industries where they are not needed at all year round.

Cyclical Unemployment

• It is the component of overall unemployment that results directly from cycles of economic upturn and downturn.

Flagship Programmes of Government of India

• Flagship schemes of the government of India are those schemes that are declared so by the Union cabinet or the Development Evaluation Advisory Committee (DEAC) of the planning commission. • Some of the schemes are given: Programme/Measure Twenty Point Programme Indira Awaas Yojana Jawahar Rozgar Yojana Nehru Rozgar Yojana Swarna Jayanti Shahri Rozgar Yojana Pradhan Mantri Gramodaya Yojana Pradhan Mantri Gram Sadak Yojana Sampoorna Grameen Rozgar Yojana Bharat Nirman Jawaharlal Nehru National Urban Renewal Mission Prime Minister Employment Generation Programme Mahatma Gandhi National Rural Employment Programme National Rurual Livelihood Mission (NRLM) Nirmal Bharat Abhiyan Swachh Bharat Abhiyan Pradhan Mantri Jan Dhan Yojana Atal Pension Yojana Digital India Programme National Skill Development Mission HRIDAY (Heritage City Development and Augmentation Yojana) Smart City Mission Amrut (Atal Mission for Rejuvenation and Urban Transformation) Pradhanmantri Jeevan Jyoti Beema Yojana Pradhanmantri Suraksha Beema Yojana Pradhan Mantri Krishi Sinchayee Yojana Start-up and Stand-up Yojana Pradhan Mantri Fasal Bima Yojana Ujala Yojana SWAYAM (Study Webs of Active-Learning for Young Aspiring Minding Pradhan Mantri Garib Kalyan Yojana Pradhan Mantri Vaya Vandana Yojana Pradhan Mantri Matritva Vandana Yojana Pradhan Mantri Sahaj Bijli Har Ghar Yojana Rashtriya Vayoshri Yojana Saubhagya Yojana UDAN Scheme

Year of Launch 1975 1985 1989 1989 1997 2000 2000 2001 2005 2005 2008 2009

Programme/Measure Ayushman Bharat Pradhan Mantri Kisan Samman Nidhi Pradhan Matri Shram Yogi Mandhan Yojna SVAMITVA Scheme One Nation One Card Scheme Ghar Tak Fibre Scheme PM SVANidhi Atmanirbhar Bharat Scheme Kisan Suryodya Yojana Pradhan Mantri Matsya Sampada Yojana PM-WANI Scheme PM POSHAN Shakti Nirman Abhiyaan PM Gati Shakti PM Daksh Yojana NIPUN scheme PM-SHRI schools Agnipath Scheme PM VIKAS PM PRANAM PM VISHWAKARMA Yojana

Year of Launch 2018 2019 2019 2020 2020 2020 2020 2020 2020 2020 2020 2021 2021 2021 2021 2022 2022 2023 2023 2023

Programme/Measure

Year

Mid-Day Meal Scheme

1995

Swadhar

1995

Swayam Sidha

2001

SSA Support to Training and Employment Programme for Women (STEP)

2001 2003–04

Ujjwala

2007

2011 2012 2014 2015 2015 2015 2015 2015

Dhanlaxmi

2008

2015 2015 2015 2015 2015 2016 2016 2016 2016 2016 2017 2017 2017 2017 2017 2017

Integrated Child Protection Scheme

2009–10

Sabia Scheme

2010

National Mission for Empowerment of Women

2011

Bal Bandu Scheme

2011

Nai Roshni

2012

Beti Bachao, Beti Padhao

2015

PM Ujjwala Yojna

2016

PM Matri Vandana Yojna

2017

Suposhit Maa Abhiyan

2020

Agriculture

• As of 2018, agriculture employed more than 50% of the Indian workforce and contributed 17–18% to country’s GDP. • In 2016, agriculture and allied sectors like animal husbandary, forests and fi shers accounted for 15.4% of the GDP (Gross Domestic Product) with about 41.45% of the workforce in 2020. • Its importance to the Indian economy can be gauged from the following facts: • Contribution to GDP • Contribution to Employment • Contribution to Industry • Contribution to Trade

Green Revolution

• Green Revolution was a part of the New Agricultural Strategy which included, the Intensive Agriculture District Programme (IADP) and later the High Yielding Variety Programme (HYVP).

79

GENERAL AWARENESS : INDIAN ECONOMY

• In the 1960s, this was launched and was the brainchild of Norman Borlaug, though in India, it was made successful by Dr. MS Swaminathan. • High Yielding Variety (HYV) seeds, chemical fertilizer and pesticides, and improved irrigation facility were the key pillars of this revolution. • In 2004, the National Commission of farmers was appointed, under the Chairmanship of Dr. MS Swaminathan suggested an Agricultural Renewal Action Plan (ARAP).

Major Agricultural Revolutions Revolution Blue Golden Fibre Pink Red White Yellow

Production Fish Production Jute Meat Tomato and Meat Milk Oilseed

Tricolor Revolutions

The tricolor revolution has 3 components • Saffron revolution–Solar energy • White revolution–Cattle welfare • Blue revolution–fi sherman’s welfare.

Industries

Industrial Policies

• Industrial policies were launched in 1948, 1956, 1977, 1980, and 1991. • Industrial Policy 1956 is called the Economic Constitution of India and gave the public sector the strategic edge. • Industrial Policy 1991 opened up the economy. Its main aims were (a) to end license-permit raj; (b) to integrate the Indian economy with the outer world; (c) to remove restrictions on FDI; and (d) to reform public sectors.

Public Sector Enterprises (PSEs)

• Industries requiring compulsory licensing (a) distillation and brewing of alcoholic drinks; (b) cigar and cigarettes and tobacco; (c) electronic aerospace and defense equipment; (d) industrial explosives; (e) specifi c hazardous chemicals. • Areas reserved for the public sector are (a) atomic energy– production, separation, and enrichment of fi ssionable materials and (b) railways. • Present Policy on PSEs is to (a) not to privatize profi t-making companies and to modernize and revive sick companies; (b) not to bring government stake in PSEs below 51%; (c) to adopt initial public offering route to disinvestment.

Maharatnas, Navratnas, and Miniratnas

• To impart greater managerial and commercial autonomy to the PSEs, the concept of Maharatna, Navratna, and Miniratna was started. • At present, there are 61 in category 1 and 12 in Category II. • As of 2023, there are 13 Maharatnas. These are ONGC, SAIL, IOC, NTPC, Coal India Ltd, BHEL, GAIL (India) Ltd, and BPCL, HPCI, Power Grid Corporation, Rural Electrifi cation Corporation Ltd., Oil India Ltd. (OIL), and Power Finance Corporation. • As of 2023, there are 16 Navratnas. These are IRCON International Ltd., Rashtriya Ispat Nigam Ltd., Rashtriya Chemicals & Fertilizers Ltd., ONGC Videsh Ltd., Rail Vikas Nigam Ltd., RITES Ltd. Bharat Electronics Ltd, HAL, MTNL, NALCO, National Mineral Development Corporation, Neyveli Lignite Company Ltd, Shipping Corporation of India Ltd, CCIL, EIL, and NBCCL.

• Miniratnas Public Sector Enterprises (PSEs) have made a profi t continuously for the last three years and have a positive net worth.

Small Scale Industry

• A new thrust in favour of small-scale industries was given in the industrial policy resolution of 1977. • The MS MED Act, 2006 clearly defi nes, for the time, not only the medium enterprises but also extends it to the services sectors too.

Classification of MSMEs Category Micro

Annual turnover

Small

Between `5 crores to `75 crores `75 to `250 crores

Not exceeding `5 crores

Medium

Major Industries in India Iron and Steel

• First Steel Industry at Kulti, West Bengal–Bengal Iron Works Company was established in 1874. • First large scale steel plant—TISCO at Jamshedpur (1907) was followed by IISCO at Burnpur (1919), West Bengal. • The fi rst public owned steel plant was Rourkela integrated steel plant. Presently, India is the 3rd the largest producer of steel and comes 1st in the production of sponge iron. Location (Plants)

Assistance

Rourkela (Odisha)

Germany

Bhilai (Chattisgarh)

Russia

Durgapur (West Bengal)

Britain

Bokaro (Jharkhand)

Russia

Visakhapatnam (Andhra Pradesh)

Russia

Jute Industry

• India ranks fi rst in jute production and second in raw jute exports after Bangladesh. • More than two-third the jute industry is concentrated in West Bengal.

Cotton and Textile Industry

It is the organized and broad-based industry accounting for 4% of GDP, 20% of manufacture value-added and one-third of total exports earning.

Cement Industry

The fi rst cement producing unit was setup at Chennai in 1904 but the modern manufacturing unit of cement started at Porbandar (Gujarat) in 1914. India is the second-largest producer of cement in the world.

Sugar Industry

India is the second largest producer of sugar in the world with a 22% share. It is the second-largest agro-based industry in the country. AC Shah Committee

Non-Banking Financial Company

Bimal Jalan Committee Market Infrastructure Instruments Malegam Committee

Functioning of Micro Finance Institutions

Birla Committee

Corporate Governance

Kirit Parikh Committee Rationalization of Petroleum Product Prices Chaturvedi Committee Improving National Highways in India SR Hashim Committee Urban Poverty Abhijit Sen

Wholesale Price Index

Abid Hussain Committee Development of Capital Markets

80

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Damodaran Committee Customer Service in Banks

Courier Services

100%

Khandelwal Committee Human Resource in Commercial Banks

Single Brand Product Retail Trading

100%

Patil Committee

Defense Sector

100%

Airlines

100%

Corporate Debt

VK Sharma Committee Credit to Marginal Farmers Sarangi Committee

Non-performing Assets

Khanna Committee

Regional Rural Banks

Dantawala Committee

Lead ank Scheme

Gadgil Committee

Financial inclusion

Indian Money Market

• The Money Market in India is a correlation for short-term funds with maturity ranging from overnight to one year in India including fi nancial instruments that are deemed to be close substitutes of money.

Indian Capital Market

Foreign-Trade

• Foreign Trade is a trade between two or more nations. It is also called international trade. • In 1991, the liberalization of the economy and adaption of export promotion measures has led to substantial growth in exports and diversifi cation of our exports.

• It is a market where medium and long-term funds are borrowed and rented. • It does not deal with capital goods but is concerned with the raising of money the capital for investment purpose.

Features of FTP

• The capital market is the market for long-term funds while the money market is the market for short-term funds. • Capital market of India is regulated by SEBI (Securities and Exchange Board of India, 1988). • A Stock Exchange provides services for brokers and traders to trade stocks, bonds, and other securities. • The Bombay Stock Exchange (BSE) is a stock exchange located on Dalal Street, Mumbai, and is the oldest stock exchange in Asia. The BSE has the largest number of listed companies in the world established in 1875. • The National Stock Exchange (NSE) is the 16th largest stock exchange in the world. It is situated in Mumbai.

• FTP is the Foreign Trade Policy. • Merchandise Exports from India Scheme MEIS aims to promote specifi c services for specifi c markets FTP. • Make in India is an initiative of the government to encourage multinational as well as domestic companies to manufacture their products in India. It was launched in September 2014.

Balance of Trade (BoT)

The difference between a nation’s imports of goods and services and its exports of them is known as balance of Trade. There are three possibilities in the Balance of Trade (BoT) which are as follows: 1. Balance BoT i.e. Exports = Imports 2. Adverse BoT i.e. Exports < Import 3. Favourable BoT i.e. Exports > Imports

Foreign Direct Investment (FDI)

It is an investment in a foreign country through the acquisition of a local company or the establishment of an operation on a new greenfi eld site. Direct investment implies control and managerial and perhaps technical, input.

Indian Financial System

The Indian Financial System consists of two parts: (a) Indian Money Market (b) Indian Capital Market Sector/Activity

Stock Exchange of India

Share Market in India

• National Stock Exchange of India Limited is the leading stock exchange of India. • It is located in Mumbai, Maharashtra, and the oldest stock exchange in Asia which was established in 1875.

SEBI

• Securities and Exchange Board of India was setup in 1983 and made a satutory body in 1992. • SEBI was merged with Forward Market Commission (FMC).

Global Indices Index

% of FDI (Equity)

Country

Hang Seng

Hong Kong

Multi Brand Retail (food)

100%

JCI

Indonesia

Telecom Services

100%

Nikkei 225

Japan

Tea Plantation

100%

Kospi

South Korea

Asset Reconstruction Company

100%

Kuala Lumpur composite

Malaysia

Petroleum and Natural Gas

49%

TSEC Weighted index

Taiwan

Commodity Exchanges

49%

SSE Composite index

China

Insurance

74%

SET

Thailand

Power Exchanges

49%

FTSE 100

UK

Stock Exchanges/Clearing Corporations

49%

NASDAQ Composite index

US

STOXX

Europe

Credit information companies, Pharma

100% (Green Field) 74% (Brown Field)

Dow Jones

US (New York)

81

GENERAL AWARENESS : INDIAN ECONOMY

Credit Rating Agencies

• A Credit Rating Agency (CRA) is a company that assigns credit for insurance of certain types of debt obligation as well as the debt instruments themselves. • CRISIL, setup in 1988, is a credit rating agency. It undertakes the rating fi xed deposits programs, convertible and non-convertible debentures and also credit assessment of companies. • Some other Credit Rating Agencies in India are ICRA, CARE, and India Rating.

Banking and Finance

• Bank of Hindustan was the fi rst bank, established in India in 1770. • The fi rst bank with limited liability managed by an Indian Board was the Oudh Commerical Bank is 1881. • First purely Indian bank was Punjab National Bank (1894).

Nationalization of Bank

• A step towards social banking was taken with the nationalization of 14 commercial banks on 19th July 1969. Six more banks were nationalized in 1980, the total number of public sector banks is 27. • Later on, in the year 1993, the government merged the New Bank of India with Punjab National Bank. • Bhartiya Mahila Bank, India’s fi rst bank exclusively for women, headquarters in New Delhi was inaugurated on 19th November 2013. It has been merged with SBI in 2017. • IDBI Bank is an Indian fi nancial service company, formerly known as Industries Development Bank of India, headquartered in Mumbai, India. • In September 2004, the RBI incorporated IDBI as a scheduled bank under the RBI Act, 1934. • In 2019, Oriental Bank of Commerce and United Bank got merged with Punjab National Bank. Syndicate Bank was merged with Canara Bank while Union Bank of India, Andhra Bank and Corporation Bank got merged. Similarly, Indian Bank got merged with Allahabad Bank.

Reserve Bank of India (RBI)

• RBI was established in 1935, under RBI Act, 1934. RBI is the Central Bank of India. The main purpose of creating RBI was to regulate the money supply and credit in this country. RBI was nationalized in 1949 and its fi rst India Governor was CD Deshmukh. Its headquarter is in Mumbai.

Functions of the RBI

• It is responsible for the monetary policy. • It regulates and supervises the banking sector and non-banking fi nancial institutions. • It looks into debt and cash management for the centre and state governments. • It is involved in foreign exchange reserves management, current and capital account management. • It looks into currency management, and payment and settlement systems. • It has a development role

The RBI and Credit Control Quantitative Credit Control

It is used to control the volume here of credit and indirectly to control the infl ationary and defl ationary pressures. The quantitative credit control consists of • Bank Rate: It is the rate, at which the RBI gives fi nance to Commercial Banks. • Cash Reserve Ratio (CRR): Cash that banks deposit with the RBI without any fl oor rate or ceiling rate. • Statutory Liquidity Ratio (SLR): It is the ratio of liquid assets, which all Commerical Banks have to keep in the form of cash, gold and government-approved securities with themselves.

• Repo Rate: It is the rate, at which RBI lends short-term money to the banks against securities. • Reverse Repo Rate: It is the rate, at which banks park short-term excess liquidity with the RBI. This is always 100 base point, 1% less than Repo rate.

Qualitative/Selective/Direct Credit Control

Qualitative measures are used to make sure that purpose, for which loan is given is not misused. It is done through • credit rationing • regulating loan to consumption etc.

New Bank Licence

• In April 2015, Reserve Bank of India provided licence for operation to two new private banks namely Bandhan Financial Services and Infrastructure Development Finance Company (IDFC).

MUDRA Bank

• Micro Units Development and Refi nance Agency Bank (MUDRA Bank) was launched on 8th April 2015. Bank setup under SIDBI (Small Industries Development Bank of India). Bank has launched three loans instruments. • Shishu—Cover loans up to ` 50,000. • Kishore—Cover loan above ` 50,000 and up to ` 5 lakh. • Tarun—Cover loans above ` 5 lakh and up to ` 10 lakh.

Indradhanush Scheme 2015

• This is for the banking reforms in India. The seven key reforms of the Indradhanush Mission includes appointments de-stressing, capitalization empowerment, the framework of accountability, governance reforms and bank board bureau.

15th Finance Commission

• The 15th Finance Commission was constituted in accordance by Article 280 of the Indian Constitution. • The fi rst fi nance commission was headed by KC Neogi and the 15th Finance Commission is headed by NK Singh.

Insurance

• The insurance industry includes two sectors-life Insurance, and General Insurance. • LIC was established on 1st September 1956. • Insurance Regulatory and Development Authority of India (IRDAI) was setup on 19th April, 2000 to regulate the Insurance Sector. IRDA has changed its name to Insurance Regulatory and Development Authority of India in December 2014.

General Insurance Corporation of India (GIC)

• It was established on 1st January 1873 with its four subsidiaries. (i) National Insurance Company Limited, Kolkata (ii) The New India Assurance Company Limited, Mumbai (iii) The Oriental Fire and General Insurance Company Limited, New Delhi (iv) United India General Insurance Company Limited, Chennai.

Unit Trust of India

• It was set up in 1964 and bifurcated in 2003 into the specifi ed undertaking of Unit Trust of India (SUUTI) and UTI Mutual Fund (UTIMF).

Indian Currency System

• Inconvertible paper-based currency is the present monetary system of India. It is managed by RBI under the Section 24 of the Reserve Bank of India Act, 1934 Printing of securities and minting in India.

82

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Security Press India Security press (1922) Security Printing press (Estd. 1982) Currency Notes press (1928)

Station Nasik (Maharashtra) Hyderabad (AP) Nasik (Maharashtra)

Banknotes from `1 to `100

Bank Notes press (1974)

Dewas (MP)

Banknotes of `20, `50, `100, `500, `2000

Modernized currency notes press (1995)

Mysore (Karnataka ); Salbani (West Bengal)

Security Papermill (Estd. 1967–68)

Hoshangabad (MP)

Inflation

• Infl ation is the rate of increase in prices of goods and services, which leads to a fall in the currency purchasing power.

Causing of inflation

• Printing too much money. • Increase in production cost. • Rise in taxes. • Decline in exchange rates • War or other events causing instability • Increase in money supply in the economy

Deflation

• The defl ation is reduction in the general level of prices in an economy. This is often caused by the decline in the supply of money or credit. • It leads to unemployment since there is a lower level of demand in the economy, which can lead to economic depression.

Stagflation

• Stagfl ation is a situation in which the infl ation rate is high, the economic growth rates slow, and unemployment remains steadily high.

Measures to control inflation

• Increasing the bank interest rates. • Regulating fi xed exchange rates of domestic currency. • Controlling prices and wages. • Providing cost of living allowance to citizens. • Regulating black and speculating market.

Indian Tax Structure

• Direct Tax: The term direct tax generally means a tax paid directly to the government by the persons on whom it is imposed e.g. income tax, Corporate income tax, capital gain tax, stamp duty, land tax, estate duty, wealth tax, petroleum revenue tax. The government earns the maximum from corporate income tax. • Indirect tax: An indirect tax is a tax collected by an intermediary from the person who bears the ultimate economic burden of the tax e.g. sales tax or VAT, customs duty, insurance premium tax, excise duties, landfi ll tax, electricity duty, climate change levy.

Goods and Services Tax (GST) The GST as it is more commonly referred to as a system of taxation where there is a single part in the economy for goods as well as services. Indian GST was fi rst proposed in India in the Union Budget speech in 2006–07. This tax came into effect on 1 July 2017. The main feature of the GST is that there is a tax credit available at each stage of the value chain.

Finance Commission

The fi nance commission is a constitutional body for allocation of certain revenue resources between the Union and the State government. 15th Finance commissions have been set up for the fi nancial year 2020–25. NK Singh was appointed as the President of 15th fi nance commission.

Related to Posted Material, postal stamps etc. Union exercise duty stamps

Bank and currency notes paper

International Economic Organizations International Monetary Fund

IMF established in 1944 is an organization of 190 countries, working to foster global monetary cooperation, secure fi nancial stability, cooperation international trade, promote high employment, sustainable economic growth, and reduce poverty around the world.

World Bank Group

World Bank is one of the fi ve institutions created at the Bretton Woods conference in 1944. Along with IMF, it constitutes twinsister of Bretton Woods.

World Trade Organization

WTO is an intergovernmental organization established in 1995, having its head quarter in Geneva (Switzerland) that regulates and faciliates international trades between two nations. Conference

Year

Place

First

1996

Singapore

Second

1998

Geneva

Third

1999

Seattle (USA)

Fourth

2001

Doha (Qatar)

Fifth

2003

Cancun (Mexico)

Sixth

2005

Hong Kong

Seventh

2009

Geneva

Eighth

2011

Geneva

Ninth

2013

Bali (Indonesia)

Tenth

2015

Nairobi (Kenya)

Eleventh

2017

Buenos Aires (Argentina)

Twelfth

2021

Nur-sultan (Kazakhstan)

Asian Development Bank (ADB)

• ADB envisions a prosperous, inclusive, resilient, and sustainable Asia and the Pacifi c, while sustaining its efforts to eradicate extreme poverty in the region.

Demography Demographics • In 1881, the fi rst synchronized census in India took place. Since 1881 it has been taking place after every decade. • The slogan of the census of 2011 is ‘Our Census, Our future.’ • India was the fi rst country to adopt family planning in the world. Total Population

121,08,54,977

Males

623724248 (51.54%)

Females

586469174 (48.46%)

Density

382 per sq. km

Adult sex ratio

943

Child sex ratio

914

83

GENERAL AWARENESS : INDIAN ECONOMY

Census 2011 Population Trend in India Total Population

1210569573

1891–1921

Stagnant population

Male

623121843 (51.47%)

1921–1951

Steady growth

Female

587447730 (48.53%)

1951–1981

Rapid high growth (stage of population explosion)

Density

382 per sq km

1981–2001

High growth rate with defi nite signs of slowing down

Child Sex Ratio

914

National Population Policy

This policy outlined the following objectives to be achieved: • To lower down Total Fertility Rate (TFR) • Population stabilization of 2045. • Reduce Maternal Mortality Rate (MMR) to below 100 per 100000 births.

• Reduce Infant Mortality Rate (IMR) to below 30 per thousand live births. • Marking school education compulsory. • Promote delay marriage of girls. • Promote and control communicable diseases.

Largest and the Smallest States/UTs (in Population) Top States/UTs

Bottom-States/UTs

Uttar Pradesh

199281477 Lakshadweep

64,429

Maharashtra

112372972 Ladakh

2,74,000

Bihar

103804637 Andaman and Nicobar Islands

3,79,944

West Bengal

91347736 Dadra and Nagar Haveli and Daman and Diu

5,85,764

Andhra Pradesh

84665533 Sikkim

6,07,688

States/UTs (According to Literacy) Top States/UTs

Effective Literacy Rate (2001–2011)

(in %)

Bottom States/UTs

(in %)

Kerala

93.91

Bihar

63.82

Lakshadweep

92.28

Arunachal Pradesh

66.95

Mizoram

91.58

Rajasthan

67.06

Tripura

87.75

Jharkhand

67.63

Goa

87.40

Andhra Pradesh

67.66

2001 (%)

2011 (%)

Difference

Persons

64.83

74.04

10.21

Males

75.26

82.14

6.9

Females

53.67

65.46

11.8

Sex Ratio 2001

2011

Population (in mn)

Proportion (in %)

Population (in mn)

Proportion (in %)

Males

532.2

51.74

623.7

51.51

Females

496.5

48.26

586.4

48.46

51.89

82.9

48.11

75.8

Adult Sex Ratio

933

Males

85.0

Females

78.8

Child Sex Ratio

943

927

52.24 47.76 914

Objective Type Questions 1. In terms of Purchasing Power Parity (PPP), the place of the economy of India is ________in the world. (2023) (2) 5th (1) 3rd (3) 4th (4) 6th 2. The relationship between the values of a country’s imports and its exports is called. (2023) (1) Balance of Trade (2) Balance of Payment (3) Balance of currency (4) Bill of exchange 3. The rate at which banks part short-term excess liquidity with RBI, is called. (2022) (1) Repo Rate (2) Reverse Repo Rate (3) Bank Rate (4) Cash Reserve Ratio

4. Which of the following indices are released by NITI Aayog? (2022)

A. State Energy Index B. District Hospital Index C. State Health Index D. Composite Water Management Index E. SDG India Index (2) A, C, D and E only (1) C, D and E only (3) A, B and E only (4) A, B, C, D and E 5. A tyre tube manufacturer faces a –1.6 price elasticity of demand for its tubes. It is presently selling 24000 units/month. If it wants to increase quantity sold by 4%, it must lower its price by _________. (2021) (1) 2.5 per cent (2) 5.6 per cent (3) 2.4 per cent (4) 6.4 per cent

84

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

6. Which Five Year Plan had an objective of ‘Rapid industrialization with particular emphasis on development of basic and heavy industries’? (1) First (2) Second (3) Third (4) Fourth 7. During which fi ve year plan the growth rate of agricultural production was negative? (1) Third Plan (2) Second Plan (3) First Plan (4) No option is correct. 8. What was the main aim of First Five Year Plan? (1) Self-dependent (2) Industrial growth (3) Economic growth (4) Agricultural growth 9. In the last one decade, which one among the following sectors has attracted the highest foreign direct investment infl ows into India? (1) Food processing (2) Petro-chemical (3) Chemicals other than fertilizers (4) Telecommunications 10. Which of the following should always be balanced in the foreign trade of India? (1) Balance of Trade (2) Balance of Payment (3) Balance of Current Account (4) No option is correct. 11. The currency of Japan is _______. (1) Renminbi (2) Euro (3) Yen (4) Dollar 12. Which of the following is NOT a type of loan available under the Pradhan Mantri MUDRA Yojana? (1) Purush (2) Kishore (3) Shishu (4) Tarun 13. In the Interim Budget of 2019, the Union Government launched a minimum assured pension scheme for those working in the unorganised sector, after they attain the age of 60. The scheme has been named as the Pradhan Mantri_______. (1) Suraksha Vima Yojana (2) Shram Yogi Maan-dhan Yojna (3) Jeevan Jyoti Bima Yojana (4) Vishwakarma Shram Samman Yojana 14. Which one of the following industries is the biggest consumer of water in India? (1) Engineering (2) Paper and pulp (3) Textiles (4) Thermal power 15. The apex body for formulating plans and coordinating research work in agriculture and allied fi elds is:_______. (1) Indian Council of Agricultural Research (2) Regional Rural Banks (3) State Trading Corporation (4) National Bank of Agriculture and Rural Development (NABARD) 1. (1) 9. (4)

Answer Key 2. (1) 3. (2) 4. (4) 5. (1) 6. (2) 7. (1) 8. (4) 10. (2) 11. (3) 12. (1) 13. (2) 14. (4) 15. (1)



3.

4..



5. 6.

7.

Answers with Explanations 1. Option (1) is correct. India currently ranks 3rd in the world in terms of Purchasing Power Parity (PPP). This means that India’s gross domestic product (GDP) adjusted for price differences across countries allows its citizens to purchase more goods and services than those in many other countries with higher nominal GDPs. 2. Option (1) is correct. Balance of Trade (BOT): Specifi cally refers to the relationship between a country’s imports and exports of goods and services.

8.

9.

Calculated as the difference between the total value of exported goods and services (credits) and the total value of imported goods and services (debits). Balance of Payment (BOP): A broader concept encompassing all international economic transactions of a country. Includes the Balance of Trade (BOT), foreign investments, fi nancial transfers, and other transactions. Captures the overall fl ow of money into and out of a country. Balance of Currency: Does not exist as a standard economic term. Bill of Exchange: A fi nancial instrument used in international trade to facilitate payments between buyers and sellers. Not a measure of trade imbalance. Option (2) is correct. Reverse Repo Rate is the rate the central bank of a country pays its commercial banks to park their excess funds in the central bank. It is the rate at which banks earn interest when they park surplus funds with the RBI. It helps to control infl ation, and increases liquidity in the economy. The repo rate set by the RBI is always higher than the reverse repo. Option (4) is correct. The NITI Aayog serves as the apex public policy think tank of the Government of India. Planning Commission was replaced by a new institution – NITI Aayog on January 1, 2015. It has emphasis on ‘Bottom –Up’ approach to envisage the vision of Maximum Governance, Minimum Government, echoing the spirit of ‘Cooperative Federalism’. Chairperson: Prime Minister Vice-Chairperson: Dr. Rajiv Kumar Some of the indices launched by NITI Aayog are School Education Quality Index, State Health Index, Composite Water Management Index, Sustainable Development Goals India Index, India Innovation Index and Export Competitiveness Index, District Hospital Index, State Energy Index, and Multidimensional Poverty Index. Option (1) is correct. If tyre tube manufacturer wants to increase quantity sold by 4%; he must lower its price by 2.5 percent. Option (2) is correct. The Second Five Year Plan came into existence in 1956 and it was operational bill 1961. It was based on the Mohalanobis Model of rapid industrialization of heavy industries. Option (1) is correct. The Third Five Year Plan underscored the development of industries and agriculture. It desired to improve agricultural exhibition by 30%; regardless, India lost to achieve its target because of battles with Pakistan and China in 1965 and 1962 and the lousy monsoon. Option (4) is correct. The focus of the fi rst fi ve-year plan was on the nation’s agricultural revolution, the second was the nation’s industrial growth, the third plan was to drive the economy independently, and the Fourth Five Year Plan was developed with stability and accomplishment self-reliance. Option (4) is correct. Telecommunications has tempted India’s most elevated foreign direct investment infl ows in the fi nal decade.

GENERAL AWARENESS : INDIAN ECONOMY

10. Option (2) is correct. Balance of Payment is described as the report that documents all transactions completed between citizens of a nation and the world’s leftovers during any provided duration. Transactions associated with both goods and services are registered elements of BOP: Current account, Capital account, Financial account. 11. Option (3) is correct. The Yen is the authorized currency of Japan. 12. Option (1) is correct. Pradhan Mantri MUDRA Yojana is a project founded on April 8, 2015, by the Hon’ble Prime Minister. This was established for delivering loans up to 10 lakhs to the non-corporate, nonfarm tiny/micro-enterprises. Shishu, Kishore, and Tarun are the kinds of loans known underneath the Pradhan Mantri MUDRA Yojana.

85 13. Option (2) is correct. Interim Budget 2019–20 the Union Minister for Finance, Corporate Affairs, Railways and Coal, Shri Piyush Goyal, presented a mega pension yojana, i.e. ‘Pradhan Mantri Shram Yogi’ Maan-Dhan Yojana. 14. Option (4) is correct. Thermal Power is the largest buyer of water in India. Engineering is the second-biggest consumer of water in India. The Paper and pulp enterprise arrives in the third rank, and textiles reach the fourth rank for gulping water. 15. Option (1) is correct. The Indian Council of Agricultural Research is answerable for developing projects and performing research work in agriculture and its associated fi elds. It notifi es the Department of Agricultural Research and Education Ministry of Agriculture. The Union Minister of Agriculture acts as its President.

Lenses

Light

Mirrors

Electricity and Magnetism

Waves

First Level

Second Level

Trace the Mind Map

Third Level



General Properties of Matter

Gravitation

MotionLaws of Motion

Heat and Thermodynamics

Physics

Unit System

86 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

GENERAL AWARENESS : GENERAL SCIENCE

87

Pollution

Biotechnology

Ecology

The Cell

First Level

Second Level

Third Level



Human System

Trace the Mind Map

Central Nervous System

Biology

Living World

88 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

General Science

5

Chapter Analysis 2022

2023

Physics

Concept Name

2021

2

3

Chemistry

2

1

2

Biology

1

2

2

Topic-1 Physics  Revision Notes Units • Unit refers to the quantity of a constant magnitude. It is used to measure the magnitudes of other quantities of the same nature. • It is based on the following seven basic units and two supplementary units.

Basic Units Unit Meter (m) Kilogram (km) Second (s) Ampere (A) Kelvin (K) Candela (cd) Mole (mol)

Supplementary Units Name of Quantity Plane angle Solid angle

Name of unit Radian (rad) Steradian (sr)

Derived Units • These units are expressed in terms of two or more fundaments units.

Bigger Units

1 light year = 3.46 × 1015 m 1 Parsec = 3.086 × 1016 m = 3.26 ly 1 AU = 1.5 × 1011 m 1 metric tonne = 103 kg 1 quintal = 102 kg • Dimensions – All physical quantities that can be expressed in terms of powers of M, L, T are called dimensions.

Motion Distance

• The length of the path covered by an object, or body is called distance. It is always positive. • The SI unit of distance is meter.

Displacement

• Displacement ≤ Distance • The distance covered by a moving body in a unit time interval is the speed. • It is always equal to or greater than the magnitude of the velocity. Total distance travelled Average speed = • Total time taken

Velocity

SI unit system

Name of Quantity Length Mass Time Electric current Thermodynamic temperature Luminus intensity Amount of substance

Additional Questions

• The shortest distance between the final and initial position of an object is called displacement. It may be positive, negative or zero. • SI unit of displacement is metre.

• The rate of change of displacement is called velocity. • It is a vector quantity. • It’s SI unit m/s Total displacement • Average velocity = Total time taken • Relative velocity is the rate of change of displacement of one body relative to that of another and vice-versa.

Acceleration

• It is the rate of change of velocity. Its SI unit is m/s2. It is a vector quantity. • Equation of Motion – for a straight line motion with initial velocity (u), final velocity (v) and constant acceleration (a), equations of motion are (i) v = u + at (ii) s = ut + 1 at2 2 (iii) v2 = u2 + 2as • In physics, we use various physical quantities, which can be classified as scalars and vectors. • Scalar quantities – These are the physical quantities that have only magnitude e.g., – speed, mass, volume, work, time, distance, power, energy, etc. •  Vector quantities – These are physical quantities that have both magnitude and direction e.g., displacement, velocity, acceleration, momentum, force, etc.

Newton’s Laws of Motion

• First law – Every body maintains the state of rest or uniform motion on a straight line unless an external non-zero force acts on it. It is called Galileo’s law of inertia. For example – While jumping from a moving bus or train, one must follow the direction of motion for a while. • Second law – The force acting upon an object is directly proportional to the product of masses of the object and the acceleration produced on it. • Third law – Every action has equal and opposite reaction. For example – Rocket goes up due to reaction of downward ejection of gas.

Circular Motion

• When an object moves in a circular path, its motion is called circular motion.

90

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• The tangent of a circle gives the direction of motions at any point in a circular motion. • The Centripetal force is the external force required to act radially inward over the circular motion of the body. • The Centrifugal force is such a pseudo force that is equal and opposite to centripetal force.

• Where, FG is gravitational force, G is the gravitational constant, M is the mass of the fi rst particle, m is the mass of second particle and r is the distance between them. • This is called Newton’s Universal Law of Gravitational • The value of G is 6.67 × 10–11 N-m2/kg2.

Friction

• The gravitational force of the earth is called gravity. • The acceleration produced in a body due to the force of gravity is called acceleration due to gravity (g). • The value of g is 9.8 m/s2. • Escape velocity is the minimum velocity with which an object just crosses the earth’s gravitational fi eld and never returns. Escape velocity at the surface of the earth is 11.2 km/s. • The escape velocity at the moon’s surface is 2.4 km/s. • The value of g decreases with height or depth from the earth surface. • g is maximum at poles • g is minimum at equator • g decreases due to rotation of the earth • g decreases if the angular speed of the earth increases and increases if the angular speed of the earth decreases. • The g at the moon is one-sixth that of the Earth.

• Friction is a force that opposes motion. It is set-up between the surface of contact of two bodies when one body slides or rolls or tends to do so on the surface of another body. • We are able to move on the surface of Earth due to friction. • The brakes stop the automobile because of friction.

Work

• Work is said to be done if force acting on a body is able to actually move it through some distance in the direction of the force. • The SI unit of work is joule • Work = Fs cos θ Where, F = force, s= displacement and θ is the angle between the direction of force and displacement. • If θ > 90°, then work will be negative. • If θ < 90°, then work will be positive. • If θ = 90°, the work will be zero. • If a coolie carrying a load on his head is moving on a horizontal platform, then theoretically he is not doing any work θ = 90°, w = Fs cos 90° = 0 because cos 90° = 0.

Energy

• Energy is the capacity of doing work by a body. • It is a scalar quantity. • The SI unit of energy is a joule. • Mechanical energy is of two types (i) Kinetic energy (K) – It is the energy possessed by a body due to its motion. K = 1 mv2 2 Where, m is mass and v is the velocity. (ii) Potential Energy (U) – It is the capacity of doing work developed in a body due to its position or confi guration. U = mgh Where, m = mass, g = acceleration due to gravity, h = height • The law of conservation of energy states that the sum of all kinds of energies in an isolated system remains constant at all times.

Power

• Power is the rate of doing work. • The SI unit of power is Watt. Work done • Power = Time Taken • 1 watt hour = 3600 joule • 1 kilowatt hour = 3.6 × 106 joule • 1 HP = 746 watt

Gravitation

• In 1686, Newton stated that in the universe each particle of matter attracts every other particle. • This universal attractive force is called gravitational force, and this event is known as gravitation.

Gravitational Force

• Although it is long-range force but is the weakest force. It is represented as FG = GMm r2

Gravity

Satellites

• A satellite is an object (either natural or artifi cial) in space that objects or circles around a bigger object. • Earth’s natural satellite is Moon, while INSAT-1B is an artifi cial satellite of Earth. • The period by a revolution of satellite revolving near the surface of Earth is 1 hour 24 minutes (84 minutes). • Geo-stationary satellite revolves around the Earth at a height of 36000 km (approx.) • Time period of rotation of geo-stationary satellite is 24 h • Polar satellite revolves around the Earth in polar orbit at a height of 800 km (approx.)

General Properties of Matter Elasticity

• Elasticity is the property of the material of a body by virtue of which the body acquires its original shape and size after the removal of deforming force. • Deforming force is a force, which changes the confi guration of a body. • Steel is more elastic than rubber.

Pressure

• The force, acting normally on a unit area of the surface is pressure. • Pressure = Force Area • SI unit of pressure is N/m2. It is a scalar quantity. • Barometer is an instrument used to measure the atmospheric pressure. • If a gravitational attraction is negligible in equilibrium condition (approx.) pressure is the same at all points in a liquid. • The pressure exerted anywhere at a point of the confi ned liquid is transmitted equally and undiminished in all directions throughout the liquid. • Hydraulic lift, hydraulic press and hydraulic brakes are based on Pascal’s law of pressure.

Archimedes’ Principle

• When a body is immersed partly or wholly in a liquid, there is an apparent loss in the weight of the body, which is equal to the weight of the liquid displaced by the body. • The weight of water displaced by an iron ball is less than its own weight whereas water displaced by the immersed portion of a ship is equal to its weight so, a small ball of iron ball sinks in water, but a large ship fl oats.

91

GENERAL AWARENESS : GENERAL SCIENCE

Laws of Floatation

• A body fl oats in a liquid if: • The density of the material of the body is less than or equal to the density of a liquid. • When the density of the material of the body is equal to the density of liquid the body fl oats fully submerged in liquid in neutral equilibrium. • The weight of the body is equal to the weight of displaced liquid when body fl oats in neutral equilibrium. The center of gravity of the body and center of gravity of the displaced liquid should be in one vertical line for this condition.

Density

• Density is defi ned as mass per unit volume. • Hydrometer is used to measure relative density. • The density of iron is more than water but less than that of Mercury. So, a solid chunk of iron sink in water but fl oat in Mercury.

Heat and Thermodynamics Heat

• A form of energy, which measures the perception of warmness or coldness of a body or environment, is called Heat. • Its unit are calorie, kilocalorie or joule. • 1 calorie = 4.18 joule

Temperature

• The temperature is the measurement of hotness or the coldness of a body. • When two bodies are placed in contact, heat always fl ows from a body at higher temperature to the body at a lower temperature. • An instrument used to measure the temperature of a body is called a thermometer. • 37°C or 98.4°F is the normal temperature of a human body. • –40° is the temperature at which Celsius and Fahrenheit thermometer read same. • The clinical thermometer reads from 96°F to 110°F. • White roof keeps the house cooler in summer than black roof because white roof refl ects more and absorbs fewer heat rays whereas black roof absorbs more and refl ects fewer heat rays. • Ice wrapped in a blanket does not melt away quickly because woolen blanket is a bad conductor of heat. • Silver is the best conductor of heat. • Cooking utensils are made of aluminum, brass and steel because these substances have low specifi c heat, and high conductivity.

Thermal Expansion

• Thermal Expansion is the increase in size on heating. • A solid undergoes three types of expansion: (i) Linear expansion (ii) Superfi cial expansion (iii) Cubical expansion

Specific Heat

• The specifi c heat capacity of a substance is the heat capacity of a sample of the substance divided by the mass of a sample. • When the temperature of the water is increased from 0°C then its volume decreases upto 4°C, becomes minimum at 4°C, and then increases. • This behavior of water around 4°C is called the anomalous expansion of water.

Latent heat

• Latent heat is heat energy absorbed or released at constant temperature per unit mass for change of state. • Latent heat of fusion of ice is 80 cal/g. • Latent heat of vaporization of steam is 540 cal/g.

Evaporation

• Evaporation is the process of conversion of liquid into its vapor even below its boiling temperature. • Humidity is the amount of water vapor in the air. • Relative humidity is measured by a hygrometer.

Simple Pendulum

• A heavy point mass suspended from rigid support using an elastic and inextensible string, is called a simple pendulum. • If a lift falls freely under gravity then the time period of the pendulum is infi nite. • The time period of the pendulum will increase if a simple pendulum is suspended in a lift with acceleration. • If lift is ascending, then the period time of the pendulum will decrease.

Waves

• A disturbance, which propagates energy from one place to the other without the transportation of matter is called a wave. • The waves are classifi ed into two types: (i) Mechanical wave (longitudinal wave and transverse wave) (ii) Electromagnetic force.

Longitudinal wave

• In this wave, the particles of the medium vibrate in the direction of propagation of the wave. • Sound waves in air is example of a longitudinal waves.

Transverse wave

• The particles of the medium vibrate perpendicular to the direction of propagation of a wave. • Waves on strings under tension, waves on the surface of the water are examples of transverse wave.

Electromagnetic wave

• Electromagnetic waves are those waves, which do not require a medium for their propagation i.e., which can propagate even through the vacuum. • Light radio waves, X-rays, etc. are examples of electromagnetic waves.

Sound waves

• Sound waves are longitudinal mechanical waves, sound waves are grouped as follows: • The sound waves, lying between the frequency range of 20 Hz to 20000 Hz, are called audible waves. • The sound waves having frequencies less than 20 Hz are called infrasonic waves. • The sound waves having frequencies greater than 20000 Hz are called ultrasonic waves.

Speed of Sound • Speed of sound is maximum in solids and minimum in gases. • When the sound goes from one medium to another medium its speed and wavelength change, but the frequency remains unchanged. • The speed of sound remains unchanged by the increase or decrease of pressure. • The speed of sound increases with the increase of temperature of the medium. • The speed of sound is more in humid air than in dry air because the density of humid air is less than density of dry air.

92 • Echo is the repetition of sound due to the refl ection of sound waves. • Intensity is defi ned as the amount of energy passing per unit time through a unit area that is perpendicular to the direction in which sound waves are travelling. • Pitch is the sensation of a frequency. • SONAR stands for sound navigation and ranging. It is used to measure the depth of the sea, to locate under water submarines and shipwrecks.

Doppler’s Effect

• The Doppler effect is the change in frequency of a wave to an observer who is moving relative to the wave source. • When the distance between the source and observer decreases, then apparent frequency increases and vice-versa.

Light

• A form of energy, which is propagated as the electromagnetic waves, is light. • It is the radiation that makes our eyes able to see. • 3 × 108 m/s is the speed of light.

Reflection of light

• Refl ection of light occurs when the waves encounter a surface or other boundary that does not absorb the energy of the radiation and bounces the waves away from the surface.

Laws of reflection

• The incident ray, refl ected ray, and the normal to the refl ecting surface at the incident point all lie in the same plane. • The angle of incidence is equal to the angle of refl ection.

Reflection from plane mirror

• The image formed on the plane mirror is virtual and laterally inverted. • The size of the image is equal to that of an object. • If an object moves towards the plane mirror with speed v, relative to the object the image moves towards it with a speed 2v. • To see his full image in a plane mirror, a person requires a mirror of at least half of his height. • The number of images formed by two plane mirrors, inclined by

 360°  − 1 an angle θ, n =   θ .

Spherical Mirror

• These are of two types: (i) Concave mirror (ii) Convex mirror • Virtual erect and diminished images are formed by the convex mirrors. • Generally real and inverted images are formed by the concave mirrors.

Refraction of light

• The bending of a ray of light passing from one medium to another medium is called refraction. • Refractive index (m) Speed of light in vacuum = Speed of light in the medium

Optical Fiber

• Optical fi ber works on the principle of Total Internal Refl ection (TIR). It is used for telecommunication and various medical purposes like endoscopy.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Lens

• Lens are of two types: (i) Convex lens (ii) Concave lens • When a lens is dipped in a liquid higher refractive index, the focal length increases, and convex lens behave as a concave lens and vice-versa. • An air bubble trapped in water or glass appears as convex but behaves as a concave lens.

Dispersion of light

• When a ray of the white light is passed through a prism, it gets split into its constituent colors. • This phenomenon is called dispersion of light. • (VIBGYOR) Violet, Indigo, Blue, Green, Yellow, Orange, and Red are the order of the splitting of lights.

Scattering of light

• Scattering of light is the phenomenon in which light rays get deviation from their straight path on striking an obstacle like dust or gas molecules, water vapors, etc. • The colors we see in the sky are due to the scattering of light.

Human eye

• The eye is an optical instrument like the camera. It forms the real image of the object on the retina of the eye. • Least distance of distinct vision is 25 cm.

Defect of eye Myopia (short sightedness)

A short-sighted eye can see only nearer objects. Distant objects are not seen clearly. This defect can be removed by using a concave lens of suitable focal length.

Hypermetropia (longsightedness)

A long-sighted eye can see distant objects clearly but nearer objects are not clearly visible. This defect can be removed by using a convex lens.

Presbyopia

In this defect both near and far objects are not clearly visible. It can be removed by using the bi-focal lens.

Astigmatism

In this defect eye cannot see horizontal and vertical lines clearly. The defect can be removed by using suitable cylindrical lenses.

Microscope

• A simple microscope is a convex lens of a small focal length. • A combination of two lenses in which objective lens is a convex lens of large aperture and large focal length while eyepiece is a convex lens of small aperture and small focal length, is an Astronomical Telescope.

Electricity and Magnetism Charge

• The basic property associated with the matter due to which electric and magnetic effects are produced is known as charge. • Similar charges repel each other and opposite charges attract each other. • The SI unit of charge is coulomb.

Electric current

• The rate of fl ow of charge or charge fl owing per unit time is called electric current. • Its SI unit is ampere and it is a scalar quantity.

93

GENERAL AWARENESS : GENERAL SCIENCE

Ohm’s law

• Ohm’s law states that the current passing through a conductor between two points is directly proportional to applied voltage across the two points. V , where R is resistance, V = voltage, I = current • I= R • If a wire is stretched, its resistance will change but its specifi c resistance will remain unaffected. • On increasing the temperature of the metal, its resistance increases. • On increasing the temperature of semiconductors, their resistance decreases. • On increasing the temperature of electrolytes, its resistance decreases. • The reciprocal of resistivity of a conductor is called its conductivity. Its unit is (Ω–1 m–1) . • The heating effect of electric current is known as Joule’s law of heating. • Ammeter is a device that is used to measure electrical current. It is connected in series. • Voltmeter is a device used to measure the potential difference between two points in a circuit. It is connected in parallel. • Fuse wire is a small conducting wire of alloy of copper, tin, and leads having a low melting point.

Magnets

• A magnet is a material or object that produces a magnetic fi eld. • When a magnet is freely suspended, it is one pole always directs towards the North. This pole is called the North pole. The other pole is called the South pole. • Like poles of a magnet repel each other while unlike poles attract each other. • Electromagnet is a current-carrying coil containing a soft iron core, which is utilized in the electric bell, telegraph receiver, telephone, transformer, dynamo, etc.

Atomic and Nuclear Physics Cathode Rays

• Sir William Crooke discovered cathode rays. • Its properties are: • These rays travel in straight lines. • These rays produce fl uorescence. • These rays can penetrate through thin fi lms of metal and defl ect by both electric and magnetic fi elds.

X-Rays

• X-rays are electromagnetic waves with a wavelength range of 0.1 Å – 100 Å.

• Roentgen discovered the X-rays. • X-rays show a photoelectric effect.

Radioactivity

• Radioactivity was discovered by Henry Becquerel, Madame Curie, and Pierre Curie for which they jointly won Nobel Prize. • The nucleus having protons 83 or more are unstable. They emit α, β, and γ particles and become stable. The elements of such nucleus are called radioactive elements and the phenomenon of emission of α, β and γ is called radioactivity. • Robert Pierre and his wife Madame Curie discovered a few radioactive elements of radium. • The end product of all-natural radioactive elements after emission of radioactive rays is lead. • With the emission of an α-particle, the atomic number is decreased by 2 and mass the number is decreased by 4. • With the emission of a β-particle, the atomic number is increased by 1 and the mass number does not change.

Nuclear Fission

• The nuclear reaction, in which a heavy nucleus split into two nuclei of nearly equal mass is nuclear fi ssion. Ba

Kr

• Atom bomb is based on nuclear fi ssion, U235 and Pu139 are used as fi ssionable material.

Nuclear Fusion

• When two or more light nuclei are combined together to form a heavier nucleus it is nuclear fusion. • 108 K temperature is required for nuclear fusion. • Hydrogen Bomb is based on nuclear fi ssion and was made by American scientists in 1952. There are several components of a nuclear reactor which are as follows: • Fuel-U235 or U239 is used as a fuel in the nuclear reactor. • Moderator: Moderator decreases the energy of neutrons so that they can be further used for fi ssion reaction. Heavy water and graphite are used as a moderator. • Control Rod: Rods of cadmium or boron is used to absorb the excess neutrons produced in the fi ssion of the uranium nucleus so that the chain reaction continues to be controlled. • Coolant: A large amount of heat is produced during fi ssion. The coolant absorbs that heat and prevents the excessive rise in the temperature. The coolant may be water, heavy water, or like He or CO2.

Units of Measurement Quantity

Unit (SI)

Quantity

Unit (SI)

Length

meter

Viscosity

N.s/m2

Time

second

Surface tension

Newton/meter

Mass

kilogram

Heat

Joule

Area

square meter

Temperature

Kelvin

Volume

cubic meter

Absolute temperature

Kelvin

Velocity

meter/second

Resistance

Ohm

2

Acceleration

meter/second

Electric current

Ampere

Density

kilogram/meter3

Electromotive force

Volt

Momentum

kilogram-meter/second

Electrical conductivity

mho/meter

Work

Joule

Electric energy

kilowatt-hour

94

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Energy

Joule

Force

Newton

Magnetic intensity

Oersted

Pressure

Pascal or Newton/meter2

Charge

Coulomb

Frequency

Hertz

Magnetic induction

Gauss

Power

Watt

Luminous fl ux

Candela

Weight

Newton or kilogram

Intensity of sound

Decibel

Impulse

Newton-second

Power of lens

Dioptre

Angular velocity

Radian/second

Depth of sea

Fathom

Topic-2

Electric power

Chemistry

 Revision Notes • Chemistry is the branch of physical science, which studies substances i.e., elements and compounds.

Physical and chemical changes

• The changes in which only physical properties like color, hardness, density, melting point, etc. get affected, are the physical changes, but they do not affect the composition of matter. • A physical change is temporary while a chemical change is permanent. • Sublimation, boiling, melting, crystallization, vaporization, cutting of trees, dissolving of sugar or salt in water, etc. are some examples of physical change. • Chemical changes affect the chemical properties of matter along with the formation of a new substance. • Burning of fuel-burning of the candle, electrolysis of water, photosynthesis, ripening of fruits, etc. are some chemical changes.

Matter

• Anything which has some mass and occupies some space is matter. It exists in three states i.e., solid, liquid, and gas. • Nowadays there is a discussion on two more states of matter i.e., Plasma (mixed gases containing super energetic and super excited particles) and Bose-Einstien condensates or BEC (a gas at super low temperature with extremely low density).

kilowatt or watt

• Protons and neutrons reside in the nucleus, whereas electron revolves around the nucleus. • The smallest part of an element or a compound capable of independent existence under ordinary conditions, is the molecule. • Element contains only one type of atom e.g. carbon (C), sulphur (S), diamond, graphite, etc. • Oganesson, with symbol Og and atomic number 118 is a recently synthesized element.

Isotopes and Isobars

• Isotopes have the same number of protons (i.e. atomic number), but different mass number (atomic number + number of neutrons), e.g. 1H1, 1H2, 1H3. • Isobars have the same mass number but different atomic numbers e.g., 18Ar40, 19K40, and 20Ca40.

Colloids

• Colloids are heterogeneous solutions, containing two phases: a dispersed phase and a dispersion medium. • They show the tyndall effect (i.e. scattering of light by colloidal particles) and Brownian motion (zig-zag motion).

Some Colloids and their Example Dispersed Dispersion Phase Medium

Type of Colloid

Example

Liquid

Gas

Aerosol

Fog, clouds, mist

Solid

Gas

Aerosol (solid)

Smoke, automobile exhaust

• The boiling point is the temperature at which liquid converts into vapors. • 100°C is the boiling point of water. • Boiling points decrease at high altitudes. That’s why at high altitudes, the boiling point of water is less than 100°C and more time is required to cook food.

Gas

Liquid

Foam

Shaving cream

Liquid

Liquid

Emulsion Milk, face cream

Solid

Liquid

Sol

Mud, milk of magnesia

Gas

Solid

Foam

Foam, rubber, sponge, pumice

Liquid

Solid

Gel

Jelly, cheese, butter

Melting Point

Solid

Solid

Solid sol

Milky glass, colored gem stone

Boling Point

• The temperature at which a substance starts melting is its melting point. It converts from solid to liquid state. • 0°C is the melting point of ice.

Atom, Molecule and Element

• The smallest particle of the element that can exist individually and retain all its chemical properties is called an atom. • It is made up of electrons, protons, and neutrons.

Battery

• A device used to convert chemical energy into electrical energy is called battery. It is of two types: • Primary batteries (non-rechargeable) act as galvanic cells, e.g., dry cell, mercury cell, etc. • Secondary batteries (rechargeable) act as galvanic as well as electrolytic cells e.g., lead storage battery, nickel-cadmium battery, etc.

Types of Batteries Battery

Anode

Cathode

Electrolyte

Used in

Leclanche cell

Zinc

Graphite

Paste of ammonium chloride and zinc chloride

Transistors, clocks

Mercury cell

Zinc-mercury amalgam Lead

Paste of HgO (Mercuric Paste of KOH and ZnO oxide) and carbon Lead packed in lead dioxide 38% solution sulphuric acid

Lead storage battery

Hearing aids and camera Automobiles, invertors

95

GENERAL AWARENESS : GENERAL SCIENCE

Corrosion

1. Innermost part – This part is black due to the presence of unburnt Carbon particles and has the lowest temperature. 2. Middle part – This part is yellow due to the incomplete combustion of fuel. 3. Outermost part – This part is blue due to the complete combustion of fuel. It is the hottest part and used by goldsmith to heat the Gold.

• The process of corroding of metal by nature is called corrosion. It is an electrochemical process. • When iron is exposed to air, the iron surface turns red due to the formation of hydrated ferric Oxide (Fe2O3 × H2O) • Corrosion can be prevented by electroplating, oiling, greasing, painting, varnishing, and galvanization.

Renewable and Non-renewable Natural Resources

• Renewable resources are available in excess, i.e. never ends, e.g. air, sunlight, etc. • Non-renewable resources are available in limited quantity and end if used excessively after a limited period time e.g., mineral, coal, petroleum, natural gas, etc.

Coal

• It is obtained by carbonization of vegetable matter and is available in different varieties: Peat (60% C), Lignite or Brown coal (70% C), Bituminous coal (60-80% C), Anthraicte coal (90% C).

Flame

• Flame contains three parts

Fire Extinguishers

• Fire is extinguished by water because it evaporates the vapors surrounding the burning substance, cutting off the oxygen supply, thus inhibiting the burning process. • In case of electrical or oil (petrol) fi res, water can’t be used, as water is a conductor of electricity and heavier than oil. In that case, Carbon dioxide, which is generated by the reaction of baking soda with acid, is used to extinguish electrical or oil fi res.

Fuels

• Fuel is a substance which produces heat and light on combustion. • To detect the leakage of LPG, the foul-smelling substance called ethyl mercaptan C2H6SH, is added to LPG. • Calorifi c value is the amount of heat obtained when 1g of a fuel is burned in excess of Oxygen.

Some Important Fuels and their Compositions Fuel

Composition

Sources

Water Gas

Carbon monoxide (CO) + Hydrogen (H2)

Bypassing steam over red hot coke

Producer Gas

Nitrogen (N2) + Carbon monoxide (CO) (2 : 1 ratio)

Bypassing insuffi cient air over red hot coke

Coal Gas

Hydrogen + Methane + Ethylene (C2H4) + Acetylene (C2H2) + CO + Nitrogen

By fractional distillation of wood

Natural Gas

Methane (83%) + Ethane (16%)

From petroleum

Liquifi ed Petroleum Gas (LPG)

Butane (C4H10) + Propane (C3H8)

From oil wells

Compressed Natural Gas (CNG)

Methane (CH4) 95%

From petroleum

Biogas or Gobar Gas

Methane (CH4) + Carbon dioxide (CO2) + Hydrogen (H2) + Nitrogen (N2)

From organic wastes

Calorific Value of Some Substance Fuel

Calorific Value (kJ/g)

Coal

25-32

Kerosene oil

48

Petrol

50

Diesel

45

Biogas

35-40

LPG

50

Cow dung

6-8

Hydrogen

150

Natural gas

35-50

Saftey Matches

• In the safety matches, the stack consists of a mixture of Antimony Trisulphide and Potaasium Chlorate at its one end. • The sides of the box contain a mixture of powdered glass and Red Phosphorous.

Acids, Bases and Salts Acid

• Acid is substance which is sour in taste and it turns blue litmus red. • They are a good conductor of electricity in an aqueous solution. • Acids react with the metal to produce Hydrogen gas that is why pickles are always kept in a glass jar.

Bases

• These are the substances, which are bitter in taste, soapy in touch, and turn red litmus blue. • Bases like NaOH, KOH, etc. are good conductors of electricity in their aqueous solution and in molten state.

Salts

• Salts are the product of neutralization reaction between an acid and a base. • pH is the measure of acidity/basicity.

Some Important compounds in Everyday Life Carbon Dioxide

• CO2 is an acidic oxide of carbon and is used by green plants for photosynthesis. • It does not help in burning. • Air and our breath contain carbon dioxide. Thus, when lime water is kept in the air or we pass our breath into it, the lime water turns milky.

Carbon Monoxide

• It is a neutral oxide of air and has more affi nity towards Hemoglobin which is 200 times more than oxygen. • That is why in the environment of Carbon Monoxide (which is a non-poisonous gas) people die for the need for Oxygen.

Plaster of Paris

1 H O) 2 2 and is prepared by heating Gypsum which is Calcium Sulfate Dihydrate (CaSO4 . 2H2O) at 373 K.

• PoP is chemically Calcium Sulphate Hemihydrate (CaSO4∙

96

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• Plaster of Paris on mixing with water, further sets into a hard solid called Gypsum.

Portland Cement

• Portland Cement is a complex mixture of silicates and aluminates of calcium with the small amount of gypsum. • It is a composition of Calcium Oxide (50-60%), alumina (5-10%) and Magnesium Oxide (2-3%), Gypsum is added to Cement to decrease its setting rate. • Mortar a mixture of sand, cement, and water is used for joining bricks and plastering walls. • Concrete is a mixture of gravel, sand, cement, and water is used for fl ooring and making roads. • Reinforced Concrete Cement (RCC) which is concrete with steel bars, and wires is used for constructing roofs, bridges, and pillars.

Soaps

Pesticides

• Pesticides are the chemicals used to destroy the organisms that harm the crop. • Following are the types of pesticides: • Insecticides – DDT, Grammaxene, Aluminium Phosphate. • Fungicides – Bordeaux mixture, Organomercury compounds. • Herbicides – Benzipram, Sodium chlorate. • Rodenticides – Aluminum phosphide

Heavy Water

• Heavy Water is Deuterium Oxide (D2O), molecular mass = 20.028 g/mol • It is used as the moderator in nuclear reactors. It is called heavy due to the presence of Deuterium, the heavy Hydrogen.

Hard Water

Color

Substance Added

• Hard water is water containing high amounts of mineral ions. • The most common ions found in hard water are the metal cations Calcium (Ca2+) and Magnesium (Mg2+), though Iron, Aluminium, and Manganese may also be found in certain areas. • The temporary hardness of water is removed by boiling or by adding Calcium Hydroxide, Ca(OH)2 - Clark’s process. • The permanent hardness of water is removed by adding Sodium Carbonate (Na2CO3), or Calgon (Sodium Hexametaphosphate Na2[Na4(PO3)6]).

Red

Copper oxide (CuO)

Hardening of oil (Hydrogenation)

Green

Chromium oxide (Cr2O3)

Ruby Red

Goldchloride (AuCl3)

Blue

Cobalt oxide (CoO)

Brown

Iron oxide (Fe2O3)

• Soaps are Sodium and Potassium salts of higher fatty acids, e.g., Sodium Palmitate, Sodium Stearate, etc.

Glass

• Glasses are amorphous solids or super-cooled liquids, contains mainly Silica (SiO2). • Different substances are added to obtain glass of different colors e.g.,

• Oil, an unsaturated fat when heated with Nickel catalyst and Hydrogen, gets converted into solid mass called ghee, a saturated fat. This process is called hardening of oil and is carried out through hydrogenation in the presence of Nickel as a catalyst.

Medicines

• These are the chemicals used for treating diseases and reducing suffering from pain.

Different Medicines and their Examples Medicine

Used to

Example

Analgesics

Reduce pain

Aspirin, paracetamol, morphine, phenacetin

Tranquilizers

To treat stress, mild and severe mental diseases

Equanil, valium, chlorodiazoepoxide, serotonin and meprobamate

Antiseptic

Prevent the growth of micro-organisms or kill them (applied to living tissues)

Dettol (a mixture of chloroxylenol—the antiseptic and α-terpineol), savlon, iodine tincture (solution of I2 in alcohol water mixture), boric acid (antiseptic for eyes), hydrogen peroxide, iodoform

Antibiotic

Destroy microorganisms (These are obtained from microorganism)

Penicillin (discovered by A. Fleming in 1929, ampicillin, amoxicillin, ofl oxacin, chloramphenicol)

Antimalarial

Cure Malaria

Chloroquine

Sulpha Drugs

Alternative for antibiotics

Sulphanilamide, sulphadiazine

Antacids

Reduce acidity

Baking soda, magnesium hydroxide

Polymers

• A polymer is a compound of high molecular weight formed by the combination of a large number of molecules of one or two types

of low molecular weight (known as monomers). This process is called polymerization. • Polymers are the backbones of four major industries plastics, fi bers, paints, and varnishes.

Some Fibers and their Monomers Fibers

Monomers

Uses

Nylon-6.6

Adipic acid + hexamethylene diamine

In making bristles for brushes, synthetic fi bers, parachutes, as a substitute for metal in bearings.

Nylon-6 or perlon

Caprolactum

In making fi bers, plastic tyre cords and ropes.

Terylene

Ethylene glycol and Terephthalic acid

For making wash and wear fabrics, tyre cords, safety belts, tents etc.

Kevlar

Terephthaloyl chloride + 1,4-diamino benzene

For making bulletproof vests.

Lexan or polycarbonate

Diethyl carbonate + bis-phenol-A

In making bulletproof windows and safety helmets.

Polyurethanes

Toluene diisocyanate + ethylene glycol

For making washable and long lasting mattresses cushions.

97

GENERAL AWARENESS : GENERAL SCIENCE

Some Important Industrial Compounds Industrial Name

Chemical Name

Chemical Formula

Alum

Potassium Aluminum sulfate

KAl(SO4)2 . 12H2O

Alcohol

Ethyl alcohol

C2H5OH

Baking soda

Sodium bicarbonate

NaHCO3

Bleaching powder

Calcium oxychloride or calcium hypochlorite

CaOCl2

Brine (or common salt)

Sodium chloride

NaCl

Borax

Sodium tetraborate decahydrate

Na2B4O7 . 10H2O

Caustic potash

Potassium hydroxide

KOH

Caustic soda

Sodium hydroxide

NaOH

Chalk (marble) or pearl

Calcium carbonate

CaCO3

Chilli saltpetre

Sodium nitrate

NaNO3

Chloroform

Trichloro methane

CHCl3

Epsom salt

Magnesium sulfate

Glauber’s salt

Sodium sulfate decahydrate

MgSO4 . 7H2O Na2SO4 . 10H2O

Gypsum

Calcium sulfate dehydrate

Hypo

Sodium thiosulphate pentahydrate

CaSO4 . 2H2O

Na2S2O3 . 5H2O

Laughing gas

Nitrous oxide

N2O

Lunar caustic

Silver nitrate

AgNO3

Marsh gas

Methane

CH4

Quick lime

Calcium oxide

CaO

Sal ammonia (Nausadar)

Ammonium chloride

NH4Cl

Sapphire (Ruby)

Aluminum oxide

Al2O3

Slaked lime

Calcium hydroxide

Ca(OH)2

Soda ash

Sodium carbonate

Na2CO3

Spirit

Methyl alcohol

CH3OH

Washing soda

Sodium carbonate decahydrate

Topic-3

Biology

 Revision Notes • Biology is the scientifi c study of life. It was coined by Lamarck and Treviranus (1802). • It mainly includes Botany (study of plants) and Zoology (study of animals). • Aristotle is known as the father of Biology and Zoology.

Living World

• Carolus Linnaeus, in the 18th century, developed binomial nomenclature for a living organism, i.e., scientifi c name consisting of genus and species. • Living organisms are classifi ed into fi ve kingdoms by Whittaker (1969) – Monera, Protista, Fungi, Plantae and Animalia. • Monera includes bacteria and Mycoplasma, while Protista includes Protozoa (unicellular Eukaryotes). • Viruses are sub-microscopic, obligate, intracellular parasites consisting of nucleoprotein. WM Stanley fi rstly crystallized TMV (Tobacco Mosaic Virus). • Viroids are the smallest infections single-stranded RNA molecule discovered by TO Diener.

The Cell • Schleiden and Schwann (1838) proposed a cell theory that the cell is the structural and functional unit of living organism. • An organism may be composed of a single cell (unicellular) or many cells (multicellular). • There are two types of cells i.e., Prokaryotic (which lacks nucleus and membrane-bound organelles) and Eukaryotic (which have nucleus and membrane-bound organelles)

Na2CO3 . 10H2O

• In bacteria, prokaryotic cell is found, mycoplasma and blue-green algae while the eukaryotic cell in plants, animals, and fungi.

Nucleic Acids • These contain the genetic instructions used in the development and functioning of all known living organisms. These are of two types namely DNA and RNA. • Deoxyribose Nucleic Acid (DNA) is a long polymer made from repeating units called nucleotides. • It has four bases, i.e. Adenine, Guanine, Cytosine, and Thymine. • Ribonucleic Acid (RNA) is made up of a long chain of nucleosides. It contains Uracil in place of Thymine base pair.

Human Systems • Tissues are the organized cells of humans and other multicellular animals. Two or more tissues grouped together to form an organ. • An organ system is a group of organs that function together to carry out the principal activities of the body.

Digestion • Digestion is the process by which complex food is converted into simple components with the help of digestive enzymes.

Respiratory System • Respiration is an oxidative process involving the oxidation of food substances such as carbohydrates, fat and proteins to form CO2 water and to release energy. • Respiration may be anaerobic, (i.e. without O2), and aerobic (i.e. with O2).

98

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Respiratory Organs of Animals

Body surface

Protozoans, porifera and coelenterates

Organ

Animal

Book lungs

Spider and scorpion

Lungs

Reptiles and mammals

Book gills

King crab, prawn, crayfi sh and Daphnia

Skin

Frog, earthworm and leeches

Mental

Mollusca (Unio)

Gills

Fishes, tadpoles and prawns

Air bladder

Long fi sh and bony fi shes (e.g. Labeo)

Tracheae

Insects, centipedes and millipedes

Air sacs lungs

Birds

Vitamins Fat Soluble Vitamins Vitamin (Name)

Rich Food Source

Function

Deficiency Disease

A (Retinol)

Fish liver oils, dairy products, Needed for healthy epithelial cells liver, most leafy vegetables and and regeneration of rodopsin in rod carrots contain carotene that can be cells of the eye converted into retinol

Dry skin and night blindness (Nyctalopia)

D (Calciferol)

Fish oils, egg yolk and butter. It can be made by the action of sunlight on skin

Promotes absorption of calcium from intestines necessary for formation of normal bone and reabsorption of phosphate from urine

Rickets in children (soft bone that bend easily) Osteomalacia (inadequate mineralization of bone tissue) in adults

E (Tocopherol)

Vegetable oils, cereal products and many other foods

Formation of red blood cells, affects muscles and reproductive system.

Mild anemia and sterility. The defi ciency is rare in humans

K (Phylloquinone)

Fresh and dark green vegetables. Also made by gut bacteria

Formation of prothrombin (involved in blood clotting)

There is delayed clotting time. May occur in new-born babies before their gut bacteria become established

Water Soluble Vitamins Vitamin (Name)

Rich Food Source

Function

Deficiency Disease

B1 (Thiamine)

Yeast, cereals, nuts, seeds and pork

Co-enzyme in cell respiration, necessary for complete release of energy from carbohydrates.

Beri-beri (muscular dystrophy, stunted growth and nerve degeneration)

B2 (Ribofl avin)

Liver, milk, eggs and green vegetables

Co-enzyme in cell respiration. Precursor of FAD

Cracked skin and blurred vision

B3 (Niacin)

Liver, yeast, whole cereals and beans

Co-enzyme in cell respiration. Precursor of NAD/NADP

Pellagra (severe skin problems, diarrhea and dementia)

B5 (Pantothenic acid)

Animal tissue whole grain cereals and legumes

Needed to manufacture adrenal hormone

Pellagra, Dermatitis and diarrhea

B6 (Pyridoxine)

Meat, fi sh, eggs, cereals brain and Interconversion of amino acids some vegetables

Skin problems and nerve disorder

B10 (Folic acid)

Liver, raw green vegetables, yeast and gut bacteria

Formation of nucleic acids and red blood cells

Anemia (especially during pregnancy)

B12 (Cyanocobalamine)

Liver, milk, fi sh and yeast. None in plant foods

Maturation of red blood cells in bone marrow. Maintenance of myelin sheath of nerves

Pernicious anemia and nerve disorders

C (Ascorbic acid)

Blackcurrants, peppers, sprouts and citrus fruits

Formation of collagen and intercellular cement

Scurvy and poor wound healing

Major Enzymes of Digestion Enzyme

Source

Where Active

Substrate

Main Breakdown Product

Carbohydrate Digestion Salivary amylase

Salivary glands

Mouth

Polysaccharides

Disaccharides

Pancreatic amylase

Pancreas

Small intestine

Polysaccharides

Disaccharides

Sucrase, maltase, lactase

Small intestine

Small intestine

Disaccharides

Monosaccharides (e.g., glucose)

Pepsin

Stomach mucosa

Stomach

Proteins

Peptide fragments

Trypsin and chymotrypsin

Pancreas

Small intestine

Proteins and polypeptide

Peptide fragments

Carboxypeptidase

Pancreas

Small intestine

Peptide fragments

Amino acids

Protein Digestion

99

GENERAL AWARENESS : GENERAL SCIENCE

Enzyme

Source

Amino peptidase

Where Active

Main Breakdown Product

Substrate

Intestinal mucosa

Small intestine

Peptide fragments

Amino acids

Pancreas

Small intestine

Triglycerides

Free fatty acids and monoglycerides

Fat Digestion Lipase Nucleic Acid Digestion Pancreatic nucleases

Pancreas

Small intestine

DNA and RNA

Nucleotides

Intestinal nuclease

Intestinal mucosa

Small intestine

Nucleotides

Nucleotides bases and monosaccharides

Blood (Lymphatic System)

• Blood is a type of fl uid connective tissue composed of plasma and blood cells. • An adult person has 5–6 Liters of blood. • It is slightly alkaline having a pH 7.3–7.4. • Plasma is pale yellow liquid and constitutes about 60% volume of blood. • Plasma consisted of 90–92% water, 7% organic substances (albumin, globulin, and fi brinogen protein) and 1% inorganic substances.

• Red blood corpuscles (most abundant) are non-nucleated and contain haemoglobin (the respiratory pigment). • White blood cells are colorless, nucleated, and granular or agranular. • Eosinophils are also called acidophilis. They are (2–8%) are phagocytic granulocytes and play important role in hypersensitivity. • Neutrophils (65%) are phagocytic granulocytes and increase during bacterial infection. • Basophils (2%) are non-phagocytic granulocytes and increase during chickenpox.

Blood Groups, Genotypes and their Transfusion Possibility Blood Group (phenotype)

Antigen in Red Blood Cells

Antibodies in Plasma

Can Give Blood to Groups

Can Receive Blood from Groups

Genotype

O

None

Anti-a, Anti-b

O, A, B and AB

O

IO IO

A

A

Anti-b

A and AB

O and A

IA IA or IA IO

B

B

Anti-a

B and AB

O and B

IB IB or IB IO

AB

A and B

None

AB

O, A, B and AB

IA IB

• Lymphocytes (26%) are agranulocytes producing antibodies and increase during the viral injection. • Monocytes (0.5%) are agranulocytes. They are called policeman of blood and increase during tuberculosis. • Platelets (thrombocytes) are non-nucleated. Platelets have a life span of about 8 to 10 days. • Rh factor discovered by Landsteiner and Veiner in Rhesus monkey which is responsible for erythroblastosis foetalis disease.

Heart • The human heart is myogenic i.e., contraction is initiated by a pulse produced by the Sino-atrial node (SA node) located in the right atrium. It is also called a pacemaker. The fi rst heart sound is lub and the second heart sound is dub. • The contraction of the heart is called systole while relaxation is called diastole (80 mm Hg). • The Blood Pressure is usually written as 120/80 where 120 is the systolic pressure and 80 is the distolic pressure.

Excretion • It is the process of elimination of harmful waste products from the animal body to regulate the composition of the body fl uids and tissues. • Human excretory system is composed of two kidneys. The Nephron is the structural and functional unit of kidneys. • The color of urine is pale yellow. It is due to pigment urochrome. • The Human urine contains about 95% water, 2% salts, 2.6% urea, and 0.3% uric acid.

Main Excretory Organs Excretory Organ

Animal

Contractile vacuole

Amoeba

Flame cells/solenocytes

Tapeworm

Renette cell

Ascaris

Nephridia

Earthworm

Malpighian tubules

Cockroach

Coxal glands

Scorpion

Green glands

Prawn

• The pH of urine is about 6.0 (mildly acidic). • The urine on standing gives a pungent smell. It is due to the conversion of urea into ammonia. • Specifi c gravity of urine is 1.015-1.025 • Volume of urine is 1 to 2 L per day.

Main Excretory Products Product

Animal

Ammonia

Most invertebrates, fi shes etc.

Urea

Ascaris, earthworm, cartilaginous fi shes, amphibian and mammals

Uric acid

Insects, land reptiles and birds

Central Nervous System The brain is the organizing and processing center of the body. It is the site of consciousness, sensation, memory and intelligence. The brain receives impulses from the spinal cord and from 12 pairs of cranial nerves coming from it and extending to the senses and to other organs. In addition, the brain initiates activities without environmental stimuli. Three major portions of the brain are recognised as the hindbrain, midbrain, and forebrain.

100

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Important Functions of Brain

Hypothalamus

Forebrain Olfactory region Cerebrum

Control

Smell Thinking, intelligence, memory, ability to learn from experience, willpower, skilled work, reasoning knowledge, consciousness and speech Laughing, weeping, micturition (passing of urine), defecation voluntary forced breathing and voluntary muscular coordination.

Diencephalon Heat, cold and pain control center of autonomic (sensation of) nervous system, control hunger, thirst, sweating, sleeping and sex.

Regulated body temperature so ‘thermostat’ of body. Appetite and safety control emotions like love, anger, pleasure and satisfaction. Control metabolism of carbohydrate, fat and water.

Midbrain and Refl ex center of visual and auditory sensation. Hindbrain Cerebellum

Involuntary muscular co-ordination, maintain posture, orientation and equilibrium of the body.

Medulla oblongata

Regulate heart rate, involuntary breathing, respiratory center, blood pressure, (vasoconstriction and vasodilation) gut peristalsis, food swallowing and vomiting gland secretion.

Some Human Diseases Caused by Viruses and Bacteria Disease Chickenpox (Varicella) Smallpox

Pathogen Herpes zoster virus

Incubation 12–20 days

Variola virus

12 days

Poliomyelitis

Poliovirus

7–14 days

Measles Rubella virus (Rubella disease) Mumps Mumps virus

12–26 days

Rabies (Hydrophobia) Tuberculosis

M. tuberculosis

10 days to 1–3 months 2–10 weeks

Diphtheria

C. diphtheria

2–6 days

Cholera

Vibrio cholera

Leprosy

Mycobacterium leprea

6 h to 2–3 days 2–5 years

Tetanus (Lockjaw) Typhoid

Clostridium tetani

3–21 days

Salmonella pestis

2-6 days

Neisseria gonorrhoea Streptococcus pneumonia Salmonella ententdrs HINI fl u virus (Orthomy)

2–10 days 1–3 days

Gonorrhoea Pneumonia Salmonellosis Swine Flu

Rabies virus

10 days

Symptoms Dark red colored rash or pox changing into vesicles, crusts and falling Appearance of rash changing into pustules, scaps and falling pockmarks are left Damages motor neurons causing stiffness of neck, convulsion, paralysis of limbs generally legs Rubella (skin eruptions), coughing, sneezing, etc Painful enlargement of parotid and salivary glands Spasm in throat and chest muscles, fears from water paralysis and death Coughing, chest pain and bloody sputum with tuberculin Infl ammation of mucosa of nasal chamber, throat, etc, respiratory tract blocked. Acute diarrhea and dehydration Skin hypopigmentation, nodulated skin, deformity of fi ngers and toes. Degeneration of motor neurons, rigid jaw muscles, spasm and paralysis Bubonic plague affects, lymph nodes, pneumonic plague affects lungs and septicemic plague causes anemia Infl ammation of urinogenital tract Decrease in respiratory effi ciency

Prevention Vaccine Varicella vaccine Smallpox vaccine Salk vaccine and Oral Polio Vaccine (OPV) Measles-mumps-rubella-Varicella Combo (MMRV vaccine) Mumps-vaccine, isolation Immunization of dogs BCG vaccine DTaP vaccine Sanitation, boiling of water and oral cholera vaccine BCG also offers variable amount of protection against leprosy. Lepromin skin tests ATS and DPT vaccines Killing of rats and rat fl eas, plague vaccine Avoid prostitution PCV 13

48 h 1–4 days

Diarrhoea Fever with or without chill, sore throat, dyspnea, myalgia, diarrhea, vomiting and dizziness

RASV vaccine Oseltamivir (Tamifl u), Zanamivir (Relenza) are antiviral drugs vaccines are available against this disease.

Ebola Virus Disease (EVD)

Ebola virus (Filovirus)

2–21 days

Haemorrhagic fever, muscle pain, headache, sore throat, diarrhea, kidney and liver dysfunction, internal and external bleeding.

No licensed vaccine available, immune therapies are done currently.

Dengue

RNA virus of genus Flavivirus

3–14 days

Muscle pain, swollen lymph nodes, fever, headache and rash

No specifi c antiviral drug is available, however symptoms based treatment is done.

Chikunguniya

RNA virus of genus Alphavirus

1–12 days

Headache, fatigue, digestive complaints and conjunctivitis

No specifi c treatment, however supportive case through drugs like naproxen, paracetamol is done.

101

GENERAL AWARENESS : GENERAL SCIENCE

Disease COVID-19

Pathogen Novel Corona Virus

Incubation 5–14 days

Symptoms Fever, dry cough, tiredness, aches, pains, nasal congestion, headache, conjunctivitis, sore throat, diarrhea, loss of taste or smell or a rash on skin or discoloration of fi ngers or toes.

Human Diseases Caused by Fungi Disease Aspergillosis Blastomycosis Candidiasis Chromomycosis Coccidiomycosis Cryptococcosis Geotrichosis Histoplasmosis Neuritis Onychomycosis

Fungus Aspergillus fl avus, A fumigatus and A. niger Blastomyces dermatitidis Candida albicans Cladosporium carrionii Coccidiodes immitis Lipomyces neoformans Geotrichum candidum Histoplasma capsulatum Mucor pusillus Trichophyton purpureum

Animal/Human Diseases Caused by Fungi Disease Athelete foot Ringworm Mucormycosis Penichosis

Fungus Trichophyton Trichophyton, Microsporum and Epidermophyton Mucor and Rhizopus Penicillium

Important Vaccines Discoverer Vaccine Smallpox Cholera Diphtheria and

Discovered By Edward Jenner (1786) Louis Pasteur (1880) Emil Adolf Von Behring and Shibasaburo

Tetanus Tuberculosis Polio Oral polio Measles Rabies

Kitasato Leon Calmette and Camille Guerin (1992) Jonas E Salk (1954) Albert Bruce Sabin (1995) John F Enders (1960) Charles Nicolle (1969)

Some Antibiotics Developed through Biotechnology Antibiotic Penicillin Bacitracin Cephatosporin Griseofulvin Streptomycin Tetracycline Erythromycin Chloramphenicol

Microbial Source Penicillium notatum and P. chrysogenum Bacillus subtilis Cephalosponum acremonium Penicillium griseofulvum Streptomyces griseus S. erythraeus S. aureofacines S. venezuelae

Ebola Virus • According to World Health Organisation WHO’s 19th August 2015 Situation Report, there were three confi rmed cases of Ebola reported in the week up to 16th August all of which were reported from Guinea. For the fi rst time since the beginning of the outbreak in Sierra Leone, a full epidemiological week has passed with no confi rmed cases reported. A total of 72 cases remain under monitoring in Sierra Leone. • On 29th June 2015, a confi rmed case of Ebola was reported in a 17-year old male who had died in Liberia. • Apart from Africa, the ebola virus has spread to USA, Spain, Mali, and to an extent in Italy and UK.

Prevention Vaccine Covishield, Covaxin and Sputnik V

Ecology

• Ecology deals with various principles which govern the relationship between organisms and their environment. • Pyramid of number is upright in grassland and bond ecosystem while inverted in tree ecosystem. • Pyramid of biomass is upright in grassland and forest ecosystem whereas, inverted in the pond ecosystem. • Pyramid of energy is always upright.

Pollution

• Motor vehicles contribute 60% of air pollution in major cities. Photochemical smog comprising of O3, H2O2, PAN, etc. • CO has 250 times more binding affi nity with hemoglobin as compared to O2. • Acid rain is composed of H2SO4 and HNO3. • Chlorofl uorocarbons released into stratosphere release free chlorine atom that causes depletion of ozone. • Sewage is the major source of water pollution. • Bioremediation is the process of using micro-organisms to remove environmental pollutants, e.g. using oil-zapper developed by TERI to prevent oil spills. • Biomagnifi cation is the process of increase in concentration of persistent chemicals in organisms in successive trophic levels. • Chernobyl disaster occurred in Ukraine (USSR) 26th April 1986 due to explosion of a nuclear power station. • Nitrate fertilizers cause blue baby syndrome or methemoglobinemia. • Noise pollution is measured in decibels (Generally sound beyond 80 dB is termed as noise).

Biotechnology

• Biotechnology: It is a fi eld of applied biology that involves the use of living things in engineering, technology, medicine and other useful applications. • Genetic Engineering: It is the insertion of a foreign gene fragment into another DNA molecule to produce DNA clones. • Gene Therapy: It is the insertion of genes into individual cell and tissue to treat diseases, especially hereditary diseases.

Test-Tube Baby

• Test-tube baby is a fusion of ovum and sperm outside body followed by implantation in the uterus at 32 celled stage and further normal development to birth. • The IVF (In Vitro Fertilisation) technology is a boon to childless couples. • First attempt to produce a test tube baby was made by an Italian scientist Dr. Petrucci in 1959. • But this human embryo survived for only 29 days. • The World’s fi rst test-tube baby (a baby girl) named Louise Joy Brown was born on 25th July 1978 in Great Britain. • India’s fi rst test-tube baby was born in Mumbai on 6th August 1986. Her name is Harsha.

Cloning

• Cloning in biology is the process of producing similar populations of genetically identical individuals that occurs in nature when organisms such as bacteria, insects, or plants reproduce asexually. • Dolly a sheep, the fi rst mammal clone was developed by Dr Ian Wilmut, UK.

Bt Crops

• Crop plants that contain genes for Bt toxins. Bt toxin gene has been cloned from the bacteria (Bacillus thuringiensis) and been expressed in plants to provide resistance from insects without the need of insecticides e.g., Bt-cotton (fi rst GM crop), Bt-corn, golden rice, etc.

102

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• It is the starting point of agriculture and dictates the ultimate productivity of other inputs. It was organized by Dr. Swaminathan in the Jounti village of Delhi state in 1965, which was designed to convert the entire village into a highquality seed-producing center. Bacteria

Dialister Pneumosintes

• Over the years, this concept has grown and been refi ned which aims to import technocracy (technical literacy or imparting the latest skills to farmers solely) for quality seed production and thereby to make available quality seed to others at the appropriate time and affordable cost. Flower

Wolffia microscopica (Angiosperm)

Bird

Humming bird (Cuba)

Mammal

Shrew (Suncus etruscus)

Bone

Stapes

Muscles

Stapedius or arrector pill

Endocrine gland

Pituitary

Virus

Foot and mouth disease virus

Mammal (on land)

African elephant (Loxodonta Africana)

Mammal (in the biosphere)

Blue whale

Flower

Raffl esia

Flower in India

Sapria

Vertebral

Lumbar vertebrae

Bone

Femur

Bone (in frog)

Tibia-fi bula

Muscles

Gluteus maximus (buttock muscle of hip)

Tooth

Tusk of elephant (upper incisor modifi cation)

Tallest angiosperm

Eucalyptus

Tallest gymnosperm

Sequoia sempervirens (Sequoia gigantean)

Branch

Concerned Field

Cryogenics

Study concerning with the application and uses of very low temperature

Cytology

Study of cells

Dermatology

Study of skin

Floriculture

Study of fl ower yielding plants

Genetics

Study of heredity and variations

Gerontology

Study of growing old

Herpetology

Study of garden cultivation

Myology

Study of muscles

Nephrology

Study of kidneys

Obstetrics

Branch of medicine dealing with pregnancy

Ornithology

Study of birds

Phycology

Study of algae Silk industry (culture of moth and pupa)

Coral reef

In Australia, great barrier reef

Sericulture

Egg or cell

Ostrich

Serpentology

Study of snakes

Vein

Inferior vena cava

Taxonomy

Study of classifi cation of organisms

Artery

Abdominal aorta

Virology

Study of virus

Cell of the body

Neuron or nerve cell

Virus

Parrot fever virus

Pedology

Study of soils

Pathology

Study of disease causing organisms.

Physiology

Science dealing with the study of functions of various parts of organisms.

Pisciculture

Study of fi sh.

Some Important Branches of Biology Branch

Concerned Field

Agriculture

Study of producing crops from the land

Anatomy

Study of the animal forms with an emphasis on human bodies

Anthology

Study of fl owers

Anthropology

Study of apes and man

Apiculture

Honey industry (Bee keeping)

Biochemistry

Deals with the study of chemical reactions in relation to life activities

Cardiology

Study of heart

Some Important Discoveries Discovery

Made by

Country

Antibiotics

Alexender Flemming (1928)

Scotland

Antiseptics

Joseph Lister (1869)

Scotland

Blood circulation

William Harvey (1628)

Britain

Blood transfusion

Jean-Baptiste Denys (1625) France

Cholera and TB germs

Robert Kock (1883)

Germany

Electrocardiogram (ECG)

William Einthoven (1903)

Dutch

CT Scan

Godfrey Hounsfi eld (1973)

England

Sphygmomanometer

Scipione Riva-Rocci (1898)

Italy

Stethoscope

Rene Laennee (1819)

France

Thermometer

Sir Thomas Aelburt (1867)

England

Ultrasound

Ian Donald (1950)

Ireland

X-ray

WC Roentgen (1895)

Germany

Electroencephalogram (EEG)

Hans Berger (1929)

Germany

Antibiotics

Source

Action

Penicillin

Penicillium, chrysogenum, P. notatum + Phenyl Acetic Acid

Tonsilitis, Sore Throat, Gonorrhea, Rheumatic Fever, some Pneumonia types

Griseofulvin

Penicillium griseofulvum

Antifungal, especially for Ringworm

Nystatin

Streptomyces noursei

Antifungal for Candidiasis and overgrowth of Intestinal Fungi during excessive antibiotic treatment.

103

GENERAL AWARENESS : GENERAL SCIENCE

Hamycin

Streptomyces pimprei

Antifungal for Thrush

Fumagillin

Aspergillus fumigatus

Broad spectrum antibacterial especially against Salmonella and Shigella.

Bacitracin

Bacillus licheniformis

Syphilis, Lymphonema or Reticulosis

Streptomycin

Streptomyces griseus

Meningitis, Pneumonia, Tuberculosis and Local Infection. Toxic in some through eighth cranial nerve.

Chloramphenicol Chloromycetin

Streptomyces venezuelae, S. lavendulae and Now synthetic

Typhoid, Typhus, Whooping cough, Atypical Pneumonia, Bacterial Urinary Infections.

Tetracyclines/ Aureomycin

Streptomyces aureofaciens

Viral pneumonia, Osteomyelitis, Whooping Cough and Eye infections.

Oxytetracycline/ Terramycin

Chlorotetracycline → Hydrogenation Streptomyces rimosus

Intestinal and Urinary Infections (Spirochaetes, Rickettsia and Viruses)

Erythromycin

Streptomyces erythreus (= S. erythraeus)

Typhoid, Common Pneumonia and Diphtheria, Whooping Cough, etc.

Gentamycin

Micromonospora purpurea

Effective against Gram (+) bacteria

Polymixin

Bacillus polymyxa

Antifungal

Environment and Ecology

• Environment: All external conditions, factors, matter, and energy living and non-living that affect any living organism or other specifi ed systems. • Ecology: The Biological science that studies the relationships between living organisms and their environment; study of the structure and functions of nature. • Ecosystem: Its is defi ned as a unit that include all the organisms (biological components) in a given area interacting with the environment (physical component), so that the fl ow energy leads to a clear, defi ned trophic structure, biotic diversity, and material cycles. • Biome: The terrestrial regions characterized by certain types of vegetation and other forms of life. Examples include various types of deserts, grasslands, and forests. • Wetland: The land that is covered all part of the time with saltwater or freshwater, excluding streams, lakes and the open ocean. • Biodiversity: The variety of different species (species diversity), genetic variability among individuals within each species (genetic diversity), variety of ecosystems (ecological diversity) and functions such as energy fl ow and matter cycling needed for the survival of species and biological communities (functional diversity). • Biosphere: The zone of the Earth where life is found. It consists of parts of the atmosphere (the troposphere), hydrosphere (mostly surface water and groundwater), and lithosphere (mostly soil and surface rocks and sediments on the bottoms of oceans and other bodies of water) where life is found. • Wildlife: All free, undomesticated species. Sometimes the term is used to describe animals only.

• Threatened species: The wild species that is still abundant in its natural range but is likely to become endangered because of a decline in numbers. • Ozone: The colourless and highly reactive gas and a major component of photochemical smog. Also found in the ozone layer in the stratosphere and protect us from ultraviolet rays. • Smog: It is originally, a combination of smoke and fog but now used to describe other mixtures of pollutants in the atmosphere. • Acid Rain: When fossil fuel is burnt, oxides are formed in the atmosphere. The oxides formed of sulphur and nitrogen get dissolved in water and causes acid rain. • Global warming: The warming of the Earth’s lower atmosphere (troposphere) because of an increase in the concentrations of one or more greenhouse gases. It can result in irreversible climate change that can last for decades to thousands of years. • Ecomarks: The Ministry of Environment Forest and Climate change, Government of India instituted a scheme, that is operating on a national basis and provides accreditation and labeling for household and other consumer products which meet certain environmental criteria. • Coral Bleaching: It occurs when the relations between the coral host and zooxanthellae, which give coral much of their color, break down. Without the zooxanthellae, the tissue of the coral animal appears transparent and the coral’s bright white skeleton is revealed. • Sustainability: The ability of Earth’s various systems, including human cultural systems and economies, to survive and adapt to changing environmental conditions indefi nitely.

Objective Type Questions 1. 1 kilo Watt-hour (kWh) is equals to (1) 3.6 × 104 joule (2) 3.6 × 107 joule (3) 3.6 × 105 joule (4) 3.6 × 106 joule 2. Which of the following represents the Boyle’s Law? (1) V ∝ T (at constant P)

(2023) (2023)

1 (2) V ∝ (at constant T) P

(3) P ∝ T (at constant V) (4) V ∝ n (at constant T & P) 3. Beta rays emitted by a radioactive material are (2023) (1) The electrons orbiting around the nucleus (2) Charged particles emitted by nucleus (3) Neutral particles (4) Electromagnetic radiations 4. Which gland in human body is called master gland? (2023) (1) Thyroid gland (2) Mucous gland (3) Pituitary gland (4) Digestive gland

5. Which among the following enzymes is required to digest protein? (2023) (1) Disaccharides (2) Trypsin (4) Lipase (3) Amylase 6. Find the resistance (in mega Ω) of a wire of length 8 m, cross sectional area 2 cm2 and made of a material of resistivity 120 Ω m. (2023) (d) 960 (a) 1920 (b) 4.8 (c) 2.4 Choose the correct answer from the options given below: (1) (a) (2) (b) (3) (c) (4) (d) 7. In a reaction of the type A+B Þ C+D one could ensure it to be a fi rst order reaction by (2022) (1) Increasing the temperature (2) Increasing the concentration of a product (3) Increasing the concentration of a reactant (4) Adding a catalyst

104

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

8. ______ is the structural and functional unit of kidneys. (1) Nucleon (2) Ribosome (3) Nephron (4) Urochrome 9. Most abundant blood cells in the human body are: (2022) (1) WBCs (2) RBCs (3) platelets (4) plasma cells 10. In a _____ wave particles of the medium vibrate in a direction parallel to the direction of wave propagation. (2022) (1) transverse (2) longitudinal (3) surface (4) fi eld 11. Light rays enter the eye through the outer, transparent structure at the front of the eye called the _____. (2022) (1) Lens (2) Iris (3) Cornea (4) Optic nerve 12. Who discovered the rubber making process through the vulcanization method? (2022) (1) John Dunlop (2) Charles Goodyear (3) Macmillan (4) Newton 13. _____ is the primary acid present in ripe bananas. (1) Formic Acid (2) Sulphuric Acid (3) Malic Acid (4) Hydrochloric Acid 14. What does a catalyst generally do in a reaction? (1) A catalyst is nothing but the fi nal product of a reaction (2) Does not alter the rate of reaction (3) Speeds up chemical reaction (4) Slows down chemical reaction 15. Amino acids are required for the synthesis of: (1) Alkaloids (2) Lipids (3) Proteins (4) Carbohydrates 1. (4) 9. (2)

Answer Key 2. (2) 3. (2) 4. (3) 2. (2) 6. (2) 7. (3) 8. (3) 10. (2) 11. (3) 12. (2) 13. (3) 14. (3) 15. (3)

Answers with Explanations 1. Option (4) is correct. 1 Kilo Watt-Hour (kWh) is equal to 3.6 × 106 J. kWh is a unit of energy and it is a non SI unit. It is normally used to calculate the electricity bill. 2. Option (2) is correct. Boyle’s Law defi nes the relationship between pressure and volume in a closed container for an ideal gas at constant temperature. It is also referred to as Boyle–Mariotte law, or Mariotte’s law. Also, Boyle’s law is defi ned as at constant Temperature (T), where P is the Pressure exerted by the gas and V is the volume occupied by the gas. At a constant temperature, if we increase the Pressure, the volume occupied by the gas decreases and vice versa. Mathematically, Boyle’s law can be represented as P1V1 = P2V2 P1 and P2 are the initial and fi nal pressure of gas where V1 and V2 are the initial and fi nal volume of the gas. 3. Option (2) is correct. An unstable atom can attain stability by radioactive decay (alpha, beta, gamma decay). Whereas beta particle is a high energy, negative charge particle which has charge and mass equivalent to an electron/positron. Hence, if beta rays are emitted by a radioactive material then charged particles will be emitted by the nucleus. 4. Option (3) is correct. Pituitary gland is referred as the ‘master’ gland of the body as it controls the functions of many of the other endocrine glands in the body. It is a pea sized gland attached to the hypothalamus of the human brain. It is also known as Hpophysis. Some of the hormones secreted by the gland are I. Human Growth Hormone (HGH): Growth and repair of all cells II. AdrenoCorticoTropic Hormone (ACTH): Infl uences the adrenal gland to release of Cortisol or the ‘stress hormone’. III. Luteinising Hormone (LH) and Follicle Stimulating Hormone (FSH)-Control the sexual and reproductive characteristics in males and females.

IV. Prolactin (PRL): Produce breast milk V. Melanocyte-Stimulating Hormone (MSH): Stimulation of the production of melanin by skin and hair. VI. Antidiuretic Hormone (ADH): Controls the water balance of the body by affecting reabsorption of water by the kidneys VI. Oxytocin: Uterine contraction and production of milk. 5. Option (2) is correct. Protein digestive enzymes are pepsin and trypsin. They break down proteins into smaller peptides and amino acids. Pepsin is secreted by the gastric gland in the stomach where as trypsin is secreted by Pancreas. 6. Option (2) is correct. Given that: Length (L) = 8m The cross – sectional area (A) = 2 cm2 = 2 × 10–4 m2 The resistivity (P) = 120 Ω m We know that, Resistance (R) = Resistivity (P) × (Length (L)/Cross sectional area (A)). 8   120 × 0.0002    Now, 8  6 R = 120 ×   = 4.8 × 10 Ω 0 . 0002   R = 4.8 mega Ω. 7. Option (3) is correct. In a reaction A+B Þ C+D, one could guarantee it is a firstorder response by advancing the attention of a reactant. 8. Option (3) is correct. Nephron is the structural and functional unit of kidneys. 9. Option (2) is correct. • The most abundant blood cells in the human body are RBCs (Red Blood Corpuscles). • Red blood cells (erythrocytes) are an important element of blood. • Their job is to transport oxygen to the body's tissues in exchange for carbon dioxide, which they carry to the lungs to be expelled. • White blood cells help the body fi ght infection and other diseases. • Platelets play a major role in blood clotting. White Blood Cell (WBC)

5,000 – 10,000/mL

Red Blood Cell (RBC)

4.5 – 5.5×106/mL

Platelet

1.4 – 4.0×105/mL

10. Option (2) is correct. Particles of the medium shake in a path parallel to the direction of wave propagation, in a longitudinal wave. 11. Option (3) is correct. The cornea is the translucent foremost of the eye surrounding the iris, pupil, and anterior chamber. The cornea, with the anterior chamber and lens, refracts light. 12. Option (2) is correct. Charles Goodyear discovered the vulcanization of rubber. It is a process that allows rubber to withstand heat and cold. 13. Option (3) is correct. Malic acid is present in ripe bananas. It is the main principal acid present in the ripe banana. 14. Option (3) is correct. A ‘catalyst’ runs up a chemical reaction, which lowers the activation power to a chemical reaction to initiate and occur fast. It accelerates the velocity of the reaction. 15. Option (3) is correct. Amino acids are required for the synthesis of body protein and other important nitrogen-containing compounds. Amino acids are constituents of protein and act as precursors for many coenzymes, hormones, nucleic acid, etc. Adult humans are unable to synthesize all 20 amino acids needed for protein synthesis; those which can not be synthesized are referred to as essential.

Internet

Security Threats

Some Commonly used Terms

Super Computer

Networking

Computer

Computer

Software

System Software

Second Level

Trace the Mind Map

Secondary Memory

Primary Memory

First Level

Application Software

Hardware

Memory

Components of Computer

Third Level



GENERAL AWARENESS : COMPUTER SCIENCE & TECHNOLOGY

105

Chapter

6

Computer Science & Technology

Chapter Analysis Concept Name

2021

Computer Science & Technology

 Revision Notes A computer is an electronic device or machine which stores, reads and processes data to produce meaningful information as output.

Components of Computer • Input Unit is defined as an input device used to give instructions, e.g. Keyboard, Mouse, Joystick, Optical character reader, CDs Bar code reader, Touch screen, Light pen, Scanner, Magnetic Ink Character Recognition (MICR), etc. • Central Processing Unit (CPU) is the device for the manipulation of information inside the computer CPU is known as the brain of the computer, but commonly called a processor and has the following components. •  Arithmetic Logic Unit (ALU) performs all logical and arithmatical operations. • Control Unit (CU) instructs, maintains and controls the flow of information. • Output Unit is the device to display the result of processing, e.g. Visual Display Unit, Printer, Monitor, Speaker, Pen Drive, etc.

Memory  emory holds all the raw and processed data, set of instructions M and information inside the CPU.

Primary Memory

Primary Memory stores the data which is currently in use by the computer. • RAM (Random Access Memory): It is a volatile memory. It is a temporary storage. • DRAM: Dynamic Random Access Memory • SRAM: Static Random Access Memory • ROM (Read Only Memory): It is a non-volatile memory where all logical data is stored that cannot be changed. • PROM: Programmable Read Only Memory. • EPROM: Erasable Programmable Read Only Memory. • EEPROM: Electrically Erasable Progammable Read Only Memory.

Secondary Memory

It stores data, programs, instructions and information permanently.

Hardware Any peripheral device which can be seen and touched is hardware. Computer hardware includes input devices, output devices, storage devices and processing devices.

Software It is a set of instructions that directs the computer to process information. It can be classified as System Software and Application Software.

System Software

• A set of one or more programs which are designed to control the operation of a computer system including hardware component and implementations of application software, is a system software. • For example: Operating system, Device driver, Language Translator etc.

2022

2023

Additional Questions

3

8

Application software

•  It is a set of one or more programs designed to carry out operations for a specific application. • It cannot run on itself, but it is dependent on system software to get executed. •  For example: Word processors, spreadsheets, accounting programs etc.

Networking

Computer networking relates to the communication between a group of two or more computers linked together. Most common example of networking is Internet, connecting millions of people all over the world together. According to scale or size, computer network can be categorised in three ways. • Local Area Network (LAN): The geographical area spread over 1 km to 10 km or within a same building. • Metropolitan Area Network (MAN): The geographical area spread over a city or town. • Wide Area Network (WAN): The geographical area spread over countries.

Security Threats

• Worm: It is a self contained program and does not need to be a part of another program to propagate itself. • Spam: Spam is an unsolicited message sent over the internet in the form of e-mails, to a large number of users for the purpose of spreading malware, advertising phishing, etc. • Spyware: It is a type of malicious software installed on computers and collects information about users without their knowledge and may send such information to another entity. • Malware: A software which is specifically designed to disrupt or damage a computer system. It is a superset of computer viruses, worms, spyware, trojan horses and other malicious or unwanted software. • Virus: A virus is defined as a program or a piece of code that gets loaded onto the computer without users knowledge and replicates itself, e.g. Creeper, Stuxnet, Melissa, Conficker, Code red, SQL Slammer, Nimda (derived from the word ‘Admin’), etc. Antivirus Antivirus is a software consisting of computer programs that attempt to identify, detect and prevent the malware from infecting your computer.

Internet

Internet is a global computer network providing a variety of information and communication facilities, consisting of interconnected networks using standardized communication protocols.

Hyper Text Transfer Protocol (HTTP)

It sends an HTTP command to the web server directing it to fetch and transmit the requested web page. It is used by the World Wide Web (www).

Uniform Resource Locator (URL)

It specifies the address of a file and every file on the Internet has a unique address.

107

GENERAL AWARENESS : COMPUTER SCIENCE & TECHNOLOGY

Hyper Text Markup Language (HTML)

It is a computer’s language used to create hypertext documents for the World Wide Web. Web pages are created using HTML. Domain Domain specifi c names used in URL’s to identify particular web page. IP Address IP Address is a 32-bit numeric address separated by four number. Each number can be 0 to 255.

Some Commonly Used Terms

• Cache memory: It is a temporary storage, where frequently accessed data can be stored for rapid access. • Registers: These are defi ned as the special memory units used by the CPU to speed up the rate of accessing information. • Operating System: It is a system software, consisting of an integrated set of programs that control computer resources and provides common services for effi cient execution of various application software. • Compiler: It is a computer program that transforms human readable source code into the machine readable code at one go. • Interpreter: It transforms source code into the machine readable code by converting it line by line. • Assembler: It converts assembly language program into machine language program. • Modem (Modulator-Demodulator): An electronic device used to convert computer (digital) electronic signals to communication channel (analog) electronic signals and vice-versa. • Cloud Computing: It is the delivery of on-demand computing resources, everything from applications to data centres, over the internet, e.g. Google. • Dual Core Processor: It is the processing technology in which two processors are scheduled together and when one is busy the other takes over. • Internet: It is the worldwide publically access a system of interconected computer networks that transmit data by using the internet protocol. • Cryptography: It is a method of storing and transmitting data in a particular coded form so that only those can read and process it, for whom it is intended. It includes encoding and decoding of data.

Super Computers

A super computer can be defi ned as the most powerful computer in terms of performance and storage capacity. They are highly expensive and are employed for specialised applications such as for weather forecasting, several scientifi c researchers, etc.

Super Computers Developed in India Name

Year

Mft Company

Param Siddhi

2020

C-DAC

Param Shivay

2019

IIT-BHU

Pratyush

2017

IITM (Pune)

Param Kanchejunga

2016

CDAC & NIT Sikkim

Param Ishan

2016

CDAC & IIT Guwahati

Aaditya

2013

Indian Institute of Tropical Meteorology

PARAM YUVA II

2013

C-DAC, PUNE

2011

ISRO

2010–11

BARC

SAGA-220 ANUPAM-Adhya

Super Computers of the World Name

Year

Country

Operating System

Fugaku

2021

Japan

Custom-Linux

Frontera

2019

America

Linux (Cent OS)

IBM Summit

2018

America

IBM

Sunway Taihulight

2016

China

Linux

Tianhe-2

2013

China

Kylin Linux

Titan

2012

America

Linux

Sequoia

2011

America

Linux

K-Computer

2011

Japan

Linux

Mira

2010

America

Linux

Sophia In October 2017, Saudi Arabia has provided citizenship to a robot Sophia. This robot can change the facial expressions and can chat with people.

Objective Type Questions 1. PROM stands for (2022) (1) Programmable Read Only Memory (2) Programmed Random Only Memory (3) Programmable Random Objective Memory (4) Programmed Random Objective Memory 2. Which component is mainly responsible for doing calculation in computer? (2022) (1) Random access memory (2) Control unit (3) Arithmetic logic unit (4) Hard disk 3. The process of arranging data in logical sequence is called (2022) (1) Classifying (2) Reproducing (3) Summarizing (4) Sorting 4. As per data or space sizes, which one of the following series is arranged from smallest to largest? (1) KB, MB, TB, GB (2) MB, TB, GB, KB (3) GB, KB, MB, TB (4) KB, MB, GB, TB 5. If a computer has more than one processor then it is known as _________ (1) Uniprocessor (2) Multiprocessor (3) Dual processor (4) Quad Core processors

6. Who discovered North Pole? (1) Amundson (2) Robert Peary (3) John Cobot (4) Captain Cook 7. Who is the inventor of the Web? (1) Mike Sendall (2) Tim Berners-Lee (3) Bill Gates (4) Ted Nelson 8. Who invented airplane? (1) Orville Wright and Wilbur Wright (2) Sir Frank Whittle (3) Michael Faraday (4) Christian Huygens 9. Which one of the following is considered as nature’s radar? (1) Hippopotamus (2) Pigeon (3) Vulture (4) Owl 10. The Indian National Grid Computing Initiative for Scientifi c Engineering and Academic Community is named (1) Ganga (2) SAGA (3) Garuda (4) PARAM 11. What is the name of the new upgraded version of Indigenous transport aircraft SARAS? (1) SARAS PT1N (2) SARAS ML-3S (3) SARAS AL-5C (4) SARAS FF-1F

108 1. (1) 9. (2)

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

2. (3) 3. (4) 10. (3) 11. (1)

Answer Key 4. (4) 5. (2) 6. (2)

7. (2)

8. (1)

Answers with Explanations 1. Option (1) is correct. PROM stands for Programmable Read Only Memory. It is a form of digital memory where the contents can be changed once after manufacture of the device. It can be modifi ed once by a user. PROM is a way of allowing a user to tailor a microcode program using a special machine called a PROM programmer. 2. Option (3) is correct. An arithmetic-logic unit (ALU) is the part of a computer processor unit (CPU) that carries out arithmetic and logic operations on the operands in computer instruction words. It performs all arithmetic computations for example multiplication and addition and all comparison operations. 3. Option (4) is correct. In the fi eld of computing, the process of arranging data in a logical sequence is called sorting. It is a process that accepts a random sequence of numbers or any other data which can be arranged in a defi nite logical sequence as input. It arranges data in a logical sequence numerical, alphabetical, etc. by record key. 4. Option (4) is correct. A bit is the smallest unit of data in a computer. A bit is a short form of binary digit, 8 bits make 1 byte. A kilobyte (KB) is 1024 bytes, a megabyte (MB) is 1024 kilobytes, a gigabyte is 1024 MB, a terabyte is 1024 GB. The series would be KBA>B

A is greater than B but not as much as C

> Means Greater < Means Smaller = Equal

Based on Comparison

POSITIONS

S

N

S-E

E

N-E

Second Level

Third Level



Left Right Top Bottom

COUNTING FROM

Based on Position (Top/Left or Right/Bottom)

Try to find position of each element from only left end so that you may not get confused to solve quetsions based on ranking.

Approach

S-W

W

N-W

Placement of Objects/Persons

North South East West

FACING

Trace the Mind Map First Level

Ahead Before Right side Left side Adjacent right Adjacent left Above Below

Ranking

Immediate right Immediate left Top Bottom Immediate above Immediate below Adjacent means both left and right of a given element Neighbours means both left and right of element

LOGICAL REASONING: RANKING

155

Chapter

Ranking

6

Chapter Analysis Concept Name Ranking

 Revision Notes

2021

2022

1

2023

Additional Questions

2

7

Alternate method

Number Series The concept of ranking involves organising a set of elements or objects according to a specific criterion or set of criteria. The goal is to establish an order or sequence based on certain conditions or rules.

Steps to Solve Question (1) Approach the banking questions systematically, reading them sentence by sentence to construct logical steps. (2)  Determine whether the order or position will be either horizontal or vertical before starting the arrangement. (3) Consider all potential cases initially but eliminate all incorrect possibilities systematically to arrive at the correct order or ranking. (4) Failure to consider all possible cases may result in an incorrect order or ranking. (5) Utilise the ranking information to deduce the total number of objects/people, or vice versa, use the total number to evaluate the ranking. (6)  Follow the prescribed process or arrangement method accurately; relying solely on reverse engineering may not consistently yield the correct answer.

Types of Question 1. Based on position (Top/Left or Right/Bottom) In these types of questions, you are provided with the position of person(s) from either end of a row. Your task is to determine the total count of individuals in the row or calculate the number of persons situated above/below or to the left/right of a specific person. When solving these type of questions, keep the following points in mind. (i) Total number of persons in a row = (Position of a person from the left end/top + Position of that same person from right end/ bottom) – 1 (ii) Position of a person from right end/bottom = (Total number of persons in row) – (Rank of that person from the left end/top) + 1 (iii) Position of a person from top/left end = (Total number of persons in a row) – (Position of that person from right end/ bottom) + 1 Example 1: In a class of 48 students rank of Priyansh from the top is 15, then the rank of Priyansh from bottom is (1) 33 (2) 32 (3) 34 (4) 35 Solution: (3) Here, total number of students = 48 Rank from top = 15 \ Rank of Priyansh from bottom = (Total number of person in the class) – (Rank of Priyansh from top) + 1 = 48 – 15 + 1 = 34

15

th

48 Students

Total number of students = 48 Given, the rank of Priyansh from top is 15, it means there are 48 – 15 = 33 students below Priyansh in the class. \ The rank of Priyansh from bottom = 33 + 1 = 34th Type 1 (a): If the total count of elements stands in a line is provided, along with the position of an element from one end, what’s the method to determine the position of that same element from the other end of the line? Example 1: In a row of 16 students, if Mira’s position is 7th from the left end then what is her position from the right end? Solution: There are 16 students in a row i.e., Total = 16 Mira position from left end = 7 Mira position from right end = (Total + 1) – position from left end = (16 + 1) – 7 = 17 – 7 = 10 i.e., 10th from right end. Trick: Treat top as left and bottom as right for doing questions of Vertical line arrangement/ranking. Type 1 (b): How would you determine the total number of elements in a queue when provided with both the left and right end positions? Example 1: Akash ranks seventh from the top and twenty-sixth from the bottom in a class. How many students are there in the class? Solution: Here Top/Left end position = 7 Bottom/ Right end position = 26 Total students = 7 + 26 – 1 = 32 Second method By drawing figure, Clearly, the whole class consists of Akash 6 students 25 students So, total students = (6 + 1 + 25) = 32 students Trick: Position from the left end/position from top of the element + Position from the right end/position from the bottom of the same element – 1 = Total no. of elements Type 1 (c): Determine the number of elements located between two specified positions in a queue, considering their positions from both the left end/top and the right end/bottom. Example 1: If the position of Suresh from the left end is l5 in a classroom consisting 26 students while Mohan position from the right end is 6. Find how many students are present in between both of them? Solution: Convert the right end position into the left end position to make the question easy. Left end position of Mohan = (Total + 1) – right end position = 26 + 1 – 6 = 27 – 6 = 21

157

LOGICAL REASONING: RANKING

Now we got to know the position of both the persons from left end. So, the number of students in between both of them = (21 – l5) – 1 =6–l=5 Trick: Number of elements in between = (N1 – N2) – 1 where N1 = Position of one element from the left end/top, N2 = Position of another element from the left end/top and N1 > N2. OR Number of elements present in between two elements = (N2 − N1) – 1 where N1 = Position of one element from the right end/bottom. N2 = Position of another element from the right end/bottom and N 2 > N1

2. Based on Comparison

In this type of question, a comparison of different objects based on some factors like ages, marks, size, height etc. is given. The candidate is required to arrange the data in ascending or descending order and then answer the related question. Example: Rahul is taller than Sonam and Geet, while Geet is taller than Sonu. Also, Sonam is taller than Geet. Who amongst the following is the shortest? (1) Sonu (2) Sonam (3) Rahul (4) Geet Solution: (1) According to the question, Rahul > Sonam and Rahul > Geet Geet > Sonu Sonam > Geet

Rahul

Sonam

Geet

Sonu

Arranging the above data, we get Rahul > Sonam > Geet > Sonu Hence, Sonu is the shortest. Type 2 (a): Age-Based Ranking/Age-Based Comparison These kinds of questions involve comparing different individuals based on statements presented in a passage-like format. An example can help to illustrate this more effectively. Example 1: B is older than C but not as old as D. E is not as old as B. Who is the oldest of all? Solution: Step 1 : Draw a vertical line and count the total number of elements i.e. 4 Step 2: Place/arrange those elements initially having fixed confirm/ definite information. Step 3: Connect the remaining elements with placed elements on the line. 1

1

D

1

D

2

2

B

2

B

3

3

C

3

C

4

4

4

E

Step 1

Step 2

Step 3

Hence, from step 3 figure, we can say D is the oldest. So, by drawing a line and placing on it is a good approach to solve these type of questions. Alternative Method: From the first statement we get D>B>C ........(1) From the second statement we get B>E ........(2) By merging both statements (1) and (2), we get D>B>C>E Hence, D is the oldest among four given elements.

Type 2 (b): Height Based Ranking/Compare This is a method of arranging or ordering items, individuals, or entities based on their relative heights or positions. Example 1: A is taller than B. C is taller than A, D is taller than E but shorter than B. Who is the tallest? Solution: A > B ...(1) C>A ...(2) B>D>E ...(3) Merging all the statements together We get, C>A>B>D>E Hence, C is the tallest Type 3: Interchange the position in a queue. This pattern involves altering the sequence or order of individuals or items in a line or queue based on certain conditions or rules. Example 1: In a row of girls, Kamya is fifth from the left and Preeti is sixth from the right. When they exchange their positions, then Kamya becomes thirteenth from the left. What will be total number of girls in a row? Solution: By drawing an exact positional figure of both girls i.e., K P 5 4 K 12 Here, Kamya is at the 5th position means there are 4 girls before Kamya in a row in the similar way Preeti is 6th from the right means there are 5 girls before Preeti. After interchange Kamya came at the place of Preeti, it is given Kamya is at 13th place from left end means there are 12 girls before Kamya and after Kamya there are 5 girls also so, the total will be sum of 12,5 and herself Kamya i.e., 1. Thus, the row consists total of (12 + 1 + 5) = 18 girls. Tip: Focus on position value of given element after interchange i.e., Here it is Kamya whose position is given after interchange so pick only new and old position value of given element in the question.

Type 4: Ahead of / Below of Ranking. In this type of question you will be getting frequently used word which is ahead of and below of. Ahead of means = “Before someone/something”. Below of means = “Lower than”. Example 1: Ravi is 7 ranks ahead of Sumit in a class of 39. If Sumit’s rank is 17th from the last, what is Ravi’s rank from the start? Solution: To deal with such type of question, draw a horizontal line and place each element position. Ravi Sumit 6 16 ? So, Ravi is 24th (16 + 1 + 6 + 1) from the last. Number of students ahead of Ravi in rank = (39 – 24) = 15 So, Ravi is 16th from the start. Type 5: If the shifting of places of elements occurred in a row/queue then how to find the total number of elements in a queue and many more. Example 1: Hemant is 5th from the left and Peter is 12th from the right end in a row of children. If Peter shifts by three places towards Hemant, he becomes tenth from the left end. How many children are there in the row? Solution: Just read the questions carefully clearly, Hemant lies towards the left end while Peter lies towards the right end of the row. So, when Peter shifts towards Hemant, he shifts 3 places to the left. Thus, Peter is now 15th from the right end. But, Peter is 10th from the left end. Hence, the total number of children in the row is = (14 + 1 + 9) = 24.

158

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Type 6: Puzzled based linear seating arrangement/ranking It is a situation based questions in which plenty of information have been given in a mix way so, we need to arrange all the elements in a queue/row by using the given statements. Example 1: Vishu, Pooja, Vishakha, Rani and Ram are sitting in a line. Pooja is third to the extreme right end. Vishu is second to the left of Pooja. Vishakha is to the right of Pooja. Rani is third to the right of Ram, who is the immediate neighbour of Vishu. Who is sitting in the middle? Solution: Step 1: Draw a horizontal line. Mark the positions of each elements i.e., Total number of elements. Here there are five persons. Note: If the direction of all elements are facing is not given in the question then just take North direction for solving question. Extreme left

Extreme right

Step 2: Try to place those persons who have fixed positions. Here Pooja has the fixed position by this statement “Pooja is third to the extreme right end”. So place Pooja at first then place other persons also who are directly related to Pooja.

Pooja

Vishu

Vishaka

Step 3: Now you have placed three person already at their places so now place the remaining with the help of other already placed. Hence,

Ram

Vishu

Pooja

Vishaka

Rani

At middle

So, from the last figure Pooja is at the middle.

Objective Type Questions 1. Ram ranks 7th from the top and 26th from bottom in a class. How many students are there in the class? (2023) (1) 31 (2) 32 (3) 33 (4) 34 2. Among A, B, C, D and E having a different amount of money, C has more money than only E, B and A. Who among them has the highest amount of money? (2023) (1) C (2) D (3) E (4) A 3. Dinesh is taller than Chinku and Elina. Akash is not as tall as Elina. Chinku is taller than Akash. Dinesh is not as tall as Bikash. Who among them is next to the tallest one? (2021) (4) Dinesh (1) Bikash (2) Chinku (3) Akash 4. A is taller than B but smaller than E. C is the smallest of all. D is taller than A but smaller than E. Who is the tallest of the five? (1) B (2) C (3) D (4) E 5. Farhan scored more runs than Raju. Simran scored more runs than Nitu but lesser runs than Raju. Who scored the highest runs? (1) Nitu (2) Farhan (3) Simran (4) Raju 6. In a row of students, Ankit is 26th from the left end. Poonam is 18 ranks to the right of Ankit. If Poonam is 32nd from the right end, then what will be the rank of Ankit from the right end? (1) 52 (2) 51 (3) 76 (4) 50 7. In a row of trees, a tree is 7th from left end and 14th from the right end. How many trees are there in the row? (1) 18 (2) 19 (3) 20 (4) 21 8. Four children, Akram, Bopsi, Priya and Tulsi are on a ladder. Akram is further up the ladder than Bopsi. Bopsi is in between Akram and Priya. If Tulsi is still further than Akram, who is the second person from the bottom? (1) Tulsi (2) Akram (3) Priya (4) Bopsi 9. F has less money than H but more than G. E has more than F but less than H. Who is the poorest? (1) F (2) E (3) H (4) G 10. Umesh is taller than Satish, Suresh is shorter than Neeraj but taller than Umesh. Who is the tallest among them? (1) Umesh (2) Suresh (3) Satish (4) Neeraj 1. (2) 9. (4)

2. (2) 3. (4) 10. (4)

Answer Key 4. (4) 5. (2) 6. (4)

Answers with Explanations 1. Option (2) is correct. Given that Ram’s rank is 7th from top and 26th from bottom. So, total number of students in class = (7 + 26) - 1 = 33 - 1 = 32 2 Option (2) is correct. According to the question, D C E/B/A So, D has the highest money. 3. Option (4) is correct. According to the question, Dinesh > Chinku …(i) Dinesh > Elina …(ii) Akash < Elina …(iii) Chinku > Akash …(iv) Bikash > Dinesh …(v) Using all above conditions Bikash>Dinesh>Elina and Chinku>Akash As we can see, Bikash is tallest among them and Dinesh is next to Bikash. 4. Option (4) is correct. According to the question, E>A>B

...(i)

E>D>A

...(ii)

C ® Smallest

...(iii)

Arranging equation (i), (ii) and (iii) together, E>D>A>B>C Hence, E is the tallest of the five. 5. Option (2) is correct. According to the question, Farhan > Raju Raju > Simran > Nitu

7. (3)

8. (4)

Arranging equations (i) and (ii) together, Farhan > Raju > Simran > Nitu Hence, Farhan scored highest run.

...(i) ...(ii)

159

LOGICAL REASONING: RANKING

8. Option (4) is correct. According to the question,

6. Option (4) is correct. According to the question,

Top

Ankit

Poonam

26

44

Tulsi Akram

From left

+18

+31

75

Ankit 25

Bopsi Priya

50

From right

Total students = 26 + 18 + 31 = 75 Ankit’s rank from right = 75 – 25 = 50 7. Option (3) is correct. According to the question,

Bottom

Hence, Bopsi is secound person from the bottom. 9. Option (4) is correct. According to the question, H>F>G H>E>F From equation (i) and (ii) H > E > F > G

7

Left

Second from Bottom

...(i) ...(ii)

↑ Poorest

Right Tree 14

Total number of trees in the row = 14 + 7 – 1 = 20 trees.

Hence, G is the poorest. 10. Option (4) is correct. According to the question, Umesh > Satish Neeraj > Suresh > Umesh Form equations (i) and (ii) Neeraj > Suresh > Umesh > Satish ↑ Tallest

Hence, Neeraj is the tallest.

...(i) ...(ii)

Linear arrangement based blood relation. Circular based blood relation. Age based blood relations. Combination of blood relation with any of the above seating arrangement.

9.

10.

Husband–Wife

Brother-Brother

Sister–Sister

Brother–Sister

3.

4.

5.

6.

11.

8.

Female

7.

S.No.

2.

Meaning

Male

Symbols

1.

S.No.

Daughter

Mother

Son

Mother

Daughter

Father

Son

Father

Symbols

Meaning

Unknown Gender

Mother–Daughter

Mother–Son

Father–Daughter

Father–Son

Following standard symbols can be used to draw a family diagram.

Apply the question in yourself. Do not assume gender. Draw a family tree diagram clearly. Keep all relation’s name in your mind. Solve through options if specific relation is not given. While checking options, keep in mind the gender. It has been seen that you can eliminate maximum options out of four given options 6. In puzzle-based qustions, a series of relations can be formed, so do not solve such questions in a haste. 7. If the statement says X in the son of Y, the gender of Y cannot be determined unless mentioned in the question. 8. In the case of coding-decoding blood relation, use a pictorial description to solve the question. This will make the symbols and relation more clear.

1. 2. 3. 4. 5.

TIPS/TRICKS

1. 2. 3. 4.

1. Solve through operations if relations have to identified and eliminate option one by one. 2. Form a diagram if relation is given.

1. Operation based (+, –, ×, ÷) 2. Symbolic based (&, *, #, $, !, %, etc.) 3. Number based (relations coded as 1, 2, 3, etc.)

Coding-Decoding

Dialogue/Conversation based

Tag line

Sister Uncle Aunt

Mother’s or Father’s Brother Mother’s or Father’s Sister

First Level

Second Level

Trace the Mind Map

Third Level



Niece

Nephew

Brother’s or Sister’s Son Brother’s or Sister’s Daughter

Sister-in-law Brother-in-law

Sister’s Husband

Husband’s or Wife’s Brother Brother’s Wife

Sister-in-law Brother-in-law

Husband’s or Wife’s Sister

Daughter-in-law Son-in-law

Daughter’s Husband

Grandmother Son’s Wife

Mother’s or Father’s Mother

Grandfather

Brother

Mother’s or Father’s Daughter

Mother’s or Father’s Father

One self

Father’s or Mother’s only Son/Daughter Mother’s or Father’s Son

Some of the blood relations are summarised below :

Blood Relation

Generation Tree OR Family Tree

Do not assume gender (Male/ Female) if not given in question.

TYPES

Puzzles

1. Pointing to man/woman. 2. Indicating to man/woman.

160 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

Blood Relation

7

Chapter Analysis Concept Name

2021

2022

2023

Blood Relation

4

3

3

 Revision Notes

Additional Questions

(2) Relation Puzzle Based

‘Blood relation’ involves the study and analysis of familial connections and relationships between individuals within a family or social unit. It focuses on understanding and deducing relationships and other familial connections based on the provided information or clues.

Family Tree: A family tree is a visual representation or diagram that depicts the relationships among individuals within a family. It typically resembles a tree structure, with branches representing different generations and lines connecting family members to illustrate their relationships.

 hese involve a set of statements providing clues about the T relationships between family members. The candidate need to decipher these clues to determine the relationships. Example 1: R is the mother of P and Q. If S is the husband of Q, then how is R related to S? (1) Mother  (2) Aunt  (3) Mother-in-law  (4) Sister Solution: Let us draw a family diagram from the given information R Mother P

Components of a Family Tree:

Generations: Family trees are organised into different levels or generations. The topmost level usually represents the oldest generation, such as grandparents, and subsequent levels represent their children, grandchildren, and so on. Individuals: Each person in the family is represented by a node or a box in the diagram. Their name is written inside the box along with other relevant information like birth and death dates. Connections: Lines or branches connect individuals to show their relationships. For instance, a horizontal line might represent a marriage or partnership, and vertical lines descend from parents to their children.

Types of Questions based on Blood Relation (1) Conversation Based

 C onversation-based questions typically revolve around understanding or interpreting dialogues or exchanges between individuals. These questions may test comprehension, inference, analysis, or even critical thinking skills. Example: Pointing to a lady on the stage, Monika said, “She is the sister of the son of the wife of my husband”. How is the lady related to Monika? Solution: To solve these type of questions start reading question from back (last). Step 1: In this question, be Monika then start from the end of the sentence. Step 2: Establish relation, “My husband” = “Monika’s husband”. “Wife of my husband” = is me Monika. “Son of the wife of my husband” = My son. “Sister of the son of the wife my husband” = My son’s sister = My daughter “She is the sister of the son of the wife of my husband” = the lady on the stage = the lady being pointed out = my daughter So, the lady on the stage = Monika’s daughter Trick: Reading question from the last and break the sentence.

Sister

Q

Couple

S

Clearly, R is the mother-in-law of S. (3) Coded Blood Relation. These questions use a code or symbolic representation for  relationships rather than using direct familial terms like father, mother, brother, etc. The candidate need to decode the relationships based on the given code. Example 1: If ‘P + Q’ means that P is the mother of Q, ‘P ÷ Q’ means that P is the brother of Q, ‘P ´ Q’ means that P is the son of Q, ‘P – Q’ means that P is the sister of Q, Which of the following means that C is the sister of D? (1) C – P ÷ D (2) P + D ÷ C (3) D × P – C (4) D – C × P Solution: Upon checking option (1). C – P ÷ D can be drawn diagrammatically by using the given information: ‘C – P’ means that C is the sister of P, ‘P ÷ D’ means that P is the brother of D. As C is the sister of P and P is the brother of D, we can say C is the sister of D. C–————P+————D (–) sign indicates the gender of C as female. (+) sign indicates the gender of P as male. Hence, option (1) is correct.

Important points to be kept in mind while solving these questions:

• You cannot assume yourself, the gender of the person based on the name. • If the statement say X is the son of Y, the gender of Y cannot be determined unless mentioned in the question. • In puzzle based questions, a series of relations can be formed, so do not solve such questions in a haste. •  In case of coding-decoding blood relation, use a pictorial description to solve the question. This will make the symbols and relation more clear.

Objective Type Questions 1. Pointing to a man a lady said “His mother is the only daughter of my mother.” How is the lady related to the man? (2023) (1) Aunt (2) Sister (3) Daughter (4) Mother 2. Following question is based on the information given below:  (2023) (I) ‘R ´ S’ means ‘R is father of S’ (II) ‘R - S’ means ‘R is sister of S’

(III) ‘R + S’ means ‘R is mother of S’ (IV) ‘R ¸ S’ means ‘R is brother of S’ In the expression A + B ´ C ¸ D how is C related to A? (1) Son (2) Grandson (3) Granddaughter (4) Grandmother

162

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

3. Rahul told Anand “yesterday I defeated the only brother of the daughter of my grandmother”. Whom did Rahul Defeated? (2023) (1) Son (2) Brother (3) Father-in-law (4) Father 4. Pointing towards a person in the photograph, Mohit said, “he is the husband of the only daughter of the father of my sister’s brother.” How is that person related to Mohit? (2022) (1) Uncle (2) Cousin (4) Brother (3) Brother-in-law 5. Kalam says, “Ravi’s mother is the only daughter of my mother”, How is Kalam related to Ravi? (2022) (1) Father (2) Brother (3) Maternal Uncle (4) Uncle 6. Introducing a man X, Y said “His wife is the only daughter of my father.” How is X related to Y? (2022) (1) Brother (2) Uncle (3) Husband (4) Father-in-law 7. Amit and Bobby are brothers. Chitransha and Dolly are sisters. Amit’s son is Dolly’s brother. How is Bobby related to Chitransha? (2021) (1) Father (2) Uncle (3) Grandfather (4) Brother 8. Sunil is the son of Kesav. Simran, Kesav’s sister, has a son Maruti and daughter Sita. Prem is the maternal uncle of Maruti. How is Sunil related to Maruti? (2021) (1) Cousin (2) Maternal uncle (4) Nephew (3) Brother 9. A and B are brothers. C and D are sisters. A’s son is D’s brother. How is B related to C? (2021) (1) Father (2) Brother (3) Grandfather (4) Uncle 10. A is B’s sister. C is B’s mother. D is C’s father. E is D’s mother. Then, how is A related to D? (2021) (1) Grandmother (2) Grandfather (4) Grand daughter (3) Daughter 1. (4) 9. (4)

2. (2) 3. (4) 10. (4)

Answer Key 4. (3) 5. (3) 6. (3)

7. (2)

8. (1)

4. Option (3) is correct. F Father Sister

Mohit

Husband

H+

Brother-in-law

5. Option (3) is correct. Kalam is maternal Uncle of Ravi.

Kalam

Ravi

6. Option (3) is correct. Daughter of my father is myself. So, X is the husband of Y. 7. Option (2) is correct. According to the question, Amit and Bobby are brothers and Chitransha and Dolly are sisters. We will make family diagram from the given information. We will use following conventions. Male

Female

Bobby

Amit

Answers with Explanations 1. Option (4) is correct. According to the question.

S–

Son

Generation

Dolly

Siblings

Chitransha

It is clearly seen from the family diagram that Bobby is the uncle of Chitransha. 8. Option (1) is correct. We can draw the family diagram on the basis of given information.

Lady Mother

Lady

Brother Keshav Sister Simran Prem Man

So, the lady itself is the mother of the man. 2 Option (2) is correct. – A

So, C is the grandson of A. 3. Option (4) is correct. According to the question, Grandmother Daughter

D

Brother Rahul

Hence, Rahul defeated his father.

Sunil

Maruti

Sister

Sita

Cousin

B+

C+

Son

Son

+ sign stands for male and - sign stands for female. Now, we can clearly see that Sunil is the cousin of Maruti. 9. Option (4) is correct. After carefully examining the statements, it was discovered that B is A's brother, and A's son is D's brother, implying that D is A's daughter. Because C and D are sisters, C is also A's daughter. As a result, B is C's uncle. 10. Option (4) is correct. A is B's sister, and C is their mother, based on the information provided. Their grandfather is D, and their great-grandmother is E. As a result, A is D's granddaughter.

Symbols, notations, mathematical signs. Miscell, anequal patterns.

We have to follow the relationship given through these sign, symbols given in the question.





Question based on symbol substitution Example: P + Q – R × S ÷

Trick based operations 9 × 5 × 2 = 529 7 × 8 × 3 = 837 4 × 7 × 6 = 764

Question based on interchanging mathematical signs. Example: + – × ÷

Mathematical Operations

Trick based Mathetical Operations







Symbol substitution Interchange of signs and Numbers Balancing the equation

















V means Vinculum or Bar... (The bar you _ see on the top of the value X) B means Bracket – () {} [ ] O means ‘Of’ (multiplication) [ ×]. D means Division [÷] M means Multiplication [×] A means Addition [+] S means Subtraction [–].

VBODMAS RULE

– ×

Interchange of sign 4×3–2+6=4

Second Level

Trace the Mind Map

Third Level



We have that option of interchanging which will satisfy the given equation in the question.

× +

First Level

÷

Balancing the Expression 8×8×1×7=8

LOGICAL REASONING: MATHEMATICAL OPERATIONS

163

Mathematical Operations

Chapter

8

Chapter Analysis Concept Name

2021

Mathematical Operations

 Revision Notes

2022

2023

Additional Questions

1

1

8

2. Problem solving by Substitution

Deals with various mathematical calculations or operations applied to logical statements or sequences. It involves using basic arithmetic operations like addition, subtraction, multiplication, and division to solve problems that involve logical reasoning. The candidate should be familiar with the mathematical operations. They are Addition (+), Subtraction (–), Multiplication (´), Division (÷), signs like Greater than (>), Less than ( 180° 360° – 286° = 74°

\

Objective Type Questions 1. At what angle the hour hand and minute hand of a clock are inclined at 15 minutes past 5? (2023) (1) 58

1° = 2

(2) 64°

(3) 67

1° = 2

(4) 72

1° = 2

2. Time appears in the mirror 11:09. Then, what time it will appear in the clock? (2022) (4) 1:09 (1) 1:51 (2) 12:09 (3) 12:51 3. At what time are the hands of clocks together between 6 and 7? (1) 32

8 min past 6 11

(2) 34

8 min past 6 11

(3) 30

8 min past 6 11

(4) 32

5 min past 6 7

9. The Chairman of the Selection Committee arrived at the interview at 10 minutes to 12 : 30 hour. He was earlier by 20 min than the other members of the board, who arrived late by 30 minutes. At what time was the interview scheduled? (1) 12:10 (2) 12:20 (3) 12:30 (4) 12:40 10. A clock with only dots marking 3, 6, 9 and 12 positions has been kept upside down in front of a mirror. A person reads the time in the reflection as 9 : 50. What is the actual time ? (1) 2:15 (2) 8:40 (3) 8:50 (4) 4:15 1. (3) 9. (1)

2. (3) 3. (1) 10. (2)

7. (3)

1. Option (3) is correct. Time = 15 minutes past 5

Angle = θ = [60 × hour − 11 × minutes] 2

And another angle = (360 - θ)° Hence, θ=

4. A clock gains five min every hour. What will be the angle traversed by the second hand in 1 minutes? (1) 360° (2) 360.5° (3) 390° (4) 380° 5. By looking in a mirror it appears that it is 6:30 in the clock. What is the real time? (1) 6:30 (2) 5:30 (3) 6:40 (4) 5:00 6. Reaching the place of meeting 20 min before 8 : 50 hour, Satish found himself 30 min earlier than the man who came 40 min late. What was the scheduled time of the meeting? (1) 08:20 (2) 08:10 (3) 08:05 (4) 08:00 7. A clock with only dots marking 3, 6, 9 and 12 positions has been kept upside down in front of a mirror. A person reads the time in the reflection as 6 : 10. The real time is (1) 06:50 (2) 12:40 (3) 12:20 (4) 6:10 8. After 9 O’clock at what time between 9 pm and 10 pm will the hour and minute hands of a clock point in opposite directions? (1) 15 min past 9 (2) 16 min past 9 4 1 (3) 16 min past 9 (4) 17 min past 9 11 11

Answer Key 4. (3) 5. (2) 6. (1)

Answers with Explanations

8. (3)

[60 × 5 − 11 × 15] 2

= (135/2)° = 67.5

And another angle = (360 – θ)° = (360 – 67.5)° = 292.5° Or [67 ½]° 2. Option (3) is correct. Given that, time in the mirror = 11:09 So, time in the clock = 11 : 60 – 11 : 09 00 : 51 So, time in the clock = 12 : 51 3. Option (1) is correct. Time between 6 and 7 when both hands are together. Both hands together means the angle between them is 0°. Let x be the minute past 6. 11° \ 0° = 6 × 30° – x × 2 0° = 180° − x ×

⇒ ⇒

11x = 180 2

⇒

x=

11° 2

180 × 2 360 8 = = 32 11 11 11

\ 32 8 minutes past 6. 11

4. Option (3) is correct. Since there is an addition of 5 min in every hour (60 min) in the clock i.e., in instead of showing 60 minute clock will show 65 min. So, 60 minutes = 65 minutes ⇒

1 minute = 65 minutes 60

⇒

1 minute = 65 × 60 s 60

212

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

⇒

1 minute = 65 s ↓ 60 s + 5 s ↓ ↓ 360° + 30° = 390° Note: Second hand forms 360° in 60 seconds i.e., in one second it forms 6°. 5. Option (2) is correct. 12 12

\ Interview scheduled at 12:40 – 30 minutes = 12:10 hour 10. Option (2) is correct. 6 7 5 8 4 3 2

9

3

3

Real Image 5:30

\ Its 5:30 in real. Shortcut method: Always subtract from 11:60/23:60 to time of mirror image/ real image to get real image mirror image. So, 11:60 – 06:30 = 05:30 6. Option (1) is correct. Satishs time to reach the place = 8:50 h – 20 minutes = 8:30 minutes Satish reached 30 minutes earlier than the other man \ Other man reached the place at 9:00 hour Other man reached 40 minutes late \ Scheduled time = 9:00 hour – 40 minutes = 8:20 hour 7. Option (3) is correct. 6

3

9

12 Mirror Image

Real Image

\ Actual time is 12:20 8. Option (3) is correct. If hands are in opposite direction then angle between them = 180° \

180° = hour × 30° – min × 180° = 9 × 30° – x ×

⇒

11° 2

11 2

11 x = 9 × 30° – 180° 2

x=

⇒

90°× 2 180 4 minutes ⇒ = 16 11 11 11

\ Time is 16 4 minutes past 9. 11

9. Option (1) is correct. Chairman arrived = 12:30 – 10 minutes = 12:20 h Other members come after the chairman had arrived = 20 minutes \ The other members arrive = 12:20 hour + 20 minutes = 12:40 hour The other members are late by 30 minutes

12

Real image

Calendar

 Revision Notes

6

Mirror Image will be 6:30

Mirror

11

\ Actual time is 8:40.

Topic-2 6

10 1

Mirror image

9

9

A calendar is a chart of series of pages showing the days, week and months of a particular year. The topic of calendar includes concepts such as odd days, leap year and finding the day of the week for a given data. A calendar is the record of all the days of the year. The smallest unit of a calender is a day. There are two types of years. (a) Ordinary year: An ordinary year is a year which has 365 days. Such years are not divisible by 4 e.g., 2001, 2002, 2003, etc. Ordinary years in the form of century are not exactly divisible by 400, e.g., 900, 1800, 2100, etc. (b) Leap year: A leap year is a year which has 366 days. Such years are exactly divisible by 4. e.g., 2004, 2008, 2012, etc. Leap year in the form of a century are exactly divisible by 400. e.g., 2000, 2400, 2800, etc. • The excess days beyond complete weeks are termed as “odd days.” Or, we can say, when total number of days is divisible by 7, then the remainder is called odd days. 1 ordinary year = 365 days = (52 weeks + odd 1 day) 1 leap year = 366 days = (52 weeks + odd 2 days) Year: 10th part of a century or 12 months together forms a year. Types of Question Type 1: Finding the day when another day is given. Example: If 26th January is Monday then what will be the day on 26th April of the same year? Type 2: Finding the day when another day is not given. Example: Which day falls on 26th January 2000? Type 3: Matching the calendar of a month. Example: Which year in the future will have same calendar exactly as 2009?

Concept about Year (i) Normal year/Ordinary year/Non-leap year (a) Total number of days = 365 days (b) First day and last day of the year are same day. (c) In any two consecutive normal year the same date of the year will be one day ahead of the same date of the previous year. Example: If 1st January 2001 = Monday then 1st January 2002 = Tuesday 1st January 2003 = Wednesday (d) Have one odd day. (ii) Leap year (a) Total number of days = 366 days (b) Last day of the year is one day ahead of the first day of the same year. (c) Have two odd days.

213

LOGICAL REASONING: CLOCK AND CALENDER

Century

A period of 100 years is called a century. Thus, each one of the years 1100, 1800, 2000, 2100 is a century. Given below are the important points about century: The first day of century cannot be Wednesday, Friday or Sunday. The last day of century cannot be Tuesday, Thursday or Saturday. Day Code Sun

Mon

Tue

Wed

Thurs

Fri

Sat

0

1

2

3

4

5

6

Month January

Number of Days 31

Odd Days 3

February March April May June July August September October November December

28 or 29 31 30 31 30 31 31 30 31 30 31

0 or 1 3 2 3 2 3 3 2 3 2 3

Objective Type Questions 1. If 18th February 1997 fell on Tuesday then what was the day on 18th February 1999? (2023) (1) Friday (2) Monday (3) Tuesday (4) Thursday 2. If today is Monday, after 62 days, it will be: (2023) (1) Wednesday (2) Saturday (3) Sunday (4) Thursday 3. If 1st January 2001 was Monday. Then which day of the week was on 31st December 2001? (2023) (1) Monday (2) Wednesday (3) Friday (4) Saturday 4. If 9th day of a month is Sunday, What will be the last day of the month, on which Sunday will fall? (2022) (1) 28th (2) 29th (3) 30th (4) 31th 5. On 7th March, 2005 Monday falls. What was the day of the week on 7th March 2004? (2022) (1) Monday (2) Saturday (3) Sunday (4) Friday 6. Mrs. Susheela celebrated her wedding anniversary on Tuesday, 30th September 1997. When will she celebrate her next wedding anniversary on the same day? (1) 30 September 2003 (2) 30 September 2004 (3) 30 September 2002 (4) 30 October 2003 7. If friday falls on 15th of September 2000, what will be the day on 15th of September 2001? (1) Friday (2) Saturday (3) Thursday (4) Sunday 8. If the day before yesterday was Thursday, what day will be four days after tomorrow? (1) Monday (2) Thursday (3) Sunday (4) Wednesday 9. The day before yesterday was Sunday. What will be the day after tomorrow? (1) Monday (2) Thursday (3) Friday (4) Saturday 10. If the third Friday of a month is 16th, what date is the Fourth Tuesday of that month? (2) 22nd (3) 27th (4) 29th (1) 20th 1. (4) 9. (2)

2. (3) 3. (1) 10. (3)

Answer Key 4. (3) 5. (3) 6. (1)

7. (2)

8. (2)

Answers with Explanations

Monday + 6

Sunday

1998 1999 2000 2001 2002

62 =6 7

1 odd day 2 odd day 1 odd day 1 odd day

1 odd day 2003 Total== 77 odd odd day Total day

Hence, she celebrates her next wedding anniversary on same day in year 2003, 30th September. 7. Option (2) is correct. 15th September 2000 to 15th September 2001 = 365 days 365 days = 52 week + 1 odd day. 1 odd day means 15th September 2001 will be Friday + 1 odd day = Saturday. 8. Option (2) is correct. Thursday + 1 + 1 + 1 + 4 days ↓ Yesterday

↓ Today

↓ Tomorrow

Total days after Thursday = 7 ⇒ Odd days Hence, it will be Thursday. 9. Option (2) is correct. Sunday

1. Option (4) is correct. 18th Feb 1997 = Tuesday Number of odd days = 1 + 1 = 2 (Because the number of odd days in an ordinary year is 1) Hence, 18th Feb 1999 will be Tuesday + 2 = Thursday 2. Option (3) is correct. Number of odd days in 62 days =

3. Option (1) is correct. 1st January 2001= Monday No. of days between 1st Jan 2001 to 31st Dec 2001 = 364 days. So, 364 / 7 = 52 weeks, which implies there are no odd days. Hence, 31st Dec 2001 will also be a Monday. 4. Option (3) is correct. It is given that the 9th day of a month is Sunday. So, after 21 days again it will be Sunday on 30th of the month. 5. Option (3) is correct. As there is 1 odd day. So, 1 year before the day was Sunday. 6. Option (1) is correct. 30 September 1997 30thSeptember 1 odd day

Monday

Tuesday

Wednesday

Yesterday

Today

Tomorrow

Day after tomorrow Thursday

Hence, day after tomorrow will be Thursday. 10. Option (3) is correct. Third Friday = 16th First Friday = 16 –14 = 2nd \ First Tuesday = 2 + 4 days = 6th Fourth Tuesday = 6th + 21 days = 27th day Hence, 27th will be the fourth Tuesday of that months.

Sum of opposite faces will always be equal to 7

Standard dice

Open dice/ Open cube

Dice and Cube

Ordinary dice or Non-standard dice

First Level

Second Level

Trace the Mind Map

Third Level



Sum of numbers on the opposite faces is not always seven

214 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

Dice And Cube

19

Chapter Analysis Concept Name

2021

2022

Standard Dice Problems

2023

Additional Questions

1

4

Ordinary Dice or Non-Standard Dice problems

5

Open Dice/ Open Cube

5

Introduction Dice is a three-dimensional small cubical, throwable object with marked faces, which when seen from any direction only three faces of a cube are always visible at a time. Dice may have different elements (numbers/symbols /alphabets) marked on each face. Some Properties of a Dice 1. A dice have six faces, eight vertices, and twelve edges. 2. Three pairs of opposite faces are there in a dice. 3. Only three faces of a cube can be seen at a time and these faces can never be opposite to each other. 4. Each face has one opposite face and four adjacent faces. D E

C

common 4 6

6

6 5

4

4

1

common numbers opposite

1

common

Then, faces number 1 and 5 will be opposite to each other.

H

3. If the faces of dice in two different positions show a common number face at the same place or position, then the opposite faces of the same cube will be at the same place/positions as shown below:

B A

From the above figure of dice, the six faces of the dice are– AGCB, BHDC, DHFE, FAGE, GEDC, and AFHB. 1. There are three pairs of opposing faces e.g., DHFE and AGCB are opposite to each other. 2. Four faces are adjacent to each face e.g., GEDC, DHFE, AFHB, and AGCB are adjacent to face FAGE. 3. GEDC is the upper face of the cube. 4. AFHB is the bottom face of the cube.

Example:

Some Tips and Tricks:

In both the positions of a dice, face number 6 is common.

1. Two opposite faces of any dice can never be adjacent to each other. The faces of dice in two different positions are shown below. 3 1

(A)

6 5

5

From above, two faces numbers 4 and 6 are common in both dices and 5 and 1 are remaining faces.

G

F

2. If the faces of dice in two different positions show two common number faces, then the remaining number of both the dices will be opposite to each other, as shown below figure.

3

6

Here, faces with numbers 1, 5, and 6 are adjacent to face number 3. Therefore, faces with numbers 1, 5, and 6 will never be opposite to face number 3. Therefore, if the question is about the opposite face to the face number 3 then, options having 1, 5, and 6 to be eliminated.

5

opposite

4

3

6

opposite

5

1 3

common number at same place

Therefore, 4 is opposite to 1, and 5 is opposite to 3. 4. If only a single face is common in two different positions of a dice then the face which is not visible will be opposite to the common number face. Example : 1

4

5

(B)

Therefore,

1

4

6

5

6

Here face numbered 2 is not visible

3 6

common number at same place

opposite

2

216

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

From the above two positions of a dice, the face that is common in 1

both the position is number 6 and the face number 2 is not visible. Then, face number 2 is opposite to face number 6. Types of questions

4

5

1. Standard dice problem 2. Ordinary dice problems

Standard Dice

3. Open dice problem

Topic-1

opposite of 1-----6 opposite of 2-----5 opposite of 3-----4

Standard dice Problems

 Revision Notes In such type of dice problem, the question is based on standard dice. A standard dice is a regular cube with its six faces numbered from 1 to 6. In a Standard Dice sum of the numbers on opposite faces is always 7 and the sum of the numbers on adjacent faces will never be 7.

1+6 = 7 2+5 = 7 3+4 = 7

Example 1: If the dice is a standard dice then which number will be opposite to the face number 4? (1) 1 (2) 3 (3) 2 (4) 4 Solution: Option (2) is correct. The sum of opposite faces of a standard dice is always 7. Therefore, opposite face + 4 = 7 Opposite face = 7 – 4 = 3

Objective Type Questions 1. In a dice, 1, 2, 3 and 4 are written on the adjacent faces, in a clockwise order and 5 and 6 at the top and bottom. When three is at the top, what will be at the bottom? (2022)

& (1) +

*

(

( $

*

$

#

(2) *

+

(3) #

(4) $

Answer Key 1. (1)

(1) 1

(2) 2

(3) 3

(4) 4

2. Two different positions of the same dice are shown the six faces of which are marked by the letters A, B, C, D, E and F. Which letter will be on the face opposite to the one showing ‘D’. A C

(1) C

D D

B

(2) F

F

@ (1) &

#

(2) $

#

(4) A

(3) #

% @ (4) %

4. In a dice 1, 2, 3 and 4 dots are on the adjacent faces, in a clockwise order, and 5 and 6 dots at the top and bottom. When 1 dot is at the top, then what number of dots will be at the bottom? (1) 4

(2) 6

(3) 3

5. (1)

Answers with Explanations 1. Option (1) is correct. It is given that, 1, 2, 3 and 4 are written on the adjacent faces, in a clockwise order and 5 and 6 at the top and bottom. The pair of opposite faces from given data is – (1)

1–3

(2)

2–4

(3)

5–6

Adjacent face can never be opposite face. \ Eliminate the option having adjacent face of D. \ We can say E is opposite to D. 3. Option (4) is correct. Writing each dice symbol clockwise, and same symbol should

& %

4. (3)

2. Option (3) is correct.

3. Three different positions of the same dice are shown. Select the symbol that will be on the face opposite to the one having ‘!’ $

3. (4)

So, when three is at the top, then 1 will be at the bottom.

(3) E

!

2. (3)

(4) 2

5. Three different positions of the same dice are shown below. Which symbol is on the face opposite the face showing ‘(‘?

be in same column from right to left. Dice 1: @ ! # Dice 2: $ % # Dice 3: @ % & opposite @ $ opposite Ans ! % opposite # & 4. Option (3) is correct. According to the question opposite 1 3 opposite 2 4 opposite 5 6 \ We can say when 1 is at top, then 3 dots will be at bottom.

217

LOGICAL REASONING: DICE AND CUBE

5. Option (1) is correct. Writing symbol clockwise from left to right and same symbol in same column. Dice 1: ( $ & Dice 2: ( # * Dice 3: + $ * \ We can say + is opposite to (.

Ordinary Dice or Non-Standard Dice problems

Topic-2

 Revision Notes Ordinary dice is a type of dice in which the sum of the numbers on opposite faces is not always 7 and the sum of the numbers at adjacent faces maybe 7. Faces of ordinary dice may be marked by different elements such as symbols, alphabets, numbers, and so on.

1

2 5

2

4

6

Options: (1) 3 (2) 2 (3) 5 (4) 1 Solution: Option (2) is correct. Steps to solve ordinary dice questions Step-1: Identify the common element of the dice at different positions. Step-2: Place the elements in clockwise sequence from left to right in a row, starting from the common element 1

2

3

5

2

4

6

5

4

ordinary dice 4+3 = 7

Problems based on ordinary dice can be solved by following simple steps as explained below with the help of example. Example: From the two different positions of ordinary dice. Find the number opposite to face number 5?

2 = common element e.g., (DICE-1) 2 6 5 Step-3: Place the common elements in the same column (even if there are more than one element common). And then place the remaining elements in a clockwise sequence from left to right in a row. e.g., (DICE-1) 2 6 5 (DICE-2) 2 4 1 Step-4: The numbers in the same column are opposite to each other i.e., 6 and 4 are opposite to each other. Similarly, 5 and 1 are opposite to each other. Also, the common number/ element (2) is opposite to the remaining number which is not visible in the given figure i.e., 3 and 2 are opposite to each other.

Objective Type Questions 1. Four different positions of the same dice are shown. Select the number that will be on the face opposite to the one having 5.

5

5

6

3 2

1

4

6

1 2

5

3

1

1

5 4

6

2

5

3

5

1 (1) 3

(2) 1

5 (3) 4

4

2

2

3

2

2 (1) 3

3 (2) 4

1. (4)

6

4

1 3

5

5

(3) 5

Answer Key 2. (2) 3. (4) 4. (2)

2 (4) 6 5. (3)

Answers with Explanations

1 2

6

(1) 4 (2) 1 (3) 6 (4) 3 5. Three different positions of the same dice are shown. Select the number that will be on the face opposite to the face showing the number ‘1’.

3

(1) 2 (2) 3 (3) 1 (4) 5 3. Two different positions of the same dice are shown. Assuming that the sum of opposite faces is an odd number, select the number that will be on the face opposite to the one having the number ‘2’.

6

5

1

5

3

(1) 2 (2) 1 (3) 6 (4) 4 2. Four different positions of the same dice are shown. Select the number that will be on the face opposite to the face having number ‘6’. 6

4. Three different positions of the same dice are shown. Select the number that will be on the face opposite to the one having 5.

3 (4) 5

1. Option (4) is correct. Putting numbers of each dice clockwise/anticlockwise and same number must be in same column.

218

2.

3.

4.

5.

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Dice 1: 5 3 2 Dice 2: 4 6 1 Dice 3: 5 6 2 Dice 4: 5 3 1. Option (2) is correct. Writing numbers of each dice clockwise/anticlockwise and same number must be in same column. Dice 1: 6 4 1 Dice 2: 6 5 2 Dice 3: 3 5 1 Dice 4: 3 5 2. Option (4) is correct. Writing each number of both dices clockwise/anticlockwise and same number must be in same column. Dice 1: 1 6 2 Dice 2: 1 3 5 \ 5 is opposite to face 2. Option (2) is correct. Writing each number of each dice clockwise/anticlockwise and same number must be in same column. Dice 1: 5 3 6 Dice 2: 5 4 2 Dice 3: 1 3 2. Option (3) is correct. Putting the same numbers of each dice in same column and fill the vacant space accordingly. Dice 1: 1 3 2 Dice 2: 5 3 4 Dice 3: 5 6 2 \ Face 5 is opposite to number face 1.

Open Dice/Open Cube

Topic-3

 Revision Notes

(B)

1

1↔6

2 3

2↔4

4 5 (C)

6

3↔5 1↔3

1 2 3

4↔6

4 5

2↔5

6

Example: Which of the given options will be a pair of opposite faces when the given flattened-out cube is folded again? 5 1

4

(1) 2 and 3

(2) 4 and 6

OPPOSITE

B

(3) 5 and 6

(4) 5 and 2

Solution: Option (4) is correct. Here, faces 5 and 2, 1 and 6, 4 and 3 are alternate faces and these alternate faces will be always opposite to each other. Therefore, 5 and 2 are opposite to each other. Example: Which of the following given options will be the folded cube when the given flattened-out cube is folded again? 1 2

Alternate faces are always opposites to each other

D C

6

2

OPPOSITE

F

3

6 3

5

4

E 3

OPPOSITE

6

A

3 4

5

6

(1)

Flattened out of a cube

1

(2)

(1) 1

Variations of open dice

From Cube (1) - faces 3, 4 and 6 are adjacent to each other; therefore, these adjacent faces can never be opposite to each other

(A)

1 2

3 5 6

4

Opposite faces to each other

(3) 3

(3)

1. Top face of the cube = A. 2. Bottom face of the cube = C. 3. Side faces of the cube = E, B, F, and D. 4. Alternate faces are always opposite to each other e.g., F and E; C and A, D and B are alternate faces and opposite to each other.

Open dice

(2) 2

5

2

(4) both 1 and 3

Solution: Option (2) is correct From given flattened-out cube, faces 6 and 1, faces 3 and 4, faces 2 and 5 are alternate faces and opposite to each other.

1↔5

From Cube (3) - faces 1, 2 and 5 are adjacent to each other; therefore, these adjacent faces can never be opposite to each other

2↔4

From Cube (2) - faces 3, 6, and 5 are adjacent to each other and these adjacent faces never are opposite to each other; this matches to given flattened-out cube

3↔6

Therefore; only cube (2) can be the folded cube.

219

LOGICAL REASONING: DICE AND CUBE

Objective Type Questions 1. A cube is made by folding the given sheet. In the cube so formed, which of the numbers will be on opposite faces?

5. Select the answer figure that can be formed by folding the given sheet along the lines. @

6 4

3

1

%

2

$

5 (1) 1 and 5 (2) 6 and 3 (3) 2 and 3 (4) 1 and 6 2. A cube is made by folding the given sheet. In the cube so formed, which number would be on the face opposite to the one showing ‘5’. 6

* ! @

%

(1)

*

(2)

$

1 4

2

1. (3)

5 3 (1) 2 (2) 1 (3) 4 (4) 6 3. Select the answer figure that can be formed by folding the given sheet along the lines. A C D

B

@ will be opposite of ($) # will be opposite of (!) % will be opposite (*)

#

%

A

(2)

C

C E

5. (2)

Answers with Explanations 1. Option (3) is correct. According to given figure of cube. opposite 6 opposite 4 opposite 3

5 1 2

opposite

1

opposite

4

2 5

opposite

\ 1 is the face opposite to one showing 5. 3. Option (2) is correct. Opposite face can never be adjacent face. \ Eliminating options having opposite face as an adjacent face. \ Remaining option is (2) that can be formed by folding the given sheet along the lines. 4. Option (1) is correct.

B

C

(4)

A

@

3

B

(3)

!

6

A will be opposite of D B will be opposite of F C will be opposite of E

A

*

(4)

2. Option (2) is correct.

F

F

# @

Answer Key 2. (2) 3. (2) 4. (1)

E

(1)

%

(3)

#

E

D

4. A cube of 3’ × 3’ × 3’ inch size was painted on all six sides and then cut into 27 smaller cubes of 1 inch size. How many smaller cubes will have only two sides painted.

(1) 12

(2) 8

(3)

10 (4)

6

\ The smaller cube at the middle of each edge is two side painted \ Total smaller cube having two side painted = number of edges = 12. 5. Option (2) is correct. Opposite face can never be adjacent face of a dice. \ Eliminating options having given opposite face as an adjacent face. \ Only option (2) left after elimination

1

2

3

4

5

6

n

a

o

b

p

c

q

d

r

e

s

f t

g u

h

7

v

i

8

w

j

N

A

O

B

P

C

Q

D

R

E

S

F T

G U

H

Mirror Image of Capital Letters

V

I W

J

x

X

K

k

9

Y

L

y

l

0

Z

M

z

m

Shortcut Approach Whenever you have to solve a mirror image question, imagine a mirror placed in front of the object and then try to find its inverted image. The portion of the object that is near the mirror will now be the portion of the image near to the mirror in the inverted form

The letters which have the same mirror images are A, H, I, M, O , T, U, V, W, X and Y.

Mirror Image

Real Image

Mirror Image

Real Image

The letters which have the same mirror image are — i, l, o, v, w and x.

Mirror Image

Real Image

Mirror Image

Real Image

Mirror Image of Small Letters

Numbers 0 and 8 have the same mirror images.

Mirror Image

Real Image

Placed Left Hand Side (L.H.S.) of the object

Mirror Images

Placed Vertically

Placed above the object

Placed Right Hand Side (R.H.S.) of the object

Mirror

Mirror Image

Second Level

Third Level



Placed below the object

L.H.S and R.H.S remain unchange

Top and bottom of the object interchange in image

Real Image

Trace the Mind Map First Level

Placed Horizontally

Mirror Position

L.H.S and R.H.S of the bottom of the object interchange in image

Top of the bottom remains unchange

220 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

221

9 8 7 6 5 4 3 2 1 0

0 1 2 3 4 5 6 7 8 9

LOGICAL REASONING: MIRROR AND WATER IMAGES

Mirror and Water Images

Chapter

20

Chapter Analysis Concept Name

2021

2022

Mirror Images

2023

Additional Questions

1

9

Water Images

10

Topic-1 Mirror Images

UNCHANGE

LHS

RHS

Top

 Revision Notes Introduction The figure obtained by putting a mirror in front of the real object is known as mirror image or we can say that the reflection of an object into the mirror is called its mirror image. For example, mirror image of ABC = . There are ‘11’ letters in English alphabet which have identical mirror image: A, H, I, M, O, T, U, V, W, X, Y

Position of Mirror There are two positions of mirror.

Interchange

Bottom Tricks (i) The section of an object that is closer to the mirror will have its corresponding image situated closer to the mirror as well. (ii) The portion of an object situated at a distance from the mirror results in an image that appears distant from the mirror as well. Mirror Vertical

1. Mirror is placed VERTICALLY  lacing a mirror vertically can position it to the left or right of an P object. In the mirror’s reflection, the left-hand side (L.H.S) and right-hand side (R.H.S) of the object switch places, while the top and bottom maintain their original positions. Mirror Image

Object

LHS

INTERCHANGE

Far

RHS

Top

Near

Near

Far Far

Mirror placed Horizontal

Near Near

Unchanged

Near Mirror

Bottom

Near Far

Object

Mirror

Example 1: In each of the following questions, select the alternative which exactly matches the mirror image of the word/number in the questions. TRIUMPHS T (1) HPMUIRTS (2) PSMIUR STRIUMPH (3) SHPMUIRT (4) Solution: If we put a mirror in front of the word, we will get the image like TRIUMPHS SHPMUIRT Example 2: In each of the following questions, select the alternative which exactly matches with the mirror image of the word/number in the questions. 2345 (1) 243 (2) 32 (3) 5432 (4) 53 2 Solution: If we put a mirror in front of the number, we will get the image like

45

If a mirror is positioned horizontally above or below an object, the positions of the top and bottom of the reflected image swap places.

5

2345

Mirror Image

4

2. Mirror is placed Horizontally

5432

Example 3: Choose the correct mirror image from alternatives (1)/ (2)/(3) and (4) of the figure X.

223

LOGICAL REASONING: MIRROR AND WATER IMAGES

Solution: Mirror image of figure (X) will be like (X)

(1)

(2 )

( 3)

(4 ) Object

Mirror Image

Objective Type Questions 1. Choose the correct mirror image:

(1)

(2)

(3)

(2023)

5. Select the correct mirror image of the given figure when the mirror is placed to the right side of the figure.

(4)

2. Select the correct mirror image of the given figure.

(1)

(2)

(3)

(4)

MIRROR (1)

(2)

(3)

(4)

6. Select the correct mirror image of the given figure when a mirror is placed on the line PQ. P

3. Select the correct mirror image of the given figure when the mirror is placed to its right side.

Q (1)

(2)

(3)

(4)

7. Select the correct mirror image of the given figure. (1)

(2)

(3)

(4)

4. Select the correct mirror image of the given figure when the mirror is placed to the right side of the figure.

(1)

(3)

(2)

(4)

MIRROR (1)

(2)

(3)

(4)

8. Select the correct mirror image of the given figure when the mirror is placed to the right side of the figure. GKM#$%NG24 (1) 42GN$%#MKG (3) 42GN% $#MKG

(2) 42GN%$#MKG GKM#$%NG24 (4)

224

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

9. Select the correct mirror image of the given figure when a mirror is placed to the right side of the figure.

4. Option (1) is correct. Given Image

SEA (2) AES

(3)

A

(1) AE S

(4) SE

10. Select the correct mirror image of the given figure when the mirror is placed to the right side of the figure. Object nearer to mirror remain nearer to mirror. 5. Option (3) is correct. Given Image

(1)

(2)

(3)

(4)

1. (4) 9. (2)

2. (3) 3. (3) 10. (2)

Answer Key 4. (1) 5. (3) 6. (1)

Figure nearer to the mirror remains nearer to the mirror. 6. Option (1) is correct. Given Image Mirror Image

7. (2)

8. (2)

Answers with Explanations 1. Option (4) is correct.

Part of figure nearer to the mirror remains nearer to the mirror. 7. Option (2) is correct. Given Image Mirror Image

MIRROR

Logic: The top becomes the bottom and the bottom becomes the top. Hence,

Shaded area is nearer to the mirror remains nearer to the mirror in mirror image. 8. Option (2) is correct. Object nearer to the mirror remains nearer in mirror image. GKM#$%NG24 42GN%$#MKG

2. Option (3) is correct. MIRROR

9. Option (2) is correct. Given Image SEA

Real Image

Mirror Image

Mirror Image AES

10. Option (2) is correct.

3. Option (3) is correct.

Object nearer to the mirror remain nearer to the mirror in mirror image.

225

LOGICAL REASONING: MIRROR AND WATER IMAGES

Water Image

Topic-2

 Revision Notes The reflection of an object into water is known as its Water Image. The water image of an object or a figure is the same as the mirror image of that object when a mirror is placed horizontally at the bottom of that figure/object. Water image is a vertically inverted image obtained by turning an object/a figure upside down. OBJECT TOP

BOTTOM

(3)

(4)

OBJECT

R.H.S L.H.S WATER

TOP

UNCHANGED

R.H.S WATER SURFACE

TOP

WATER IMAGE

INTERCHANGED

R.H.S BOTTOM

Therefore, the answer figure in option (1) represents the correct water image of the given question figure.

BOTTOM

In the above figure, L.H.S and R.H.S of an object/a figure remain unchanged while the top and bottom of the same object/figure gets interchanged i.e., the bottom becomes the top and the top becomes the bottom. In water-based problems of this kind, candidates need to envision how an object or figure would look when positioned close to a water surface, considering how its image would be reflected or appear within the water. Some of the examples are given below will help you understand better.

Type-2: Water Image of Numbers and Letters

In such type of water image questions, the candidates are required to find the water image of numbers or letters or a combination of both. The water image of numbers and letters is given below .This will be helpful to solve the problems based on water images of numbers and letters: Water Image of Capital letters

OBJECT

A B C D E F G H I J K L M

WATER SURFACE

A B C D E F G H I J K L M

WATER IMAGE

Identical water images

Sometimes, a water image formed of an object/ a figure is similar to the actual object/figure. This happens when the upper half and the lower half of an object/a figure are similar. Some of the examples of such objects/figures are shown below. Example identical water images of figures

NOP Q R S T U VW X Y Z NOP Q R S T U VW X Y Z

a b c d e f g h i j k l m OBJECT WATER SURFACE WATER IMAGE

Types of questions of water image

Type-1: Water Image of Symbols, Signs or Figure

In such types of questions, the candidates are required to find the water image of symbols, signs or figures given in questions. The question may consist of symbols, signs or figures or a combination of them. Example 1: Which of the given alternatives represents the correct water image of the following figure?

LETTERS WATER SURFACE WATER IMAGE LETTERS WATER SURFACE WATER IMAGE

C, D, E, H, I, K, O, and X are eight capital letters having the same water image as they actually are. Water image of small letters:

a b c d e f g h i j k l m n o p q r s t u v w x y z n o p q r s t u v w x y z

LETTERS WATER SURFACE WATER IMAGE LETTERS WATER SURFACE WATER IMAGE

c, l, o, and x are the small letters having the same water image as they actually are. Water Image of numbers:

1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

DIGITS WATER SURFACE WATER IMAGE

The water image of digits 0, 3, and 8 remain the same as they actually are. Example 7: Which of the given alternatives represents the correct water image of ‘RABBIT’. (1) T (2) R R IABBBBA R R (3) (4) A B B

TI

L.H.S

(2)

TI TIBBA

L.H.S

(1)

Solution: Option (1) is correct. While solving the water image problems, the top and the bottom of the given figures interchange their places while the L.H.S and R.H.S of that figure remain at their original positions (unchanged) as shown below.

226

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Solution: Option (3) is correct. While solving the water image problems, the Left-hand side and the Right-hand side of a letter or a number remain unchanged (remain at their original position) while the top and the bottom of the same interchanged their place.

R A B B I T R A B B I T

WORD WATER SURFACE WATER IMAGE

Tips and Tricks

1. A water image is a vertically inverted form of an object or a figure. 2. C, D, E, H, I, K, O, X, are eight capital letters having the same water images as they actually are. 3. c, l, o, x are four small letters having the same water images as they actually are. 4. 0, 8, have the same water image and mirror images as they actually are.

Hence, option (3) is correct.

Objective Type Questions Directions: In each of the following questions, you are given a combination of alphabets and/or numbers followed by four alternatives (1), (2), (3), and (4). Choose the alternative which most closely resembles the water-image of the given combination. 1. F R O G (1) (2) (3) (4) 2. F A M I L Y (2) (1) (3) (4) 3. r i s e (2) (3) (4) (1) 4. w r o t e (2) (3) (4) (1) 5. m o n d a y (2) (1) (3) (4) 6. 9 6 F S H 5 2 (2) (1) (3) (4) 7. N 4 t Q j 3 (2) (1) (3) (4) 8. N h R q S y (2) (1) (3) (4) 9. a b 4 5 C D 6 7 (2) (1) (3) (4) 10. A 1 M 3 b (2) (1) (3) (4) 1. (1) 9. (2)

2. (4) 3. (1) 10. (3)

Answer Key 4. (3) 5. (4) 6. (3)

7. (1)

8. (2)

Answers with Explanations 1. Option (1) is correct.

Water image

Analysing option (1), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 2. Option (4) is correct.

Analysing option (4), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 3. Option (1) is correct.

Water image

Analysing option (1), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 4. Option (3) is correct.

Water image

Analysing option (3), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 5. Option (4) is correct.

Water image

Analysing option (4), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 6. Option (3) is correct.

Water image

Analysing option (3), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 7. Option (1) is correct.

Water image

Analysing option (1), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 8. Option (2) is correct.

Water image

Analysing option (2), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 9. Option (2) is correct. ab45CD67 ab45CD67 Analysing option (2), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer. 10. Option (3) is correct.

Water image

Analysing option (3), L.H.S and R.H.S remains at their same place and part of object nearer to the water surface remains nearer.

g

)

Integers

A.P.

G.P.

A.P. are a, a + d, a + 2d ... G.P.

Integers

,

H.C.F.

s

n  n   n   5  +  5 2  +  53  + .....      

Counting number of zeros at the end on n! is integral value of

QUANTITATIVE APTITUDE: NUMBER SYSTEM

227

Chapter

Number System

1

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

3

7

Progressions and Series Basic Concepts of Number System

6

9

Division Algorithm, Arithmetic Progression and Geometric Progression

1

6

Problems on Numbers, H.C.F. and L.C.M.

2

10

Types of Number System

Properties of G.P.

• Progressions and Series • Basic Concepts of Number System • Division Algorithm, Arithmetic and Geometric Progressions • Problems on Numbers, H.C.F. and L.C.M.

• ak, bk, ck,…will also be in G.P.

Topic-1 Progressions and Series

Scan to know more about this topic

consecutive terms is constant then the terms are said to be in A.P. Example. 2, 5, 8, 11 or a, a + d, a + 2d, a + 3d… • If ‘a’ is the first term and ‘d’ is the common difference then the general ‘nth’ term is Tn = a + (n – 1)d.

Arithmetic and Geometric Progression 2nd

If a, b, c, d,…. are in A.P. and ‘k’ is a constant then, • a – k, b – k, c – k,… will also be in A.P. • ak, bk, ck,…will also be in A.P.

• G/a = b/G or, • G2 = ab. • G = √ab. Harmonic Progression • If a, b, c, d,.…..are unequal numbers then they are said to be in • The ‘n’ term in H.P. is 1/(nth term in A.P). Properties of H.P.

Scan to know more about this topic

If a, b, c, d,…are in H.P. then,

• a + d > b + c.

• ad > bc. Polynomial

 et H be the harmonic mean between two numbers a and b. So, a, L H, b are in H.P. This means that 1/a, 1/H, 1/b are in A.P. 1 1 1 1 2ab . a+b 1 1 2 • – = – . or, . •H= = + = ab a b a b a+b H H H

a , b , c will also be in A.P. k k k

Let us assume a and b are two numbers. And A be the arithmetic mean between two numbers. So, a A b are in A.P. • A – a = b – A or, 2A = a + b • A = (a + b) . 2

Geometric Progression • If in a succession of numbers the ratio of any term and the previous term is constant then that numbers are said to be in Geometric Progression. Example :1, 3, 9, 27 or a, ar, ar2, ar3. • The general expression of a G.P. • Tn = arn–1(where a is the first terms and ‘r’ is the common ratio). • Sum of ‘n’ terms in G.P. a ( r n − 1) Sn = 1 , if r > 1 where a1 is the first term. r −1

• Sum of terms of infinite series in G.P. • S∞ = a/(1 – r), (–1 < r < 1).

Geometric Mean

Harmonic Mean

Arithmetic Mean

•

a b c , , will also be in k k k

H.P. if 1/a, 1/b, 1/c,……are in A.P.

Properties of A.P.

•

G.P.

•

Let, G be the geometric mean between two numbers a and b. So, a G b are in G.P.

 Revision Notes Arithmetic Progression • If the difference between any two

If a, b, c, d,…. are in G.P. and ‘k’ is a constant then,

Arithmetic Geometric Series • Suppose a1, a2, a3, …. is an A.P. and b1, b2, b3, …… is a G.P. Then the sequence a1b1, a2b2, …, anbn is said to be an arithmetic– geometric progression. An arithmetic–geometric progression is of the form ab, (a + d)br, (a + 2d)br2, (a + 3d)br3, ……

• Sum of ‘n’ terms of A.G.P. series, Sn = ab/1 – r + dbr(1 – rn – 1)/ (1 – r)2 – (a + (n – 1)d)brn/1 – r.

• Sum of term of infinite series in G.P., S.∞ = ab/1 – r + dbr/(1 – r)2, (–1 < r 0 6a = 4d a =2

 Revision Notes Types of Numbers

11 11 11 11  === −− 55( 8( 8) ) 33 55 88  1 1 1 1  =  −  299( 302 ) 3  299 302 

All the terms like 1 , 1 ,......... 1 will cancel out. \

5 8 299 11 1  11 ((300 300)) 50 25 25 50 = = = The sum =  − = 3  2 302  = 33((22))((302 302)) =302 302 151 151

8. Option (3) is correct. We want to maximise the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since, a52 is 100 and all the numbers are positive integers, maximising a1 entails maximising a2, a3, ... a51 The only way to do this is to assume that a2, a3.... a52 are in an A.P. with a common difference of 1. Let the average of a2, a3.... a52 i.e., a27 be A. (Note: The average of an odd number of terms in an A.P. is equal to the value of the middle-most term.) Since, a52 = a27 + 25 and a52 = 100 ⇒ A = 100 – 25 = 75 a2 + a3 + ... + a52 = 75 × 51 = 3825 Given, a1 + a2 + ... + a52 = 52(A – 1) = 3848 Hence, a1 = 3848 – 3825 = 23 9. Option (1) is correct. For such questions, we can take value of n = 1. The right option must give the first term, i.e.,

1

a1 + a2

Only Option (1) satisfies. 10. Option (4) is correct. Both the sequences are in arithmetic progression. The common difference (d1) for the first sequence = 4 The common difference (d2) for the first sequence = 5 The first term common is 19. The common terms will also be in arithmetic progression with common difference LCM (d1, d2) = LCM (4, 5) = 20 Let there be ‘n’ terms in this sequence, then the last term would be ≤ 415 i.e., a + (n – 1)d ≤ 415 ⇒  19 + (n – 1) × 20 ≤ 415 ⇒ (n – 1) × 20 ≤ 415 – 19 ⇒ (n – 1) × 20 ≤ 396  396   (where [ ] is the greatest integer)  20 

⇒ (n – 1) = 

⇒ (n – 1) = 19

Rational numbers : A number that can be expressed as

p is q

called a rational number, where p and q are integers and q ¹ 0. e.g.,

2 −2 1 , , , 1 , etc. 3 3 7 - 1000 7 But is not a rational number, It is not defined. 0

The set of rational numbers is denoted by “Q”. Irrational numbers : The numbers which are not rational number are called irrational numbers, e.g., 2 , 3 , 2 + 3 , 5 , etc. Real numbers : These numbers include both rational as well as −5 7 irrartional numbers, e.g., 1,0, –3, , , 3 , 3 + 2 , etc. 7 9

I

a2 =

R

Q

I

W

N

Î , q ¹ 0, p , q ona l 4 , 3 100 e t c .

11 11 11 11  === −− 22( 5( 5) ) 33 22 55 

ad

ad

p q

ati

a1 =

a100 =

So, on

16

ir r

4

ad 0’ d ‘ -ve t eger in

The G.P. is 24, 6, 1.5, 1.5 , 1.5 .......

Natural numbers : The counting numbers are Scan to know called natural numbers. 1, 2, 3, 4, ……. are all more about this topic natural numbers. It is denoted by ‘N’ Whole numbers : All natural numbers and ‘0’ make the set of whole numbers. 0, 1, 2, 3, 4, ……. are whole numbers. It is denoted by ‘W’ Integers : All natural numbers, ‘0’ and Number System negative numbers make the set of integers, i.e., …… –3, –2, –1, 0, 1, 2, 3, ……., etc. It is denoted by ‘I’ or Z. Integers are divided in two parts: Positive integers : 1, 2, 3, 4, …… Negative integers : ........ –4, –3, –2, –1, …… Remember: ‘0’ is neither a positive integer nor negative integer. We call the set {0, 1, 2, 3, 4,} ……. as non-negative integers.

d

3

6. Option (3) is correct. For any n ≥ 1, an = 3(an + 1 + an + 2 + ........)

1.5 3 a5 = . = 16 32 \ 7. Option (1) is correct.

Basic Concepts of Number System

d

d

Topic-2

ad

d

Even number : Whole number which is divisible by 2 is called even number, e.g., 2, 4, 6, 8, ….., etc.,. Odd number : Natural number which is not divisible by 2 is called odd number, e.g., 1, 3, 5, 7, ……, etc. Prime number : A number that is divisible only by 1 and itself is called prime number. It is greater than ‘1’, e.g., 2, 3, 5 , ……, etc. Co-primes : The two numbers (a,b) are said to be co-prime if they are not having any common divisors other than 1, e.g., (3,4), (8,27), (2, 3) , ……, etc. Composite numbers : The natural number which is not prime is called composite number, e.g., 4, 6, 8 , ……, etc.

Points to Remember

1 is neither a prime nor composite number. There are 25 prime numbers between 1 and 100, i.e., 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.  2 is the only even prime number. To check whether a number is prime or not  Consider the given number (to be checked) as ‘l’.  Choose integer k such that k2 ≥ l. [nearest perfect square]  Find the prime numbers less than k.  If these prime numbers divide ‘l’ then given number is a composite number, if none of them divides ‘l’ then the given number is a prime number.  

231

QUANTITATIVE APTITUDE: NUMBER SYSTEM

Example : Check whether the following numbers are prime or not ? (a) 437 (b) 811 (a) Sol. 437 < 441= (21)2 Prime numbers less than 21 are 2, 3, 5, 7, 11, 13, 17, 19, and 19 divides 437. Hence, 437 is not a prime number. (b) Sol. 811 < 841 = (29)2 Prime numbers less than 29 are 2, 3, 5, 7, 11, 13, 17, 19, 23. Here, none of them divides 811. Hence, 811 is a prime number.

Tests of Divisibility Divisibility by 2 : A number is divisible by 2 if it is an even number, i.e., its last digit (unit place digit) is any of 0, 2, 4, 6, 8. e.g., • 85724 is divisible by 2. • 3433 is not divisible by 2. Divisibility by 3 : A number is divisible by 3 if the sum of its digits are divisible by 3, e.g., 964251 ⇒ 9 + 6 + 4 + 2 + 5 + 1 = 27 Sum of the digits is 27 which is divisible by 3. \ 964251 is divisible by 3. Divisibility by 4 : A number is divisible by 4 if its last 2 digits are divisible by 4, or if the last 2 digits are 0’s. e.g., • 4321724 is divisible by 4 since 24 is divisible by 4. • 4329621 is not divisible by 4 since 21 is not divisible by 4. Divisibility by 5 : A number is divisible by 5, if its last digit is ‘0’ or ‘5’. e.g., 245 and 343420 are divisible by 5. 299342 is not divisible by 5. Divisibility by 7 : A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either zero or a multiple of 7. Example : To check 8724107 is divisible by 7. Just the double of the unit digit and then subtract it from the rest of the number. i.e., 872410 – 2 × 7 = 872396 (continue the process) ⇒ 87239 – 12 = 87227 ⇒ 8722 – 14 = 8708 ⇒ 870 – 16 = 854 \ 85 – 8 = 77 Since 77 is divisible by 7, so the number 8724107 is divisible by 7. Divisibility by 8 : A number is divisible by 8 if the number formed by the last three digits of the given number is divisible by 8. e.g., 343568 is divisible by 8 because 568 is divisible by 8. Divisibility by 9 : A number is divisible by 9 if the sum of its digits is divisible by 9. e.g., 659241 is divisible by 9 Sum of the digits = 6 + 5 + 9 + 2 + 4 + 1 = 27, since 27 is divisible by 9, so the number is divisible by 9. • 345729 is not divisible by 9 Sum of the digits = 3 + 4 + 5 + 7 + 2 + 9 = 30, since 30 is not divisible by 9, so the number is not divisible by 9. Divisibility by 10 : A number is divisible by 10, if its last digit is ‘0’, e.g., 342490 is divisible by ‘10’. Divisibility by 11 : A number is divisible by 11 if the difference between the sum of its digits at odd places and sum of its digits at even places is either ‘0’ or a number divisible by 11. e.g., (a) 4685835 is divisible by 11? Sol: (Sum of its digits at odd places) – (Sum of its digits at even places) = (4 + 8 + 8 + 5) – (6 + 5 + 3) = 25 – 14 = 11, which is divisible by 11. Hence, the number is divisible by 11.

(b) Is 7686162 divisible by 11? Sol: (Sum of its digits at odd places) – (Sum of its digits at even places) \ (2 + 1 + 8 + 7) – (6 + 6 + 6) = 0, which is divisible by 11. Hence, the number is divisible by 11. (c) Is 349628 divisible by 11? \ (8 + 6 + 4) – (2 + 9 + 3) = 4, which is not divisible by 11. Hence, the number is not divisible by 11.

To Find Unit digit

Unit place digit To find the unit digit in the product of two or more numbers, we take unit digit of each number and then multiply them. If the two digits product is obtained, then consider only unit place digit and neglect the tens place digit. The unit place digit in the last number is the required digit. Example : Find the unit digit of 3867 × 248 × 67 Sol. Multiply unit digit of each number: 7 × 8 × 7 56 × 7 42 Condensible only unit digit in 42, Unit place digit is 2. Hence, the unit digit of 3867 × 248 × 67 is 2.

Last digit of a product [Powers]

Last digit of the product a × b × c ….. is the last digit of the product of last digits of a, b, c. Last digit of a product of power numbers Consider 2n, when n is 1, 21 = 2 when n is 2, 22 = 4 when n is 3, 23 = 8 when n is 4, 24 = 16 when n is 5, 25 = 32 After 4th power of 2, unit digit repeats. when n is 6, 26 = 64 The last digit of 2n repeats after four consecutive powers, in the order 2,4,8,6. Similarly, 3n also repeats after four consecutive powers, in the order 3, 9, 7, 1. For 4n, it repeats after two consecutive powers in the order 4, 6 For 5n, the last digit remains same for all, i.e., 5. For 6n, also the last digit is 6 and it is same for all the values of n. Pattern of 2n : 2, 4, 8, 6 Pattern of 3n : 3, 9, 7, 1 Pattern of 4n : 4, 6 Pattern of 5n : 5 Pattern of 6n : 6 Pattern of 7n : 7, 9, 3 Pattern of 8n : 8, 4, 2, 6 Pattern of 9n : 9, 1

Factorials

Let ‘n’ be the natural number then ‘n’ factorial is denoted by n! n! = 1 × 2 × 3 × ..... × n Example : 1 × 2 × 3 × 4 × 5 = 5! = 120 Note :To find the number of trailing zeros at the end of n!. We use the shortcut method, n  n   n   n  Number of zeros in n! =   +  2  +  3  +  4  +...., where [ ] 5 5  5  5  represents a function called greatest integer function. Example : [3] = 3, [4.3] = 4, [–1.5] = –2, etc. Example : Find the number of zeros in 136 !  136   136   136   136 

Sol. Number of zeros =  + +  + ... +  5   5 2   53   5 4  \ 27 + 5 + 1 = 33 zeros Hence, 136 ! has 33 zeros.

232

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions x = 8 and y = 4 So, product = 8 × 4 = 32

1. The difference between two numbers is 4 and there average is 6. The product of these numbers is: (CUET 2023) (1) 24 (2) 12 (3) 32 (4) 48 2 4 7 1 , is largest?(CUET 2023) 2. Which fraction among , , 3 5 11 3 2 7 4 1 (2) (3) (4) (1) 3 11 5 3

2. Option (3) is correct. 2 4 7 1 Given fractions = , , , 3 5 11 3

3. The sum of the numerator and the denominator of a fraction is 11. If 1 is added to the numerator and 2 is subtracted from the 2 (CUET 2023) denominator, it becomes . The fraction is: 3 5 6 3 8 (2) (3) (4) (1) 6 5 8 3

So, the largest fraction is

3

4. The value of ( 625) is: (CUET 2023) (1) 110 (2) 25 (3) 125 (4) 115 5. Which one of the following numbers is not a prime number? (CUET 2023) (1) 241 (2) 337 (3) 391 (4) 571 6. If 2433x = 27(4x – 1), then the value of 4x is: (CUET 2023) 1 (1) 4 (2) ×4 (3) (4) 1 4 16 7. If ‘a’ is a natural number then (7a2 + 7a) is always divisible by (1) 7 and 14 both (2) 7 only (3) 14 only (4) 21 only 8. If the number 59a44b is divisible by 36, then the maximum value of a + b is (1) 16 (2) 12 (3) 14 (4) 10 9. What should be the value of N to make 396258N divisible by 8 ? (1) 2 (2) 8 (3) 4 (4) 6 10. If the 8-digit number la765b12 is to be divisible by 72, the least value of 2a + 3b is (1) 10 (2) 9 (3) 12 (4) 11 11. Which of the following number is divisible by both 7 and 11 ? (1) 16,324 (2) 12,235 (3) 16,257 (4) 16,425 12. Which number is divisible by both 9 and 11 ? (1) 10,089 (2) 10,098 (3) 10,108 (4) 10,087 13. If the 8-digit number 43A5325B is divisible by 8 and 9, then the sum of A and B is equal to (1) 12 (2) 18 (3) 14 (4) 15 14. If 8-digit number 4432A43B is divisible by 9 and 5, then the sum of A and B is equal to (1) 12 (2) 5 (3) 7 (4) 8 15. If the 8-digit number 342x18y6 is divisible by 72, then what is the value 9x + y , for the largest value of y ? 4

(1) 2 7

(2) 4 7

(3) 8

(4) 6

Answer Key 1. (3)

2. (3)

3. (3)

4. (3)

5. (3)

6. (3)

7. (1)

9. (3)

10. (4) 11. (1) 12. (2) 13. (3) 14. (3) 15. (4)

8. (3)

Answers with Explanations 1. Option (3) is correct. Let the numbers be x and y.

According to the question, x−y=4 x+y and =6 2 ⇒ x + y = 12 By equation (1) and (2),

(1) (2)

LCM of denominators = 3 × 5 × 11 = 165 Now fractions =

110 132 105 55 , , , 165 165 165 165 132 4 or 165 5

3. Option (3) is correct.

Let the denominator and numerator of given number are x and y. According to the question, x + y = 11 (1) x +1 2 = and (2) ⇒ 3x − 2 y = −7 y−2 3 from equation (1) and (2), x = 3, y = 8 3 So, fraction = 8

4. Option (3) is correct.

Given expression =

4

( 625 )

(

3

= ( 625)3

)

1/4

(

= 5 4×3

)

1 4

= 53 =125

5. Option (3) is correct. A number that can be divided exactly only by itself and 1 is called a prime number. Here, 391 is not a prime number as it is divisible by 1, 17, 23 and 391. 6. Option (3) is correct. Given: (243)3x = (27)(4x – 1)

( )

⇒ 35 ⇒

3x

( )

= 33

( 4 x − 1)

315 x = 312 x − 3

Comparing both sides, 15x = 12x – 3 ⇒ 3x = −3 1 4 x = 4 −1 = ⇒ x = −1 So, 4 7. Option (1) is correct. Given: 7a2 + 7a, a is a natural number. 7a2 + 7a = 7a(a + 1) \ It is always divisible by 7. Suppose ‘a’ is even, then it is divisible by ‘2’. i.e., 7 × 2 = 14, The number is divisible by 14. Suppose ‘a’ is odd, then ‘a + 1’ becomes even once again and it is divisible by 14. Therefore, the given number is divisible by 14 and 7. 8. Option (3) is correct. Given: 59a44b is divisible by 36. 36 = 9 × 4, (9, 4) are co-primes 5 + 9 + a + 4 + 4 + b = 22 + a + b, which is divisible by 9. Now, if a + b = 16, then 16 + 22 = 38 which is not divisible by 9. a + b = 14, then 14 + 22 = 36 which is divisible by 9. \The maximum value of a + b is 14. 9. Option (3) is correct. Given: 396258N is divisible by 8. The number is divisible by 8 if its last three digits are divisible by 8.

233

QUANTITATIVE APTITUDE: NUMBER SYSTEM

∴ 58N must be divisible by 8 for N = 4 Since 584 is divisible by 8, therefore, N = 4 10. Option (4) is correct. Given: 1a765b12 is divisible by 72. 72 = 9 × 8, so the number is divisible by 9 and 8 also. Now, 1 + a + 7 + 6 + 5 + b + 1 + 2 = 22 + a + b must be divisible by 9. 22 + a + b = 27 (nearest divisible by 9 number) ∴ a+b=5 Since, the number is divisible by 8, last three digit of number must be divisible by 8. We can choose b as 1, 3 or 5, since 112, 312 and 512 are divisible by 8. Least value of (2a + 3b) so choose b = 1, a = 4 \ (2a + 3 b) = (8 + 3) = 11 11. Option (1) is correct. Shortcut Method : To check divisibility by 11 (Sum of digits in odd places) – (Sum of digits in even places) Option (1): 16324 ⇒ (4 + 3 + 1) – (6 + 2) = 0 It is divisible by 11. To check divisibility by 7 1632 – 4 (2) = 1624 ⇒ 162 – 4 (2) = 154 ⇒ 15 – 8 = 7, which is divisible by 7. \16324 is the number which is divisible by both 7 and 11. 12. Option (2) is correct. (1 + 0 + 0 + 9 + 8) = 18 which is divisible by 9. (Sum of digits in odd places) – (Sum of digits in even places) (8 + 0 + 1) – (0 + 9) = 0, which is divisible by 11 \10098 divisible by both 9 and 11. 13. Option (3) is correct. Given: 43A5325B is divisible by 8 and 9. Since, the number is divisible by 8. 25B must be divisible by 8. The only option for B is 6. Also, given that the number is divisible by 9. Sum of digits = 4 + 3 + A + 5 + 3 + 2 + 5 + 6 = 28 + A The minimum number to add is 8 i.e., 28 + 8 = 36, which is divisible by 9. Therefore, A = 8 ⇒ A + B = 8 + 6 = 14 14. Option (3) is correct. Given: 4432A43B is divisible by 9. Sum of the digits = 4 + 4 + 3 + 2 + A + 4 + 3 + B = 20 + A + B The minimum number added to 20 to make it divisible by 9 is 7. Therefore, A + B = 7 15. Option (4) is correct. Given: 342x18y6 is divisible by 72. So, it is divisible by factors of 72 also. Since, 8 and 9 are the factors of 72. Last three digit number is divisible by 8 i.e., 8y6 ÷ 8 Then by inspection possible values of y are 1, 5, and 9. Choosing the largest value for y that is 9. To find x, The number is divisible by 9 if its sum is divisible by 9: \ (3 + 4 + 2 + x + 1 + 8 + 9 + 6) ÷ 9 (33 + x) ÷ 9 \ x=3 Now,

9 x + y = 9 ( 3 ) + 9 = 27 + 9 = 36 = 6

Division Algorithm, Arithmetic and Geometric Progressions

Topic-3

 Revision Notes Division algorithm Let ‘a’ and ‘b’ be two integers such that b ≠ 0. On dividing a by b, we get ‘q’ and ‘r’ as quotient and remainder, respectively. Then, the relationship between a, b, q and r is established as a = bq + r. i.e., Dividend = Divisor × Quotient + Remainder For example, 37 ÷ 4 4 Divisor

9 37 36 1

Quotient Dividend Remainder

37 = (4) × (9) + (1).

Results (an – bn) is divisible by (a – b) for all value of n. (an – bn) is divisible by (a + b) for all even value of n. (an + bn) is divisible by (a + b) for all odd values of n. If any number is divisible by two co-prime numbers then it is divisible by their product. If any two numbers are divisible by ‘n’ then the sum of those two numbers and difference of those two numbers are divisible by ‘n’.

Arithmetic Progression Definition : A sequence of terms in which each term differs from the preceding one by a constant quantity. The constant quantity is called common difference. A.P. can be represented as a, a + d, a + 2d, ..., etc., where a is called as first term and d is called as common difference.

Formula

Tn = a + (n – l)d, where Tn is the nth term. Sn =

n  2 a + ( n − 1 ) d  , where Sn is the sum of first terms. 2

Sn =

n [ a + l ] , where last term ‘l’ is given. 2

Geometric Progression

Definition : A sequence of terms in which the ratio between any two successive terms is the same. G.P. can be represented as, a, ar, ar2, …, etc. where, a is the first term and r is called the common ratio.

Formula Tn = arn – 1 Sn = Sn =

(

a 1 − rn 1−r

(

a rn − 1 r −1

Note: S¥ =

) , where r < 1 ) , where r > 1

a where –1 < r < 1 1−r

Points to remember

• Sum of first n natural numbers = 1 + 2 + 3 + … + n =

∑n =

n ( n + 1) 2

• Sum of the squares of first n natural numbers = 12 + 22 + 32 + … + n2 = ∑ n2 =

n ( n + 1) ( 2n + 1) 6

234

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Arithmetic mean

• Sum of the cubes of first n natural numbers = 13 + 23 + 33 + … + n3 = ∑n = 3

n2 ( n + 1)

• If a, b, c are in A.P., then

2

4

• Algebraic Identities (i) (a + b)2 = a2 + b2 + 2ab (ii) (a – b)2 = a2 + b2 – 2ab (iii) a2 – b2 = (a + b) (a – b) (iv) (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (v) (a3 – b3) = (a – b) (a2 + ab + b2) (vi) (a3 + b3) = (a + b) (a2 – ab + b2) (vii) (a + b)3 = a3 + b3 + 3ab (a + b) (viii) (a – b)3 = a3 – b3 – 3ab(a – b) (ix) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) (x) (x + a) (x + b) = x2 + (a + b)x + ab

a+c =b 2

• For n terms a1, a2, a3, …, an, then A.M. will be a1 + a2 + a3 + ... + an . • A.M. of a, b is a + b .

n

2

Geometric mean • If a, b, c are in G.P. then b2 = ac • For n terms a1, a2, a3, …, an, the G.M. will be n a1 ⋅ a2 ⋅ a3  an . • G.M. of a, b, is ab . • A.M. ≥ G.M.

Objective Type Questions 1. The smallest five digit number which is exactly divisible by 12, 15 and 18 is: (CUET 2023) (1) 10000 (2) 10020 (3) 10080 (4) 10260 2. If a number is divided by 3, the remainder will be 2. If the number is added by 5 and then divided by 3, then what will be the remainder ? (1) 3 (2) 1 (3) 0 (4) 0 3. The value of l + 3 + 5 + 7+ ................ (2n – 1) is (1) (2n – 1) × (2n – 1)

(2)

(3) n × n

(4)

n 2 n ( n + 1) 2

4. In a question on division, the divisor is 6 times the quotient and 3 times the remainder. If the remainder is 40, then find the dividend. (1) 2455 (2) 2450 (3) 2440 (4) 2445 5. Given that 220 + 1 is completely divisible by a whole number, which of the following is completely divisible by the same number ? (1) 215 + 1 (2) 5 × 230 (3) 290 + 1 (4) 260 + 1 6. If 2941 + 3741 is divided by 33, then the remainder is (1) 2 (2) 3 (3) 1 (4) 0 7. 225 + 226 + 227 is divisible by (1) 6 (2) 7 (3) 5 (4) 9 Answer Key 1. (2)

2. (2)

3. (3)

4. (3)

5. (4)

6. (4)

7. (2)

Answers with Explanations 1. Option (2) is correct.

If a five digit number is divisible by 12, 15 and 18. Then it must be divisible by 4, 9 and 5. From given options, only (3) and (4) are divisible by 9, as there sum of digits is 9. Here, option (3) has smaller number of four digit and also divisible by 4 and 5. 2. Option (2) is correct. Let the number be x. Then, x divided by 3 will leave remainder 2. Now, x + 5 is divided by 3. We get, 5+2=7 When we divide 7 by 3 the remainder is 1.

3. Option (3) is correct. The given series is in A.P. n Sn = ( a + l ) [here, last term l is given] 2 ⇒ Sn =

n 1 + ( 2n − 1 )  2

\ Sn = n × n 4. Option (3) is correct. We have, Dividend = Divisor × Quotient + Remainder Given: d = 6q and d = 3r Also given r = 40, then d = 3 × 40 = 120 120 \ = q = 20 6 Now, Dividend = 120 × 20 + 40 = 2400 + 40 = 2440 5. Option (4) is correct. Given: 220 + 1 is divisible by a whole number Let 220 = x Then, 220 + 1 = x + 1 (adding 1 both side) Let x + 1 be completely divisible by the natural number N, then ⇒ (260 + 1) = [(220)3 + 1] ⇒ (x3 + 1) = (x + 1) (x2 – x + 1) ⇒ (260 + 1) = (220 + 1) [(220)2 – 220 + 1] is completely divisible by N, since (x + 1) is divisible by N. 6. Option (4) is correct. Remember the result (a + b) divides an + bn provided ‘n’ is odd. Here, (29 + 37) divides 2941 + 3741 i.e., 66 divides 2941 + 3741 We know that 66 = 33 × 2 \ 33 divides 2941 + 3741 \ Remainder is 0. 7. Option (2) is correct. Given: 225 + 226 + 227 = 225 + 225 . 2 + 225 . 22 = 225 [1 + 2 + 4] = 225 [7] \The number is divisible by 7

Topic-4

Problems on Numbers, H.C.F. and L.C.M.

 Revision Notes In number problems, you are given some clues about one or more numbers and you use these clues to build an equation. Then you have to solve the equation to get the solution.

235

QUANTITATIVE APTITUDE: NUMBER SYSTEM

H.C.F. and L.C.M.

Least Common Multiple (L.C.M.)

If ‘k’ divides ‘l’ and leaves remainder zero. Then we say ‘k’ is a 7 factor of ‘l’. Also ‘l’ is a multiple of ‘k’. Example : 21 ÷ 3

The least number which is exactly divisible by each one of the given numbers are called their L.C.M.

3 21 21 0

Here 3 is a factor of 21 and 21 is a multiple of 3.

Highest Common Factor (H.C.F.) or Greatest Common Divisor (G.C.D.) The H.C.F. of two or more numbers is the greatest common number that divides each number and leaves reminder as zero. Example : H.C.F. (3, 27) = 3 The H.C.F. of 3 and 27 is 3. H.C.F. (12,20) = 4 The H.C.F. of 12 and 20 is 4.

Methods of finding H.C.F.

(i) Factorisation Method : In this method, write the number as a product of prime factors. The product of least powers gives the H.C.F. of two given numbers. (ii) Division Method : In this method, we divide the largest number by the smallest number. Then, we repeat the process of dividing divisor by the remainder until remainder is zero. The last divisor is the required H.C.F. Example : Find the H.C.F. of 64 and 324. (i) Factorisation Method : 2 324 2 64 \ 2 162 2 32 3 81 2 16 3 27 2 8 3 9 2 4 3 3 2 2 1 1 324 = 22 × 34 And 64 = 26 \ H.C.F. (324, 64) = 22 = 4

Methods of finding L.C.M.

(i) Factorisation Method : Express the number in the form of product of primes. Then, L.C.M. is the product of the highest power of all the factors. (ii) Common Division Method : Write down the numbers in a row, then divide by a prime number which exactly divides at least two of the given numbers and carry forward the numbers which are not divisible. Repeat until we obtain 1. The product of the divisors and undivided numbers is the required L.C.M. Example : Find the L.C.M. of 64 and 324. (i) Factorisation Method : 324 = 22 × 34 and 64 = 26 L.C.M. = 26 × 34 = 5,184 (ii) Common Division Method : 2 324, 64 2 162, 32 2 81, 16 81, 8 2 81, 4 2 2 81, 2 81, 1 81 1, 1 = 2 × 2 × 2 × 2 × 2 × 2 × 81 ∴ L.C.M. of 324 and 64 is 5,184 Note : 1. If a and b are two numbers, then a × b = H.C.F. (a, b) × L.C.M. (a,b) H.C.F. of fractions =

3.

L.C.M. of fractions = L.C.M. of Numerators

4.

H.C.F. (a, b, c) =

a × b × c × L.C.M.( a, b, c ) L.C.M.( a, b) × L.C.M.( b, c ) × L.C.M.( a, c)

5.

L.C.M. (a, b, c) =

a × b × c × H.C.F.( a, b, c ) H.C.F.( a, b) × H.C.F.( b, c ) × H.C.F.( a, c)

H.C.F. of Denominators

(ii) Division Method : 5 64 324 320 4

16 4 64 64 0

H.C.F. of Numerators L.C.M. of Denominators

2.

Last divisor is 4. Therefore, H.C.F. (324, 64) = 4

Objective Type Questions 1. Find the L.C.M. of (1) 45

3 4 5 , and is 5 9 8

(2) 55

(3) 60

(CUET 2022) (4) 65

2. The L.C.M. and H.C.F. of two numbers are 35 and 15, respectively. The product of these numbers is ______. (1) 625

(2) 525

(2) 425

(CUET 2022) (4) 325

3. The difference between a number and one-third of that number (2) 58.9

(3) 61.8

(4) 72.5

4. The proportion among three numbers is 3 : 4 : 5 and their L.C.M. is 1800. The second number is (1) 150

(2) 30

(3) 120

(1) 56

(4) 90

(2) 284

(3) 72

(4) 26

(3) 4 , 9

(4) 2 , 9

7. Two fractions are such that their product is 9 and sum is 77 . 10 40 The two fractions (1) 2 , 9

is 228. What is 20% of that number ? (1) 68.4

5. What is the H.C.F. (highest common factor) of 77 and 275 ? (1) 12 (2) 11 (3) 7 (4) 25 7 9 7 6. If of of a number is 126, then of that number is 2 4 2

5 4

(2) 1 , 9

5 2

5 8

5 4

8. lf 6 of 8 of a number is 192, then 3 of that number is 7 4 5 (1) 105

(2) 77

(3) 36

(4) 80

236

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

9. The sum of a non-zero number and thrice its reciprocal is Find the number. (1) 8 (2) 9

(3) 7

10. lf Gaganjyot’s salary is

52 . 7

(4) 6

7 times of Hafiz’s and Sayed’s is 8 6 7

times of Hafiz’s, what is the ratio of Gaganjyot’s salary of Sayed’s salary? (1) 49 : 48 (2) 3 : 4 (3) 4 : 3 (4) 48 : 49 11. In a fraction when 3 is added to its numerator and denominator it becomes 4 . And it becomes 1 when 2 is subtracted from both 5

2

⇒

= x

Then, the second number = 4x = 4 × 30 = 120 5. Option (2) is correct. 77 = 11 × 7 and 275 = 11 × 5 × 5 Hence, H.C.F. of 77 and 275 is 11. 6. Option (1) is correct. Let the number be x. According to the question, 9 of 7 of x = 126 4 2 9 of 7 x = 126 ⇒ 9 × 7 x = 126 4 2 4 2

⇒

the numerator and denominator. Find the fraction. (1)

15 16

(2)

14 16

(3)

11 16

(4)

9 16

⇒

12. What is the value of (81 + 82 + 83 + ......... + 130) ? (1) 5275 (2) 10550 (3) 1585 (4) 21100

1. (3)

2. (2)

3. (1)

4. (3)

9. (3)

10. (1) 11. (3) 12. (1)

5. (2)

⇒ Now,

Answer Key 6. (1)

7. (3)

1. Option (3) is correct. Given numbers are 3/5, 4/9, 5/8. As we know, L.C.M. of Numerator L.C.M. of fraction = H.C.F. of Denominator L.C.M.( 3, 4 , 5) 60 = = = 60 H.C.F.( 5, 8 , 9 ) 1

2. Option (2) is correct. Given that, 1st number = 35 2nd number = 15 As we know, Product of two numbers = L.C.M. × H.C.F. So, L.C.M. × H.C.F. = 35 × 15 = 525 3. Option (1) is correct. Let x be the number. According to the question, 1 x − x = 228 3 2 ⇒ x = 228 3 228 × 3 ⇒ x= 2 \ x = 342 Now, 20% of x = 20% of 342 20 × 342 = 68.4 = 100 4. Option (3) is correct. Given: Three numbers are in the ratio 3 : 4: 5 Then, the three numbers are 3x, 4x and 5x and their L.C.M. is 1800. So, L.C.M. of 3x, 4x and 5x = 1800 60x = 1800

63x = 126 8 126 × 8 x= = 16 63 7 7 7 of x = × x = × 16 = 56 2 2 2

Shortcut Method : Let the number be x. 126 × 2 × 4 7 ∴x= × = 14 × 4 = 56 9×7 2

8. (1)

Answers with Explanations

1800 = 30 60

7. Option (3) is correct. Let the fractions be x and y. According to the question, 9 10 77 x+y = 40

x×y = and

...(i) ...(ii)

From equation (i) and (ii), we get 9 9  77   77 − 40 x  ⇒ x = x  − x =  40   40  10 10

⇒ ⇒ ⇒ ⇒ ⇒

40x2 – 77x 40x2 – 32x – 45x + 36 8x(5x – 4) –9(5x – 4) (8x – 9) (5x – 4)

= –36 =0 =0 =0

x=

⇒ Shortcut Method : From the option (iii) and

9 4 , 8 5

4 9 36 9 × = = 5 8 40 10

4 9 32 + 45 77 + = = 5 8 40 40

8. Option (1) is correct. Let x be the number then Given: ⇒ ⇒ ⇒ ⇒ Now,

6 of 8 of x = 192 7 5 6 of 8 x = 192 7 5 6 × 8 x = 192 7 5 48 x = 192 35 192 × 35 6720 = = 140 x= 48 48 3 3 of x = ×140 = 3 × 35 = 105 4 4

...(iii)

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QUANTITATIVE APTITUDE: NUMBER SYSTEM

9. Option (3) is correct. Let x be a non-zero number, 1

⇒

52

Multiplying through out by 7x ⇒ 7x2 + 21 = 52x ⇒ 7x2 – 52x + 21 = 0 ⇒ 7x2 – 49x – 3x + 21 = 0 ⇒ 7x (x – 7)–3(x – 7) = 0 ⇒ (7x – 3)(x – 7) = 0 x = 7 or x =

Choosing answer from the options \ x = 7 is the required number. 10. Option (1) is correct.

16 3 11 a 3 11 = = b 16 16 3

⇒

b=

3 7

Given: Gaganjyot's salary = 7 Hafiz's salary and 6

Sayed's salary =

11 3

From equation (i) Now, 5a – 4b = – 3 55 ⇒ − 4b = −3 3

Given: x + 3   = x 7

⇒

a=

8 Hafiz's salary 7

7 Gaganjyot's salary 6 7 7 49 = = × = ⇒ 8 6 8 48 Sayed's salary 7

\ Gaganjyot’s salary : Sayed’s salary = 49 : 48 11. Option (3) is correct. Let a be the required fraction. b a+3 4 a−2 1 = and According to the question, = b+3 5 b−2 2

⇒ 5a + 15 = 4b + 12 and 2a – 4 = b – 2 ⇒ 5a – 4b = –3 ….(i) and 2a – b = 2 ...(ii) Multiplying equation (ii) by 4, then subtract from equation (i), 5a − 4 b = −3 8a − 4b = 8 − + − −3a = −11

\ The required fraction is

11 . 16

12. Option (1) is correct. Given: 81 + 82 + 83 +… + 130. We have (1 + 2 + 3 + …+ 130) – (1 + 2 + … + 80) = [81 + 82 + … + 130] (Sum of 130 numbers) – (Sum up to 80) = (Sum from 81 to 130) Sum of first ‘n’ numbers is n ( n + 1) 2

= 130 ( 130 + 1) − 80 ( 80 + 1) 2

2

= 8515 – 3240 = 5275 81 + 82 + … + 130 = 5275.

∴

Shortcut Method : Here, ∴

n = 50 [ 81 + 82 +....... 130] a = 81, l = 130 (l = Last term) Sn = n (a + l) 2

= 25(81 + 130) = 5275

• Recurring decimals • Non - recurring decimals. • Decimal fractions

Decimals

• Proper fractions, improper fractions, like fractions, unlike fractions, mixed fraction, continuous fraction

Types of Fractions

a ; a, b ∈I b and b ≠ o

Fractions

•

•

•

1 m

a = an

(m p )

n

=p

First Level

Second Level

Trace the Mind Map

•

•



Rules of Surds

am m−n • n a a

m n m+n • a ×a = a

Third Level

Method of finding • Prime factorisation method

Methods of finding • Prime factorisation method • Division method

Rules of Indices

l = k3 l is cube of k k is cube root of l

l = k2 l is square of k k is square root of l

Square Root and Cube Root

5 + 11 + 19 + 29 + 49 = 3

_ 1.414 2∼ _ 3 ∼ 1.732

Some Root Value

Simplification

V means Vinculum or bar. (The bar you see on the top of the values x) B means Brackets - ( ) {} [ ] O means of [i.e., Powers and square roots, etc.] D means Division [÷] M means Multiplication [×] A means Addition [+] S means Subtraction [−] a, a ≥ 0 Modulus: a =  −a, a < 0

V BODMAS Rule

238 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

Simplification

2

Chapter Analysis Concept Name BODMAS Rule, Square Root and Cube Root

2021

2022

2023

1

2

1

Additional Questions

Fraction and Decimal Fraction, Surds and Indices

Types of Simplification

• BODMAS Rule, Square Root and Cube Root • Fraction and Decimal Fractions, Surds and Indices

= 2 × 5 of 6 × 1 × 51  3  = 2 × 5 of [2 × 51]

Topic-1 BODMAS Rule, Square Root and Cube Root

= 2 × 5 × 102 = 1020

 Revision Notes

Modulus of Real Number

BODMAS Rule, Square Root, Cube Root, Fraction and Decimal Fraction, Surds and Indices.

BODMAS Rule

This is the rule to simplify the problems with many operations (+, –, ×, ÷ of, etc.) BODMAS rule helps us to solve the maths problems easily without error. Method of Solving : Whenever we solve math problems with many operations, we strictly follow the order. V → Vinculum (or Bar) B → Bracket [first of all remove all the brackets in the order ( ), { }, [ ] O → of [i.e., we multiply while solving ‘×’] D → Division M → Multiplication A → Addition S → Subtraction Note : To solve ‘of ’, it is replaced with ‘×’ sign, then we do multiplication of the numbers, e.g., 2 of 8 = 2 × 8 = 16 To solve the division (÷), we replace it by ‘×’ and we write the reciprocal of the second number. Then we multiply those two numbers. e.g., 28 ÷ 7 1 28 × = 4 7

Solving the Problem with BODMAS

{

(

)}

Example: 2 × 5 of 6 ÷ 3 6 × 10 + 21 ÷ 7 14 − 24 + 7    (First solve the bar ‘–––‘) 2 × 5 of [6 ÷ 3 {6 × 10 + 21 ÷ 7 (– 10 + 7)}] (Now solve brackets one by one)   = 2 × 5 of [6 ÷ 3 {6 × 10 + 21 ÷ 7 × (–3)}] [If no symbol is found between number and brackets then it is considered as ‘×’]

=

 1   2 × 5 of 6 ÷ 3 6 × 10 + 21 × × ( −3) 7   

[To perform ‘÷’, write reciprocal of the second term and replace ‘÷’ by ‘×’. Then multiply the numbers] = 2 × 5 of [6 ÷ 3 {6 × 10 + 3 × (– 3)}] = 2 × 5 of [6 ÷ 3 {60 – 9}] = 2 × 5 of [6 ÷ 3 × 51]

6 10

= 2 × 5 of [102][‘of ’ says we multiply the numbers]

Modulus of real number ‘a ’ is defined as a , if a ≥ 0 i.e., a = −a , if a < 0 5 = 5 ( 5 > 0 )

Properties



−5 = 5 ( −5 < 0 , − ( −5 ) = 5 )

• |a| = |– a| • |ab| = |a| |b| • a = a b b

Square Roots and Cube Roots Square : If a number is multiplied with itself then the product is called as the square of a number, e.g., 5 × 5 = 25, 25 is the square number of 5. Perfect Square : If l = k2 where ‘k’ is any natural number, then ‘l’ is called a perfect square, e.g., 49, 64, 81, etc. List of Squares from 1 to 20 List of Cubes from 1 to 20 12 = 1 112 = 121 13 = 1 113 = 1331 2 2 3 12 = 144 2 = 8 123 = 1728 2 = 4 2 2 3 13 = 169 3 = 27 133 = 2197 3 = 9 2 2 3 14 = 196 4 = 64 143 = 2744 4 = 16 152 = 225 53 = 125 153 = 3375 52 = 25 162 = 256 63 = 216 163 = 4096 62 = 36 172 = 289 73 = 343 173 = 4913 72 = 49 182 = 324 83 = 512 183 = 5832 82 = 64 192 = 361 93 = 729 193 = 6859 92 = 81 2 2 3 20 = 400 10 = 1000 203 = 8000 10 = 100 Square Root : If l = k × k = k2, then we say k is the square root of 2 ‘l’ denoted as l = k , e.g., 5 = 25, 25 = 5 We say square root of 25 is 5.

Finding Square Root of a Number

Prime factorisation method : • Write the number in the form of the product of primes. • Pair the same prime factors. • Now, we have to find the product of prime numbers taken one number from each pair of prime numbers. • The product is the required square root of a number.

240

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Example : Find the square root of 4096. Sol. 4096 = \

Cube Root : If l = k3, where k is any natural number, then we say ‘k’ is the cube root of ‘l’. e.g.,27 = 3 × 3 × 3 = 33 So, we say ‘3’ is the cube root of 27.

2×2×2×2× 2×2 ×2×2 ×2×2 ×2×2 ↓ ↓ ↓ ↓ ↓ ↓ 2 × 2 × 2 × 2 × 2 × 2 = 64

Method of Finding Cube Roots

64 is the square root of 4096

Example : Find the square root of 784. Sol. 784 =

Prime factorisation method • Write the numbers in the form of product of primes.

2×2×2×2× 7×7 ↓ ↓ ↓ 2 × 2 × 7 = 28

• Make a group of three same prime numbers [prime factors].

We say, 28 is the square root of 784.

Division Method

• Pair a number using Bar’s from unit place. If single digit left (without pair) we call it as period. • Now, find a perfect square number which is less than or equal to the first period. • Write the quotient and remainder for chosen divisor as in division. • Now add unit place value of divisor to next divisor then shift it to left leaving right side vacant. Now search for a number which is largest perfect square near to remainder. • Now repeat the same process for other period. • Repeat until remainder is 0. • Now the number in the quotient place is the required square root of the number. Example : Find the square root of 4096 Sol. 64 6 4096 6 36 124 496 496 0

Example : To find the square root of 784 Sol. 28 2 784 2 4 48 384 384 0

Cube: If a number is multiplied thrice then the product is a cube of that number. i.e., a × a × a = a3 e.g., 2 × 2 × 2 = 8 = 23 8 is the cube of 2

• Find the product of prime numbers taken one number from each group of prime numbers. • The product is required cube root of the given number. Example : Find the cube root of 1728. 1728 =

2×2×2×2×2×2× 3×3×3 ↓ ↓ ↓ 2 2 × × 3 = 12

\ 12 is the cube root of 1728. Note : • Square of a number ending with 5. Suppose p is the left number of 5 (5 is in unit place) Then, we write the square as p × (p + 1) 25 e.g., (275)2 = (27 × 28) 25 = 75625 • Square of any number can be calculated by using (a + b)2, (a – b)2 and a2 – b2 formulae. Example : Find the square of 86. Sol. (80 + 6)2 = (80)2 + (6)2 + 2 × 80 × 6 = 6400 + 36 + 960 = 7396 • Suppose a and b are two consecutive factors of x such that ab = x and a < b, then (a) If y =

x + x + x +  ∞ , then y = b

(b) If y =

x − x − x −  ∞ , then y = a

(c) If y = x i x i x ∞ , then y = x (d) If y

x i x i x  k , then y = ( x )

2 k −1 2k

Objective Type Questions 1. Which should replace the question mark in below term?

2. 3. 4. 5. 6.

7.

(4.25 + 2.75)2 + ? = 53 – (9 × 8) (1) 3 (2) 4 (3) 5 The value of 3 – 3 ÷ 3: (1) 2 (2) 4 (3) 0 The value of xa – b × xb – c × xc – a is : (1) 1 (2) 0 (3) –1 What is the value for 25 – [9 – {6 + (10 – 8)}] (1) 20 (2) 21 (3) 22 8.2 × 7.5 × 9.3 = _________ (1) 175.95 (2) 375.95 (3) 571.95 What will come at place of x, (x < 10) for (132 ÷ 12 × x − 3 × 3 ) =1 52 − 6 × 4 + x 2 (1) 2 (2) 4 (3) 3 Evaluate : [7 + 7 × (7 + 7 ÷ 7)] + 7 ÷ 7. (1) 10 (2) 5 (3) 63

 289  3 2 2  1 18 17  3 10. If A =  of 4 ÷ +  of  ÷  +   , then the value 5 25 24  7  16  4 3  

(4) 24 (CUET 2021) (4) 751.95

Answer Key

(4) 1 (4) 64

8. If x is the square of the number when  2 of 6 1 ÷ 3  of 1 2 is 4 7 7 5

(2) 16

(3) 9

(1) 14

(2) 15

(3) 17

(4) 4

(4) 16

of 8A is (1) 231

(2) 321

1. (2)

2. (1)

9. (4)

10. (1)

3. (1)

4. (4)

(3) 132

5. (2)

6. (4)

(4) 213

7. (4)

Answers with Explanations 1. Option (2) is correct. Given: ( 4.25 + 2.75)2 + ? = 53 − ( 9 × 8 )

⇒ 7 2 + ? = 125 − 72 ⇒ ? = 4 2. Option (1) is correct. 3–3÷3=3–1=2

1 4

divided by 11 , then the value of 81x is : (1) 36

9. Simplify the expression 25 – [16 – {14 – (18– 8 + 3 )}].

(CUET 2023) (4) 6 (CUET 2022) (4) 1 (CUET 2022) (4) x

8. (1)

241

QUANTITATIVE APTITUDE: SIMPLIFICATION

3. Option (1) is correct. xa b xb c xc a xa x0

1

4. Option (4) is correct. 25 – [9 – {6 + (10 – 8)}] = 25 – [9 – {6 + 2]] = 25 – [9 – 8] = 25 – 1 = 24 Hence, option (4) is correct. 5. Option (2) is correct. The given expression will be solved by multiplying numbers one by one. There are 2 total decimal places in first numbers. Ignore the decimal places and complete the multiplication as operating on two integers. Thus, 82 × 75 = 6150 Hence, rewrite the product with 2 total decimal places = 61.50 Now, we will multiply 61.50 with 9.3 There are 3 total decimal places in both numbers. Ignore the decimal places and complete the multiplication as if operating on two integers. Thus, 6150 × 93 = 571950 Rewrite the product with 3 total decimal places. Therefore, 61.50 × 9.3 = 571.950 Hence, 8.2 × 7.5 × 9.3 = 571.950 6. Option (4) is correct. Given: ⇒

(132 ÷ 12 × x − 3 × 3 )

( 52 − 6 × 4 + x 2 )

=1

(11 × x − 3 × 3 ) 25 − 24 + x 2

⇒

(11x − 9 ) 1 + x2

4

Now,

81x = 81 ×

4 = 36 9

9. Option (4) is correct.

{ (

)}

Given: 25 − 16 − 14 − 18 − 8 + 3    = 25 – [16 – {14 – (18 – 11)}] = 25 – [16 – {14 – 7}] = 25 – [16 – 7] = 25 – 9 = 16 10. Option (1) is correct. 3 7

1 5

Given: A =  of 4 ÷

18 17  + of 25 24 

 3 21 18 17  ⇒ A =  × ÷ +  of  7 5 25 24 

 289  3 2 2   ÷ +    16  4 3  

 289  9 + 8 2   ÷    16  12  

 9 25 17   289 289  +  of  ÷   5 18 24   16 144 

⇒ A= × 5 2

⇒ A= +

17   289 144  of × 24   16 289 

 60 + 17   of 9  24 

=1

77  231 × 9 = 8  24 

⇒ A = 

=1

Now, 8 A = 8 ×

7. Option (4) is correct. Given: [7 + 7 × (7 + 7 ÷ 7)] + 7 ÷ 7  1   = 7 + 7 ×  7 + 7 ×   + 7 ÷ 7 7    = [7 + 7 × 8] + 7 ÷ 7

Topic-2

231 = 231 8

Fraction and Decimal Fractions, Surds and Indices

 Revision Notes Number Analogy

a

Fraction: A number expressed in the form b , where a and b are integers, b ≠ 0. Here a and b are numerator and denominator, respectively.

Types of Fraction

= [ 7 + 56] + 7 ÷ 7 = 63 + 7 ÷ 7

• Proper fraction : When numerator is less than denominator,

= 63 + 1 = 64

then the fraction is called proper fraction, e.g., 2 , 8 , 98

5 10 125

1 3 2 2 Consider  of 6 ÷  of 1 4 7 7 5 2

2

2

x =   = 9 3

∴

⇒ A=

⇒ 11x – 9 = 1 + x2 ⇒ x2 + 1 – 11x + 9 = 0 ⇒ x2 – 11x + 10 = 0 ⇒ (x – 10) (x – 1) = 0 ⇒ x = 10 or x = 1 but given (x < 10). Therefore, x must be 1.

8. Option (1) is correct.

15 45 15 4 2 ÷ = × = 2 4 2 45 3 2 Given: x is the square of 3

=

b b c c a

• Improper fraction : When numerator is greater than denominator, then the fraction is called improper fraction. 25

3

9

=  of ÷  of 4 7 7 5  2 25 3  9 =  × ÷  of 5 4 7 7

5 7 9 =  ×  of = 35 × 9 = 15 2 3 7 6 7 2 1 Now, divide the answer by 11 4 15 1 = ÷ 11 2 4

etc. e.g., 28 , 39 ,etc . 17 20

• Like fraction : A number of fractions having same denominator etc. is called like fraction, e.g., 2 , 5 , 11 ,etc . 9 9 9

• Unlike fractions : A number of fractions having different etc. denominator are called unlike fractions, e.g., 7 , 5 , 7 ,etc . 9 11 17

• Mixed fraction : Those fractions which contain a whole number and a proper fraction are known as mixed fraction. e.g., 1 2 , 17 14 etc , etc. . 9

45

242

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• Continued fraction : It contains an additional fraction in the numerator or in the denominator, e.g., 1 +

1

1+

Operations on Fractions:

Laws of Indices : n

a an •   = n b

• am × an = am + n

1 2

a

•

m

an

 

= am −n

( )

Addition : Make fractions like fractions by taking L.C.M. of the denominators, then add the numerators of the fractions,

m • a

e.g., 2 + 8

• (a × b) = a × b

7

n

= amn n

(7, 11) = 77

∴

22 + 56 78 = 77 77

n

• a −m =

n

1 am

Method of Arranging Surds in Increasing or

11

L.C.M.

b

• a0 = 1 • ax = ay then x = y

Decreasing Order Let

m

p

a , n b and c be three surds. Find the L.C.M. of m, n and p

Subtraction : Make fractions like fractions, then subtract the

and make the surds of the same roots. Then arrange the numbers in

numerators of the fraction, e.g., 2 − 6 5 7

order, e.g.,

5, 2 2, 4 5

Now, consider 3, 2 and 4. L.C.M. (3, 2, 4) = 12.

L.C.M. (5, 7) = 35 14 − 30 −16 = 35 35

∴

3

1

Multiplication : To multiply two fractions, multiply the numerator of one with numerator of another and multiplying the denominator of one with denominator of another fraction, e.g., 2 × 6 = 12 5 7 35 Division : To divide first fraction with second fraction, we write the reciprocal of the second fraction and change the sign ‘÷’ by ‘×’, 2 6 2 7 1 e.g., √ = × = 21 7 21 6 9 Decimals : A decimal number can be defined as a number whose whole number part and the fractional part are separated by a decimal point. e.g., 38.7 is a decimal number. 38 is whole number and 7 is the fractional part.

Types of Decimal Numbers

1 4

1

1

(5)3 , (2)2 , (5)4

We can write

1 6

1 3

( 5 ) 3× 4 , ( 2 ) 2× 6 , ( 5 ) 4 × 3 4

6

3

( 5 )12 , ( 2 )12 , ( 5 )12 12 4 12 6 12 3

5 ,

12

2 ,

5

625 , 12 64 , 12 125

Now, increasing order of the surds is 12

64 , 12 125 , 12 625

Operation on Surds Addition and Subtraction • To add two surds, its nth root and number inside the root must be

1. Recurring decimal numbers : Repeating or non-terminating decimals, e.g., 3.121212 2. Non-recurring decimal numbers : Non-repeating or terminating decimals, e.g., 3.237654…

Decimal fraction

But

Surds and Indices

3

5 + 73 5 = 83 5

5 and 8 5 , cannot be added as two surds have different

roots. 3

In decimal fractions, the denominator is a power of 10 such as 100, 1000, etc.= e.g., 1 0= .3333 0.3 3

3

same, i.e.,

5 + 8 3 3 , cannot be added as these two surds have different

numbers inside the root. • Subtraction is same as addition. Multiplication and Division of Surds To multiply two surds, we make root nth root the same. Then, we

Surds are the roots of numbers which cannot be simplified into a whole or rational number, e.g., 2 , 5 , 7 , etc

multiply the numbers inside the root.

Note:

Example :

3

5 × 3 7 = 3 5 × 7 = 3 35

Example :

3

3 × 5 2 we can’t multiply directly.

9 is not a surd because

We write

n

9 = 3.

1

a = a n , it is called nth order surd.

It is read as nth root of ‘a’, e.g.,

1

3

5 is read as cube root of 5.

7

17 is read as 7th root of 17.

Indices: It is defined as the power of exponent which is raised to a number or variable.

Laws of Surds: • • •

n

1

a = (a)n

(n a ) n

n

a na = b nb

•

n

 1 =  an  = a    

•

(n a )

n

•

n

n

ab = a × b

m

mnp

= (a)

= a

= a a

1 5

1 3

( 3 ) 3× 5 × ( 2 ) 5× 3

5

3

=

15 5

15 3

=

15

= ( 3 ) 15 × ( 2 ) 15 3 ×

2

243 × 15 8 = 15 243 × 8 = 15 1944

Remember = 2 1= .414 , 3 1= .732 , 5 2.236 , 7 = 2.646 Important identities.

m n

mnp

2

But ( 3 ) 3 × ( 2 ) 5 , LCM (3, 5) = 15

• (a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c) (c + a) 1 mnp

• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca. • a3 + b3 + c3 – 3abc = (a + b + c) [a2 + b2 + c2 – ab – bc – ca]

243

QUANTITATIVE APTITUDE: SIMPLIFICATION

Objective Type Questions 1. lf the numerator of a fraction is increased by 15% and denominator is decreased by 20%, then the fraction, so

Given:

obtained, is 17 . What is the original fraction ? 65

(1)

281 1495

(2)

278 1495

(3)

267 1495

(4)

272 1495

⇒

2. By adding 3 and 5 in numerator and denominator, respectively 2 . If 1 and 3 are subtracted and added 3 from numerator and denominator, respectively, it becomes 2 . 5

of a fraction, it becomes

Find the fraction. (1)

5 7

(2)

6 7

(3)

7 6

(4)

⇒

+ 0.036

(3) 36

(1) 2.1014

(2) 2.1973

Let

(4) 30

(3) 1.9876

(4) 1.9996

−6

−3

3 x +7 )  5  2 ( 2 x +3 )  4  3 (   2 2     3     =    5. If         3 3                

then the value of 2 − 42 x is : (1) 5 (2) 6 (3) 3 6. Find the value of

(2) 4 a+3

a+6

(3) 6

(4) 5

a +1

8. Find the value of x in 3 15625 − x = 4 . (1) 625 (2) 343 (3) 441 9. Evaluate the following.

(4) 81

3

9. (1)

10. (3)

4. (2)

5. (1)

b

7

(0.1)3 +(0.2)3 +(0.3)3 +3(0.005+0.016+0.027)+0.036 (0.1)2 +(0.2)2 +(0.3)2 +0.04+0.06+0.12

3

6. (3)

7. (3)

Answers with Explanations 1. Option (4) is correct. a Let be the fraction.

\ The required fraction is 5 .

We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca ...(i) (a + b + c)3 = a3 + b3 + c3 + 3(a + b) (b + c) (c + a) = a3 + b3 + c3 + 3[a2(b + c) + b2(a + c) + c2(a + b)] + 6abc …(ii) Comparing formula and given problem a = 0.1, b = 0.2 and c = 0.3 [Numerator is in the (ii) form and denominator is in (i) form]

Answer Key 3. (3)

a =5 Substituting a = 5 in (i), we get 3(5) – 2b = 1 ⇒ 15 – 2b = 1 ⇒ 2b = 14 ⇒ b =7

A=

(1) 20 (2) 25 (3) 30 (4) 15 10. If x = 3 − 5 ,then the value of x2 – 16x + 6 is : (1) 0 (2) 4 (3) 2 (4) – 2

2. (1)

5(a – 1) = 2(b + 3) 5a – 5 = 2b + 6 5a – 2b = 11 …(ii)

3. Option (3) is correct.

(12.5 ) + ( 7.5 ) (12.5 )2 + ( 7.5 )2 − 12.5 × 7.5

1. (4)

a −1 2 = b+3 5

and

⇒ 3(a + 3) = 2(b + 5) and ⇒ 3a + 9 = 2b + 10 and ⇒ 3a – 2b = 1 …(i) and Solving (i) and (ii), we get,

(4) 4

× 25 4 , then the value of a + 9 is : 7. If 3 × 4 = 27 a −1 × 8 a − 2 × 125a + 4 1526 (1) 4 (2) 6 (3) 5 (4) 8

3

a+3 2 = b+5 3

3a − 2b = 1 5a − 2b = 11 ( − )( + ) ( −) −2 a = −10

30 + 30 + 30 + 30 + ⋅⋅⋅... ... ... ... ∞

(1) 3

a be the fraction. Then b

According to the question,

1 . x

4. If x = 1 + 2 , then find the vahe of x +

15 a 100 = 17 20 65 b− b 100 100 a + 15a 17 100 = 100b − 20b 65 100 115a 17 = 80b 65 115 a 17 × = 80 b 65 a 17 80 = × b 65 115 a 272 = b 1495 a+

\ 2. Option (1) is correct.

( 0.1)2 + ( 0.2 )2 + ( 0.3 )2 + 0.04 + 0.06 + 0.12

then the value of 60A is : (1) 20 (2) 60

⇒ ⇒

7 5

( 0.1)3 + ( 0.2 )3 + ( 0.3 )3 + 3 ( 0.005 + 0.016 + 0.027 ) 3. If A =

⇒

a + 15% of a 17 = b − 20% of b 65

8. (3)

Now, A = ( 0.1 + 0.2 + 0.3 ) = 0.1 + 0.2 + 0.3 = 0.6

( 0.1 + 0.2 + 0.3 )2

Now, 60A = 60 × 0.6 = 36 4. Option (2) is correct. Given: x = 1 + 2 ∴ 1 = 1 × 1 − 2 = 1 − 2 = −1 + 2 x 1+ 2 1− 2 −1

244

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)



Consider,  x + 

2

1   = x

=x+

( x)

2

1 + 2 = 1+ 2 −1+ 2 + 2 = 2 + 2 2 x 2

(

1    x+  = 2 1+ 2 x 

∴

2

1  1  +  +2 x× x  x

)

Given:

Taking square root on both the sides, we get x+

1

x

(

= 2 1+ 2

)

= 2 + 2 ( 1.414 ) = 4.828 = 2.1973

5. Option (1) is correct. −

3

6

3 x +7 )  5 ( 2 x +3 )  − 4   2 (    2 3  2       3  Given:     =        3    3            

−3( 2 x + 3 )  2 3  4

    3  

⇒

−9( 2 x + 3 ) 2 4

⇒

3  

−6( 3 x +7 ) 2   5  2  3 

=     3    

−12( 3 x +7 ) 2 15

=  3

We have, ax = ay then x = y ∴

−9 ( 2 x + 3 )

⇒ ⇒ ⇒ ⇒

4

=

−12 ( 3x + 7 ) 15

135(2x + 3) 270x + 405 270x – 144x 126x

= 48(3x + 7) = 144x + 336 = 336 – 405 = –69 −69 −23 x= = 126 42

⇒ Now,

 −23  2 − 42 x = 2 − 42    42 

6. Option (3) is correct.

= 2 + 23 = 25 = 5

Given: A = 30 + 30 + 30 +  ∞ We know that, 30 = 5 × 6 5 and 6 are two consecutive factors of 30 such that 5 < 6 Here, ‘+’ sign is there \ The answer is 6. OR Let A = 30 + 30 + 30 +  ∞ ⇒ A = 30 + A Squaring on both sides, A2 = 30 + A ⇒ A2 – A – 30 = 0

⇒ A2 – 6A + 5A – 30 = 0 ⇒ (A – 6) (A + 5) = 0 ⇒ A = 6 or A = – 5 (not possible) \ A=6 7. Option (3) is correct. 3a + 3 × 4 a + 6 × 25a +1

27 a −1 × 8 a − 2 × 125a + 4

=

4 1526

3a + 3 × 2 2 a +12 × 52 a + 2 22 ⇒ 3a −3 3a −6 3a +12 = 26 26 3 ×2 ×5 5 ×3 ⇒3a + 3 – 3a + 3 × 22a + 12 – 3a + 6 × 52a + 2 – 3a – 12 = 22 × 5 –26 × 3–26 3–2a + 6 × 218 – a × 5 – a – 10 = 22 × 5–26 × 3–26 Comparing base (any one). –2a + 6 = –26 ⇒ – 2a = –32 ⇒ a = 16 Now, a + 9 = 16 + 9 = 25 = 5

8. Option (3) is correct. Given: 3 56 − x = 4 15625 = 56 Now,

3 6 5 − x =4 6

⇒

5 36 − x = 4

5532 − x = 4 ⇒ ⇒ 25 − x = 4 ⇒ x = 21 Squaring on both sides ⇒ x = 441 9. Option (1) is correct.

Given:

(12.5 )3 + ( 7.5 )3 (12.5 )2 + ( 7.5 )2 − 12.5 × 7.5

…(i)

Consider a = 12.5 and b = 7.5 a3 + b 3

i.e.,

2

a + b 2 − ab

= (a + b)

…(ii)

By comparing (i) and (ii) a + b = (12.5 + 7.5) = 20 10. Option (3) is correct. Given: ⇒ ⇒ ⇒ ⇒

x = 3− 5 x=

(

3− 5

)

2

= 3 + 5 − 2 15

x = 8 − 2 15

x − 8 = −2 15 ( x − 8 )2 = ( −2 15 )2

⇒ x 2 − 16 x + 64 = 60 ⇒ Now, \

x 2 − 16 x + 4 = 0

x 2 − 16 x + 6 = x 2 − 16 x + 4 + 2 = 0 + 2 = 2

x2 – 16x + 6 = 2

•

•

•

•

•

Constants fixed value e.g., 3, 4, 5 etc.

Degree of the equation  highest power of x in the equation

Identities:

•

•

•

•

•

Identities:

•

• Two variable •

Solving  Elimination method

One variable

Linear equations (Degree one)

First Level

Second Level

Third Level



b a c a

Roots of quadratic equation - solutions of a quadratic equation are called roots

Trace the Mind Map

Solving  Factor method

Identities:

Factor method

Identities:

Quadratic equation (Degree two)

•

•

•

•

Long division method

Division

Subtraction of like terms

Multiplication of each and every term

Operation on Algebraic term

Algebraic Expressions

Variables not fixed values e.g., x, y2 etc.

Addition of like terms

QUANTITATIVE APTITUDE: ALGEBRAIC EXPRESSIONS

245

Chapter

3

Algebraic Expressions

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

Algebraic Identities

2

3

10

Fundamental Operations and Equations

4

Types of Algebraic Expressions

11

Division : First arrange the terms in decreasing order, then proceed as shown in the following example.

• Algebraic Identities • Fundamental Operations and Equations

x+3 2

x+2

Topic-1 Algebraic Identities Algebraic Expression:

Operations on Algebraic Expressions

Addition : Only like terms, we can add Example : (2x2 + 5y2) + (2x + 4x2 + 7y + 8y2) (2x2 + 4x2) + (5y2 + 8y2) + 2x + 7y 6x2 + 13y2 + 2x + 7y Subtraction : This is similar to addition. We subtract only like terms. Example : (7x2 + 8y + 3y2) – (9x2 + 7y + 8x + 5y2) = (7x2 – 9x2) + (8y – 7y) + (3y2 – 5y2) – 8x = 2x2 + y – 2y2 – 8x Multiplication : (i) We multiply each and every term of the first expression with the second expression. (ii) Multiplying numerical co-efficients with numerical and variables with variables. Example : (2x2 + 5y + 2) × (3x2 + 2x) = 2x2(3x2 + 2x) + 5y(3x2 + 2x) + 2(3x2 + 2x) = 6x2+2 + 4x2+1 + 15x2y + 10xy + 6x2 + 4x = 6x4 + 4x3 + 15x2y + 10xy + 6x2 + 4x

x 2 + 2x 3x + 6 3x + 6 0

 Revision Notes An algebraic expression is an expression which is made up of constants and variables along with algebraic operations. Example : 2x + 5, x2 + xy + 6y, etc. Constants : Constants are fixed values. Example : 15, 53, e, p, etc. Variables : Variables are not fixed value. It may vary with problems. (x, y, etc.) Example : x, 2y, 3y2, etc.

x + 5x + 6

x 2 + 5x + 6

(x + 2) ( x + 3) ( x + 2)

= = OR (x + 3) ( x + 3 )2 ( x + 3 )2 Important Identities : (a + b)2 = a2 + b2 + 2ab (a – b)2 = a2 + b2 – 2ab a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab a2 – b2 = (a + b) (a – b) (a + b)2 + (a – b)2 = 2(a2 + b2) (a + b)2 – (a – b)2 = 4ab (a + b)2 = (a – b)2 + 4ab (a – b)2 = (a + b)2 – 4ab (a + b)3 = a3 + b3 + 3ab (a + b) (a – b)3 = a3 – b3 – 3ab (a – b) a3 + b3 = (a + b)3 – 3ab (a + b) a3 + b3 = (a + b) (a2 – ab + b2) a3 – b3 = (a – b)3 + 3ab (a – b) a3 – b3 = (a – b) (a2 + ab + b2) (a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – ca) If a + b + c = 0, then a3 + b3 + c3 = 3abc (a + x) (b + x) = x2 + (a + b) x + ab ab (a + b) + bc (b + c) + ca (c + a) = (a + b) (b + c) (c + a) A function or expression f(x) is divisible by x – a, if f(a) = 0 [To find f(a), simply put x = a in the given expression or function]

Objective Type Questions 1. If arithmetic mean and geometric mean of roots of a quadratic equation are 8 and 5, respectively, then the quadratic equation is: (2) x2 – 8x + 25 = 0 (1) x2 + 8x + 5 = 0 2 (3) x – 16x + 25 = 0 (4) 2x2 – 8x + 25 = 0  (CUET 2023) 2. If

x y z = = ( b + c )( b + c − 2 a ) ( c − a )( c + a − 2b ) ( a − b )( a + b − 2 c )

then

value of x + y + z is: (CUET 2023) (1) a + b + c (2) a2 + b2 + c2 (3) 0 (4) 1 3. Linear equations 3x + 5y = 19 and 10x – 3y = 24 have solution β α and y = , then the value of α + β is: (CUET 2023) x= 2 3 (1) 5 (2) 13 (3) 10 (4) 7

4. If x −

1 1 = 3 then the value of x 2 + 2 is x x

(CUET 2022)

(1) 9 (2) 11 (3) 7 (4) 3 5. Find the mean of 12, 15, 7, 8 and ‘x +13’, if x = 2  (CUET 2022) (1) 11.2 (2) 9.2 (3) 11.4 (4) 12 6. If the difference between two numbers is 6 and the difference between their squares is 60, what is the sum of their cubes? (1) 678 (2) 945 (3) 894 (4) 520 7. The sum of two numbers is 59 and their product is 840. Find the sum of their squares. (1) 2961 (2) 1754 (3) 1801 (4) 1875 8. If a and b are two positive real numbers such that a + b = 20 and ab = 4, then the value of a3 + b3 is (1) 7760 (2) 240 (3) 8000 (4) 8240



247

QUANTITATIVE APTITUDE: ALGEBRAIC EXPRESSIONS

9. The sum of two numbers is 47 and their product is 550. Find the sum of their squares. (1) 1109 (2) 986 (3) 876 (4) 1209 10. If a + 3b = 12 and ab = 9, then the value of (a – 3b) is (1) 4 (2) 8 (3) 6 (4) 9 11. If a + 2b = 10 and 2ab = 9, then |a – 2b| is equal to : (1) 4 (2) 6 (3) 8 (4) 2 1 1 12. If a 2 + = 98 , a > 0 , then the value of a3 + 3 will be : a a2 (1) 960 (2) 950 (3) 970 (4) 870 13. If a3 + b3 = 62 and a + b = 2, then the value of ab is (1) – 9 (2) 9 (3) – 6 (4) 6 14. If a = 2b, then the value of a + b is a−b

(1) 5 (2) 4 (3) 3 15. (a + 2b)2 – (a – 2b)2 is equal to : (1) 10ab (2) 6ab (3) 8ab

(4) 6 (4) 4ab

Answer Key 1. (3)

2. (3)

3. (2)

4. (2)

5. (3)

6. (4)

7. (3)

9. (1)

10. (3) 11. (3) 12. (3) 13. (1) 14. (3) 15. (3)

8. (1)

Answers with Explanations 1. Option (3) is correct. Let the roots of quadratic equation are α and β. Given that: Arithmetic mean of roots = 8 α+β ∴ = 8 or α + β = 16 2 Geometric mean of roots = 5 ∴ αβ = 5 ⇒ αβ = 25 Using, formula for quadratic equation ⇒ x 2 − ( α + β)x + αβ = 0 ⇒ x2 − 16x + 25 = 0 2. Option (3) is correct. Given:

x y = ( b − c )( b + c − 2 a ) ( c − a )( c + a − 2b ) z = =k ( a − b )( a + b − 2 c )

So,

x = (b – c)(b + c – 2a)k = k(b2 – c2 –2ab + 2ac) Similarly y = k(c – a)(c + a – 2b) = k(c2 – a2 +2ab – 2bc) and z = k(a – b)(a + b – 2c) = k (a2 – b2 + 2bc – 2ac) Now x + y + z = k[b2 – c2 – 2ab + 2ac + c2 – a2 + 2ab – 2bc + a2 – b2 + 2bc – 2ac] =k×0=0 3. Option (2) is correct. Given that 3x + 5y = 19 10x – 3y = 24 β Now putting x = α and y = 2 3 Now, 3 α + 5 β = 19 and 10 α − 3 β = 24 3 2 3 2 ⇒ 2α + 5β = 38 and 20 α – 9β = 144 By solving equation (i) and (ii), α = 9 and β = 4 So, α + β = 9 + 4 = 13

(i) (ii)

4. Option (2) is correct. As we know that, (x – 1/x)2 = x2 + 1/ x2 – 2 × x × 1/ x So, 32 = x2 + 1/ x2 – 2 x2 + 1/ x2 = 9 + 2 = 11 5. Option (3) is correct. Given numbers = 12, 15, 7, 8 and x +13 where x =2 So, last number = 15 Now, mean = sum of all numbers/5 = 57 / 5 = 11.4 6. Option (4) is correct. Let x and y be the two numbers. Given: x–y =6 ...(i) and x2 – y2 = 60. We have, (x2 – y2) = (x – y) (x + y) ⇒ 60 = 6 × (x + y) ⇒ x + y = 10 ...(ii) Solving (i) and (ii), x + y = 10 x−y =6 2 x = 16 x = 8, y = 2 Now, x3 + y3 = (x + y) (x2 – xy + y2) ⇒ x3 + y3 = 10 (64 – 16 + 4) ⇒ x3 + y3 = 520 or x3 + y3 = (8)3 + (2)3 = 512 + 8 = 520 7. Option (3) is correct. Let x and y be the two numbers Given: x + y = 59 and xy = 840, x2 + y2 = ? We have, (x + y)2 = x2 + y2 + 2xy (59)2 = x2 + y2 +2(840) ⇒ x2 + y2 = (59)2 – 1680 ⇒ x2 + y2 = 1801 8. Option (1) is correct. Given: a + b = 20, ab = 4 then a3 + b3 = ? We have, (a + b)2 = a2 + b2 + 2ab 400 = a2 + b2 + 8 ⇒ 392 = a2 + b2 Now, a3 + b3 = (a + b) (a2 + b2 – ab) = 20(392 – 4) = 20 × 388 = 7760 9. Option (1) is correct. Let x and y be the two numbers. Given:x + y = 47 and xy = 550 to find x2 + y2 = ? We have, (x + y)2 = x2 + y2 + 2xy (47)2 = x2 + y2 + 2(550) ⇒ 2,209 = x2 + y2 + 1100 ⇒ x2 + y2 = 1,109 10. Option (3) is correct. Given: a + 3b = 12, ab = 9 to find (a – 3b) Consider, (a + 3b)2 = a2 + 9b2 + 6ab 144 = a2 + 9b2 + 6(9) ⇒ 144 – 54 = a2 + 9b2 ⇒ a2 + 9b2 = 90 Now, (a – 3b)2 = a2 + 9b2 – 6ab ⇒ (a – 3b)2 = 90 – 6(9) ⇒ (a – 3b)2 = 90 – 54

248

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Example x3 + x2 + 3x, here the highest degree is 3. \ The equation is cubic equation.

⇒ (a – 3b)2 = 36 ⇒ a – 3b = 6 11. Option (3) is correct. Given: a + 2b = 10, 2ab = 9 We have, (a + 2b)2 = a2 + 4b2 + 4ab 100 = a2 + 4b2 + 2(9) ⇒ 82 = a2 + 4b2 Now, (a – 2b)2 = a2 + 4b2 – 2(2ab) = 82 – 2(9) = 64 ⇒ (a – 2b) = ± 8 ⇒ |a – 2b| = 8 12. Option (3) is correct. Given:

2

a +

1 a2

We have, 

2

⇒

2

Linear Equation

An equation of the type ax + b = 0 is a linear equation with one variable (x only) An equation of the type ax + by + c = 0 is a linear equation with two variables (x and y) Finding solution of the equation [Solving for x] Example 2x + 5 = 0 ⇒ 2x = – 5

⇒

= 98 , a > 0 ,

1 1 1 2  a + a  = a + 2 + 2 ( a )  a  = 98 + 2     a

Now,

a+ a3 +

1 = 10 a

[

∴

⇒

1   a + a  = 100  

a>0]

1  1  1   =  a +   a2 + − (a)   2 a  a    a a 1

3

= 10(98 – 1) = 970 13. Option (1) is correct. Given: a3 + b3 = 62 and a + b = 2 We have, (a + b)2 = a2 + b2 + 2ab 4 = a2 + b2 + 2ab ⇒ a2 + b2 = 4 – 2ab ..... (i) Consider, a3 + b3 = (a + b) (a2 + b2 – ab) 62 = 2(4 – 2ab – ab) 62 ⇒ = 4 − 3ab 2 3ab = 4 – 31 −27 ab = = −9 3

⇒ ⇒ 14. Option (3) is correct. Given:

a = 2b a + b 2 b + b 3b = = =3 a − b 2b − b b

Now,

15. Option (3) is correct. We know that (a + b)2 – (a – b)2 = 4ab Applying above, (a + 2b)2 – (a – 2b)2 = 4(a) (2b) = 8ab

Topic-2

Fundamental Operations and Equations

 Revision Notes Equation In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. Example 5x2 + 2x + 17 = 0, x2 + y2 + 2xy + 2y3 = 0, etc. Degree of the equation : Degree of the equation is the highest power of x in the given equation.

x=−

5 2

We can say that linear equation with single variable like ax + b = b 0, its solution x = − a Two Variables : To solve linear equations with two variables, we need equations. Example 2x + 5y = 6 4x + 2y = 10 We can use Elimination method, Substitution method, Graphical method, etc. to find x and y. In this chapter we use Elimination method. Now 2x + 5y = 6 ..... (i) 4x + 2y = 10 ..... (ii) Multiply equation (i) by 2, we get 4x + 10y = 12 ..... (iii) 4x + 2y = 10 ..... (ii)

(−) (−)

(−)

8y = 2 2 1 y= = 8 4

[(iii) – (ii) means we change the sign of the second equation] Substituting y =

1 in (ii), we get 4

⇒ ⇒ ⇒ ⇒ Quadratic equation

4 x + 2 y = 10 1 4 x + 2   = 10 4 1 4 x + = 10 2 8x + 1 = 10 2 19 x= 8

An equation of the form ax2 + bx + c = 0, a ¹ 0 is called quadratic equation. In quadratic equation, degree of the equation is 2. Example 2x2 + 5x + 6 = 0, x2 + 5x + 6 = 0, etc. Factors : The linear expression which divides completely ax2 + bx + c and leaves remainder zero is called a factor of the equation ax2 + bx + c. Example x2 + 7x + 12 = x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3) (x + 4) Thus, x + 3 and x + 4 are the factors of the quadratic polynomial. Solution. The value of x which satisfies the equation ax2 + bx + c = 0 is called the solution of the equation. • Linear equation has one solution. • Quadratic equation has two solutions.

Finding solution of the quadratic equations • First find the factors of the equation. • Write the equation as product of factors. • Equate the factors to zero and find x. • The two x values are also called roots of the equation.

249

QUANTITATIVE APTITUDE: ALGEBRAIC EXPRESSIONS

Points to Remember • Consistency Test (linear equation with 2 variables).

a + b = Sum of the roots = −

If a1 ≠ b1 , then system will have exactly one solution and will a2

b2

be consistent. a1 b1 c1 If = , then system is consistent and has infinitly = a2 b2 c2 many solution. a1 b1 c1 , then system has no solution and it is ≠ If = a 2 b2 c 2 inconsistent.

• Let a and b be the two roots of the equation ax2 + bc + c = 0. Then we have,

ab = Product of the root =

b a

c a

• If px + q is a factor of the equation

ax2 + bx + c = 0 When ax2 + bx + c is divided by px + q, remainder is zero. q i.e., while substituting x = − in ax2 + bx + c, we get ax2 + bx + p c=0 Suppose kx + l, is not a factor of (ax2 + bx + c), then remainder is not zero when ax2 + bx + c is divided by kx + l l To find the remainder put, x = − in ax2 + bx + c. k

Objective Type Questions 1. If m – n = 16 and m + n = 400, the value of mn is = (CUET 2022) 2

2

(1) 72 (2) 25 (3) 144 (4) 192 2. The value of ‘y’ in the question 2x + 3y – 7 = 0 if x = –3/2 (CUET 2022) (1) 10 (2) 10/3 (3) 4/3 (4) 13/3 3. If lengths of two diagonals of rectangle are (2x – 3) cm and (x + 2) cm, then the value of x is : (CUET 2022) (1) 1 (2) 5 (3) 5/2 (4) 5/3 4. The value of x in the equation 2x – 3 = 7 – 3x (CUET 2022) (1) 10 (2) 2 (3) –2 (4) 5 5. If the value of 3x y + 2 y x 3x y − 2 y x

−

3x y − 2 y x is same as that of 3x y + 2 y x

x y , then

which of the following relations between x and y is correct ? (1) 9x – 4y = 36 (2) 9x + 4y = 24 (3) 9x + 4y = 36 (4) 9x – 4y = 24 6. If 1 + 9r2 + 81r4 = 256 and 1 + 3r + 9r2 = 32, then find the value of 1 – 3r + 9r2. (1) 8 7. If A = (1)

(2) 4

(3) 16

(4) 12

x −1 1 , then find the value of A − is: x +1 A

x2 − 1 (2) −4 x −4 ( 2 x + 1 ) x2 − 1

(3)

x2 − 1 (4) −4 ( 2 x − 1) −4 ( 2 x − 1 ) x2 − 1

8. A man buys 2 apples and 3 kiwi fruits for `37. If he buys 4 apples and 5 kiwi fruits for `67, then what will be the total cost of 1 apple and 1 kiwi fruit ? (1) `20 (2) `28 (3) `18 (4) `15 5x 1 . = 8 , then find the value of 2 x + 1 − 6x x (1) 2.5 (2) 6 (3) 5 (4) 6.5 10. If a and b are two roots of the quadratic equation ax2 + bx + c = 1 1 0 where a, b, c are constants and a ≠ 0, then the value of + is α β (1) −b (2) c (3) c (4) b b a c c

9. If x +

11. Find the values of x for the given equation 3x2 + 5x – 2 = 0 (1) –3 and –2 (2) 3 and − 1 (3) 2 and –3 (4) –2 and 1 3 2

12. If kx3 + 4x2 + 3x – 4 and x3 – 4x + k leave the same remainder when divided by (x – 3), then the value of k is (1) 2 (2) –1 (3) 1 (4) 0 13. If x3 – 6x2 + ax + b is divisible by x2 – 3x + 2, then the values of a and b are : (1) a = –11 and b = 6 (2) a = –6 and b = –11 (3) a = 11 and b = –6 (4) a = 6 and b = 11 14. If 4x2 + y2 = 40 and xy = 6, (x > 0, y > 0), then the value of 2x + y is (1) 24 (2) 16 (3) 4 (4) 8 15. If p +

1 1 15 = 112 , find ( p − 112 ) + 15 . p p

(1) 10

(2) 0

(3) 15

(4) 1

Answer Key 1. (1)

2. (2)

3. (2)

4. (2)

5. (4)

6. (1)

7. (2)

8. (4)

9. (1) 10. (1) 11. (4) 12. (2) 13. (3) 14. (4) 15. (2)

Answers with Explanations 1. Option (1) is correct. As we know that, (m – n)2 = m2 + n2 – 2mn 162 = 400 – 2mn 2mn = 400 – 256 2mn = 144 mn = 72 2. Option (2) is correct. Given that, 2x + 3y – 7 = 0 if x = –(3/2) –2(3/2) + 3y –7 = 0 –3 + 3y –7 = 0 3y – 10 = 0 3y = 10 y = 10/3 3. Option (2) is correct. Given that, Lengths of two diagonals of rectangle are (2x – 3) cm and (x + 2) cm. We know that the length of diagonals of rectangle is equal. So, 2x – 3 = x + 2 2x – x = 2 + 3 x =5

250

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

4. Option (2) is correct. Given that, 2x – 3 = 7 – 3x 2x + 3x = 7 + 3 5x = 10

Given:

−

3x y − 2 y x

Given:

x =2

5. Option (4) is correct. 3x y + 2 y x

The cost of one apple = `8 and the cost of one kiwi fruit = `7 \ Total cost of 1 apple and 1 kiwi fruit = 7 + 8 = `15 9. Option (1) is correct.

3x y − 2 y x 3x y + 2 y x

Taking L.C.M. and simplifying,

( 3x y + 2 y x ) − ( 3x y − 2 y x ) ( 3x y − 2 y x ) ( 3x y + 2 y x ) 2

Consider, = xy

(

( 3x

)( 2 y x ) = 2 2 y ) − (2y x ) 24 xy xy

⇒

9x 2 y − 4 y 2 x

= xy

=

10. Option (1) is correct. Given: ax2 + bx + c

xy

= xy

(

)

1 − 3r + 9 r 2 =

7. Option (2) is correct.

256 =8 32

x − 1 , then 1 x + 1 = x +1 A x −1 1 x −1 x +1 A− = − A x +1 x −1

Given: Now,

A=

=

=

( x − 1 )2 − ( x + 1 )2 ( x + 1) ( x − 1) x 2 + 1 − 2x − x 2 − 1 − 2x 2

=

−4 x

x −1 x2 − 1 8. Option (4) is correct. Let x and y be the cost of one apple and one kiwi fruit respectively. According to the equation, 2x + 3y = 37 …(i) 4x + 5y = 67 …(ii) Using Elimination method 2 ( 2 x + 3 y = 37 ) 4 x + 6 y = 74 4 x + 5 y = 67

(−) (−) (−) y=7

Substituting in (i), we get 2x + 3(7) ⇒ 2x + 21 ⇒ 2x ⇒ x

= 37 = 37 = 16 =8

5 5 = 1 8−6  x + x  − 6  

5 = 2.5 2

1 1 β+α α +β + = = αβ α β αβ b Now, sum of the roots = α + β = a Product of the roots = αβ = c a −b 1 1 −b \ + = a = c α β c a 11. Option (4) is correct. Given: 3x2 + 5x – 2 = 0 2 ⇒ 3x + 6x – x – 2 = 0 ⇒ 3x(x + 2) – 1(x + 2) = 0 ⇒ (3x – 1) (x + 2) = 0 ⇒ x + 2 = 0 or 3x – 1 = 0 1 ⇒ x = – 2 or x = 3 12. Option (2) is correct. Given: kx3 + 4x2 + 3x – 4 and x3 – 4x + k leave same remainder when divided by x – 3. Put x = 3 in both the equation and equate both. k (3)3 + 4(3)2 + 3(3) – 4 = (3)3 – 4(3) + k ⇒ 27k + 36 + 9 – 4 = 27 – 12 + k ⇒ 26k = – 26 ⇒ k=–1 13. Option (3) is correct. Given: x3 – 6x2 + 9x + b is divisible by x3 – 3x2 + 2 Now, x3 – 6x2 + 9x + b ..… (i) x2 – 3x + 2 = x2 – 2x – x + 2 = (x – 2) (x – 1) Since x – 2 and x – 1 are the factors of equation (i), ∴ x – 2 and x – 1 will divide x3 – 6x2 + x + ax +b and leave reminder 0. Now, Put x = 1 in equation (i) a+b–5=0 ..… (ii) Also, Put x = 2 in equation (i) 2a + b – 16 = 0 ….. (iii) Solving (iii) and (ii), we get

Consider,

⇒ 24xy = xy(9x – 4y) ⇒ 24 = 9x – 4y ⇒ 9x – 4y = 24 6. Option (1) is correct. Given:1 + 9r2 + 81r4 = 256 and 1 + 3r + 9r2 = 32 We have (a + b) (a – b) = a2 – b2 So we can write (1 + 9r2 + 3r) (1 + 9r2 – 3r) = (1 + 9r2)2 – (3r)2 = 1 + 81r4 + 18r2 – 9r2 = 1 + 81r4 + 9r2 Now, (32) (1 – 3r + 9r2) = 256 ⇒

5x 1   x x + − 6  x  

=

24 xy xy = xy 9 x 2 y − 4 y 2 x

⇒

=

2

4 3x y

∴

5x x 2 + 1 − 6x

1 =8 x

2

We know that (a + b) – (a – b) = 4ab. 2

x+

=

2 a + b = 16 a+b = 5

(−) (−) (−) a = 11

251

QUANTITATIVE APTITUDE: ALGEBRAIC EXPRESSIONS

15. Option (2) is correct.

From (ii), 11 + b – 5 = 0

⇒ b=–6 \ a = 11 and b = – 6 14. Option (4) is correct. Given: 4x2 + y2 = 40 and xy = 6, x>0, y>0 Consider, (2x + y)2 = (2x)2 + y2 + 2(2x) (y) ⇒(2x + y)2 = 4x2 + y2 + 4xy ⇒ (2x + y)2 = 40 + 4(6) ⇒ (2x + y)2 = 64 ∴ 2x + y = 8

Given: p + Now p +

⇒

1 = 112 p

1 = 112 p

p − 112 =

−1 p

⇒ ( p − 112 )15 +

1 p

15

15

 −1  =   p 

+

1 p15

=0

4

n  n  1 2

(n+ 1)  2n 1  6

•

•

•

•



n1 A1  n A   .....  nk A k n1  n  .......  nk

k number of observations with n1, n, ... nk numbers having A1, A2, …… Ak averages respectively. Then, Combined average

Average (Arithmetic Mean)

Average and Problems on Ages

Average of first n terms =

n1 2

First Level

Second Level

Trace the Mind Map

Third Level



Solving using known Mathematical tools

Puzzle type Questions Based on Ages of a Person

If the average of n1 number is a1 and the average of n2 numbers is a2, then the average of total numbers n1 and n2 , n1 a1  n2 a2 Average = n1  n2

Problems on Ages

Sum of Observations A Total number of Observations

then the age after n years will be (x + n) years. If the current age is x, then the age years before will be (x – n) years. If the age is given in the form of a ratio, for example, p : q, then the ages shall be considered as px and qx, repectively. If the current age is x, then n times the current age will be (x × n) years. If the current age to be x, 1 x then of the age will be equal to   years. n n

• If the current age is x,

The Average of first n consecutive odd natural numbers, i.e., 1, 3, 5, …..(2n – 1) = n

The Average of first n consecutive even Natural numbers, i.e., 2, 4, 6, …. 2n = n+ 1

• Average of 13, 23, 33 …… n3 =

• Average of 12, 22, 32, 42, 52 … n2 =

252 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Average and Problems on Ages

Chapter

4

Chapter Analysis Concept Name

2021

Average

2022

2023

Additional Questions

3

4

7

Problems on Ages

5

Types of Analogies

• If 0 is one of the observation of a given data, then that 0 will also be included while calculating average.

• Average • Problems on Ages

Points to Remember :

• The average of first n natural number is n + 1 . 2

Topic-1 Average  Revision Notes Types of Numbers An average of a given data is obtained when the sum of the given observations is divided by the number of observations. Average = Sum of the given observations Number of observations Example : Find the average of 3, 4, 9, 10, 16 and 18 60 Average = 3 + 4 + 9 + 10 + 16 + 18 = = 10 6 6

Properties of Averages • If the number of quantities in two groups be n1 and n2 and their averages are x and y, respectively. Then the combined average is

n1 x + n2 y . n1 + n2

• If the average of n1 quantities is x and the average of n2 quantities out of them is y, then the average of the remaining groups is

n1 x − n2 y n1 − n2

. Thus, if there are k groups having number of observations n1, n2, n3……nk with averages A1, A2, A3……Ak, respectively.

Then weighted average, Aw =

n1 A 1 + n2 A 2 + n3 A 3 + ......nk A k n1 + n2 + n3 + ......nk

Properties of Average

• If all the given numbers are added, subtracted, multiplied or divided by a non-zero number a, then their average will also be added, subtracted, multiplied or divided by a. • Average of the given data is less than the greatest observation and greater than the smallest observation.

• Average is also called as Arithmetic Mean. • If ‘0’ is one of the observations of a given data then that ‘0’ will also be included while calculating average.

Example: Average of 3, 6 and 0 is 3 + 6 + 0= 9= 3 3

3

• If the average of ‘n1’ numbers is a1 and the average of ‘n2’ numbers is a2, then the average of total numbers n1 and n2,

n a +n a

2 2 Average = 1 1 n + n 1 2 • If A goes from P to Q with speed x km/h and returns from Q to P with speed y km/h, then the average speed of total journey, 2xy Total distance = Average speed = Total time taken x+y

• If a distance is travelled with three different speeds a km/h, b km/h and c km/h, then Average speed of total journey =

3abc km/h ab + bc + ca

Example 1: The average of first 100 positive integers is

Sol. We know that, average of first n natural numbers is n + 1 . 2 Here, n = 100 100 + 1 101 ∴ Required average = = = 50.5 2 2 Formula 1. Average of even natural numbers from 1 to n is n   2 + 1.   Example 2 : The average of even numbers from 1 to 50 is n Sol. As we know, average of even numbers from 1 to n = + 1 2 Here,

n = 50 50

  ∴ Required average =  + 1  = 26  2 

Objective Type Questions 1. If the average weight of 10 students is 25 kg and that of other 10 students is 35 kg, then the average weight of these 20 students is: (CUET 2023) (1) 30 kg (2) 35 kg (3) 25 kg (4) 20 kg 2. The mean of scores obtained by 50 students is found to be 79.5. Later on, it was found that the score of one student was read as

94 in place of 49 and the score of another student was read as 69 in place of 89. Find the correct mean. (CUET 2023) (1) 85.52 (2) 79.35 (3) 79 (4) 78 3. If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then value of x: (CUET 2023) (1) 7 (2) 43 (3) 51 (4) 8

254

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

4. Eight persons with an average weight of 49 kg were in a lift. Two more persons Anita and Kamal boarded the lift and the average weight of the persons in the lift increased by 2.8 kg. What is the average weight of Anita and Kamal? (CUET 2023) (1) 51.8 kg (2) 54.6 kg (3) 59 kg (4) 63 kg 5. The average age of group of 5 friends is 32 years. The youngest friend amongst them is 4 years old. What was the average age of the group at the time of birth of the youngest friend? (CUET 2022) (1) 36 years (2) 35 years (3) 34 years (4) 33 years 6. Find the average of following numbers. 12, 15, 18, 14, 16, 13, 25, 28, 23, 27 (CUET 2022) (1) 18.25 (2) 12.5 (3) 19.1 (4) 16.1 7. Please read the following carefully and answer the questions. The marks of 7 students in a unit test are given below: (CUET 2022)

9, 10, 7, 6, 9, 3, 5

Mean of the data is : (1) 9 (2) 10 (3) 7 (4) 6 8. The average of 40 numbers is 36. The average of the first 25 numbers is 31 and the average of last 16 numbers is 43. Find the 25th number. (1) 23 (2) 24 (3) 21 (4) 22 9. The numbers 24, 45, a, 35, 59, 83, 46, b, 29, 74 are serially numbered as they appear in the sequence. When each number is added to its serial number, then the average of the new numbers formed is 55. The average of the missing numbers (a and b) is (1) 38 (2) 58 (3) 62 (4) 50 10. The sum of 17 consecutive numbers is 289. The sum of another 10 consecutive numbers, whose first term is 5 more than the average of the first set of consecutive numbers is (1) 315 (2) 285 (3) 265 (4) 300 11. In a set of three numbers, the average of the first two numbers is 7, the average of the last two numbers is 10, and the average of the first and the last numbers is 14. What is the average of the three numbers ? (1) 25 4

(2) 37 3

(4) 31 3

(3) 29 4

12. The average of 24 numbers is 26. The average of the first 15 numbers is 23 and that of the last 8 number is 33. Find the 16th number. (1) 15 (2) 16 (3) 17 (4) 18 13. A library has an average of 265 visitors on Sundays and 130 visitors on other days. The average number of visitors per day in a month of 30 days beginning with a Monday is (1) 148 (2) 135 (3) 165 (4) 129 14. A player has a certain average for 15 innings. In the 16th inning he scores 120, thereby his average increases by 6 runs. What is the new average ? (1) 20 (2) 8 (3) 30 (4) 24 Answer Key 1. (1)

2. (3)

3. (1)

4. (4)

5. (2)

6. (3)

7. (3)

8. (1)

9. (4) 10. (3) 11. (4) 12. (1) 13. (1) 14. (3)

Answers with Explanations 1. Option (1) is correct. Given that average weight of 10 students = 25 kg and average weight of 10 students = 35 kg



∴ Average weight of 20 students = =

10 × 25 + 10 × 35 10 + 10 250 + 350 600 = 30 = 20 20

2. Option (3) is correct. Given that: Average score of 50 students = 79.5 ∴ Total score of 50 students = 79.5 × 50 = 3975 According to the question, Correct mean =

3975 − ( 94 + 69 ) + ( 49 + 89 ) 50

= 79

3. Option (1) is correct. We have, Mean of x, x + 2, x + 4, x+ 6 and x + 8 = 11 x+x+2+x+4+x+6+x+8 ∴ = 11 5 ⇒ 5x + 20 = 55 ⇒ x = 7 4. Option (4) is correct. Average weight of 8 persons = 49 kg ∴ Total weight = 49 × 8 = 392 kg. After adding Anita and Kamal, new average = 49 + 2.8 = 51.8 kg ∴ Total weight of 10 persons = 51.8 × 10 = 518 kg So, total weight of Anita and Kamal = 518 – 392 = 126 126 = So, Average weight of Anita and Kamal = 63 kg 2 5. Option (2) is correct. Given that, The average age of 5 friend = 32 years So, sum of their ages = 32 × 5 = 160 years The youngest friend is 4 years old, that means youngest was born 4 years ago. Now 4 years ago, Sum = 160 – (4 × 5) = 160 – 20 = 140 years So, the average age of group at the time of the birth of youngest one = 140/4 = 35 years 6. Option (3) is correct. Given numbers are 12, 15, 18, 14, 16, 13, 25, 28, 23, 27. As per the average formula, Average = (12 + 15 + 18 + 14 + 16 + 13 + 25 +28 + 23 + 27) / 10 = 191 /10 = 19.1 7. Option (3) is correct. Given that, Data = 9, 10, 7, 6, 9, 3, 5. As we know, Mean = Sum of all numbers

Total numbers 9 + 10 + 7 + 6 + 9 + 3 + 5 49 = =7 = 7 7

8. Option (1) is correct. Given: Average of 40 numbers is 36 and average of the first 25 numbers is 31 Also, average of last 16 numbers is 43. In this problem, 25th number is considered in the first 25 group as well as last 16 group. Consider the 25th number as x So, we can write ⇒ 40 × 36 = (25 × 31) + (16 × 43) – x [ 25th term is added twice] ⇒ 1440 = 775 + 688 – x ⇒ 1440 = 1,463 – x ∴ x = 23

255

QUANTITATIVE APTITUDE: AVERAGE AND PROBLEMS ON AGES

9. Option (4) is correct. Given: The average of the number added to its series number is 55. i.e.,

( 24 + 1) + ( 45 + 2 ) + ( a + 3 ) + ( 35 + 4 ) + ( 59 + 5 ) + ( 83 + 6 ) + ( 46 + 7 ) + ( b + x ) + ( 29 + 9 ) + ( 74 + 10 )

⇒ ⇒ ⇒ ∴ Average,

10

= 55

439 + a + b + 11 = 55 10 a + b + 450 = 550

b+c = 10 ⇒ b + c = 20 2 a+c = 14 ⇒ a + c = 28 2 Adding equations (i), (ii) and (iii), 2(a + b + c) = 62 ⇒ a + b + c = 31 a + b + c 31 = ⇒ 3 3

The problems contain clues regarding the ages. Using the known mathematical tool (like solving linear and quadratic equation), we have to find the answer.

….(i) ….(ii) ….(iii)

31 ∴ Average of a, b, c is 3 12. Option (1) is correct. Let 16th number be x. Now, sum of 24 numbers = 24 × 26 = 624 Sum of 24 numbers = Sum of first 15 numbers + 16th numbers + Sum of last 8 numbers ⇒ 624 = (15 × 23) + 16th number + (8 × 33) ⇒ 624 = 345 + 16th number + 264 ⇒ 16th number = 624 – 609 ⇒ 16th number = 15 13. Option (1) is correct. There are 4 Sundays and 26 other days in a month of 30 days. ( 265 × 4 ) + (130 × 26 ) ∴ Average = 30

=

1060 + 3380 4440 = = 148 30 30

Thus, on average 148 visitors per month. 14. Option (3) is correct. Let batsman’s average before 16th innings be x. According to the question, ( x × 15) + 120 = x + 6 16

Problems on Ages

Puzzle Type Problems on Ages

a + b = 100 a + b 100 = = 50 2 2

a + b = 14

Topic-2

 Revision Notes

10. Option (3) is correct. Sum of 17 consecutive number is 289. 289 = 17 Then average = 17 The first term of the second series is 5 more than the average of the first i.e., 17 + 5 = 22 Now, 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 = 265 11. Option (4) is correct. Let a, b, c be the three numbers. a+b =7 Given: 2 ⇒

⇒ 15x + 120 = 16x + 96 ⇒ x = 120 – 96 ⇒ x = 24 runs New average after 16th innings is x + 6 = 24 + 6 = 30 runs

Tips and Tricks to Solve Problems on Ages Candidates who are not much familiar with the concept and tend to either skip the age problems or answer them incorrectly can refer to the tips given below. These tips may help you answer the question following a set pattern and then finding the answer. • The most important thing is to read the question carefully and gradually form the equation which will help you answer the question. • Basic things like addition, subtraction, multiplication and division will help a candidate reach the answer and no complicated calculations are required to answer such questions. • Arrange the given values by placing them correctly in an equation by giving variables to the unknown values. • Once the equation has been formed, solve the equation to find the answer. • The final step is to recheck the answer obtained by placing it in the equation formed to ensure that no error has been made while solving. The ‘problems on ages’ is one such topic which is not just asked in the first or the preliminary phase of examination but questions from this topic may also be asked in the mains phase of examination in a rather complex manner.

Problems on Ages

Example : The product of the ages of Anu and Scan to Sonu is 240. If twice the age of Sonu is more than know more about Anu’s age by 4 years, find the Sonu’s age. this topic Sol. Let Anu’s age be x years, then Sonu’s age = 240 years. x According to the question, Problems 240 ⇒ 2× −x = 4 on Ages x 480 − x 2 = 4x ⇒ x2 + 4x – 480 = 0 ⇒ (x + 24) (x – 20) = 0 x = 20 or x ¹ –24 [not possible because age cannot be negative] \ x = 20  240  Hence, Sonu’s age   = 12 years  20  Example : The average age of a family of 10 members is 20 years. If the age of the youngest member of the family is 10 years, then the average age of the members of the family just before the birth of the youngest member was approximately? Sol. Sum of the present age of 10 members = 20 × 10 = 200 years 10 years before, total age of 9 members = 100 years 100 1 ∴ Required average = = 11 years 9 9

256

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Example : 5 years ago, the average age of P and Q was 15 years. Average age of P, Q and R today is 20 years. How old will R be after 10 years? Sol. Let P and Q be the present ages of P and Q. Sum of ages of P and Q 5 years ago =

( P − 5) + (Q − 5) = 15 2

⇒ P + Q = 30 + 10 = 40 years ...(i) Sum of (P + Q + R)‘s present age = 20 × 3 = 60 years From (i) & (ii), R’s present age = 60 – 40 = 20 years ∴ R’s age after 10 years = 20 + 10 = 30 years Example : The average age of a husband and his wife was 23 years at the beginning of their marriage. After five years they have a oneyear old child. The average age of the family of three, when the child was born, was: Sol. After five years of marriage, sum of the ages of husband, wife and child = 46 + 10 + 1 = 57 years At the time of birth of child, the sum of the ages of husband, wife and child = 57 – 3 = 54 years 54 ∴ Required average age = = 18 years 3 Example : From a class of 42 boys, a boy aged 10 years goes away and in his place a new boy is admitted. If on account of this change, the average age of the boys in that class increases by 2 months, the age of the newcomer is: 2 × 42   Sol. Age of new boy =  10 + = 17 years 12   Example : The average age of Ram and his two children is 17 years and the average age of Ram’s wife and the same children is 16 years. If the age of Ram is 33 years, the age of his wife is (in years): Sol. Sum of the ages of Ram and two children = 17 × 3 = 51 years ...(i)

Also, the sum of Ram’s wife and two children = 16 × 3 = 48 years ...(ii) From (i) and (ii), difference between Ram and his wife age, ∴ Ram – wife = 3 years ⇒ 33 – wife = 3 years ⇒ Wife = 33 – 3 = 30 years ∴Age of wife is 30 year. Example: The average age of A and B is 20 years. If A is to be replaced by C, the average would be 19 years. The average age of C and A is 21 years. The ages of A, B and C in order (in years) are: Sol. Sum of ages of A and B, A + B = 2 × 20 = 40 years ...(i) Sum of ages of B and C, C + B = 2 × 19 = 38 years ...(ii) Sum of ages of A and C, C + A = 2 × 21 = 42 years ...(iii) Adding all three question, 2 (A + B + C) = 40 + 38 + 42 = 120 ⇒ A + B + C = 60 ∴ A = (A + B + C) – (B + C) = 60 – 38 = 22 years B = (A + B + C) – (A + C) = 60 – 42 = 18 years C = (A + B + C) – (A + B) = 60 – 40 = 20 years Example : ln a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was: Sol. Sum of the present age of family members = 33 × 5 = 165 years, 9 years ago. Sum of their ages = 165 – 9 × 5 = 120 years 120 ∴ Required average age = = 30 years 4

Objective Type Questions 1. The ratio of a man’s age to his father’s age is 4 : 5, and the ratio of his age to his son’s age is 6 : 1. Four years ago these ratios were 11 : 14 and 11 : 1, respectively. The ratio of the age of the grandfather to that of the grandson 12 years from now will be (1) 12 : 5 (2) 14 : 3 (3) 18 : 5 (4) 18 : 7 2. The ratio of present ages (in years) of a father and son is 15 : 8. Six years ago, the ratio of their ages was 13 : 6. What is the father’s present age ? (1) 65 years (2) 58 years (3) 45 years (4) 78 years 3. The average age of a man and his son is 60 years. The ratio of their ages is 13 : 7, respectively. What is the son’s age ? (1) 40 years (2) 41 years (3) 42 years (4) 43 years 4. The average of the ages of a group of 65 men is 32 years. If 5 men join the group, the average of the ages of 70 men is 34 years. Then the average of the ages of those 5 men joined later (in years) is (1) 50 (2) 55 (3) 65 (4) 60 5. The ratio of the present ages of A and B is 6 : 5. Four years ago, this ratio was 5 : 4. What will be the ratio of the ages of A and B alter 12 years from now ? (1) 3 : 2 (2) 8 : 7 (3) 9 : 8 (4) 7 : 6 Answer Key 1. (3)

2. (3)

3. (3)

4. (4)

5. (3)

Answers with Explanations 1. Option (3) is correct. Let M, F, S be the ages of man, his father and his sons, respectively. According to the question, M = 4 ⇒ M = 4 F ….(i) F 5 5 M 6 = ⇒M= 6S S 1

….(ii)

4 years ago,

M − 4 11 = F − 4 14

….(iii)

and

M − 4 11 = S−4 1

….(iv)

From equation (iv), M – 4 = 11(S – 4) Substituting (ii) in (iv) we get ⇒ ∴ From (ii), From (i),

= 11S − 44 = 40 = 8 years = 6 × 8 = 48 years 4 5 × 48 = 60 years M = F⇒F= 5 4

6S − 4 5S S M

Ages of Grandfather and Grandson after 12 years from now become 60 + 12 = 72 years And, 8 + 12 = 20 years.

257

QUANTITATIVE APTITUDE: AVERAGE AND PROBLEMS ON AGES Grandfather's age = 72 = 18 20 5 Grandson's age

∴ Ratio becomes 18 : 5 2. Option (3) is correct. Let F and S be the ages of father and son, respectively. According to the question, F 15 = ...(i) S 8 F−6 13 and = ...(ii) S−6 6 8 F From (i), S= ….(iii) 15 From (ii), C(F – 6) = 13(S - 6) ⇒ 6F – 13S = – 42

⇒

 8  6F − 13  F  = – 42  15 

(using (iii))

104 F = – 42 15 90F − 104F ⇒ = – 42 15 −14F ⇒ = – 42 15 42 × 15 ∴ = 45 years F= 14 3. Option (3) is correct. Let F and S be the ages of father and son, respectively. Given:

⇒

6F −

and

From (i),

(using (ii))

20S = 120 7

⇒

S=

120 × 7 = 42 years 20

⇒ ⇒

5x + 2080 = 34 × 70 5x = 300

⇒

x=

300 = 60 years 5

∴ Average age of the last 5 men who joined the group is 60 years. 5. Option (3) is correct. Let the ages of A be x and B be y. 6 x = 5 y

...(i)

x−4 5 = y−4 4

...(ii)

x=

6 y 5

From (ii),

13 S + S = 120 7

⇒

2080 + 5x = 34 70

From (i),

...(ii)

13S + 7S = 120 7

We have,

and

F + S = 120 13 F= S 7

⇒

65

∴ Sum of the ages of 65 men = 65 × 32 = 2080 Also, sum of 5 men who joined the group = 5x

Given:

F 13 F+S = 60 years and = S 7 2

∴

∴ The son’s age is 42 years. ∴ B’s present age is 90 years. 4. Option (4) is correct. Let x be the average age of the last 5 men who joined Now, sum of the ages of 65 men = 32

...(i)

44xx −−16 16 == 55yy −− 20 20 44= xx 55yy −− 44 ⇒ = Substituting (iii) in (iv), we get 6 4 × y = 5y – 4 5

⇒ ⇒ Then,

….(iii) ….(iv)

24 y − 5y = – 4 5 y = 20 6 x= × 20 = 24 years 5

24 + 12 36 9 = = 20 + 12 32 8 i.e., 9 : 8 is the ratio of the ages of A and B after 12 years from now. = After 12 years, ratio

= S.P – C.P = C.P – S.P P = × 100% C.P L = × 100% C.P

x − y  × 100 • Gain/Loss% =   y    gain    100 ±  loss      × C.P  = • S.P   100     where x = S.P, y = C.P • If an object is sold on r% loss, [100 − Loss%] × C.P then, S.P = 100 100 or C.P = × S.P P [100 − Loss%]

• Loss %

• Profit %

• Profit • Loss

• Cost Price (C.P) • Selling Price (S.P) • Marked Price (M.P) Loss (P-L) (P L) • Profit Profit-Loss

x2 % 100

If a man sells two similar objects, one at a loss of x% and another at a gain of x%, then he always incurs loss in this transaction and loss% is

ndred un ‘Per’ hundred d by % denoted

Depreciation p on 3 × 100 = 75% 4

Fraction to %

% tto fraction 50 50 50% = 100

Population P Po aaf ft n years after n R   = P1 +   100 

R   Value after n years = P  1 −   100 

n

Second Level

Trace the Mind Map

Third Level



xy  xy  Single g discount =  x + y − % 100   where x and y are two successive discounts.

More/less Percentage • A is R % more than B, then B is less than A by  R  × 100 %  R + 100   • A is R% less than B, then B is more than th h A by  R  × 100   %  100 − R 

First Level

• If a% and b% are two succeissive losses then (negative sign shown ab loss and positive sign shows profit). Total loss % = [−a − b + ] 100 • If a% profit and b% loss occur, simultaneously then overall loss or ab profit % is [a − b − ]% 100 ab  loss  % • If a% loss and b% profit occur then, total is −a + b − 100  profit  (−ve sign for loss, +ve sign for profit)

S.P P × 100  100 − D  S.P = MP   or MP = 1000 − D  100 

If an article is sold at D% discount, ou unt, then n

Discount

Percentage, Profit–Loss and Discount

x1− x ×100 x

= Marked Price – Selling Price Discount 00 × 100 Discount % = Marked price

Discount

Profit / Loss

then, Increment x1 % =

x − x0 then Reduction %= × 100 x • If x is increased to x1,

• If x is reduced to x0,

258 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Percentage, Profit–Loss and Discount

Chapter

5

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

Percentage

4

2

8

Profit–Loss

2

6

2 10

Discount, Tax

Types of Analogies

Sol.

• Percentage • Profit–Loss • Discount

Topic-1

20 × 100% = 25% 100 − 20

Results on Depreciation:

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum then, n R   Value of the machine after n years = P  1 −   100 

Percentage

 Revision Notes

The word “Per Cent” means per hundred. It is denoted by the symbol ‘%’. Scan to Example : 30 per cent is written as 30% know more 30 about meaning 30 out of 100 = this topic 100

a a as a per cent, We have × 100% b b i.e., we can express the ratio in terms of per cent by multiplying it by 100%

To express

Profit and

Loss Example: 4 = 4 × 100%= 80% 5 5 We can express the percentage in terms of fraction by dividing it by 100. 25 1 Example: 25% = = 4 100

Percentage on base value: a ×b Suppose we want to find a% of b = 100 (here b is called base value) Percentage increase =

Increase × 100 Base value

Percentage decrease =

Decrease × 100 Base value

Percentage change =

Change × 100 Base value

More/Less Percentage Problems : • If A is R% more than B, then B is less than A by  R   R + 100 × 100  %  

•

Required percentage =

If A is R% less than B, then B is more than A by  R   100 − R × 100  %  

Example: If A’s salary is 20% less than B’s salary, by how much per cent is B’s salary more than A’s ?

Value of the machine n years ago =

P R    1 −  100 

n

Example: The value of a mobile depreciates at the rate of 20% per annum. The present value of the mobile is `40,000. What will be the worth of mobile two years from now ? Sol. Depreciation value of a mobile 2 20   2 years from now = 40,000  1 −   100 

= 40,000 ×

80 × 80 = ` 25,600 100 × 100

Problems on Population:

If P is the population of a town and suppose it increases (decreases) at the rate of R% per annum, then, n R  Population after n years = P  1 +   100  n

R   [for decrease P  1 −  ]  100 

Population n years ago =

P R   1+   100 

n

[for decrease

P R    1 −  100 

n

]

Example: The population of a town 2 years ago was 62,500. Due to migration to cities, it decreases every year at the rate of 4%. The present population of the town is ? n R  Sol. Present population = P  1 −   100  

4 

= 62,500  1 −   100 

2

24 24 × 25 25 62, 500 × 576 = = 57,600 625

= 62,500 ×

∴ The present population of the town is 57,600

260

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions 1. A shop and a car were sold for ` 2,70,000 each. Due to increased scope of business, shop was sold at a profit of 20% but due to depreciation, car could be sold at a loss of 10%. How much profit or loss resulted in the entire transaction? (CUET 2023) (1) Loss ` 5,000 (2) Loss ` 10,000 (3) Profit ` 15,000 (4) Profit ` 15,480 2. The price of rice increases by 30%. A family reduced its consumption so that the expenditure of the rice is up only by 10%. If the total consumption of the rice before the price rise was 10 kg per month, then the consumption of rice per month at present (in kg ) is: (CUET 2023) 1 6 6 5 (1) 7 (2) 8 (3) 8 (4) 8 15 15 13 13 3. If P = x% of y and Q = y% of x, then which of following is true? (CUET 2022) (1) P = Q (2) P > Q (3) P < Q (4) Relationship between P and Q cannot be established 4. A boy secures 70%, 40% and 60% marks in test papers with 100, 50 and 150, respectively as maximum marks. The percentage of his aggregate is : (CUET 2022) (1) 55 (2) 60 (3) 56.66 (4) 70 5. What per cent decrease in salaries would exactly cancel the 20 per cent increase ? (CUET 2022) 1 2 2 1 (1) 15 % (2) 14 % (3) 16 % (4) 33 % 3 3 3 3 6. If 20% of a number is 30, then the number is : (CUET 2022) (1) 6 (2) 150 (3) 60 (4) 15 7. Sachin’s income is 25% more than Dileep’s income. By how much percentage is Dileep’s income less than Sachin’s income ? (1) 15% (2) 20% (3) 18% (4) 22% 8. The sum of the number of male and female students in an institute is 100. If the number of male students is x, then the number of female students becomes x% of the total number of students. Find the number of male students. (1) 50 (2) 65 (3) 45 (4) 60 9. In an election, candidate X got 70% of the overall valid votes. If 20% of the overall votes were declared invalid and the total numbers of votes is 6,40,000, then find the number of valid votes polled in favour of the candidate. (1) 3,58,400 (2) 4,50,000 (3) 4,00,000 (4) 3,58,000 10. If 40% of a number is less than its 60% by 30, then the 20% of that number is : (1) 60 (2) 40 (3) 50 (4) 30 11. Ravinder invests `3,750 which is equal to 15% of his monthly salary in a medical insurance policy. Later he invests 25% and 8% of his monthly salary on a child education policy, and mutual funds, respectively. The total amount left with him is : (1) `15,000 (2) `12,000 (3) `8,000 (4) `13,000 12. Bala decided to donate 10% of his salary to PM Care Fund. On the day of donation, he changed his mind and donated `1,800 which was 60% of what he had decided earlier. How much is his salary ? (1) `36,000 (2) `32,000 (3) `30,000 (4) `40,000 13. A shopkeeper has certain number of apples of which 10% are found to be rotten. He sells 85% of the remaining good apples and still has 405 good apples. How many apples did he originally have ? (1) 3500 (2) 3000 (3) 2500 (4) 2000 14. Sudha spends 80% of her income. When her income is increased by 30%, she increases her expenditure by 25%. Her (1) increased by 5% (2) decreased by 30% (3) decreased by 5% (4) increased by 50%

Answer Key 1. (3)

2. (3)

3. (1)

4. (2)

5. (3)

6. (2)

9. (1)

10. (4) 11. (4) 12. (3) 13. (2) 14. (4)

7. (2)

8. (1)

Answers with Explanations 1. Option (3) is correct. Given that the selling price of shop and car each = ` 2,70,000 According to the question, The cost price of shop = and the cost price of car =

2 , 70 , 000 × 100 = ` 2,25,000 120

2, 70, 000 × 100 = ` 3,00,000 90

So, total cost price of both articles = 2,25,000 + 30,000 = 5,25,000 and selling price of both = 2 × 2,70,000 = 5,40,000 So, profit = 5,40,000 –5,25,000 = ` 15,000 2. Option (3) is correct. Given price of rice increased by 30%, let the family reduced the consumption by x % Now percentage increase in expenditure =10% xy +y Using successive percentage change = x + 100 30 x ⇒ 10 = 30 − x − 100 200 130 x ⇒ = 20 ⇒ x = % 100 13 According to the question, present consumption of rice 200    100 − 13  110 6 = = 8 kg. = 10 ×   13 13  100    3. Option (1) is correct. Here, P = x% of y = xy/100 and Q = y% of x = xy/100 So, P = Q 4. Option (2) is correct. 70% of 100 = (70/100) of 100 = 70 marks 40% of 50 = (40/100) of 50 = 20 marks 60% of 150 = (60/100) of 150 = 90 marks So, marks secured = 70 + 20 + 90 = 180 marks And total marks = 100 + 50 + 150 = 300 marks So, required percentage = (180/300) × 100 = 60% 5. Option (3) is correct. Let the salary amount in starting = ` 100 Then after 20% increment the salary amount = ` 120 Required percentage decrease = [(120 – 100) / 120] × 100 = 2000 / 120 = 50/3% =16(2/3)% 6. Option (2) is correct. Let the number is x. As per the question, 20% of x = 30 1% of x = 30/20 100% of x = 1.5 × 100 = 150 7. Option (2) is correct. Let x and y be the income of Sachin and Dileep, respectively. If A is R% more than B, then B is less than A by

261

QUANTITATIVE APTITUDE: PERCENTAGE, PROFIT-LOSS AND DISCOUNT

12. Option (3) is correct. Let Bala’s salary be x. Then, 60% of 10% of x = 1,800

 R   R + 100 × 100  %  

So, Dileep’s income less than Sachin’s income  25   25  × 100  % = 20% × 100  % =  =   125   25 + 100  8. Option (1) is correct. Given: i.e., x + x% of 100 = 100  x  ⇒ x+ × 100 = 100  100 

⇒ 2x = 100 ⇒ x = 50 9. Option (1) is correct. Given: Overall votes = 6,40,000 and invalid votes are 20% of the overall votes. Then, valid votes are 80% of the overall votes 80 × 6, 40, 000 = 5,12,000 = 100 Now, number of vote’s candidate ‘x’ got 70 = × 5,12, 000 = 3,58,400 100 10. Option (4) is correct. Let the number be x. 60% of x − 40% of x = 30 ⇒

 60   40  × x −  × x  = 30    100  100

⇒

4 6 x  −  = 30 10 10  

⇒

1 x   = 30 5

x = 150 20 × 150 = 30 20% of x = 100

∴

Shortcut Method  60 40  x −  = 30  100 100  ⇒

3, 750 × 100 15

= 25,000

Now, 25% of 25,000 + 8% of 25,000 + 3,750

=

60 x × 100 10

= 1,800

6x = 1,80,000 x=

⇒

1, 80, 000 6

= 30,000

∴ Bala’s salary is `30,000 13. Option (2) is correct. Let the original number of apples shopkeeper has be x. Given: 10% are rotten, so good apples are 90% of x. Also, he sells 85% of good apples. so, he is left with 15% of good apples. Now, 15% of [90% of x] = 405  90  × x  = 405 ⇒ 15% of   100 

15 9 x = 405 × 100 10

⇒

x=

⇒

405 × 1, 000 135

= 3,000

∴ Originally, the shopkeeper has 3,000 apples. 14. Option (4) is correct. Let Sudha’s income be `100 Therefore, amount she expends is 80% of income, i.e., `80 units. Increment is 30%, i.e., her income becomes 100 + 30 = `130 Expenditure = 25% of 80 = `20 ∴ Total expenditure = 80 + 20 = `100 Hence, her savings now is 130 − 100 = `30 Her earlier savings is 20 ∴ Difference in savings is 30 − 20 = `10 10 So percentage increases in savings = × 100 = 50% 20

Topic-2

Profit and Loss

Some Important Definitions:

 20  x  = 30  100 

x =

⇒

 Revision Notes

∴ 20% of x = 30 11. Option (4) is correct. Let Ravinder’s salary be x. 15 × x = 3,750 Then, 100 ⇒



× x  = 1,800 60% of   100 

⇒

male students + female students = 100

⇒

 10

⇒

25 8 × 25, 000 + × 25,000 + 3,750 100 100

= 6,250 + 2,000 + 3,750 = 12,000 ∴ Amount left with Ravinder = `25,000 − `12,000 = `13,000

Cost Price : The price at which an article is purchased is called its Cost Price and denoted as C.P. Selling Price : The price at which an article is sold, is called its Selling Price and denoted as S.P. Marked Price : The price on the label of an article is called the Marked Price. Profit : If Selling Price is greater than Cost Price, the seller is said to have a profit. Profit = S.P. − C.P. Loss : If Cost Price is greater than Selling Price, the seller is said to have loss. Loss = C.P. − S.P. Formulae List : Gain/Profit = S.P. − C.P. Loss = C.P. − S.P. P × 100% Profit Per cent = C.P.

262

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Poss Per cent =

L × 100% C.P.

 100 + Gain Per cent  S.P. =   × C.P. 100  100 − Loss Per cent  S.P. =   × C.P. 100

C.P. =

100 × S.P. 100 + Gain%

C.P. =

100 × S.P. 100 − Loss%

If the Cost Price of x articles is equal to Selling Price of y article, then x − y Gain/Loss =   × 100  y  [If sign of the answer is ‘+’ then profit and If sign of the answer is ‘−‘then loss.] We can also solve the problems by rewriting the problems and solving it by cross multiplication method. Example: A person incurs 5% loss by selling a watch for `1140. What is the cost price of the watch ? Sol. Suppose, Cost Price of the watch is `100. ( 5% Loss) Then person sells it at 100 − 5 = `95 Now, the person sold it at `1140. ∴ Cost Price of the watch C.P. S.P. 100 × 1,140 = 95 100 95 1140 ? = `1,200 Example: A man buys a cycle for `1,400 and sells it at a loss of 15%. What is the selling price of the cycle ? Sol. Suppose, C.P. of the cycle is 100 Then S.P. of the cycle become 85 Now, C.P. of the cycle is 1,400 1, 400 × 85 C.P S.P Then S.P. of the cycle is = 100 85 100 1,400 ? = `1,190 Similar way, one can solve the problems on profit also, building a problem like above is useful when formulas are confused or forgotten. Example: A person purchases an article at `3,000, sells it at a profit of 15%. What is selling price of the article ?  100 + gain per cent  Sol. S.P =   × C.P.  100

⇒

S.P = S.P =

115 × 3, 000 100

3, 000 × 115 100

= `3,450 OR

= `3,450

C.P 100 3,000

S.P 115 ?

Objective Type Questions 1. The difference between the cost price and the selling price of an article is ` 240. If the profit is 20%, then the selling price is: (CUET 2023) (1) ` 1,200 (2) ` 1,240 (3) ` 1,440 (4) ` 1,600 2. If the cost price of 12 pens is equal to the selling price of 8 pens, then what is the gain percent? (CUET 2023) (2) 50% (3) 66 2 % (4) 70% (1) 33 1 % 3 3 3. A shop and a car were sold for ` 2,70,000 each. Due to increased scope of business, shop was sold at a profit of 20%

but due to depreciation, car could be sold at a loss of 10%. How much profit or loss resulted in the entire transaction? (CUET 2023) (1) Loss ` 5,000 (2) Loss ` 10,000 (3) Profit ` 15,000 (4) Profit ` 15,480 4. A trader sold half of his articles at 60% profit, half of the remaining at 20% loss and the rest was sold at the cost price. In total transaction his gain % or loss % will be: (CUET 2023) (1) Gain 20% (2) Loss 5% (3) Gain 40% (4) Gain 25% 5. By selling on book for `250 , loss percentage is 10%. What is the cost price? (CUET 2023) (1) 277.7 (2) 250 (3) 285 (4) 315 6. A wholesaler sells 20 pens at the marked price of 16 pens to a retailer. The retailer sells them at the marked price. Determine the gain percent of the retailer. (CUET 2023) (1) 30% (2) 23% (3) 25% (4) 28% 7. A man makes a profit of 20% after selling a product on 28% discount on the printed price. Find the ratio of printed price to cost price (CUET 2022) (4) 8 : 3 (1) 7 : 3 (2) 5 : 3 (3) 4 : 3 8. Radha sells her scooty for `50,000 at a loss of 20%. At what price she should sell her scooty to make a profit of 10% ? (CUET 2022) (1) ` 65,750 (2) ` 64,750 (3) ` 66,750 (4) ` 68,750 9. If a saree is sold for `3,060, the seller will face 15% loss, at what price should he sell the saree to gain a 20% profit ? (1) `4,650 (2) `3,600 (3) `3,440 (4) `4,320 10. A man sold his furniture at a 25% gain. Had he sold it at 15% loss, he would have received `800 less. Find cost price of the furniture. (1) `2,500 (2) `1,500 (3) `2,000 (4) `3,000 Answer Key 1. (3)

2. (2)

9. (4)

10. (3)

3. (3)

4. (4)

5. (1)

6. (2)

7. (2)

8. (4)

Answers with Explanations 1. Option (3) is correct. Let the cost price of the article = `x According to the question, 120 × x − x = 240 100 20 x = 240 × 100 ⇒ x = 1,200 120 So, selling price = 1, 200 × = 1,440 100 ⇒

2. Option (2) is correct. According to the question, 12 C.P = 8 S.P. C.P. 2 C.P. 8 ⇒ = ⇒ = S.P. 3 S.P. 12 ∴ Required profit percentage =

( 3 − 2 ) × 100 2

= 50%

3. Option (3) is correct. Given that the selling price of shop and car each = ` 2,70,000 According to the question, The cost price of shop = 2 , 70 , 000 × 100 = ` 2,25,000 120

and the cost price of car =

2, 70, 000 × 100 = ` 3,00,000 90

So, total cost price of both articles = 2,25,000 + 30,000 = 5,25,000

263

QUANTITATIVE APTITUDE: PERCENTAGE, PROFIT-LOSS AND DISCOUNT

and selling price of both = 2 × 2,70,000 = 5,40,000 So,

profit = 5,40,000 –5,25,000 =` 15,000

4. Option (4) is correct. Given, Half of his article sold at 60% profit. Half of the remaining at 20% loss. And rest at cost price only. Let there was total 4 articles. Using weighted average, Total Profit =

( 2 × 60 ) + (1 × ( −20 ) ) + (1 × 0 )

2+1+1 120 − 20 = = 25% 4

=

Hence, the total profit = 25% 5. Option (1) is correct. Given that: S.P. of book = ` 250 loss % = 10% Using, C.P. ×  100 − loss%  = S.P.   100  

250 × 100 = 277.7 90

6. Option (3) is correct. According to the question: For wholesaler, wholesale, S.P. of 20 pens = M.R.P. of 16 pens. Let the M.R.P. of 1 pen = `100 1, 600 ∴ Cost price of 1 pen for retailer = = `80 20

∴ Profit % for retailer =

120 × 3, 600 100

= `4,320

10. Option (3) is correct. Suppose the cost price is `100. Then, selling price with 25% gain= `125 and selling price with 15% loss is= 85 Now, 125 − 85 = 40 800 × 100 = `2,000 ∴ Cost price of the furniture = 40

 100 − 10 %  C.P. ×   = S.P. = 250 100   C.P. =

8. Option (4) is correct. Given that, The S.P. of scooty = ` 50,000 Loss percentage = 20% So, C.P. = (50,000 × 100) / 80 = ` 62,500 Now, new profit percentage = 10% So, new S.P. = (62,500 × 110) / 100 = ` 68,750 9. Option (4) is correct. Let the cost price of the saree be x. C.P. S.P. By given data 100 85 3, 060 × 100 = 3,600 C.P = ? 3,060 85 The selling price of the saree with 20% profit

(100 − 80 ) × 100%

80 20 = × 100% = 25% 80

Topic-3

Discount

 Revision Notes Discount is a concession given by a seller to the customer while selling goods. Discount is always given on Marked Price. Discount = Marked price − Selling price Remember : Discount ×100 Rate of discount(Discount %) = Marked price The formula for total discount in case of successive discounts. If the first discount is x% and the second discount is y%, then xy   % Total discount =  x + y −  100 

7. Option (2) is correct. Let the C.P. of product = X Then S.P. = 1.2X

Example: The successive discount of 10% and 20% are given on the purchase of a bag. Write the single discount equivalent to this. xy   % Sol. We have total discount =  x + y −  100 

Now let the printed price = Y Then S.P. = 0.72 Y Now, comparing both S.P.,

⇒

0.72 Y = 1.2X Y/X = 1.2 / 0.72 = 5/3

 (10 × 20 )  Total discount =  10 + 20 − %  100   

200 

=  30 −  % = 28%  100 

So, the ratio of printed price to cost price = 5/3

Objective Type Questions 1. The marked price and cost price of a book are `850 and `748, respectively. The discount in percentage is : (1) 12% (2) 10% (3) 15% (4) 8% 2. If the marked price of a television set is `24,500, then, its selling price after a 12% discount on it is : (1) `21,460 (2) `21,650 (3) `21,640 (4) `21,560 3. In a grocery store, an item with an MRP of `1,100 is on a discount counter with a special price of `979. What is the percentage of discount given for that item ? (1) 9% (2) 10% (3) 12% (4) 11%

4. A dealer marks his goods at 30% above the cost price. Then he allows 35% discount on it. What would be his loss percentage ? (1) 16.5% (2) 17.5% (3) 18.5% (4) 15.5% 5. A product, whose MRP is `978, is sold for `925 by a wholesale shop owner. What is the percentage of discount given by him ? (1) 6.5% (2) 9.2% (3) 5.4% (4) 7.8% 6. The marked price of a shirt was `1,800. A man bought the same shirt for `1,200 after getting two successive discounts. If the first discount was 12%, what was the second discount rate ? (Correct to two decimal place) (1) 25.25% (2) 22.22% (3) 24.24% (4) 20.20%

264

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

7. A man gets a discount of 30% and then 20% on his food bill. How much equivalent single discount does he get ? (1) 44% (2) 50% (3) 40% (4) 35% 8. A man gets a discount of 30% and then 20% on his food bill of `1,250. How much amount does he has to pay ? (1) ` 700 (2) ` 550 (3) `350 (4) `500 9. In a 15% discount sale, the cost of a book is `2,150. What was the original price of the book ? (Correct to two decimal places) (1) `2,529.41 (2) `2,500.00 (3) `1,527.00 (4) `2,250.50 10. A shopkeeper pays 12% of the cost price as tax while purchasing an item whose cost is `500. He wants to earn a profit of 20% after giving a discount of 16% on the marked price. So, the marked price should be : (1) `840

(2) `780

(3) `960

(4) `800

Let the second discount be x%. Then x × 1, 584 = 384 100 x=

1. (1)

2. (4) 10. (4)

3. (4)

4. (4)

5. (3)

6. (3)

7. (1)

8. (1)

Answers with Explanations 1. Option (1) is correct. We have Discount % =

=

Total discount = =

12 y  100  %=  12 + y − % 3 100   Solve it to get y = 24.24 % 7. Option (1) is correct. We have,

xy   % Total discount =  x + y −  100 

Given: Marked price = `24,500 and discount = 12% 88 × 24, 500 = `21,560 ∴ S.P = 100

= (50 − 6)% = 44%

8. Option (1) is correct.

xy   % Total discount =  x + y −  100   

=  30 + 20 −

Now, amount to pay =

Now, loss% = 100 − 84.5 = 15.5% 5. Option (3) is correct.  978 − 925  × 100 Discount% =   978 

=

5, 300 978

= 5.4%

6. Option (3) is correct. The amount after first discount of 12% 88 × 1, 800 = `1,584 = 100 The second discounted amount = 1,584 − 1,200 = 384

(100 − 44 ) × 1, 250 100

56 × 1, 250 = `700 = 100

= 11%

4. Option (4) is correct. Let the cost price of the goods be `100. Then, marked price = 100 + 30 = `130 and selling price after 35% discount 65 = 130 × = 84.5 100

600  % 100 

= (50 − 6)% = 44%

3. Option (4) is correct.

=

600 



2. Option (4) is correct.

1,100 − 979 × 100 1,100

1, 800 − 1, 200 100 × 100 = % 1, 800 3

=  30 + 20 − %  100 

850 − 748 10, 200 × 100 = = 12% 850 850

Discount % =

M.P. − S.P. × 100 M.P.

xy   % Also, total discount =  x + y − 100  

M.P. − S.P. × 100 M.P

 M.P. − S.P.   M.P.  × 100

= 24.24%

Shortcut Method

Answer Key 9. (1)

38, 400 1, 584

9. Option (1) is correct. 85 We have, × C.P. = 2,150 100

2150 × 100 = `2,529.41 85 10. Option (4) is correct. The amount paid by shopkeeper Cost price = 500 + 12% of 500 = 500 + 60 = `560 120 × 560 = `672 Now, selling price with 20% profit = 100 Now, to find the marked price, given that he gives 16% discount, then 84 × M.P. = 672 100 ⇒ C.P. =

⇒ M.P =

672 × 100 = `800 84

a or a : b b

a:b3a : 3a

Properties

Componendo – Dividendo a c ab cd    b d ab c d

a : b :: c : d d is forth proportion

ab is mean proportion of a and b

a : b :: b : c, c is third proportion

Inversely proportional k If x  where k is a constant. y 1 It is written as x  y

Directly proportional If x  ky, where k is a constant. It is written as x  y

Proportion a c a : b  c : d or  b d

If a : b :: c : d then ad = bc

Second Level

Third Level



Componendo a c ab cd    b d b d

Simple Partnership

Compound Partnership

Trace the Mind Map First Level

Dividendo a c ab cd    b d b d

Partnership Partnership Share in a Business, Joint venture, etc.

Alternendo a c a b    b d c d

Invertendo a c b d    b d a c

Sleeping Partner

Partners Who invest in a Business

Alligation

Active Partner

C.P. of dearer Mean price Quantity of cheaper  Quantity of dearer Mean price  C.P. of cheaper

Ratio & Proportion/ Alligation & Partnership

Mixed ratio Let x:y and P:Q be two ratio then, Px:Qy is Called mixed ratio.

Sub-triplicate ratio

Triplicate ratio a : b  a3 : b 3

Ratio and Proportion

Ratio:

Duplicate ratio a : b  a2 : b 2

a:b a : b

Sub-duplicate ratio

QUANTITATIVE APTITUDE: RATIO AND PROPORTION, ALLIGATION & PARTNERSHIP

265

Chapter

6

Ratio and Proportion, Alligation & Partnership

Chapter Analysis Concept Name

2021

Ratio and Proportion, Alligation and Partnership

 Revision Notes Ratio and Proportion Ratio : A ratio is a way to show a relationship or compare two numbers of same kind. The ratio of two amounts is equal to a fraction. It shows how much less or more time an amount is in comparison to another. Denoted by a : b i.e., a b

2022

2023

Additional Questions

1

5

14

(ii) Alternendo : a = c then a = b b d c d Scan to know more about this topic

(iii) Componendo : a = c then a + b = c + d b d b d (iv) Dividendo : a = c then a − b = c − d b d b d a+b c+d a c = = then a−b c−d b d Example: Find the third proportional of 16 and 36 Sol. Given: 16 : 36 :: 36 : x Then, 16 × x = 36 × 36 36 × 36 ⇒ x= = 81 16 ∴ The third proportion is 81. Example: Two numbers are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is Sol. Given: Two numbers are in the ratio 3 : 5. Let x be the common ratio, then 3x − 9 12 = 5x − 9 23 ∴

(v) Componendo - Dividendo : If

Ratio and

Proportion • In a : b we call ‘a’ as first term or anticedent and ‘b’ as second term or consequent. • The multiplication or division of each term of a ratio by the same non−zero number does not affect the ratio. Example : 3 : 5 or 3 5 Now, multiplying both the side by 5 we get 3 × 5 = 15 ratio is 3 : 5 only 5 × 5 25 i.e., 3 : 5 = 15 : 25 Remember : (i) Duplicate Ratio : If a : b is a ratio, then a2 : b2 is a duplicate ratio of a : b (ii) Sub-duplicate Ratio : If a : b is a ratio, the a : b is sub− duplicate ratio of a : b (iii) Triplicate Ratio : If a : b is a ratio then a3 : b3 is triplicate ratio of a : b (iv) Sub-triplicate Ratio : If a : b is a ratio, then 1 1   a : 3 b  or a 3 : b 3  is a sub-triplicate ratio of a : b  Proportion : Proportion says that two ratios are equal. 3

a c = b d Proportion is the equality of two ratio as a : b :: c : d here ‘a’ and ‘d’ are called extremes and ‘b’ and ‘c’ are called mean terms. e.g., 2 : 7 = 6 : 21, then we write 2 : 7 : :6 : 21 Remarks • If a : b :: c : d, then we have, Product of extremes = Product of means i.e., a × d = b × c • Fourth Proportional : If a : b = c : d then d is called fourth proportion to a, b, c. • Third Proportional : If a : b = b : c then c is called the third proportional to a and b • Mean Proportional : Mean proportional between a and b is ab • Continued Proportion : The non-zero numbers a, b, and c will a b continued proportion, if = i.e., b2 = ac b c be in Here, b is known as mean proportional and c is known as third proportional.

i.e., a : b = c : d or

Properties of Proportion (i) Invertendo : a = c then b = d [or a : b = c : d then b : a = d:c] b d a c

⇒ ⇒ ⇒ ⇒

23(3x − 9) = 12(5x − 9) 69x − 207 = 60x − 108 69x − 60x = 207 − 108 9x = 99 x = 99 = 11

⇒ ∴

9

Smaller number = 3x = 3 × 11 = 33

Alligation or Mixture

Alligation is the simplified, faster technique to solve the problems based on weighted average. • Mean Price : The cost price of a unit quantity of the mixture is called the mean price. • If two ingredients are mixed then Quantity of cheaper C.P. of dearer − Mean price = Quantity of dearer Mean price − C.P. of cheaper Above ratio can be explained as follows C.P. of a unit quantity of cheaper C.P. of a unit quantity of dearer. (c)

(d)

(Mean Price) (m) (d - m)

(m - c)

[Note mean price value higher matter’s or lower] (Cheaper quantity) : (Dearer quantity) = (d − m) : (m − c)

267

QUANTITATIVE APTITUDE: RATIO AND PROPORTION, ALLIGATION & PARTNERSHIP

Example: Cost of two types of pulses is `15 and `20 per kg, respectively. If both pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg. Sol. Let x be the cost of mixed variety of pulse, Now, 2 : 3 = (20 − x) : (x − 15) ⇒ 2x + 3x = 60 + 30 ⇒ 5x = 90 ⇒ x = 18 Weighted average =

n1 A1 + n2 A 2 , n1 + n2

where, n1 and n2 are numbers in a group one and two respectively. A1 and A2 are averages of group one and two respectively. Example: 20 kg of rice costing `24 per kg and 30 kg of rice costing `40 per kg are mixed. Find the average cost of the mixture per kg. Sol. Let Aw be the average cost of the mixture. n1 A 2 − A w We have, = n2 A w − A 1 20 40 − A w = 30 A w − 24

⇒ ⇒ ⇒ ⇒

20(Aw – 24) = 30(40 – Aw) 20Aw + 30 Aw = 1200 + 480 50Aw = 1680 Aw = 1680 = 33.6 50

⇒

(ii) Sleeping Partner : A partner who provides capital but does not involve directly in the business is known as a sleeping partner. Types of Partnership: (i) Simple Partnership : If all the partners invest capital for the same time period. This is known as simple partnership. (ii) Compound Partnership : If all the partners invest capital for different period of time, then this is called compound partnership. Remark : If a1 : a2 : a3 be the ratio of investment and b1 : b2 : b3 be the ratio of profits then ratio of time periods is given by b1 b2 b3 : : a1 a2 a3

Example: A, B, C enter into a partnership investing `35,000, `45,000 and `55,000, respectively. Then find the shares of A, B and C on profit of `40,500. Sol. Given: A, B and C are partners. and, A : B : C = 35,000 : 45,000 : 55,000 = 7 : 9 : 11 Also, 7 + 9 +11 = 27 7 A’s share = 40 , 500 × = `10,500 27

Shorcut Method

• Weighted average = n1 A1 + n2 A 2 = 20 × 24 + 30 × 40 n1 + n2

20 + 30

1680 = = 33.6 50

Partnership

When two or more people invest in the same business, then this is called Partnership and the people who invest in the business are called Partners. There are two types of partners: (i) Active Partner : A partner who manages the business is known as an active partner or working partner.

B’s share = 40 , 500 ×

9 = `13,500 27

C’s share = 40 , 500 ×

11 = `16,500 27

Example: Arun, Kamal and Vijay invested `8,000, `4,000 and `8,000, respectively in a business. Arun left after 6 months. If after eight months, there was a gain of `4,005, then what will be the share of Kamal? Sol. Given: Arun : Kamal : Vijay = (8,000 × 6) : (4,000 × 8) : (8,000 × 8) = 48 : 32 : 64 = 3 : 2 : 4 2

Kamal’s share = `  4 , 005 ×  = ` 890 9 



Objective Type Questions 1. Two numbers are in the ratio 3: 5. If 10 is added to each number, then the ratio becomes 5: 7. Find the numbers. (CUET 2023) (1) 30, 50 (2) 45, 60 (3) 15, 25 (4) 36, 60 2. The fourth proportional to 5, 8, 15 is: (CUET 2023) (1) 18 (2) 24 (3) 19 (4) 20 3. The difference between two numbers P and Q (P > Q) is 40% of their sum. The ratio P : Q is: (CUET 2023) (1) 6: 4 (2) 7: 3 (3) 8: 5 (4) 5: 4 4. Joseph gifted ` 20,000 to his wife and some money to his three children aged 12,14 and 16 years in the ratio of their ages. If he gave ` 3,000 to his youngest child, then how much money he gifted to his family? (CUET 2023) (1) ` 30,000 (2) ` 30,150 (3) ` 30,225 (4) ` 30,500 5. A sum of ` 1,250 is divided among A, B and C such that A gets 2 of B’s share and C gets 3 of A’s share. Find the shares of A, 4 9 B and C (in `). (CUET 2023) (1) 200, 900, 150 (2) 200, 800, 250 (3) 150, 800, 300 (4) 190, 850, 210

6. Ratio between age of P and Q is 5 : 7. If 2 years ago Q’s age was 2 years more than the age of P after 4 years, then find their total age (CUET 2022) (1) 30 years (2) 35 years (3) 42 years (4) 48 years 7. If a : b = 3 : 5 and b : c = 2 : 3 then the proportion a : b : c is: (1) 6 : 10 : 15 (2) 3 : 5 : 3 (3) 3 : 10 : 3 (4) 3: 10 : 15 8. Dividing the amount `18,144 among three people A, B, C in the ratio 3 : 5 : 8, the amount B gets more than A is: (1) `2,178 (2) `2,268 (3) `2,464 (4) `2,386 9. If an amount of `800 is distributed between Ravi, Mohan and Govind in the proportions 2 : 5 : 3, then the sum of the shares of Mohan and Govind is: (1) `560 (2) `400 (3) `640 (4) `240 10. In a bag, white marbles and red marbles are in the ratio of 3 : 5. If the number of red marbles are 150, then how many white marbles are there? (1) 90 (2) 60 (3) 30 (4) 120 11. The ratio of the incomes of A and B is 2 : 3 and that of their expenditures is 1 : 2. If 90% of B’s expenditure is equal to the income of A, then what is the ratio of the savings of A and B? (1) 1 : 1 (2) 9 : 8 (3) 8 : 7 (4) 3 : 2

268

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

12. What number must be added to each of the numbers 8, 13, 26 and 40 so that the numbers obtained in this order are in proportion? (1) 2 (2) 3 (3) 1 (4) 4 13. The ratio of the number of men and women in a factory is 14 : 19. If the total number of employees in the factory is 2,145, then the number of women in the factory is: (1) 1,367 (2) 1,976 (3) 1,645 (4) 1,235 14. When 50% of a number A is added to B, the second number B increases by 25%. The ratio between the numbers A and B is: (1) 3 : 2 (2) 2 : 3 (3) 1 : 2 (4) 3 : 4 15. What is the third proportional to 16 and 24? (1) 36 (2) 28 (3) 34 (4) 32 16. The ratio of tables and chairs in a room is 7 : 9. If there are 560 tables and chairs in the room, then what is the number of chairs? (1) 397 (2) 315 (3) 463 (4) 489 17. How much should be added to each term of 4 : 7 so that it becomes 2 : 3? (1) 3 (2) 4 (3) 2 (4) 1 18. Two numbers A and B are in the ratio 5 : 2. If 4 is added to each number then this ratio becomes 9 : 4. If 5 is subtracted from each of the original numbers, then the ratio of A and B will be: (1) 3 : 1 (2) 8 : 3 (3) 7 : 2 (4) 4 : 1 1 2 3 19. Three numbers are in the ratio : : . The difference between 2 3 4 the greatest and the smallest number is 27. The smallest number is: (1) 81 (2) 40 (3) 72 (4) 54 20. An oil merchant wants to make a minimum profit of `2,100 by selling 50 litres of oil he purchased at `236 per litre. For this, he adds a few litres of duplicate oil whose cost price is `180 per litre and sells at `250 per litre. How many litres of duplicate oil is needed for this purpose? (1) 22 litres (2) 16 litres (3) 18 litres (4) 20 litres Answer Key 1. (3)

2. (2)

3. (2)

4. (4)

5. (1)

6. (4)

7. (1)

4. Option (4) is correct. Given: Joseph gifted his wife = `20,000 Ratio in which he gifted amount to his children = 12 : 14 : 16 =6:7: 8 = 6 x 3000 = x 500 According to question, So, total amount gifted to children = 500 × 21 = `10,500 ∴ Total amount he gifted to his Family = 20,000 + 10,500 = `30,500 5. Option (1) is correct. Let B’s share = 36x then A’s share 2 × 36 x = 8 x 9 3 and C’s share × 8 x = 6 x 4 So, ratio of shares of A, B and C = 8x : 36x : 6x = 4 : 18 : 3 So, A’s share 4 × 1, 250 = 200 25

B’s share 18 × 1, 250 = 900

25 3 C’s share × 1, 250 = 150 25

6. Option (4) is correct. Let the age of P = 5x Then the age of Q = 7x As per the question, 7x – 2 = (5x + 4) + 2 2x = 8 x =4 Total sum = 12x = 48 7. Option (1) is correct. Given a : b = 3 : 5 and b : c = 2 : 3 To make ‘b’ as same ratio a 3 and b 2 = = b 5 c 3

8. (2)

9. (3) 10. (1) 11. (3) 12. (1) 13. (4) 14. (3) 15. (1) 16. (2)

⇒

17. (3) 18. (1) 19. (4) 20. (4)

Answers with Explanations 1. Option (3) is correct. Let the numbers be 3x and 5x. According to the question, 3x + 10 5 = 5x + 10 7 ⇒ 21x + 70 = 25x + 50 ⇒ 4x = 20 ⇒ x = 5 So the numbers = 3 × 5 and 5 × 5 = 15 and 25 2. Option (2) is correct. Given numbers = 5, 8, 15 let fourth number = x Using, A : B = C : D 5 15 = ⇒ 5 : 8 = 15 : x ⇒ 8 x ⇒ 5x = 15 × 8 ⇒ x = 24 3. Option (2) is correct. According to the question, 40 (P − Q) = (P + Q) 100 ⇒ 100 P − 100Q = 40 P + 40Q ⇒ 60 P = 140Q

⇒

P 140 P 7 ⇒ = = Q 3 Q 60

or P : Q = 7 : 3

a 33 2 and b 2 × 5 = × = b 55 2 c 3×5

a 6 and b 10 = = b 10 c 15 ∴ a : b : c ⇒ 6 : 10 : 15 8. Option (2) is correct. Let x be the common ratio Given: A : B : C is 3 : 5 : 8 i.e., 3x + 5x +8x = 18,144 ⇒ 16x = 18,444 18 , 144 ⇒ = x = 1, 134 16 Amount B gets more than A = 5x − 3x = 2x 2x = 2 × 1,134 = `2,268 9. Option (3) is correct. Given: Amount of `800 is distributed to Ravi, Mohan and Govind in the proportion 2 : 5 : 3. Let x be the common ratio Then 2x + 5x + 3x = 800 ⇒ 10x = 800 ⇒ x = 80 Sum of shares of Mohan and Govind = 5x + 3x = 8x = 8 × 80 = `640 10. Option (1) is correct. Given: Ratio of white marble to red marbles is 3 : 5. Let x be the common ratio. Red marbles = 150 (given) i.e., 5x = 150

⇒

269

QUANTITATIVE APTITUDE: RATIO AND PROPORTION, ALLIGATION & PARTNERSHIP

⇒

= x

Then, a : b :: b : c

150 = 30 5

Number of white marbles = 3x = 3 × 30 = 90 11. Option (3) is correct. Given: Ratio of incomes of A and B is 2 : 3. Ratio of expenditures of A and B is 1 : 2. Let income of A = 2x and income of B = 3x. Also expenditure of A = y and expenditure of B = 2y. We have, 90% of B’s expenditure = income of A i.e., ⇒

90 × 2 y = 2x 100 18 y = 2x 10

….(i)

Ratio of savings = (2x – y) : (3x – 2y) Substituting (i) in (ii), we get

….(ii)

 3 18  18 Ratio of savings =  y − y  :   y  − 2 y   10   2  10  

=  8 y  :  27 y − 2 y  10 10 

   7 y 8     = = 8:7  10 y  :  10     

12. Option (1) is correct. Let x be the number to be added Then, (8 + x) : (13 + x) = (26 + x) : (40 + x) We have (8 + x) × (40 + x) = (13 + x) × (26 + x) ⇒ 320 + 8x + 40x + x2 = 338 +13x + 26x +x2 ⇒ 320 + 48x = 39x + +338 ⇒ 48x − 39x = 338 − 320 ⇒ 9x = 18 18 ⇒ = x = 2 9 13. Option (4) is correct. Given: Ratio to men and women in the factory is 14 : 19. Total number of employees in the factory is 2145. Let x be the common ratio then 14x + 19x = 2,145 ⇒ 33x = 2,145 2 , 145 ⇒ = x = 65 33

∴ Number of women in the factory = 19x = 19 × 65 = 1,235. 14. Option (3) is correct. Given: A and B are two numbers, such that 50% of A + B = 25% of B + B 50 25 ⇒ A+ B = B+ B 100

⇒ ⇒

100 1 1 A= B 2 4 A 2 1 = = B 4 2

∴ A : B = 1: 2 15. Option (1) is correct. Let c be the third proportion. Here, a = 16, b = 24

ac = b2 16c = 24 × 24 24 × 24 c= 16 c = 36

⇒ ⇒

⇒ 16. Option (2) is correct. Given: There are 560 tables and chair in the room. Ratio of tables to chair is 7 : 9 i.e., 7x + 9x = 560 ⇒ 16x = 560 560 ⇒ = x = 35 16 ∴ Number of chairs in the room = 9x = 9 × 35 = 315 17. Option (3) is correct. Let x be the number to be added. According to the question, 4 + x = 2 7+x

3

⇒ 3(4 + x) = 2(7 + x) ⇒ 12 + 3x = 14 + 2x ⇒ x=2 18. Option (1) is correct. Given: Ratio of A and B is 5 : 2 Let A and B be 5x and 2x, respectively. Then,

5x + 4 9 = 2x + 4 4

⇒ 4(5x + 4) = 9(2x + 4) ⇒ 20x + 16 = 18x + 36 ⇒ 2x = 20 ⇒ x = 10 ∴ A = 5x = 5 × 10 = 50 and B = 2x = 2 × 10 = 20 Now, when 5 is subtracted from A as well as B, we get 50 − 5 45 3 = = 20 − 5 15 1 ∴ Required ratio = 3 : 1 19. Option (4) is correct. Let the common ratio be x 3 1 Then, (given) x − x = 27 4 2 3x − 2 x ⇒ = 27 4

x = 27 × 4 x = 108 1 ∴ The smallest number = × x = 1 × 108 = 54 2 2 20. Option (4) is correct. Let x be the quantity of duplicate oil added Given: Cost of oil = 50 × 236 = `11,800 Cost of duplicate oil = x × 180 = `180x Merchants profit = `2,100 Now, Cost of oil + Cost of duplicate oil + Profit = Selling price ⇒ 11,800 + 180x + 2,100 = 250(50 + x) ⇒ 13,900 + 180x = 12,500 + 250x ⇒ 13,900 − 12,500 = 250x − 180x ⇒ 1,400 = 70x ⇒ ⇒

⇒

x=

1, 400 = 20 litres 70

Amount = Principal + S.I.

Simple Interest

A × 200 then, equal installments = T  200 + ( T − 1 ) r  where P = Principal T = Time (in years) A = Amount R, r = Rate of interest per annum

If a sum is to be deposited in equal installments,

2n

4n

12 n

S.I. × 100 PT

PTR 100

Rate = Ra

Simple Interest =

S.I. × 100 Time = PR

Simple Interest and Compound Interest

R   Compounded Monthly, Amount = P  1 +   12 × 100 

R     Compounded Half Yearly, Amount = P  1 + 2  100    

R/4  Compounded Quarterly, Amount =P  1 +   100 

n

 – 1  100   r

n

S.I. × 100 TR

First Level

Second Level

Trace the Mind Map

Principal =

Third Level



2 P r 2 ( 300 + r )  r  and 3 years is P  (100 )  100 

The difference between C.I. and S.I. obtained on ` P at r % per annum for 2 years is

Amount = Principal + C.I. or C.I. = Amount – Principal

 C.I = P 1+ 

Compound Interest

R  Compounded Annually, Amount = P  1 +   100 

270 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Simple Interest & Compound Interest

Chapter

7

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

2

2

6

2

8

Simple Interest Compound Interest •

Types of Interest • Simple Interest • Compound Interest

Topic-1

•

Simple Interest

When an amount becomes n times of itself in t years at the rate of interest R% per annum, then R × T = (n − 1) × 100 If rate of interest becomes R2 % from R1% and `a is earned more in T years, then Principal (P) =

 Revision Notes • • • • •

Interest : It is the monetary charge for the Scan to privilege of borrowing money. know more Principal : If A borrows money from B, about then the borrowed money is called principal. this topic Rate of Interest : This is the rate at which the extra money we have to pay for using or keeping borrowed money. Time : The period for which the money is lent / borrowed is called time. Simple Simple Interest : If the principal remains Interest & the same for whole loan period, then the Compound interest is called Simple Interest. Interest PTR S.I. = 100 where P is Principal, T is Time or Period and R is Rate of Interest.

Amount : The sum of principal and interest is known as amount. Amount = Principal + S.I. ⇒

•

•

•

•

•

If a principal P becomes n times of itself in T1 years and m times in T2 years at the rate of R per annum, then n −1 m−1 = T1 T2 Principal P becomes n times of itself at the rate of R1 per annum and m times at the rate of R2 per annum in T years, then n −1 m−1 = R1 R2 If the difference of interests obtained from two sources on `P in time T years be `a, then the difference of rates of interest will be R=

a × 100 P× T

1 1 : R1 T1 R 2 T2

If there are distinct rates of interest for distinct time periods, i.e., Rate for 1st T1 years → R1% Rate for 2nd T2 years → R2% Rate for 3rd T3 years → R3% Then, total S.I. for (T1 + T2 + T3) years =

 a If the S.I. is   part of the principal, P at the rate R% per  b annum in t year in place of year also T = R, then

 a   × 100 b

If a sum is divided in two parts such that the simple interest got by first part in T1 years at R1% rate of interest is equal to the simple interest got by second part in T2 years at R2% rate of interest, then

If the sum is divided in more than two parts, then ratio is found accordingly.

RT   A = P1 +   100 

T =R=

)

− R1 × T

aT2 − bT1 ( b − a) × 100 and Rate (R) = aT2 − bT1 T2 − T1

Ratio of the two parts of the sum =

Points to remember : •

2

If a certain sum of money amounts to `a in T1 years and to `b in T2 years, then Principal (P) =

•

(R

a × 100

(

P R1 T1 + R 2 T2 − R 3 T3

)

100 •

If a certain sum becomes n1 times of itself at R1% rate and n2 times of itself at R% rate in same time then, n2 − 1 R1 R2 = n1 − 1

( (

•

If a certain sum becomes n1 times of itself in T1 years and n2 times of itself in T2 years at the same rate, then n2 − 1 t1 T2 = n1 − 1

( (

•

) )

) )

If a certain sum P amounts to `A1 in T1 years at rate of R% and the same sum amounts to `A2 in T2 years at same rate of interest R%. Then,



A −A



 A 2 T1 − A 1 T2  T1 − T2 

1 2 × 100 (ii) P =  (i) R =   A 2 T1 − A 1 T2  

272

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions 1.

2.

3. 4.

5.

6.

7. 8.

9.

1 A certain sum on 13 % per annum simple interest amounts to 2 ` 2,502.50 in 4 years. Find the sum. (CUET 2023) (1) ` 1,500 (2) ` 1,625 (3) ` 1,650 (4) ` 1,720 For a certain period, simple interest on ` 1,000 at 10% per annum is ` 100 more than that on ` 800. Find the period (in years). (CUET 2023) (1) 4 (2) 10 (3) 5 (4) 2 An amount becomes its 3 times in 20 years. What is the rate of simple interest per annum? (CUET 2022) (1) 15% (2) 10% (3) 6.67% (4) 6% What will be the simple interest earned on an amount of ` 1 (CUET 2022) 22,000 in 8 months at 8 % per annum? 4 (1) ` 1,013 (2) ` 1,012 (3) ` 1,210 (4) ` 1,215 The difference of simple interest on a sum of money for 8 years and 10 years is `200. If the rate of interest is 10% p.a. then what is the sum of money ? (1) `1,000 (2) `1,400 (3) `1,600 (4) `1,200 Suresh lent out a sum of money to Rakesh for 5 years at simple 9 interest. At the end of 5 years, Rakesh paid of the sum to 8 Suresh to clear out the amount. Find the rate of simple interest per annum. (1) 3% p.a. (2) 2% p.a. (3) 3.5% p.a. (4) 2.5% p.a. In how many years will a sum of `5,000 yield a simple interest of `2,000 at an interest rate of 10% p.a. ? (1) 5 years (2) 3 years (3) 4 years (4) 6 years A person invested a total of `9,000 in three parts at 3%, 4% and 6% per annum on simple interest. At the end of a year, he received equal interest in all the three cases. The amount invested at 6% is: (1) `2,000 (2) `3,000 (3) `4,000 (4) `5,000 If the total simple interest on a sum of `1,400 for 4 years at rate of interest x% p.a. and on the same sum for two years at the same rate, is `672, then the value of x is: (1) 9% (2) 8% (3) 6% (4) 10%

10. At simple interest, Gaurav borrows `1,500 from Sandeep at the rate of 14% per annum. What amount of money should Gaurav pay to Sandeep after 1 year to clear the debt ? (1) `1,700 (2) `1,710 (3) `1,705 (4) `1,715 Answer Key 1. (2)

2. (3)

3. (2)

4. (4)

5. (1)

6. (4)

7. (3)

9. (2) 10. (2)

Answers with Explanations 1. Option (2) is correct. 1 2

= 13 = % Given: Rate of interest

27 % 2

Total amount = ` 2,502.5 Time = 4 years. Let the principal amount = `P Using,

SI =

PRT 100

P × 27 × 4 2 × 100 2 , 502.5 = 1.54 P ⇒ P = 1, 625 ( 2 , 502.5 − P) =

⇒

Hence, principal amount `1,625

8. (1)

2. Option (3) is correct. Let the time period = T years. P × R×T Using, S.I. = 100 According to the question: 1000 × 10 × T 800 × 10 × T − = 100 100 100 ⇒ 20 T = 100 ⇒ T = 5. Hence required time period =5 years. 3. Option (2) is correct. Given that, Time duration = 20 years Let the principal amount = x So, interest earned in this duration = 2x As per the simple interest formula, SI = (P × R × T) / 100 2x = (x × R × T) / 100 R = (2 × 100) / 20 R = 10% p.a. 4. Option (3) is correct. Given that, P = `22,000 T = 8 month = 2/3 years 1 33 = R 8= % % 4 4 As we know, S.I. =

P × R×T 100

S.I. =

22 , 000 × 33 × 2 3 × 4 × 100

S.I. = `1,210 5. Option (1) is correct. Let the principal amount be P. We have,

Simple Interest =

PRT 100

According to the question, (S.I.)10 years − (S.I.)8 years = 200 ⇒ ⇒ ⇒ ⇒

P × 10 × 10 P × 8 × 10 − = 200 100 100 P − 0.8P = 200 0.2P = 200 200 = `1,000 P= 0.2

Shortcut Method : 10 × 10 − 10 × 8 = 20% Given, 20% = 200 100% = 200 × 5 = `1,000 6. Option (4) is correct. Let P be the principal amount T = 5 years and R = ? S.I. =

P×T×R , also A = P + S.I. 100

S.I. = A − P =

9 P−P 8

=

1 P 8

…(i)

273

QUANTITATIVE APTITUDE: SIMPLE INTEREST & COMPOUND INTEREST

∴ ⇒ ⇒

1 P×5×R P = 8 100 P × 100 R= 8×5×P R = 2.5% p.a

using (i)

Shortcut Method : 1 Rakesh paid of sum as interest, 8

Shortcut Method :

100 then rates = 2.5% 8×5

S.I. per year =

7. Option (3) is correct. Given: P = 5,000, S.I. = 2000, R = 10% and T = ? PTR We have, S.I. = 100 2,000 =

⇒

500T = 2,000 T=

Rate =

2, 000 = 4 years 500

Shortcut Method : Time =

2, 000 × 100 = 4 years 5, 000 × 10

8. Option (1) is correct. Given:(S.I.)3% = (S.I.)4% = (S.I.)6% ….(i) Let x be the amount invested at 3% per annum and y be the amount invested at 4% per annum. Then [9,000 − (x + y)] is the amount invested at 6% per annum. Now, from (i) 4y 3x = 100 100 9, 000 − ( x + y )  × 6 100 ⇒ 3x = 4y = [9,000 − x − y]6 ⇒ 3x = 4y = 54,000 − 6x − 6y Consider first two terms of (ii) ⇒ 3x = 4y 4 ⇒ x= y 3 Consider last two terms of (ii), ⇒ 54,000 − 6x − 6y = 4y Substituting (iii) in (iv), we get 4  4y ⇒ 54,000 − 6  y − 6 y = 3 

=

⇒ ⇒ ⇒ ∴

….(ii)

112 × 100 = 8% 1, 400

Shortcut Method : Amount to pay =

114 × 1, 500 100

= 1.14 × 1,500 = `1,710

Topic-2

Compound Interest

 Revision Notes The interest for which principal changes with every conversion period is called compound interest. The sum can be compounded, annually, quarterly or monthly. Amount : Amount is the sum of principal value and interest. In other words, amount is the money which is paid to the lender at the end of period/time. Amount = Principal + Interest

Points to remember : ….(iii) ….(iv)

54,000 − 8y − 6y = 4y 18y = 54,000 54, 000 = 3,000 y= 18 x=

672 = 112 6

10. Option (2) is correct. Given: P = `1,500, R =14% and T = 1 year PTR We have, S.I. = 100 1, 500 × 1 × 14 = = `210 100 ∴ Amount to pay to clear the debt = `1,500 + `210 = `1,710

5, 000 × T × 10 100

⇒

⇒

9. Option (1) is correct. Given: (S.I.)4 years + (S.I.)2 years = 672 1, 400 × 4 × x 1, 400 × 2 × x + ⇒ = 672 100 100 ⇒ 56x + 28x = 672 ⇒ 84x = 672 672 ⇒ x= = 8% 84

4 × 3, 000 3

⇒ x = 4,000 Therefore, amount invested at 6% = 9,000 − (3,000 + 4,000) = `2,000 Shortcut Method L.C.M. of (3, 4, 6) = 12 3 × 4, 4 × 3, 6 × 2 2 × 9,000 = `2,000 Amount inverted at 6% = (4 + 3 + 2)

• When interest is compounded annually R   Amount = P  1 +   100 

n

• When interest is compounded half-yearly 2n   R      2 Amount = P  1 +  100     • When interest is compounded quarterly, then 4n   R      4 Amount = P  1 +  100     • When interest is compounded yearly, but period is in fraction, we 3 can use the following formula for the period of 4 years 5 3   4 R R    5  then Amount = P  1 + 1 +     100   100   

274

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• When rates are different for consecutive years. That is, R1%, R2%, R3% for 1st, 2nd and 3rd respectively, then  R  R  R  Amount = P  1 + 1   1 + 2   1 + 3   100   100   100  • If compound interest on an amount for 2 years is C.I. and simple interest on that money is S.I, then Principal (P) =

( S.I.)2

4 ( C.I. − S.I.)

• If a certain sum at compound interest becomes a times in T1 year and b times in T2 year, then 1

( a) T

1

1

= ( b ) T2

• A − B Formula: If a sum is increased by A% and again by B%, A×B then, effective rate of increase = A + B + 100 • At compound interest, a sum becomes `A in ‘a’ years and `B in ‘b’ years then, B  (i) If b − a = 1, then, R% =  − 1  × 100% A 

(ii) If b − a = 2, then,

 B  − 1  × 100% R% =    A 

1    B n  (iii) If b − a = n then, R% =   − 1 × 100%  A     where ‘n’ is a whole number. • If a sum becomes ‘n’ times of itself in ‘t’ years on compound  1  interest, then R% = n t − 1 × 100%  

• The simple interest for a certain sum for 2 years at an annual R   interest rate R% is S.I., then C.I. = S.I.  1 + 200   • A certain sum at C.I. becomes x times in n1 year and y times in n2 1

1

years then x n1 = y n2 • The difference between C.I. and S.I. obtained on `P at r% per 2 P r 2 ( 300 + r )  r  and for 3 years is . annum for 2 years is P   3  100  (100 )

Objective Type Questions 1. ` 800 becomes ` 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what will ` 800 amount to in 3 years: (CUET 2023) (1) ` 1,02,080 (2) ` 1,025 (3) ` 1,052 (4) ` 1,050 2. A certain sum becomes 36 of itself after 2 years on compound

25

3.

4.

5.

6.

7.

8.

9.

interest. Find the rate of interest per annum. (CUET 2023) (1) 15 % (2) 20 % (3) 25 % (4) 10 % The difference between the compound interest on a sum of `8,000 for 1 year at the rate of 10% per annum, interest compounded yearly and half yearly is: (1) `20 (2) `40 (3) `30 (4) `10 The compound interest on `4,000 after 3 year is `630.50. Then the rate of interest compounded yearly is: (1) 7% (2) 5% (3) 8% (4) 6% In how many years will a sum of `320 amount to `405 if interest is compounded at 12.5% per annum ? 1 1 (1) 2 years (2) 1 year (3) 2 years (4) 1 years 2 2 There is a 60% increase in an amount in 5 years at simple interest. What will be the compound interest on `6,250 for two years at the same rate of interest, when the interest is compounded yearly ? (1) `1,590 (2) `1,560 (3) `1,500 (4) `1,480 If the difference between the compound interest and simple interest on a certain sum of money for three years at 10% p.a. is `558, then the sum is: (1) `18,500 (2) `18,000 (3) `16,000 (4) `15,000 A sum of `10,000 amounts to `11,664 in 2 years, at a certain rate per annum, when the interest is compounded yearly. What will be the simple interest on the same sum for 5 2 years at the 5 same rate ? (1) `4320 (2) `4160 (3) `3840 (4) `4040 The compound interest on a certain sum for 3 years at 15% p.a., compounded yearly, is `4,167. What is the simple interest on 4 the same sum in 4 years at the same rate ? 5 (1) `6,144 (2) `6,000 (3) `4,800 (4) `5,760

10. What is the rate of interest (in %) if simple interest earned on a certain sum for the 3rd year is `2,000 and compound interest earned in 2 years is `4,160? (1) 8 (2) 10 (3) 12 (4) 6 Answer Key 1. (3)

2. (2)

3. (1)

4. (2)

5. (1)

6. (1)

7. (2)

9. (4) 10. (1)

Answers with Explanations 1. Option (3) is correct. Given that: Principal = ` 800, Amount = ` 956 and Time = 3 years Simple Interest = 956 – 800 = ` 156 P×R ×T SI = 100 800 × R × 3 100

⇒

⇒ 156 =

⇒

R=

156 24

⇒ R = 6.5% New Rate of Interest = 6.5 + 4 = 10.5% Now, Principal = ` 800, R = 10.5% and Time = 3 year P×R ×T SI = 100 SI =

800 × 10.5 × 3 = 252 100

∴ Amount = 800 + 252 = ` 1,052 2. Option (2) is correct. Given that: Time = 2 years. Let the principal amount = p According to the question: 36 r   p = p1 + 25 100  

Þ

⇒

36  r  = 1+ 25  100 

2

n

8. (1)

275

QUANTITATIVE APTITUDE: SIMPLE INTEREST & COMPOUND INTEREST

Þ

⇒

6 r =1+ 5 100

r 1 = = r = 20% 100 5 3. Option (1) is correct. Given P = `8,000, n = 1 year, R = 10% Compounded yearly :

Þ

⇒

 405   1125   320  =  1000     

⇒

 81   45  =  64   40 

⇒

R   Amount = P  1 +   100 

n

∴ 6. Option (1) is correct. Given:

1

Compounded half yearly : R  1+  2 Amount = P  100     

We have,

2n

5   = 8, 000  1 +  100 

8, 000 × 105 × 105 = `8,820 100 × 100 Now, the difference between yearly and half yearly = `8,820 − `8,800 = `20

=

Shortcut Method : C.I. when compounded annually

10 = 8,000 × = 800 100 C.I. when compound half yearly 5× 5 = 10.25 Effective rate = 5 + 5 + 100 10.25 = 8,000 × = 820 100 Difference = 820 − 800 = `20

2

⇒ ⇒ ⇒

R   A = P 1 +   100 

R   4,630.5 = 4, 000 1 +   100 

⇒

46, 305 100 + R  =   40, 000  100 

46, 305 × 100 × 100 × 100 = 40, 000

⇒

We have,

3

n

12   A = 6,250  1 +   100 

R   100 

2

6, 250 × 112 × 112 = `7,840 100 × 100 C.I. = Amount − Principal = 7,840 − 6,250 = `1,590 A=

⇒ Now, Shortcut Method :

Rate of interest =

60 = 12% 5

Effective rate when compounded annually = 12 + 12 + 12 × 12 100 = 25.44% 25.44 = `1,590 100

n    PTR R  − P − = 558 P  1 +    100   100

⇒

3

 

S.I. = 60% of P. PTR S.I. = 100 P×5×R 60% of P = 100 60 P×5×R P = 100 100 60 = R = 12% 5

7. Option (2) is correct. Let P be the principal Given: n = 3 years and R = 10% per annum According to the question, (C.I.) − (S.I.) = 558 ⇒ [A − P] − [S.I.] = 558

25

A = P1 +

n = 2 years

3

[105] = [100 + R] 100 + R = 105 R = 5% 5 46305 3 9261 7 3087 7 441 7 63 3 9 3 3 1 5. Option (1) is correct. P = `320, A = 405, n = ?, R = 12.5% per annum ⇒ ⇒ ⇒

n

C.I. = 6,250 ×

[100 + R ]3

3

n

⇒

5 × 33 × 73 × 52 = [100 + R]3 46.305

n

n

R   A = P1 +   100 

n

⇒

125   1000 

For compound interest,

4. Option (2) is correct. Given: P = 4,000, n = 3 years, C.I. = 630.5, R = ? We have,

2

 9  9   =   8 8

⇒

10   = 8, 000  1 + = `8,800  100 

⇒

 

405 = 320  1 +

⇒

n



⇒ P  1 + 

⇒ ⇒

3  P × 3 × 10 10  = 558  − 1 − 100 100 

P[0.331] − 0.3P = 558 P[0.031] = 558 P=

⇒

558 = `18,000 0.031

∴ Principal value is `18,000 Shortcut Method : Effective rate for S.I. = 10 × 3 = 30% Effective rate for C.I. = 10 + 10 + = 21% 21 × 10 21 + 10 + = 33.1% 100

100 100

276

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Difference = 3.1% 3.1% = 558 558 100% = × 100 = `18,000 3.1

⇒

⇒

8. Option (1) is correct. Given: P = `10,000, A = `11,664, n = 2, R = ? R   A = P1 +   100 

⇒ ⇒

11, 664 ( 100 + R ) = 10, 000 10, 000

Now,

S.I = 2

Effective rate for C.I.=

By number crunching we get 8% 27 = `4,320 Now, S.I.= 10,000 × 8 × 500 9. Option (4) is correct. We have, Then, ⇒ ⇒ ⇒

Shortcut Method : Effective rate for C.I. = 15 + 15 +

2

1, 664 × 100 = 16.64% 10, 000

R   A = P1 +   100 

n

n   R  C.I = P  1 + − 1   100  

 115  3  4,167 = P   − 1   100   23  3  4,167 = P   − 1   20   ( 23 ) 3 − ( 20 ) 3   4,167 = P   ( 20 )3 

8, 000 × 24 × 15 100 × 5

S.I = `5,760

⇒

32.25 + 15 +

(108)2 = (100 + R)2 R = 8% PTR 10, 000 × 27 × 8 = = `4,320 Simple Interest = 100 100 × 5

Shortcut Method :

4,167 × 8, 000 4,167

P = `8,000 [Using a3 − b3 = (a − b) (a2 + b2 + ab)] PTR Now, Simple Interest = 100

n

R   11,664 = 10,000  1 +   100 

⇒ ∴

P=

⇒

15 × 15 = 32.25 100

32.25 × 15 = 52.0875%, 100

Sum =

4167 = `8,000 0.520875

S.I. = 8,000 × 15 ×

24 5 × 100

= `5,760 10. Option (1) is correct. S.I. for 2 years = `4,000 PR = 2,000 100 ∴ ⇒ PR = `2,00,000 For 2 years, C.I. − S.I. = 4,160 − 4,000 = `160 ∴

PR 2 = 160 10, 000

⇒

2, 00, 000 × R = 160 10, 000

⇒

20R = 160 R=

⇒

….(i)

160 = 8% per annum 20

Shortcut Method According to the question, S.I. on `2,000 for 1 year = `160 ∴Rate=

160 × 100 = 8% per annum 2, 000

Working together A & B need x days B & C need y days and A & C need z days to complete the work together 2xyz A, B & C need days xy  yz  zx

A and B together need x days to complete the work and A needs y days to complete the work. xy Then B needs days 4x

days

Then A and B together need

x 1 k

A needs x days, B is k times efficient than A

Outlet treated as –ve work

Tank

First Level

Second Level

Third Level



nD m Trace the Mind Map

x

m part of the work in n D days, then total time taken to finish the work

If x can finish

1m3  1000 liter

then work done by A in 1 one hour  x

‘A’ can finish a work in x hours,

Fills tank in x hours, then part filled 1 in one hour = x

When two pipes fill and one pipe empty, then (F1 + F2 – E)

Inlet treated as +ve work

Pipes & Cisterns

M1 D1 M D  2 2 W1 W2

If M1 men can finish W1 work in D1 days and M2 men can finish W2 work in D2 days then, Relation is

M1  D1  T1 M  D 2  T2  2 W1 W2

If M1 men can finish W1 work in D1 days, working T1 time each day and M2 men finish W2 work in D2 days, working T2 time then, Relation is

Total work done  Number of days  Efficiency

Time & Work

Similar to time & work, deals with problems on filling and emptying the tank

 less efficiency, efficiency  less work done

Work man Man-Day

Work done  Time taken  Rate of work Efficiency  more efficiency,  more work done

A, B and C complete the work in x, y and z days. Together they complete the work in xyz days xy  yz  zx

xy need days xy

A and B need x and y days to complete the work, A & B together

Efficiency and Time are inversely proportional to each other

1 Time taken

1 Rate of work or

Rate of work 

Time taken 

QUANTITATIVE APTITUDE: TIME & WORK AND PIPES & CISTERNS

277

Time & Work and Pipes & Cisterns

Chapter

8

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

1

3

4

12

Time & Work and Pipes & Cisterns

 Revision Notes

Then, they will complete the work together in

Time and Work: Time and Work deal with the time taken by an individual or a group of individuals to complete a piece of work and the efficiency of the work done by each of them.

Concept of Efficiency: Efficiency means rate of doing work. That is if the efficiency is more than one number of days required to complete a certain work will be less. Vice versa. Work done : A can do a work in x days. Then, the amount of work it does is called work done by A. Suppose, A can complete a work in x days, then the work done by one day is 1 . x A in Example : A is twice as efficient as B, and together they finish a piece of work in 18 days. In how many days will A alone finish the work? Sol. (A’s one day work) : (B’s one day work) = 2 : 1 (A + B)’s one day work = 1



18

⇒ ratio]

A’s one day’s work =  1 × 2  = 1 [2 out of 3 from  18 3    27

Hence, A alone can finish the work in 27 days.

Important Formula:

• Work done = Time taken × Rate of work • Rate of work =

1 Time taken

• If a work is completed in x number of days. Then the work done in one

day is 1 . x

• Time = Work done Efficiency or Total work = Efficiency × Number of days • If X number of people can do W1 work in D1 days, working T1 hours each day and Y number of people can do W2 work in D2 days, working T2 hours each day. Then we can write

X × D1 × T1 = Y × D 2 × T2 W1 W2

Some Shortcut Formula:

• If A can do a piece of work in x days and B can do a piece of work in y days, Then both of them working together will do the xy days. x+y

same work in • A, B and C working alone can complete a work in x, y and z days, respectively

xyz days. xy + yz + zx

• Two persons A and B working together, can complete a piece of work in x days. If A working alone, can complete the work in y xy

days, then B, working alone, will complete the work in y−x days. • If A and B working together can finish a piece of work in x days, B and C in y days, C and A in z days. Then A, B and C working together will finish the job in

2xyz days. xy + yz + zx

• If A can finish a work in x days and B is k times efficient than A, then the time taken by both A and B working together to complete the work in

x days. 1+ k

• If A and B working together can finish a work in x days and B is k times efficient than A, then the time taken by (i) A working alone will take (k + 1) x days.  k +1 (ii) B working alone will take   x days.  k  • If A working alone takes ‘a’ days more than A and B, and B working alone takes ‘b’ days more than A and B. Then, the number of days, taken by A and B working together to finish the job is ab . Example : A can finish a piece of work when working alone in 6 days and B while working alone can finish the same work in 12 days. If both of them work together, then in how many days, the work will be finished? 1 6

Sol.

Work done by A in one day =



Work done by B in one day = 1

12

∴ Work done by A and B together in one day =

1 1 2+1 3 1 + = = = 6 12 12 12 4

Hence, A and B working together, can finish the work in 4 days. Shortcut Method A and B working together can complete the work in = xy = 6 × 12 = 72 = 4 x+y

6 + 12

18

Example : P and Q working together takes 15 days to complete a work. If P alone can do this work in 20 days, how long would it take to complete the same work? Sol.

Work done by P in one day =

1 20

Work done by P and Q together in one day =

1 15

279

QUANTITATIVE APTITUDE: TIME & WORK AND PIPES & CISTERNS

∴

Work done by Q in one day = =

Example : A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in how many hours? Sol. Given: Cistern is filled in 8 hours. ∴ Portion of a cistern filled in 1 hour = 1

1 1 − 15 20 4−3 1 = 60 60

Hence, Q working alone can complete the work in 60 days. Shortcut Method Number of days =

=

8

Let x be the time taken by the leak to empty the full tank. According to the question,

xy 15 × 20 = ( y − x ) ( 20 − 15 )

⇒

15 × 20 = 60 5

⇒

Example : Anu is twice as efficient as Sunita and can finish a piece of work in 25 days less than Sunita. Sunita can finish this work in how many days? Sol. Given: Anu is twice as efficient as Sunita, i.e., A = 2S Let x be the number of days Anu takes to complete the work. ∴ Sunita takes 2x days to complete the work. According to the question, 2x − x = 25 ⇒ x = 25 Hence, Sunita takes 25 × 2 = 50 days to complete the work.

Pipes and Cisterns:

Pipes and Cisterns is another format of time and work based questions. Water filling or emptying the tank is treated as work.

Inlet and Outlet:

Inlet is a pipe connected with a tank to fill it. Outlet is a pipe connected with a tank to empty it. Inlet is treated as positive work and outlet is treated as negative work. Example : Pipes A and B can fill a tank in 5 and 6 hours, respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in how many hours? Sol. By given data, we have 17 1 1 1  Net part of the tank filled in 1 hour =  + −  = 60  5 6 12 

1 1 1 − = 8 x 10 1 1 1 = − x 8 10 1 5−4 1 = = x 40 40

⇒ x = 40 Hence, leak will empty the cistern in 40 hours. Points to remember : • If a pipe can fill a tank in x hours, then part filled in one hour is 1 . x

• If a pipe can empty a tank in y hours, then part emptied in one hour is

1 . y

• If a pipe can fill a tank in x hours and another pipe can empty the tank in y hours. If both the pipes are opened, then, the net part filled in one hour = • If a pipe fills

1 1 − x y

1 part of the tank in 1 hour, then the time taken x

by pipe to fill the full tank = x • If three pipes can fill a tank separately in x, y and z hours, then time taken to fill the tank by working together =

xyz xy + yz + zx

hours.

Hence, the tank will be filled in 60 = 3 9 hours 17

17

Objective Type Questions 1. A is twice as efficient as B, and together they finish a piece of work in 14 days. The number of days that A alone will take to finish the work is: (CUET 2023) (1) 11 (2) 21 (3) 28 (4) 42 2. A and B working together can do a work in 10 days. If A can do the work in 15 days, then how many days will B take to do the same work? (CUET 2023) (1) 40 (2) 30 (3) 20 (4) 25 3. A completes 7 of a work in 10 days. Then he completes the 20 remaining work with the help of B in 5 days. The time required for A and B together to complete the entire work is: (CUET 2023) 100 100 100 (1) days (2) days (3) days (4) 200 days 13 17 13 29 7 of a work in 21 days. How many more days will 4. Rajan does 11 he take to complete the work? 1 (1) 9 (2) 12 (3) 15 4

(CUET 2023) (4) 13

1 2

5. A and B can complete a piece of work in 30 days. B and C in 10 days, which C and A in 5 days. If all of them work together, the time taken in completing the work is (CUET 2022) (1) 4 days (2) 5 days (3) 6 days (4) 7 days 6. P, Q and R can complete a work in 12, 9 and 15 days, respectively. Working together, they will complete the same work in (CUET 2022) 39 15 38 39 (1) 3 days (2) 5 days (3) 3 days (4) 5 days 47 42 47 47 7. If 36 farmers can do a piece of work in 24 hours, in how many hours will 18 farmers do it? (CUET 2022) (1) 36 hours (2) 42 hours (3) 48 hours (4) 56 hours 8. 21 workers can make 1500 breads in 18 days. How many workers are required to make 1000 breads in 21 days? (CUET 2021) (1) 10 (2) 12 (3) 15 (4) 16 9. P and Q can finish a work in 10 days and 5 days, respectively. Q worked for 2 days and left the job. In how many days can P alone finish the remaining work? (1) 4 days (2) 10 days (3) 6 days (4) 8 days

280

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

10. 5 men and 8 women can complete a task in 34 days, whereas 4 men and 18 women can complete the same task in 28 days. In how many days can the same task be completed by 3 men and 5 women? (1) 64 (2) 72 (3) 56 (4) 36 11. If 27 people, working 8 hours a day, can complete a task in 12 days, then in how many days will 18 people finish the task working 9 hours a day? (1) 18 days (2) 15 days (3) 16 days (4) 20 days 12. A and B together can complete a piece of work in 15 days. B and C together can do it in 24 days. If A is twice as good as C, then in how many days can B alone complete the work? (1) 40 days (2) 60 days (3) 52 days (4) 45 days 13. Kamal and Anil can dig a pond in 8 days and 14 days, respectively. If the total expense of digging is `4,400, then how much money will Anil earn when they both work together? (1) `1,500 (2) `1,300 (3) `1,600 (4) `1,400 14. P can do a work in 10 days and Q can do the same work in 15 days. If they work on it together for 3 days, then the fraction of the work that left is : 1 2 1 4 (2) (3) (4) (1) 3 2 3 3 15. Ravi can complete a task in 6 days and Mohan can complete the same task in 9 days. In how many days can Ravi and Mohan together complete the same task? (1) 5 days

(2) 3

2 days 5

(3) 3

3 days 5

(4) 4 days

16. A alone can complete a task in 3 days and B alone can complete the same task in 6 days. In how many days can A and B complete it together? (1) 6 (2) 1 (3) 3 (4) 2 17. Tap A can fill a tank in 20 hours and tap B can fill the same tank in 30 hours. If both taps are opened together, then how much time will be taken to fill the tank? (1) 10 hours (2) 16 hours (3) 12 hours (4) 24 hours 18. Tap A can fill a tank in 6 hours and tap B can empty the same tank in 10 hours. If both taps are opened together, then how much time (in hours) will be taken to fill the tank? (1) 15 (2) 18 (3) 20 (4) 16 19. The volume of a tank is 72 cubic metres. Water is poured into it at the rate of 60 litres per minute. How much time will it take to fill the tank? (1) 6 hours (2) 20 hours (3) 12 hours (4) 10 hours 20. Pipes A and B can fill a tank in 8 hours and 12 hours, respectively whereas pipe C can empty the full tank in 6 hours. A and B are opened for 3 hours and then closed and C is opened instantly. C will empty the tank in (1) 4

1 1 1 3 hours (2) 4 hours (3) 3 hours (4) 3 hours 4 2 4 2

Answer Key 1. (2)

2. (2)

3. (2)

4. (2)

5. (3)

6. (1)

7. (3)

8. (2)

9. (3) 10. (3) 11. (3) 12. (2) 13. (3) 14. (3) 15. (3) 16. (4) 17. (3) 18. (1) 19. (2) 20. (4)

Answers with Explanations 1. Option (2) is correct. According to the question, Ratio of efficiency of A and B = 2 : 1 Let total work = 14( 2 + 1) = 42 unit So, time taken by A alone to finish the work =

42 = 21 days. 2

2. Option (2) is correct. Given that

L.C.M.

Efficiency of B = 3 − 2 = 1 Total time taken by B alone to complete the task alone = 30 1 = 30 days 3. Option (2) is correct. 7 Given: A can do of work in = 10 days 20 7 13 part. = So, remaining work = 1 − 20 20 13 Also given of the work, A and B can do in = 5 days. 20 So, total time taken by them together to finish the complete 5 5 × 20 100 days. = work = 13 = 13 13 20

4. Option (2) is correct. 7 Given: Time taken by Rohan to complete of a work in 21 days. 11

So, time taken by Rohan to complete the whole work = 21 ×

11 = 33 days. 7

Hence, the required number of days = 33 − 21 = 12 days. 5. Option (3) is correct. It is given that, Time taken by A and B = 30 days Time taken by C and B = 10 days Time taken by A and C = 5 days Let total work be 30 units (LCM of 30, 10 and 5). (A+B)'s one day work = 30/30 = 1 units (B+C)'s one day work = 30/10 = 3 units (C+A)'s one day work = 30/5 = 6 units Adding all three equations, 2A+2B+2C = 10 2(A+B+C) = 10 A+B+C = 5 So, (A+B+C)'s one day work = 5 units So, time taken by them together to finish the task = 30/5 = 6 days. 6. Option (1) is correct. Given that, Time taken by P to finish the work = 12 days Time taken by Q to finish the work = 9 days Time taken by R to finish the work = 15 days By L.C.M. method of time and work, Total work = L.C.M.(12, 9, 15) = 180 So, per day work of P = 180 / 12 = 15 and per day work of Q = 180 / 9 = 20 and per day work of P = 180 / 15 = 12 So, per day work of P, Q and R together = 15 + 20 + 12 = 47 39 So, the required days 180 = = / 47 3 days 47 7. Option (3) is correct. Given that, time taken by 36 farmers to complete the work = 24 h Let the time taken by 18 farmers to complete the same work = x h

281

QUANTITATIVE APTITUDE: TIME & WORK AND PIPES & CISTERNS

Then by unitary method, 36 × 24 = 18 × x x = 864 / 18 = 48 h 8. Option (2) is correct. 1500 breads are made in 18 days by 21 workers.

21 × 18 workers 1500 1000 bread would be made in 1 day by 21 × 18 × 1000 workers 1500 21 × 18 × 1000 1000 bread would be made in 21 days by 1500 × 21

1 bread would be made in 1 day by

= 12 workers

So, needed workers = 12 workers. 9. Option (3) is correct. Give: P and Q can finish the work in 10 days and 5 days, respectively. 1 Work done by P in one day = 10 Work done by Q in one day =

1 5

Work done by Q in 2 day = 2 × Remaining work = 1 −

x = 2592 = 16 days.

∴

162

12. Option (2) is correct. (A + B)’s one day work =

1 15

(B + C)’s one day work =

1 24

Ratio of efficiency of A and C is = A : C = 2 : 1 Now,

1 2 = 5 5

2 3 = 5 5

Now remaining work is done by 'P'

Let x be the number of days required for 3 men and 5 women to complete the task. Then, x (3m+5w) = 952w 952 w 3m + 5 w

⇒

x =

952 w 3 ( 4 m) + 5w

= 952 w 17 w x = 56 days.

⇒

2C + B = 1

….(i)

and

C+B= 1

….(ii)

15 24

B= 1

60

Shortcut Method :

∴ Time taken by P to complete the remaining work is 6 days. 10. Option (3) is correct. Let the work done by 1 man in 1 day be m Work done by 1 woman in 1 day be w Total work done by 5 men and 8 women in 34 days = 34(5m + 8w) ….(i) = 170m + 272w Also, total work done by 4 men and 18 women in 28 days = 28(4m + 18w) = 112m + 504w ….(ii) From (i) and (ii), 170m + 272w = 112m + 504w ⇒ 58m = 232w ⇒ m = 4w Now, total work = 170m + 272w ⇒ 170(4w) + 272w = 952w

x =

1 15

∴ B alone can finish the work in 60 days.

3 10 = × = 6 days. 5 1

⇒

A+B=

Solving (i) and (ii), we get

3 5 Time taken by P to complete the work = 1 10

⇒

11. Option (3) is correct. Given: 27 people working 8 hours a day can complete the task in 12 days. ∴ Total working hours = 27 × 8 × 12 = 2592 man-hours Let x be the required number of days to complete the task for 18 people working 9 hours a day, According to the question, 18 × 9 × x = 2592 ⇒ 162 × x = 2592

Let the total work done be 120 units, then work done in one day by A + B = 120 = 8 units

….(i)

B + C = 120 = 5 units

….(ii)

15

24

From equ. (i) and (ii) A − C = 3 units Also given, A = 2C ∴ 2C − C = 3 ⇒ C = 3 units ∴ B = 5 − 3 = 2 units Now, time taken by 3 to finish the work =

= 60 days 13. Option (3) is correct. Given: Kamal and Anil dig the pond in 8 and 14 days, respectively. Kamal’s one day work = Anil’s one day work =

[∴ m = 4ω]

120 2

(K + A)’s one day work = =

1 8 1 14 1 1 + 8 14 7+4 11 = 56 56

282

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

17. Option (3) is correct. Tap A can fill the tank in 20 hours

1 K Ratio of Kamal’s work and Anil’s work = = 8 1 A 14

Tap B can fill the tank in 30 hours.

K 14 7 = = A 8 4

⇒

i.e., K:A According to the question, 4x + 7x ⇒ 11x x Hence, Anil’s share

If Tap A and Tap B together opened of then the part of the tank they fill in one hour

= 4,400 = 4,400 = 400 = 4 × 400 = `1,600

=

Shortcut Method : Since efficiency and time taken is inversely proportional. ∴ Ratio of efficiency of Kamal : Anil = 14 : 8 8 = ` 1600 22

If tap A and tap B are opened together, then they need 12 hours to fill the tank. 18. Option (1) is correct. Tap A can fill the tank in 6 hours.

∴ Tank emptied by tap B in 1 hour = 1

15

1

=  −   6 10  =

Q’s one day work = 1

1 10

∴ Tank filled in 1 hour when both the tap’s are opened

10

10 − 6 4 1 = = 60 60 15

∴ When both tap A and tap B are opened, then 15 hours needed to fill the tank.

1  1 Then, (P + Q)’s 3 days work = 3  +   10 15  5

1

= 3 = =  30  10 2 The remaining work = 1 −

1 6

∴ Tank filled by tap A in one hour =

P’s one day work = 1

 3 + 2

1 1 3+2 5 1 + = = = 60 20 30 60 12

Tap B can empty the tank in 10 hours.

14. Option (3) is correct. Given: P and Q do the same work in 10 and 15 days, respectively.

∴

1 30

∴ Tap B can fill the tank in 1 hour =

=7:4

Hence, share of Anil = 4,400 ×

1 20

∴ Tap A can fill the tank in 1 hour =

Shortcut Method : Let the capacity of the tank be 30 litres [L.C.M. of (6, 10)] tap A fills

and

15. Option (3) is correct. Given: Ravi and Mohan complete the task in 6 and 9 days, respectively. 1 Ravi’s one day work = 6 1 Mohan’s one day work = 9

30 = 5 litres/hour 6 30 = 3 litres/hour tap B empties = 10

Then,

1 1 = . 2 2

∴ Time taken to fill the tank when both the taps are opened =

3+2 1 1 5 Then, Ravi and Mohan’s one day work = + = = 6 9 18 18 18 If Ravi and Mohan work together, they need days to 5

∴

5

16. Option (4) is correct.

=

30 = 15 hours 2

19. Option (2) is correct. Given: Volume of the tank = 72 m3 We have, 1 m3 = 1000 litres. ∴ Capacity of the tank = 72 × 1000 = 72,000 litres Now, water poured into it at the rate of 60 litres per minute. i.e.,

complete the task. i.e., 3 3 days to complete the task.

30

( 5 − 3)

72, 000 = 1,200 min 60

Conversion to hour =

1200 = 20 hours. 60

20. Option (4) is correct.

1 A’s one day work = 3

Tank filled by tap A in one hour =

1 8

B’s one day work = 1 6

Tank filled by tap B in one hour =

1 12

Work done by (A + B) in one day = 1 + 1 3

6

= 2 + 1 = 1 6

2

∴ If A and B worked together, they need 2 days to complete the work

∴ Tank filled by tap A and tap B in one hour 1

1

5

=  +  =  8 12  24 Water level in the tank after 3 hours (Tap A + Tap B) = 3 ×

5 5 = 24 8

Average speed =

Opposite direction Relative Speed = (x + y)km/h

Relative speed Train 1 − x km/h Train 2 − y km/h Same direction Relative Speed = (x − y) km/h

Distance travelled to cross a bridge/platform of length y m, by a train of x m is (x + y)

Speed in still water

Downstream Downstrea eam m

Speed of Boat in still water x km/h Speed of stream y km/h

Upstream

Boat and Stream

Downstream speed = (x + y) km/h

Upstream speed = (x− y) km/h

1 (Downstream speed 2 − Upstream speed)

Downstream speed × Upstream speed

Speed of stream =

a pole/man by a train of length x.

Distance Speed = Time

A+B

Total distance Total time

, where A and B

First Level

Second Level

Trace the Mind Map

Third Level



1 2 (Downstream speed + Upstream speed)

Speed of Boat in still water =

Speed = SB + SS

If a : b is the ratio of speed, then b : a is the ratio of time

Average speed =

2 are two speeds

same =

Average speed when time is

Average speed when distance is same 2AB Avg. Speed = A +B (where A and B are two speeds)

1 km /h =

5 m/s 18 18 1 m /s = km / h 5

Conversion of units of speed

Speed, Time and Distance

Speed d = SB – SS

Problem on Trains

Distance travelled to cross

Length of a train = Speed × Time

Total Distance = Distance + Length of the train

Trains having length a and b and Speeds x and y crossing each other then,   a + b  (Opposite     x + y  direction) Time =    a + b  (Same   x − y  direction)   

Distance = Speed × Time or Trains of equal lengths crossing each otherr  2t1 t2 (Opposite  direction)  t1 + t2 Time =  2t1 t2 (Same   t2 − t1 direction)

QUANTITATIVE APTITUDE: SPEED, TIME AND DISTANCE

283

Speed, Time and Distance

Chapter

9

Chapter Analysis Concept Name Speed, Time and Distance & Boat and Stream

2021

2022

2023

Additional Questions

1

3

1

10

1

2

7

Problems on Trains

Topic-1

a of the original time, then the change in b b  speed to cover the same distance is given by =  − 1  × Original a 

Speed, Time and Distance & Boat and Stream

(i) If the new time is

 Revision Notes

speed.

Speed:

Speed is the distance travelled by an object per unit time. We get speed by dividing distance travelled by time taken to cover the distance. i.e.,

Speed = Distance Time

This shows that speed is directly proportional to distance and inversely proportional to time. Similarly, Distance = Speed × Time and

1m=

(ii) ∴ (iii) ∴ (iv)

1 1 minute = hours 60

=

d1 + d2 + d3 + .... + dn t1 + t2 + t3 + .... + tn

In these problems, we are dealing with speeds of boat and stream in a system. Speed of boat in still water : If the speed of a boat is given, then that particular speed is the speed of boat in still water. Speed of the stream : The moving water in a river is called stream. The speed of the water is called speed of the stream. (i) Upstream : The boat moves against the stream i.e., Upstream Speed = Speed of the boat in still water − Speed of the stream

1 hour = 3600 seconds

1 m/s =

Total time taken

Speed of the boat in moving water :

1 hour = 60 minutes

1 km/h =

a

Boat and Stream:

1 km 1000

1 hours 3600

1 second =

a

If speed decreases by , then time increases by or viceb b−a versa.

Speed, Time and Distance

Units : The S.I. units of time and distance are second and metre, respectively. Conversion of Units : (i) 1 km = 1000 m ∴

vice-versa.

a a , then time decreases by or a+b b

• Average speed = Total distance travelled

Distance Speed

Time =

(ii) If speed increases by

Scan to know more about this topic

Upstream Speed = SB − SS (ii) Downstream : The boat moves along the flowing water

5 m/s 18

i.e., Downstream Speed = Speed of boat in still water + Speed of stream

18 km/h 5

Downstream Speed = SB + SS

Points to remember :

• When a certain distance is covered at speed A and the same distance is covered at speed B, then the average speed during the whole journey is given by 2AB

(iii) Still water : While solving problems, water is considered to be stationary and the speed of the water is zero.

Points to remember: •

A+B

Speed of boat in still water = 1 2

• If the ratio of the speed of A and B is a : b, then the ratio of the time taken by them to cover the same distance is b : a or

1 1 : . a b

a • If the new speed is b of the original speed, then the change in b  time taken to cover the same distance is given by =  − 1 × a  Original time

[Downstream speed + Upstream speed]

•

Speed of stream = 1 [Downstream speed − Upstream speed]

•

Average speed of boat =

2

Upstream speed × Downstream speed Speed of boat in still water

285

QUANTITATIVE APTITUDE: SPEED, TIME AND DISTANCE

Objective Type Questions 1. Find the distance covered by a man who walked for 15 minutes at a speed of 3 km/h. (CUET 2023) (1) 700 m (2) 750 m (3) 600 m (4) 1200 m 2. Alice is faster than Bob, Alice and Bob each walk 30 km. The sum of their speeds is 8 km/h and sum of time taken by them is 16 hours. Then Alice’s speed is equal to (CUET 2022) (4) 8 (1) 5 (2) 4 (3) 6 3. A and B are two points in a river. Point X and Y divide the line AB in three equal parts. Let the flow of the river is from A to B. The ratio of the time taken to row from point A to Y and from B to Y is 3 : 4. Find the ratio of the speed of the boat upstream to the flow of the river. (CUET 2022) (1) 5 : 6 (2) 6 : 5 (3) 4 : 5 (4) 5 : 4 4. Suresh covers a distance by a cycle at 10 km/h. He returns to the starting point in a car at a speed of 50 km/h. Find the average speed for the entire journey. (CUET 2022) (1) 36.66 km/h (2) 16.66 km/h (3) 26.66 km/h (4) 46.66 km/h 5. A bus covers the first 39 km of its journey in 45 minutes and the remaining 25 km in 35 minutes. What is the average speed of the car? (CUET 2021) (4) 54 km/h (1) 30 km/h (2) 48 km/h (3) 50 km/h 6. How much time will a horse take to run around a square field of side 175 m, if it runs at the speed of 15 km/h ? (1) 180 s (2) 175 s (3) 155 s (4) 168 s 7. Two cars start from the same place at the same time at right angles to each other. Their speeds are 54 km/h and 72 km/h, respectively. After 20 seconds, the distance between them will be : (1) 480 m (2) 540 m (3) 720 m (4) 500 m 8. Mohan covers a distance of 2.5 km by scooter at the rate of 30 km/h. The time taken by him to cover the given distance in minutes is : (1) 10 (2) 5 (3) 6 (4) 8 9. Rakesh walking at the speed of 8 km/h crosses a bridge in 30 minutes. What is the length of the bridge in kilometres ? (1) 4 km (2) 2 km (3) 3 km (4) 5 km 10. A person walks a distance from point A to B at 15 km/h, and from point B to A at 30 km/h. If he takes 3 hours to complete the journey, then what is the distance from point A to B ? (1) 25 km (2) 10 km (3) 15 km (4) 30 km 11. A man travelled a distance of 35 km in 5 hours. He travelled partly on foot at the rate of 4 km/h and the rest on bicycle at the rate of 9 km/h. The distance travelled on foot is : (1) 8 km (2) 12 km (3) 10 km (4) 15 km 12. A man is walking at a speed of 12 km/h. After every km, he takes rest for 3 minutes. How much time will he take to cover a distance of 6 km ? (1) 42 minutes (2) 40 minutes (3) 48 minutes (4) 45 minutes 13. The wheel of a car has 210 cm diameter. How many revolutions per minute must the wheel make so that the speed of the car is kept at 120 km/h ? (1) 326.42 (2) 245 (3) 303.03 (4) 289 14. A boat covered a distance of 15 km upstream in 5 hours and a distance of 42 km downstream in 6 hours. The speed of the stream in km/h is : (1) 3 (2) 2 (3) 2.5 (4) 1.5 15. A motorboat goes 24 km in 2 hour along the stream and 10 km in 1 hour against the stream. The speed of the motorboat in kilometres per hour, is : (1) 14 (2) 10 (3) 11 (4) 12 Answer Key 1. (2)

2. (1)

3. (2)

4. (2)

5. (4)

6. (4)

7. (4)

9. (1) 10. (4) 11. (1) 12. (4) 13. (3) 14. (2) 15. (3)

8. (2)

Answers with Explanations 1. Option (2) is correct. Given that, Speed of a man =

3 km 5 = 3 × m/s h 18

Time = 15 min = 15 × 60 s . Using, Distance = Speed × time Distance = 3 × 5 × 15 × 60 18 Hence, required distance = 750 m 2. Option (1) is correct. Explanation: Let Alice speed = x km/h Then, Bob’s speed = 8 – x km/h As we know that, Distance / Speed = Time 30 30 So, + = 16 x 8−x 30(8 – x) + 30x = 16x (8 – x) 240 – 30x + 30x = 128x2 – 16x x2 – 8x + 15 = 0 (x – 5) (x – 3) = 0 x = 3 and x = 5 As per the question, Alice speed = 5 km/h 3. Option (2) is correct.

A

||

I

X

||

I

Y

||

I

B

AX = XY = YB Ratio of time taken by boat to go from A to Y and B to Y = 3 : 4 As we know that, Speed of the boat going downstream = (Speed of the boat + Speed of the current) Speed of the boat going upstream = (Speed of the boat − Speed of the current) Time = Distance/Speed Let AX = XY = BY = d Let the speed of the boat = a km/h Let the speed of the current = b km/h As per the question: 2d/(a + b) ÷ d/(a – b) = 3/4 ⇒ 2(a – b)/(a + b) = 3/4 ⇒ 11b = 5a ⇒ a/b = 11/5 Speed of the boat upstream = (11 – 5) = 6 Flow of the current = 5 ⇒ The required ratio = 6 : 5 4. Option (2) is correct. Let the distance = x km Time taken to cover the distance by a cycle = x/10 h And time taken to cover the distance by a car = x/50 h Total distance Total time 2x = x x + 10 50 1000 = = 16.66 km/h 60

So, average speed =

286

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

9. Option (1) is correct. Given: Walking speed of Rakesh = 8 km/h Time taken by him to cross the bridge = 30 minutes Then, length of the bridge is nothing but distance travelled. We have, Distance = Speed × Time

5. Option (2) is correct. Since, we know that Average speed = \ and

Total distance Total time

Total distance = 39 + 25 = 64 km total time = 45 + 35 = 80 min 80 H = 60 H =

 30 

Distance = 8 ×   = 4 km  60  ∴ The length of the bridge is 4 km. 10. Option (4) is correct. Given: Speed of the person

4 3

15 km/h

64 64 × 3 = Average speed = = 48 km/h 4 4 3

 

5 75 = m/s 18 18

Average Speed =



Perimeter of the square field = 4a = 4 × 175 m, when a is the side of the square field = 700 m Time = Distance = 700  75  Speed   18

Now,

=

= Now,

[Distance = Time × Speed] (Unit conversion) Car 1 = 3 × 5 × 20 = 300 m Also, speed of car 2 is 72 km/h (given) Distance travelled by car 2 in 20 seconds

t=

Starting Point

= 72 × 5 × 20 = 400 m 18

By Pythagorean theorem, d2 = (300)2 + (400)2 ⇒ d2 = 90,000 + 1,60,000 ⇒ d2 = 2,50,000 ⇒ d = 500 m Hence, the distance between the cars after 20 seconds is 500 m. Shortcut Method By pythagorean triple (3, 4, 5). (500)2 = (300)2 + (400)2 ∴ Distance = 500 m. 8. Option (2) is correct. Given: Speed of scooter = 30 km/h Distance travelled by Mohan = 2.5 km  Distance   Time 

2.5 = 30 Time

⇒

Time =

Converting into minutes,

t=

2

Given: Let the distance form A to B be d; then, Total time,

d

⇒

AB = 20 × 3 = 30 km

Shortcut Method : Car 2

 5

Speed =  

900 = 20 km/h 45

∴ The distance between the points is 30 km.

7. Option (4) is correct. Given: Speed of car 1 is 54 km/h. Distance travelled by car 1 in 20

We have,

2xy 2 × 15 × 30 = x+y 15 + 30

Distance = Speed × Time 2AB = 20 × 3

⇒

700 × 18 = 168 s 75

seconds = 54 ×   × 20  18 

B

30 km/h

When he walks from A to B, x = 15 km/h Speed of the person when he walk from B to A, y = 30 km/h He covers distance A to B twice (return journey) in 3 hours.

6. Option (4) is correct. Given: Speed of the horse = 15 km/h Conversion to m/s = 15 ×

A

2.5 h 30 2.5 × 60 = 5 m 30

d d + , 15 30

⇒

3 =

d d + 15 30

⇒

3 =

d  1 1+  15  2

⇒

3 =

d 3 × 15 2

⇒

d = 30 km.

11. Option (1) is correct. Let x km be the distance travelled on foot. Then, (35 − x) is the distance travelled by bicycle. Given: Speed of walk = 4 km/h And speed of bicycle = 9 km/h Time = Distance

We have

Speed

According to the question, Total time = t1 + t2 = 5 ⇒

x  35 − x  +  =5 4  9 

9 x + 140 − 4 x = 5 36 ⇒ ⇒ 5x = 40 ⇒ x = 8 km ∴ Distance travelled on foot is 8 km. 12. Option (4) is correct. Given: Walking speed = 12 km/h And distance travelled = 6 km Also, 3 minutes rest after every km.

Time =

Distance + (5 × 3) m Speed

287

QUANTITATIVE APTITUDE: SPEED, TIME AND DISTANCE

Topic-2

= 6 hours + (5 × 3) minutes = 45 m 12

[3 minutes rest after 6th km can be ignored since he already reached the destination] 13. Option (3) is correct. Given: Diameter, d = 210 cm = 2.1 m Perimeter of wheel = πd =

22 × 2.1 = 6.6 metres 7

A wheel covers 6.6 metres in one revolution. Revolutions needed to cover 120 km = 1, 20, 000 = 18,181.81 6.6

∴ Revolutions per minute =

18,181.81 = 303.03 60

 Revision Notes Problems on trains are based on the basic concept of speed, time and distance.

Relative Speed :

• If two trains are moving in the same direction with speed x km/h and y km/h, then relative speed = (x − y) km/h. • If two trains are moving in the opposite direction with speed x km/h and y km/h, then relative speed = (x + y) km/h.

Distance :

• While considering distance, length of the train is also included. Total Distance = Distance covered + Length of the train D + L = Distance travelled Train

14. Option (2) is correct. Let x be the speed of boat and y be the speed of stream.

Note:

3

6

….(i) ….(ii)

x−y= 3 x + y = 10

x = 5 km/h x−y =3 5−y =3 ∴ y = 2 km/h ∴ Speed of the stream, y = 2 km/h

Now,

Shortcut Method :

Total distance covered by the train

• Speed of the train = Time taken to cover the distance

• Let train 1 has length ‘a’ m and speed x m/s and train 2 has length ‘b’ m and speed y m/s and (i)When trains move in opposite direction, time taken by them to a+b . x+y

(ii)When trains move in the same direction, time taken by them to cross each other =

a+b . x−y

• If two trains of equal lengths and different speeds take t1 and t2 time to cross a pole, then the time taken by them to cross each other, (i) When trains are moving in the opposite direction.

Given: Speed of stream = 1 2 [Downstream speed − Upstream speed] = 1  42 − 15 

2  6 3  = 1 [7 − 3] = 2 km/h 2

15. Option (3) is correct. Let x be the speed of boat and y be the speed of stream. Downstream speed = x + y = 24 = 12 km/h 2 10 Upstream speed = x − y = = 10 km/h 1

x + y = 12 x−y= 10

2x = 22 x = 11 ∴ Speed of the boat is 11 km/h

(ii) When trains are moving in the same direction.

1 Speed of Boat = 2

[Downstream speed + Upstream speed] =

1  24 10  +  2  2 1

= 22 = 11 km/h 2

2t1 t2 t1 + t2

2t1 t2 , (t2 > t1) t2 − t1

Example 1 : A train 240 m long passes a pole in 24 s. How long will it take to pass a platform of 650 m long. Sol. Given: Length of the train = 240 m Speed =

240 24

= 10 m/s

∴Time taken to cross 650 m long platform is 240 + 650 890 = = = 89 seconds 10

10

Example 2 : The distance between two railway stations is 1176 km. To cover this distance, an express train takes 5 hours less than a passenger train while the average speed of the passenger train is 70 km/h less than that of the express train. Time taken by the passenger train to complete the travel is : Sol. Let ‘t’ be the time taken by the passenger train to cover the distance 1176 km. Then, time taken by the express train to cover the distance becomes (t − 5) h. 1176 1176 − t−5 t

= 70 ⇒ 1176 t − 1176(t − 5) = 70 t (t − 5) ⇒ 70t2 − 350t − 5,880 = 0 ⇒ t2 − 5t − 84 = 0 2 ⇒ t − 12t + 7t − 84 = 0 ⇒ (t − 12)(t + 7) = 0 ⇒ t = 12 or t = −7 (negative is not possible) ∴ t = 12 hours

According to the question,

Shortcut Method : Given:

Points to remember :

cross each other =

2 x = 10

So, we have

End Point

Distance travelled = D + L

When a train crosses a stationary object the distance travelled by the train is equal to the length of the train.

And speed in downstream = x + y = 42 x − y = 3 km/h x + y = 7 km/h Solving (i) and (ii), we get

Train D Starting Point

Speed in upstream = x − y = 15

So, we have

Problem on Trains

Hence, time taken by the passenger train to complete the travel is 12 hours.

288

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions 1. A train running at a speed of 45 km/h crosses a 100 m long platform in 60 seconds. The time taken by the train to cross an electric pole is: (CUET 2023) (1) 52 seconds (2) 8 seconds (3) 1 minute (4) 1 minute 2 seconds 2. A 110 m long train takes 3 seconds to pass a pole. How long is the platform if the train passes it in 15 seconds? (CUET 2023) (1) 440 m (2) 400 m (3) 550 m (4) 450 m 3. A 400 m long train passes a railway platform in 20 seconds with speed 90 km/h. What is the length of platform? (CUET 2022) (1) 105 m (2) 102 m (3) 99 m (4) 100 m 4. A train covers a distance in 30 min if it runs at a speed of 54 km/h on an average. The speed at which the train must run to reduce the tune of the journey to 20 min is : (1) None of these (2) 81 km/h (3) 60 km/h (4) 75 km/h 5. A train travels the distance between stations P and Q at a speed of 126 km/h, while it comes back at 90 km/h. Another train travels at the average speed of the first train, then the time taken by the second train to travel 525 km is : (1) 5 hours (2) 4 hours (3) 5 hours 20 min (4) 4 hours 20 min 6. Two trains start at same time from stations A and B, respectively, 1800 km apart, and proceed towards each other at an average speed of 44 km/h, and 46 km/h, respectively. Where will the trains meet ? (1) 920 km from station A (2) 900 km from station B (3) 880 km from station A (4) 880 km from station B 7. A train covers a distance of 12 km in 12 minutes. If its speed is decreased by 5 km/h, then the time taken by it to cover the distance of 22 km will be : (1) 24 minutes (2) 20 minutes (3) 22 minutes (4) 18 minutes 8. A 360 m long train running at a uniform speed, crosses a platform in 55 seconds and a man standing on the platform in 24 seconds. What is the length (in metres) of the platform ? (1) 480 (2) 445 (3) 410 (4) 465 9. Two trains of equal length travelling in opposite directions at 72 km/h and 108 km/h cross each other in 10 seconds. In how much time (in seconds) does the first train cross a platform of length 350 m ? (1) 30 (2) 32 (3) 36 (4) 24 10. A train covers a certain distance in 45 minutes. If its speed is reduced by 5 km/h, it takes 3 minutes more to cover the same distance. The distance (in km) is : (1) 64 (2) 60 (3) 54 (4) 80

(100 + l ) = 45 ×

⇒ ⇒ l = 650 m So, time taken by train to cross the pole 650 650 × 18 pole = = = 52 seconds 5 45 × 5 45 × 18 2. Option (1) is correct. Given that: Length of the train = 110 m time = 3 second ∴ speed of train = 110 3 Let the length of platform = x m According to the question, (110 + x ) ⇒ = 15 110 3 ⇒ 330 + 3x = 1650 ⇒ 3x = 1320 ⇒ x = 440 3. Option (4) is correct. Given that, Length of train = 400 m Time taken by the train to cross the platform = 20 s Speed of the train = 90 km/h = 90 × (5/18) m/s = 25 m/s Let the length of platform = L As per the time formula, 20 = (400 + L) / 25 500 = 400 + L L = 500 – 400 = 100 m 4. Option (2) is correct. Given: Average speed of the train = 54 km/h Time taken by the train to cover distance = 30 minutes Then, Distance = Speed × Time

2. (1)

3. (4)

4. (2)

5. (1)

t1 = 20 minutes

Reduced time,

Speed = =

Distance Time

27 = 81 km/h  20    60

5. Option (1) is correct. Given: Train travels PQ at 126 km/h And returns at 90 km/h 126 km/h

6. (3)

7. (1)

8. (4)

P

9. (1) 10. (2)

Q 90 km/h

Answers with Explanations 1. Option (1) is correct. Given that speed of train = 45 km/h = 45 × Length of platform time taken Let the length of train Using, distance

30 = 27 km 60

= 54 ×

Answer Key 1. (1)

5 × 60 18

= 100 m = 60 s =l = speed × time

5 m/s 18

2xy

Average speed of train 1= x + y =

2 × 90 × 126 = 105 km/h 90 + 126

Given, speed of train 2 is average speed of train 1 ∴

Speed of train 2 = 105 km/h Distance = 525 km

∴ Time taken by the second train to travel the given distance = D = 525 = 5 hours S

105

289

QUANTITATIVE APTITUDE: SPEED, TIME AND DISTANCE

6. Option (3) is correct. Given: Distance between two stations A and B = 1800 km Average speed of train 1 (Starting point A) = 44 km/h Average speed of train 2 (Starting point B) = 46 km/h Both trains are moving in opposite direction Relative speed = 46 + 44 = 90 km/h Time (after which they meet) =

D 1800 = = 20 hours S 90

Distance travelled by train 1 from station A in 20 hours = 44 × 20 = 880 km Distance travelled by train 2 from station B in 20 hours = 46 × 20 = 920 km By choosing correct option from the given options. 880 km from station A. 7. Option (1) is correct. Given: Train covers a distance of 12 km in 12 minutes.

9. Option (1) is correct. Let the length of the train be ‘l’ m Speed of train 1 = 72 km/h Speed of train 2 = 108 km/h Given: Trains cross each other in 10 seconds. Relative speed = (72 + 108) = 180 km/h  

D = S × T =  180 ×

⇒ 2l = 500 m ⇒ l = 250 m Now, S = 72 km/h and D = (250 + 350) = 600 m T=

600 Distance = = 600 = 30 s 5  Speed 20 72 ×   18 

∴ Train 1 crosses the platform in 30 seconds. 10. Option (2) is correct. Let the distance between A and B be x km Speed =

Conversion minutes to hour = 12 = 1 hours 60

Then,

5

12 Distance speed of the train = = = 60 km/h  1 Time   5

Also, speed is decreased by 5 km/h (given) Then,

new speed = 60 − 5 = 55 km/h

=

Conversion to minutes =

Distance 22 2 = = h Speed 55 5 2 × 60 = 24 m 5

8. Option (4) is correct. Let the length of the platform be ‘l’ m. Given: Length of the train = 360 m Speed of the train = 360 = 15 m/s 24 ⇒

( l + 360 ) 55

= 15

⇒

l = 15 × 55 − 360

⇒

l = 825 − 360

⇒

l = 465 m

∴ Length of the platform is 465 m.

Distance = x  45  Time   60

60 4 x km/h = x km/h 45 3

The speed of the train is reduced by 5 km/h New speed =

4 4 x − 15 x−5= km/h 3 3

Distance = Speed × Time

Time taken by train to cover a distance of 22 km with speed 55 km/h Time =

5  × 10 18 

x =

4 x − 15 48 × 3 60

⇒ 60x = 16(4x − 15) ⇒ 15x = 4(4x − 15) ⇒ 15x = 16x − 60 ⇒ x = 60 ∴ Distance between two points is 60 km. Shortcut Method : Given: Time increases by ∴ Speed will reduce by

3 1 = 45 15

1

(15 + 1)

=

1 16

1 of S = 5 16

⇒ S = 80 km/h Distance travelled in 45 minutes = 80 ×

45 = 60 km 60

P = 2r +

Sector

r

πr 2 2 C = r (π + 2 )

A=

1 d1 d2 2 P = 4a a A=

Rhombus

1 A = bh 2 P = a+b+c

3 3 2 A= a 2 P = 6a

Regular Hexagon

b

Room

P = 2 (a + b)

A = B× h

a Parallelogram

A=

(

1 5 + 5 +2 2 a 2 4

)

1 (a + b) h 2 c P = a+b+c+d

b Trapezium

A=

l

Sphere 4 V = πr 3 3 TSA = 4 πr 2

h

CSA = 2 πr 2

Hemisphere 2 V = πr 3 3 TSA = 3πr 2

Slant height (l) =

h2 + (R − r )

2

h

R

First Level

Second Level

Third Level



l = Slant height

TSA = π r (l + r )

1 2 l πr h 3 CSA = π rl

V=

Cone

V = lbh

TSA = 2 (lb + bh + lh )

Trace the Mind Map

Curved surface area = π(R + r)l Total surface area π( R + r )l + π( R 2 + r 2 ) πh 2 Volume = (r + R2 + rR) 3 Where, r = radius of top, R = radius of base h = height, l = slant height

● ●

●

Cuboid

CSA = 2(l + b)h

If a cone is cut by a plane parallel to the base, so as to divide the cone into upper part and lower part, then the lower part is called frustum. r

Frustum of Cone

h

TSA = 2 πrh + 2 πr 2

V = πr 2 h CSA = 2 πrh

Cylinder

3D Mensuration

3

Cube

V=a TSA = 6 a 2 LSA = 4 a 2

Mensuration

Regular pentagon

d

A=a P = 4a

2

Square

3 2 a 4 P = 3a A=

Equilateral triangle

= 2(l + b) × h • Total volume of the room =l×b×h

2D Mensuration

Semi-circle

πr θ 180°

Sector of a circle θ A= πr 2 360

A = πr 2 C = 2 πr

Circle

P = 2 ( l + b)

A = lb

Rectangle

A=

1 bh 2 P = a+b+c

Scalene triangle

Right angle triangle

LSA = Lateral Surface Area CSA = Curved Surafec Area

P=Perimeter C=Circumference

• Total area of wall

TSA =Total Surface Area

A= Area

h

290 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

Mensuration

10

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

2D Mensuration

2

1

12

3D Mensuration

1

8

6 OR



Types of Mensuration • 2D Mensuration • 3D Mensuration

• Area = b 4 a 2 − b 2

Topic-1 2D Mensuration

• Height, h =

 Revision Notes

• Perimeter = a + a + b = 2a + b

4

Mensuration is a science of measurement of the lengths of lines, areas of surfaces and volumes of solids. Conversion of unit : 1 m = 10 dm = 100 cm = 1000 mm 1dm = 10 cm = 100 mm 1 cm = 10 mm 1 km = 1000 m

2D Mensuration :

2D Mensuration is nothing but a measurement which deals with a problems on area, perimeter, etc., asked on a plane figure. Area : Area is the quantity that expresses the extent of a twodimensional region or shape. Perimeter : A perimeter is a path that surrounds a two-dimensional shape. Therefore, sum of the sides of a plane figure is the perimeter of the plane figure. In a plane figure, we are defining a mensuration for three or more vertices figures. Because one vertex is just dot and two vertex is line, it is not a closed area to find area of the region. We will start by triangle. Triangle : Vertices : Three Sides : Three Sum of the interior angles = 180° Note : In a triangle, sum of any two sides is more than the third side.

C

3

a

2 • Area = 4 a

• Height =

3 a 2

A

h D a

a B

• Perimeter = a + a + a = 3a (where a = side) (ii) Isosceles Triangle • Two sides are equal • Two angles are equal [∠A = ∠B]



(iii) Scalene Triangle • All the three sides are unequal • All the three angles are unequal

a

h D b

C b

 = 1 × Base × Height

h

2

• Area = 1 × AC × AB sin A

A

D

S ( S − a )( S − b )( S − c )

• Area =

a B

c

2

(Heros’s formula) 

where, S is the semi-perimeter and S = a + b + c 2

• Perimeter = a + b + c (iv) Right Angled Triangle • One of the angle is 90° • Sum of other two angles is 90° • Three sides are in the relationship, c2 = a2 + b2 • AC itself is a height • Area = 1 × AB × AC

C c

b

2

• Perimeter = a + b + c

Types of Quadrilaterals :

1 • Area = d 2  2

C

1 × AB × DC 2 A 1 = × Base × Height 2

• Area =

2

Square : • All sides are equal. • All angles are equal (each 90°). • Diagonals are also equal • Diagonals bisect each other • Area = a × a = a2 • Perimeter = 4a • Diagonal = 2a

Different Types of Triangles (i) Equilateral Triangle • All sides are equal. • All angles are equal (60°).

 b a2 −    2

a

A

a

A a D

B

d

B a

a

C

(Using diagonals)

Rectangle : • Opposite sides are equal. • All angles are equal (each 90°).

a B

• Two adjacent sides are called length and breadth denoted by ‘l’ and ‘b’.

292

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Circle : • Set of all points in a plane which are at a fixed distance from a fixed point in the same plane is called a circle. • Fixed point is the centre. d O r • Fixed distance is the radius, 2 times radius = diameter Centre • Area = πr2 OR

• Diagonals are equal. • Diagonals of a rectangle bisect each other • Area = l × b l

D

C

d

b A

b B

l

• Perimeter = l + l + b + b = 2(l + b) • Diagonals d =

l2 + b2

Area = π

Parallelogram : • Opposite sides are equal. D C • Opposite sides are parallel. • Opposite angles are equal. h b • Area = Base × Height AB × DM A M B a • Perimeter = 2(a + b) • Diagonals divide the parallelogram into triangle of equal area. • Diagonals bisect each other. Trapezium : • One pair of opposite sides are parallel. D

d

h

b

C

c

d2 4

• Perimeter (Circumference) c = 2πr or πd Semi-Circle : • Area =

πr 2 2

O

• Circumference = r(π + 2)

θ πr 180° πrθ • Perimeter = 2r + 180°

• Length of an arc =

Segment of a Circle :

• Perimeter = a + b + c + d Rhombus : • Parallelogram whose all the four sides are equal is called Rhombus. • Opposite angles are equal (not 90°). • Diagonals bisect each other at right angle. a

a

d1

B d2

a

C

A a

a

• All sides are equal.

D

• All angles are equal (each 108°).

a

B

a

)

1 5 + 5 + 2 5 a2 • Area = 4

• Vertices : Six

a

A

a

F

B

• All sides are equal.

a

a

• All angles are equal (each 120°).

E

C

• Sum of the interior angles = 720° • Area = 3 3 a 2 2 where a = side

 πθ sin θ  − 2   360°

Area = r2 

Important Results (i) Circular Ring :

R r

Area of ring = πR2 − πr2

Regular Hexagon : • Sides : Six

πr θ θ + 2r sin , 180° 2

Region enclosed between two concentric circles of different radii in a plane is called a ring.

C

• Sum of the interior angles = 540°

(

B Small Segment

A segment of a circle is a region enclosed by a chord and an arc of the circle : Perimeter =

a

E a

• Vertices : Five

D

C

D

• Perimeter = 4a Regular Pentagon : • Sides : Five

r sector

A

A

1 × d1 × d2 2 1 2 d1 + d22 • Side, a = 2

O

q

θ πr 2 • Area of the sector = 360°

A B a 1 1 • Area = (sum of parallel sides + height) = ( a + b ) h 2 2

• Area =

diameter

Sector of a Circle : • Sector of a circle is the portion of a circle enclosed by two radii and an arc of the circle.

a

D

a

= π(R2 − r2) (ii) Cyclic Quadrilateral : A cyclic quadrilateral is a quadrilateral which has all its four vertices lying on a circle. A Area =

( S − a)( S − b )( S − c )( S − d )

and S = a + b + c + d 2

a B

d b C

c

D

Theorem 1 : In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary. Theorem 2 : The product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.

293

QUANTITATIVE APTITUDE: MENSURATION

(iii) Tangential Quadrilateral : D

a A

a 2

Radius of circumscribed circle =

c

r b

Then radius of in-circle =

C

d

2a

a 2

(v) Regular Polygon All sides and interior angles are equal.

B

A tangential quadrilateral is a convex quadrilateral whose sides are all tangent to a single circle with in the quadrilateral new line. Area = r. S (S is semi-perimeter) Property: a + c = b + d (iv) Radius of Circumscribed Circle : Let a be the side of the square.

• Number of diagonals of a polygon = • Each interior angle = 180° −

a

n (n − 3) 2

360° n

where n is the number of sides.

Objective Type Questions 1. The circumference of a circular field is 396 m and that of the other circular field is 132 m. Find the area (in m2 ) of the third circular field whose radius is the sum of the radii of the first two 22   fields.  Take π =  (CUET 2023) 7   (1) 13860 (2) 19536 (3) 22176 (4) 23984 2. The diameter of a circle whose area is numerically 110 more than its circumference is (CUET 2022) (1) 5 units (2) 7 units (3) 14 units (4) 10 units 3. If a square has a diagonal of length 6 2 cm, then the area of square is : (CUET 2022) (1) 48 cm2 (2) 72 cm2 (3) 16 cm2 (4) 36 cm2 4. A triangle has sides 25, 39, 34 units. If the area of a square exceeds the area of this triangle by 21 units, then the side of the square is : (1) 22 units (2) 21 units (3) 18 units (4) 25 units D 5. If in the following figure (not to scale), ∠DAB + ∠CBA = 90°, BC = AD, AB = 20 cm, CD = 10 cm then the area of C the quadrilateral ABCD is : (1) 120 cm2

(2) 150 cm2

B

A

(3) 100 cm2 (4) 75 cm2 6. In the given figure, the ratio of the area of the largest square to that of the smallest square is : (1) 4 : 1

(2) 2 : 1

(3) 3 : 1

(4) 3 : 2

60

(2) 1

2

11. What is the area of an equilateral triangle of side 4 3 cm ? (1) 15 3 cm2 (2) 16 3 cm2 (3) 20 3 cm2 (4) 12 3 cm2 12. What will be the area of a circle whose radius is 2

2

(1) 30 cm2 (2) 34 cm2 (3) 36 cm2 (4) 32 cm2 14. The length and breadth of a rectangle are in the ratio 5 : 3. If the length is 8 m more than the breadth, what is the area of the rectangle ? (1) 240 m2 (2) 380 m2 (3) 360 m2 (4) 400 m2 15. The length and breadth of a rectangle are in the ratio 3 : 2. If its perimeter is 730 cm, what is the area of the rectangle ? (1) 31,974 cm2

(2) 24,452 cm2

(3) 20,567 cm

(4) 28,976 cm2

2

Answer Key 2. (2)

3. (4)

4. (2)

5. (4)

6. (1)

7. (1)

9. (4) 10. (3) 11. (4) 12. (4) 13. (3) 14. (1) 15. (1)

A E B

(3) 25 49

G C

F

(4) 1 3

8. The perimeter and breadth of a rectangle are in the ratio of 8 : 1, and the area of the rectangle is 1323 cm2. Find the breadth of the rectangle. (1) 22 cm (2) 23 cm (3) 25 cm (4) 21 cm 9. Find the area of an equilateral triangle whose sides are 12 cm. (1) 38 cm2

(2) 29 5 cm2

(3) 45 2 cm2

(4) 36 3 cm2

5 cm ?

(1) 3π cm (2) π cm (3) 2π cm (4) 5π cm2 13. If M is the mid-point of the side BC of ∆ABC, and the area of ∆ABM is 18 cm2, then the area of ∆ABC is : 2

1. (3)

D

N

(1) 31

of the rectangle is 1764 cm2, then what is the length of the rectangle ? (1) 44 cm (2) 56 cm (3) 84 cm (4) 21 cm

Answers with Explanations

7. In the given figure, if AD = 3, DE = 4, AB = 12, BF = 2, FG = 6, BC = 10, then the value of M is : (Assume : M is the area of the quadrilateral FGDE and N is the area of the triangle ABC.)

10. The length of a rectangle is four times of its breadth. If the area

1. Option (3) is correct. The circumference of first circle = 396 m Circumference of first circle = 2πr1 ⇒ 396 = 2 × 22 r1 7 ⇒ r1 = 63 m The circumference of second circle = 132 m So, circumference of second circle = 2πr2 = 132 ∴ r = 132 = 132 × 7 2 2π 2 × 22 = 21 m So, radius of the new circle = 63 + 21 = 84 m Now, area of the new circle = πr2 22 = × 84 × 84 7 = 22176 m 2

8. (4)

294

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

2. Option (2) is correct. As we know that, area of circle = πr2 And circumference = 2πr As per the question, πr2 = 2πr + 110 2 πr – 2πr = 110 πr( r – 2) = 110 r( r – 2) = 7 × 5 So, r =7

6. Option (1) is correct.

a Let the side of the largest square be a.

3. Option (4) is correct. Given that, Length of the diagonal of square = 6 2 cm As we know, Side of square =

Diagonal 2

=

6 2 = 6 cm 2

∴ Side of the smallest square = Area of the smallest square

7. Option (1) is correct.

S = 98 = 49 units 2

S ( S − a )( S − b )( S − c )

 a   2

2

2

]

S [ r = ] 2

=

1 4 = = 4:1  1 1   4

= 5 sin B

….(i)

= 36 sin B

….(ii)

1 Area of ∆DBG = × 9 × 8 × sin B 2

= 49 ( 49 − 25)( 49 − 39 )( 49 − 34 )

A

49 × 24 × 10 × 15

3

= 7 3600

D 4

= 7 × 60 = 420 square units. Area of the square exceeds the area of this triangle by 21 units. Area of the square = 420 + 21 = 441 units

E 5 B

Hence, side of the square = A = 441 = 21 units. 5. Option (4) is correct. Given CD = 10 cm, AB = 20 cm and BC = AD a E

x 90°

A

10 C

20

a+

b B

….(i) ….(ii)

1  =  × a × x  + 1 ( a + b + x ) × b   2  2

1  ax + ab + b 2 + bx  2

=

1 × 150 = 75 2

G 2 C

1 × 12 × 10 × sin B 2

= 60 sin B = N

Hence,

….(iii) ⇒ b 2 + ab + bx + ax = 150 Now, area of ABCD = area of ∆DEC + area of ∆ABE

=

6

F

Area of ∆ABC =

Join BC to AD at E. Let CE = x, CB = a + b, AE = b and ED = a In ∆DEC, a2 + x2 = 100 In ∆ABE, (a + b + x)2 + b2 = 400 ⇒ a2 + b2 + x2 + 2ab + 2bx + 2ax + b2 = 400 Substituting (i) in (ii), we get ⇒ 2b2 + 2ab + 2bx + 2ax = 300

∴ Area of ABCD = 75 cm2

2

Area of FGDE = 36 sin B – 5sin B = 31 sin B = M

D

b

a2

S

1 × 5 × 2 × sin B 2

Area of ∆EBF =

Now, area of the triangle

=

a 2 2

a 2

∴ Area of the largest square =

[ r =

2

Now, radius of the smaller circle =

4. Option (2) is correct. Given: Sides of the triangle are 25, 39, 34 units. ∴ 2S = 25 + 39 + 34 ⇒ 2S = 98

Area =

a

Now, side of the middle square =

So, area of square = 6 × 6 = 36 cm2

⇒

a 2

The radius of big circle is

M 31 = N 60

8. Option (4) is correct. Given: Perimeter (P) and breadth (b) of the rectangle are in ratio 8 : 1. i.e., P: b = 8 : 1 Let the breadth be x, then perimeter be 8x, Now, 2(l + b) = P ⇒ 2(l + x) = 8x ⇒ l = 3x Area of the rectangle = 1323 (given) l × b = 1323 ⇒ 3x × x = 1323 ⇒ x2 = 441 ⇒ x = 21 cm 9. Option (4) is correct. Given: Side of the equilateral triangle is 12 cm. Area of equilateral triangle =

[from (iii)]

3 2 a 4

3 × 12 × 12 4

∴

A=

Thus,

A = 36 3 sq. cm

295

QUANTITATIVE APTITUDE: MENSURATION

10. Option (3) is correct. Given: Area = 1764 cm2 And, the length of the rectangle is four times of its breadth ∴ l = 4b We have, A = l × b = 1764 Area = 4b × b = 1764 ⇒

b2 =

⇒

b=

1764 = 441 4 441 = 21 units

Hence, the length (l) = 4l = 4 × 21 = 84 units. 11. Option (4) is correct. Given: Side of the equilateral triangle is 4 3 cm. Area of an equilateral triangle =

3 2 a 4

3 ×4 3 ×4 3 4

⇒

Area =

∴

Area = 12 3 cm

2

12. Option (4) is correct. Given:

Radius = 5 cm Area of a circle = πr2

x Length (l) Breadth (b) Area of rectangle

⇒ ∴

Topic-2

Area = π × 5 × 5 = 5π cm2 13. Option (3) is correct. Given: M is the mid-point of BC. Then, AM divides the ∆ABC in two equal parts. A

 Revision Notes Volume :

The amount of space that a substance or object occupies is called volume of that object. Now, we will see volume, surface area, etc. of 3D − figures. Cube : • All sides are equal. • 6 faces • 8 vertices • 12 edges a a • Volume of the cube = a3 a • Total surface area = 6a2 • Lateral surface area = 4a2

• Total surface area = 2(lb + bh +lh)

d = diagonal C

Area of ∆ABC = Area of ∆ABM + Area of ∆AMC Area of ∆ABC = 2 Area of ∆ABM [  Area of ∆ABM = Area of ∆AMC] ∴ Area of ∆ABC = 2 × 18 = 36 cm2 14. Option (1) is correct. Given: The length and breadth of the rectangle are in the ratio 5 : 3. Also, the length is 8 m more than the breadth i.e., l =8+b l 5 = b 3 8+b 5 = b 3

3(8 + b) 24 + 3b 2b b l

= 5b = 5b = 24 = 12 m =8+b = 8 + 12 = 20 m ∴ Area of the rectangle = l × b = 20 × 12 = 240 m2 15. Option (1) is correct. Given: Ratio to length and breadth = 3 : 2 and perimeter = 730 cm Perimeter of a rectangle = 2(l + b) i.e., 730 = 2(3x + 2x) ⇒ 730 = 6x + 4x ⇒ 730 = 10x ⇒ ⇒ ⇒ ⇒ Then,

l 2 + b2 + h2

l = length, b = breath, h = height,

i.e.,

⇒

b l

• Volume = l × b × h

cm 18

Now,

d h

• 12 edges

• Diagonal =

M

3a

Cuboid : • 6 faces • 8 vertices

2

B

3D Mensuration

• Diagonals =

∴

= 73 = 3 × 73 = 219 cm = 2 × 73 = 146 cm = l × b = 219 × 146 = 31,974 cm2

Solid Right Circular Cylinder : • Volume = πr2h

h

• Curved surface area = 2πrh • Total surface area = 2πrh + 2πr

2

r

r = radius, h = height Right Circular Cone : • Volume = 1 × πr 2 × h = 1 πr 2 h 3 3 Also, volume = 1 × area of the base × height 3

• Slant height (l) =

2

r +h

2

• Curved surface area = πrl • Total surface area = πr(l + r) Sphere : • Volume of the sphere = 4 πr 3 3

• Total surface area = 4πr2

r

O

l

h r

296

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Hemisphere : • It is the half sphere. • Volume =

Right Prism : • Volume = Area of the base × Height • Lateral surface area = Perimeter of the base × Height

r

2 3 πr 3

r

• Total surface area = 3πr

2

• Curved surface area = 2 πr 2

• Total surface area of Prism = Lateral surface area of prism + 2 × Area of the base

Hollow Sphere : • Volume = 4 π ( R 3 − r 3 ) 3

r

• Internal surface area = 4 πr 2 R

• External surface area = 4 πR 2

Objective Type Questions 1. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of same height and same diameter is carved out. The total surface area of the remaining solid is:

22   (CUET 2023)  Use π = 7    (1) 18 cm2 (2) 17.6 cm2 (3) 17 cm2 (4) 16.7 cm2 2. The volumes of 3 solid cubes made of metal are 125 cm3, 64 cm3 and 27 cm3, respectively. After melting all the three cubes a solid cube is made. Find the edge of the new cube. (CUET 2023) (1) 2.5 cm (2) 4 cm (3) 6 cm (4) 5 cm 3. A cuboidal slab of copper having dimensions 22 cm × 10 cm × 5 cm is melted and recasted in the form of a wire of 1 mm diameter. The wire is rubber coated at the rate of ` 1.50 per metre. Find the cost of rubber coating (in `) 22  .  (CUET 2023) Use π =  

7 

(1) 525 (2) 650 (3) 2100 (4) 2600 4. The area of four walls of a rectangular hall having length 18 m and height 8 m is 448 m2. What is the breadth of the hall (in m)? (CUET 2023) (1) 10 (2) 9 (3) 8 (4) 7 5. 64 solid iron spheres of radius r are melted to form a sphere of radius R. Find R : r. (CUET 2023) (1) 2 : 1 (2) 3 : 1 (3) 8 : 1 (4) 4 : 1 6. The surface area (in m2) of a sphere of radius 7 cm is:

22    Use π = 7   

(CUET 2023)

(1) 0.0616 (2) 6.16 (3) 61.6 (4) 0.616 7. The volume of a sphere of radius r is obtained by multiplying its surface area with: (CUET 2023) 4 r 4r (2) (3) (4) 3r (1) 3 3 3 8. A toy is made in the shape of a hemisphere of diameter 7 cm surmounted by a cone. If this 15.5 cm high toy is polished at 20 paise per cm2, then find the cost of polishing.  Take π = 22  7   (CUET 2023) (1) ` 39 (2) ` 40.50 (3) ` 42.90 (4) ` 45 9. A volume of a wall is 128 cm3. If the height of the wall is 6 times its breadth and the length is 9 times its breadth, find the breadth. (CUET 2022) (1)

3

3.45 cm (2)

3

4.74 cm (3)

3

2.37 cm (4)

3

2.38 cm

10. A cylindrical vessel with radius 6 cm and height 5 cm is to be made by melting a number of spherical metal balls of diameter 2 cm. The minimum number of balls needed is (1) 125 (2) 135 (3) 115 (4) 105 11. If each edge of a cube is increased by 10%, then the percentage increase in its surface area is : (1) 21% (2) 19% (3) 22% (4) 20% 12. A sphere is placed in a cube so that it touches all the faces of the cube. If ‘a’ is the ratio of the volume of the cube to the volume of the sphere, and ‘b’ is the ratio of the surface area of the sphere to the surface area of the cube, then the value of ab is (1)

π2 36

(2)

36 π2

(4) 1

(3) 4

13. The length and breadth of a cuboid are increased by 10% and 20%, respectively, and its height is decreased by 20%. The percentage increase in the volume of the cuboid is 4 5

(1) 5 %

1 5

2 5

(2) 5 %

(3) 5 %

3 5

(4) 5 %

14. The radius of a circular cone is R and its height is H. The volume of the cone is 2 (1) πR H

3

(2)

2 3πR 2 H

(3) pR 2 H

(4) πRH 2

15. If 3.96 cubic dm of lead is to be drawn into a cylindrical wire of diameter 0.6 cm, then the length of the wire (in metres) is (1) 130 m (2) 125 m (3) 140 m (4) 120 m Answer Key 1. (2)

2. (3)

3. (3)

4. (1)

5. (4)

6. (1)

7. (2)

9. (3) 10. (2) 11. (1) 12. (4) 13. (4) 14. (1) 15. (3)

Answers with Explanations 1. Option (2) is correct. Given that: height of cylinder = 2.4 cm diameter of cylinder = 1.4 cm or radius of cylinder = 0.7 cm 0.7 ) Slant height of cone = ( 2.4 ) + (= 2

2

= 6.25 2.5

So required Surface area = 2 πrh + πrl + πr 2 = πr( 2 × 2.4 + 2.5 + 0.7 ) 22 cm 2 = × 0.7 × ( 8 ) = 17.6  7

2. Option (3) is correct. Given: Volume of cube1 = 125 cm3 Volume of cube2 = 64 cm3 Volume of cube3 = 27 cm3

8. (3)

297

QUANTITATIVE APTITUDE: MENSURATION

After melting, volume of new cube = 125 + 64 + 27 = 216 cm3 Using, volume of cube = (side)3 So, (side)3 =216 = 3= 216 6 cm ∴ Side 3. Option (3) is correct. Given that the dimension of cuboid= 22 cm × 10 cm × 5 cm ∴ Volume of cuboid = 22 × 10 × 5 = 1100 cm3 1 Also given that radius of= wire 1= mm cm 2 20 Let the length of wire = l using, volume of cylinder = πr2h According to the question, 22 1 1 1100 = × × ×l 7 20 20

l = 140000cm or 1400 m So, the cost of rubber coating = 1400 × 1.5 = ` 2100 4. Option (1) is correct. Given that: length of wall = 18 m height of wall = 8 m let the breadth of wall = x According to the question: 2(18 × 8 ) + 2( x × 8 ) = 448 ⇒ 36 × 8 + 16 x = 448 ⇒ 16x = 160 ⇒ x = 10 So, breadth of wall = 10 m 5. Option (4) is correct. Volume of 64 iron sphere = 64 × 4 πr 3 3 4 and volume of one big sphere = π R 3 3 According to the question: 4 3 4 ⇒ πR = 64 × πr 3 3 3

7 2

= 3.5 cm Height of cone = 15.5 – 3.5 = 12 cm 12 2 =+ 3.12 5 2 2 + 3.5 2 So, slant height of= cone = 144 =+ 12144 .25 +=12 .5 cm 12.25 = 12.5 cm

Total surface area of toy = 2πr2 + πrl = π × 3.5 [ 2 × 3.5 + 12.5] =

22 × 3.5 × 19.5 = 214.5 cm 2 7

So, required cost = 214.5 ×

20 = `42.90 100

9. Option (3) is correct. Given that, The volume of the wall = 128 cm3 Let the breadth = x cm So, height = 6x cm And length = 9x cm As we know that, Volume = L × B × H 128 = 9x × 6x × x 128 = 54x3 x=

3

2.37

10. Option (2) is correct. Metal balls : Diameter, d = 2 cm ∴ Radius, r = 1 cm Volume of the one ball = 4 πr 3

5 cm

3

3

R 64 ⇒ ⇒ R = 3 64 = r3 1 r 1 R 4 ⇒ = ⇒ R:r=4:1 r 1 6. Option (1) is correct. Given that radius of sphere = 7 cm Using, surface area = 4 πr2 22 = 4 × × 7 × 7 = 616 cm2 or 0.0616 m2 7 7. Option (2) is correct. Given that the radius of sphere = r 4 2 we have, volume of sphere = pr 3

and surface area of sphere = 4 πr 4 3 pr r So, required multiplier = 3 2 = 3 8. Option (3) is correct. 4 pr According to the question: 2

15.5 cm

7 cm

Radius of hemisphere =

= 4 × 22 × (1)3 3

6 cm

7

= 88 cm3 21

Cylinder : Radius, r = 6 cm, height, h = 5 cm Volume of the cylinder = πr 2 h = 22 × 36 × 5 7

= 3960 cm3 7 Number of sphere balls required =

Volume of the cylinder Volume of one ball

3960 3960 21 = 7 = 7 × 88 88 21

= 45 × 3 = 135 ∴ The minimum number of balls needed is 135. 11. Option (1) is correct. Let the edge of the cube be 100 units increased by 10% ∴ New edge = 110 units Surface area of the cube = 6a2 = 6 × 100 × 100 = 60,000 sq. units Surface area of the cube with new edge = 6 × 110 × 110 = 72,600 sq. units ∴ % increase in surface area =

[ 72, 600 − 60, 000 ] × 100 60, 000

= 21%

298

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Shortcut Method : Surface area of the cube = 6a2 When edge is increased by 10% then, the surface area = 6 × (1.1 a)2 = 1.21 × 6a2 which is 21% increase Hence, option (1) is correct. 12. Option (4) is correct. Let a and r be the side and radius of the cube and sphere. According to the question, Volume of the cube a = Volume of the sphere

=

b =

a3 4 3 πr 3

Surface area of sphere Surface area of cube

4 πr 2 6a2 [ Sphere is placed in the cube of side a] 3a 3 4 πr 2 ab = × 4 πr 3 6 a 2 a a = =1 [ a = 2r] = 2r a

=

∴

13. Option (4) is correct. Volume of the cuboid = l × b × h Given: length is increased by 10% = 1.1 l Breadth is increased by 20% = 1.2 b

Height is decreased by 20% = 0.8 h Now, new volume = 1.1 l × 1.2 b × 0.8 h = 1.056 lbh

3 5

Hence, volume is increased by 5.6%, i.e., 5 % . 14. Option (1) is correct. We have, volume of the cone = Given: r = R and h = H ∴ Volume of the cone =

1 2 πr h 3

πR 2 H 3

15. Option (3) is correct. Volume of the lead = 3.96 cubic dm We know, 1 decimetre = 10 centimetre 1 dm3 = 1000 cm3 ∴ Volume of the lead = 3.96 × 1000 = 3960 cm3 Given: Diameter of the wire, d = 0.6 cm r = 0.3 cm Volume of the lead = Volume of the wire Volume of the cylinder = πr2h 3960 = 22 × 0.3 × 0.3 × h ⇒

7 3960 ×7 h= 22 × 0.09

⇒ = 14000 cm ∴ Converting to metre = 14000 = 140 metres 100

2

On the Basis of Angles • Acute angled triangle • Right angled triangle • Obtuse angled triangle

Tangents • Touches at one point • ⊥r to radius • Two tangents drawn from m external point are equal.

g2 + f 2 − c

Secant • Touches in 2 pts • Properties of secant

and r =

Cyclic Quadrilateral • Quadrilateral inscribed in a circle • Opposite angles are supplementary

Circless

Angle

Equation of circles x2 + y2 = a2 • • (x – h)2 + (y – k)2 = a2 2 2 • x + y + 2gx + 2fy + c = 0 where centre (–g, –f)

Geometry

Right angle, x = 90°° Acute angle, x < 90° 90° Obtuse angle, x > 90°° •

•

•

•

•

•

•

•

•

1

2

2

1

2

Second Level

Trace the Mind Map First Level

[(

Third Level



Area of 1 x1 ( y 2 − y3 )+ x 2 ( y3 − y1 )+ x 3 ( y1 − y 2 )] = 2 Points A, B and C are collinear if (i) AB + BC = AC (ii) Area of ABC = 0

 x + x 2 y1 + y 2  , Midpoint =  1  2   2 Centroid:  x + x 2 + x 3 y1 + y 2 + y 3  , G=  1  3 3  

Section formula:  mx + nx1 my 2 + ny1  , A(x, y) =  2  m+n   m+n

2

(x − x ) + ( y − y )

Distance formula: PQ =

y − y1 x − x1 = y 2 − y1 x 2 − x 1

Slope-intercepts: y = mx + c x y Intercept: + = 1 a b General form: Ax + by+ c = 0

Two points:

Equation of straight line in different form • Slope-point: y – y1 = m(x – x1)

• Parallel lines • Transverse lines

Co-ordinate Geometry •

Line

Polygons

Sum of interior angles with ‘n’ sides = (n – 2)180°

Sum of exterior angles 360°

Regular Polygon Measure of (n − 2 ) 180 ° each angle g = n

Circumcircle Incircle

Triangles

Types of triangle On the Basis of Sides • Scalene triangle • Isosceles triangle • Equilateral triangle

Congruency • All sides are equal • All angles are equal

area  1  S1  =  area  2  S 2 

Similar Triangle • All angles are equal • Sides are proportional

Equilateral, Isosceles, Scalene, Right Angled

Median: Joining vertex to the midpoint of the opposite side. Centroid: Intersection of medians

Altitude: ⊥r drawn from vertex to opposite side. Orthocentre: Intersection of all altitudes

QUANTITATIVE APTITUDE: GEOMETRY

299

Chapter

Geometry

11

Chapter Analysis Concept Name

2021

Basic Concepts and Coordinate Geometry

2022

2023

4

Triangles

Additional Questions

5

6

4

11

Circle

15

Types of Geometry • Basic Concepts and Coordinate Geometry • Triangles • Circle

Topic-1  Basic Concepts and Coordinate Geometry  Revision Notes Point :  A point is a geometrical element which has no dimensions. A point represents position only. Line :  Line is a collection of endless points which has only length, neither breadth nor thickness. Line segment :  A line with two end points. Ray :  A ray is a line which has a fixed starting point but no end point. Angle :  The figure made by the intersection of two lines or line segments is called angle.

Vertically Opposite Angles :  When two straight lines intersect each other, four angles are formed. The pair of angles which lie on the opposite sides of the point of intersection are called vertically opposite angles. (Vertically opposite angles are always equal.) Linear Pair of Angles :  Two adjacent angles sum is 180°, then we say, those two angles are linear pairs. Parallel Lines :  Two lines in a plane are said to be parallel if they do not intersect, when extended infinitely in both the direction. Also, the distance between the two lines is the same throughout. These are denoted by ||.

l m   Here, l || m.

Transversal Line :  A straight line which intersects two or more given lines at distinct points is called a transversal of the given lines. r

4

A

3

Angle B

C

 B and BC are two line segments angle is formed here at vertex B. A Here, AB and BC are two line segments which form an angle at vertex B. Angle is denoted by symbols, ∠B.

Types of Angles 

(i) Acute Angle : An angle whose measure is more than 0° but less than 90° is called an acute angle. (ii) Right Angle : An angle whose measure is 90° is called a right angle. (iii) Obtuse Angle :  An angle whose measure is more than 90° but less than 180° is called an obtuse angle. (iv) Straight Angle :  An angle whose measure is 180° is called a straight angle. (v) Reflexive Angle :  An angle whose measure is more than 180° but less than 360° is called a reflex angle. (vi) Complete Angle :  An angle whose measure is 360° is called a complete angle. Note :  Complementary Angles :  Two angles ∠A and ∠B are said to be complementary angles, if ∠A + ∠B = 90° Supplementary Angles :  Two angles ∠A and ∠B are said to be supplementary angles, if ∠A+ ∠B = 180° Adjacent Angles :  Two angles are called adjacent angles, if :  (a) They have the same vertex (b) They have a common arm, and (c) Uncommon arms on both side of the common arm.

7

8 5 6

1 l 2

m

Exterior Angles :  ∠1, ∠4, ∠6 and ∠7. Interior Angles :  ∠2, ∠3, ∠5 and ∠8. Corresponding Angles :  Two angles on the same side of a transversal are known as corresponding angles if both lie either above the lines or below the lines. ∠1 and ∠5; ∠2 and ∠6; ∠3 and ∠7; ∠4 and ∠8 are the pairs of corresponding angles. Alternate Angles :  ∠2 and ∠8, ∠3 and ∠5, ∠1 and ∠7, ∠4 and ∠6 are alternate angles. Consecutive Angles :  The pair of two interior angles on the same side of the transversal are called the pairs of consecutive interior angles. ∠2 and ∠5, ∠3 and ∠8 are the pairs of consecutive interior angles. Properties :  If a transversal intersects two parallel lines, then, (a) Each pair of corresponding angles is equal. (b) Each pair of alternate angles is equal. (c) Each pair of interior angles on the same side of the transversal is supplementary. Bisector :  In geometry, bisection is the division of something into two equal or congruent parts, usually by a line, which is then called a bisector.

301

QUANTITATIVE APTITUDE: GEOMETRY

Polygon : A polygon is a closed geometric plane figure bounded by straight line segments. Triangle : Triangle is a polygon of three sides. Quadrilateral : Quadrilateral is a polygon with four sides, e.g., Square, Rectangle, Parallelogram, Trapezium, etc. Pentagon : Pentagon is a polygon with five sides. Regular Polygon : All sides are of the same length.. Equiangular Polygon : All the vertex angles are equal.

OM being X-co-ordinate or abscissa and MP is the Y-co-ordinate or ordinate. Y

P(x, y) y

Polygon Angle Sum Theorem

The sum of the measures of interior angles of n side polygon is (n − 2) 180°.

( n − 2 ) 180° . For a regular polygon, the measure of each angle is n

The sum of the measure of the exterior angles of a polygon is 360°. For a regular polygon, the measure of each exterior angle is

360° . n

Note : In a regular polygon, all sides and interior angles are equal.

Regular Polygons: Name

Sides

Interior Angle

Triangle

3

60°

Quadrilateral

4

90°

Pentagon

5

108°

Hexagon

6

120°

Heptagon

7

128.571°

Octagon

8

135°

Nonagon

9

140°

Decagon

10

144°

Hendecagon

11

147.273°

Dodecagon

12

150°

y -positive

y -positive

x ’ y plane

xy plane

x’

O IV quadrant

x-negative

x-positive

y -negative

y -negative

x ’ y ’ plane

xy ’ plane

y’

Distance formula : Let P(x1, y1) and Q(x2, y2) be two points on a plane, then the distance between them is

(x

PQ =

2

− x1

) + (y 2

Distance of a point P(x, y) from origin is OP =

2

− y1

)

2

x12 + y12

Section formula : Let P(x, y) and Q(x2, y2) be two points, and let A(x, y) be the point which divides PQ in the ratio m : n internally, then

 x 1 + x 2 y1 + y 2  , 2 2 

• If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle, then the co-ordinates of centroid are  x 1 + x 2 + x 3 y1 + y 2 + y 3  ,  3 3

G=  

• If A(x1, y1), B(x2, y2) and C(x3, y3) are vertices of a triangle, then the area of triangle ABC : Area of ∆ABC =

1 x1 y 2 − y3 + x 2 y3 − y1 + x 3 y1 − y 2  2

(

)

(

)

(

)

• Collinear : If A(x1, y1), B(x2, y2) and C(x3, y3) are collinear points (lie in one line), then Area of ∆ABC =

1 x1 y 2 − y3 + x 2 y3 − y1 + x 3 y1 − y 2  = 0 2

(

)

(

)

(

)

Also, if AB + BC = AC, then the points A, B and C are collinear.

Equation of a Straight line: Point–Slope formula : To find the equation of a line when one point (x1, y1) and slope of the line ‘m’ is given (y − y1) = m(x − x1)

Two–point formula : To find the equation of a line when two points (x1, y1) and (x2, y2) are given :

x

III quadrant

Points to Remember :

Midpoint =  

Let P be any point on the plane. Draw PM perpendicular to X-axis. The lengths OM and MP are called the co-ordinates of the point P. y

x-positive

We denote OM by x and MP by y and the point P by (x, y)

Midpoint : Let P(x1, y1) and Q(x2, y2) be two points, then the midpoint of PQ :

To simplify the calculation, we divide the plane into 4 quadrants using X-axis and Y-axis such that X-axis and Y-axis intersect at origin.

I quadrant

X

m

 mx 2 + nx1 my 2 + ny1  , m+n m + n 

Cartesian Coordinate System : Co-ordinate system is used to locate the position of a point in a plane using two perpendicular lines. Points are represented in the form of coordinates, (x, y) with respect to X and Y-axis in two dimensions.

x-negative

x

A(x, y) =  

Co-ordinate Geometry:

II quadrant

O

y − y1 x − x1 = y 2 − y1 x 2 − x 1

[Slope of a line, m = tan θ =

y 2 − y1 ] x 2 − x1

Slope–intercepts formula : To find the equation of a line when slope and y-intercept is given y = mx + c, where ‘m’ is slope and ‘c’ is y-intercept Intercept form : To find the equation of a line when x-intercept ‘a’ and y-intercept ‘b’ are given x y + =1 a b

302

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

General Equation : General equation of a line is Ax + By + C = 0

Equation of a Circle:

• Equation of a circle, whose centre is at origin (0, 0) and radius ‘a’ is x2 + y2 = a2 • Equation of a circle, whose centre is at (h, k) and radius ‘a’ is (x − h)2 + (y − k)2 = a2

• Equation of a circle, in general form is x2 + y2 + 2gx + 2fy + c = 0 g2 + f 2 − c

where centre is (−g, −f) and radius is r =

• Let P(x1, y1) be any point on a circle and O(x, y) be the centre of the circle, then radius is OP

(x

OP =

) ( 2

1

− x + y1 − y

)

2

Objective Type Questions 1. The distance between two parallel sides of a trapezium is 15 m and its area is 480 m2. If one of the parallel sides is 20 m long, then length of the other side is: (CUET 2023) (1) 42 m (2) 44 m (3) 40 m (4) 46 m 2. An interior angle of a regular polygon is 135°. The polygon is a/an: (CUET 2023) (1) hexagon (2) square (3) pentagon (4) octagon 3. If CE|| DB, what is the value of x: (CUET 2023) B A 30 110 E D

60

75

x

10. In the given figure, if AC, DE are parallel and ∠CAB = 38°, then the value of ∠ABC + 5∠CBD is : C

D 2a

A

a b B E

(1) 158° (2) 178° (3) 218° (4) 196° 11. In the given figure, AD is bisector of ∠CAB and BD is bisector of ∠CBF. If the angle at C is 34°, the ∠ADB is : D

C

C

E

(1) 45° (2) 30° (3) 75° (4) 85° 4. A quadrilateral has vertices in the order (0, −1), (−2, 3), (6, 7) and (8, 3). The quadrilateral is a: (CUET 2023) (1) trapezium (2) square

(3) rhombus

(4) rectangle

3 5. An angle is times its supplementary angles. Find the angle. 7 (CUET 2023)

B

A

in the figure. The value of

(1) 54° (2) 63° (3) 27° (4) 126° Direction for questions (6-8): From the given diagram, answer the following question. C

∠ABC + 2 ∠EGD + 3∠BAJ is : 6

J

I B A

C

F

G

F

(1) 34° (2) 32° (3) 17° (4) 16° 12. ABCDE is a regular pentagon. Its sides are extended as shown

E

H

D

3 cm

D

B

G

10 cm

(1) 45° (2) 30° (3) 75° (4) 66° 13. In the given figure. AP is perpendicular to BC, and AQ is the bisector of angle PAC. What will be the measure of angle PAQ ?

7 cm

F

E

A

9 cm 11 cm

A 6. 'AC' is 23 cm in the adjoining figure. Consider the figure and answer the questions (CUET 2022) The area of DGFE is

(1) 133 cm2 (2) 140 cm2 (3) 70 cm2 7. The area of ∆DBC if DB is a straight line

(4) 66.5 cm2 (CUET 2022)

(1) 32.5 cm2 (2) 65 cm2 8. The total area of figure is

(4) 15 cm2 (CUET 2022)

(3) 25 cm2

(1) 184.5 cm2 (2) 205 cm2 (3) 175.5 cm2 (4) 153 cm2 9. If two adjacent angles of a parallelogram are (2x + 30)° and (3x – 15)°. Then the value of x is : (CUET 2022) (1) 36

(2) 39

(3) 33

(4) 35

60°

30°

B

P

Q

C

(1) 45° (2) 30° (3) 50° (4) 60° 14. In the given figure, AP is perpendicular to BC, and AQ is the bisector of angle A. What will be the measure of angle PQA ? A

60° B

(1) 50°

30° P

(2) 60°

Q

(3) 30°

C

(4) 75°

303

QUANTITATIVE APTITUDE: GEOMETRY

15. The equation of circle with centre (1, −2) and radius 4 cm is : (1) x 2 + y 2 + 2 x − 4 y = 16 (2) x 2 + y 2 − 2 x + 4 y = 16 (3) x 2 + y 2 + 2 x − 4 y = 11

Similarly slope of BC × Slope of DC = ( −2 ) ×

(4) x 2 + y 2 − 2 x + 4 y = 11

So, these two are also perpendicular and AB will also be parallel to DC.

Answer Key 1. (2)

2. (4)

3. (4)

4. (4)

5. (1)

6. (4)

7. (1)

8. (3)

9. (3) 10. (3) 11. (3) 12. (4) 13. (2) 14. (4) 15. (4)

Answers with Explanations 1 (sum of parallel sides) × height 2

1 ⇒ 480 = × ( 20 + x ) × 15 2

A

⇒ 20 + x = 64 ⇒ x = 44 2. Option (4) is correct. B Given: An interior angle of a regular polygon = 135° Let the number of sides in this polygon = n Using,

20

D

15m x

Also, BC = ( 8 − 0 )2 + ( 3 + 1)2 = 70 units DC = ( 6 − 8 )2 + (7 − 3)2 = 20 units

All four sides are not equal. So, we can say that the given quadrilateral is a rectangle. 5. Option (1) is correct. Let first angle = x So, the required angle = 3 x 7

1. Option (2) is correct. Given that area of trapezium = 480 m 2 Using, area =

C

( n − 2 )180° = 135° n

Using, sum of supplementary angles = 180° 3 ⇒ x + x = 180° 7 ⇒ 10 x = 180° × 7 ⇒ x = 126° 3 ∴ Required angle = × 126° = 54  7

6. Option (4) is correct. The area of DGFE (here, GF = h = 7 cm)

⇒ ( n − 2 )180 ° = 135n ⇒ 180n − 360 = 135n ⇒ 45n = 360 ⇒ n = 8

So, given polygon is an octagon. 3. Option (4) is correct. A B 11030 D

75 60

C

Slope of BC = Slope of DC =

3 − ( −1) 8

= −2 =

1 2

7−3 = −2 6−8

As, slope of AB × slope of BC = ( −2 ) × So, both lines are perpendicular.

=

1 × 19 × 7 = 66.5 cm2 2

=

ÐDBC

−2 − 0

1 1 (a + b) × h = (10 + 9) × 7 2 2

Area of triangle BDC =

x

∠ADB = 180° – (110 + 30°) = 40° ∠BDC = 75° – 40° = 35° ∠DBC = 180 − ( 60° + 35° ) = 180 − 95° = 85° Hence, by interior alternate angle property of parallel line: ∠DBC = ∠BCE = 85° 4. Option (4) is correct.

3 − ( −1)

=

7. Option (1) is correct. E

In ΔABD, So, in ΔBDC,

Slope of AB =

1 = −1 2

1 = −1 2

1 ×B×H 2 1 × 13 × 5 = 32.5 cm2 2

8. Option (3) is correct. Area of figure = area ∆DBC + area ∆AGB + area ∆EFA + area of trapezium DGFE =

1 ( B×H + B ×H + B ×H + (a+b)H) 2

=

1 ( 13×5 + 18 ×3 + 9 ×11 + (10 + 9)7) 2

=

1 ( 65 + 54 + 99 + 133) 2

= 175.5 cm2 9. Option (3) is correct. Given that, The angles two adjacent angles of a parallelogram are (2x + 30)° and (3x – 15)°. We know that the sum of two adjacent angle of parallelogram is 180°. So, (2x + 30) + (3x – 15) = 180° 5x + 15 = 180° 5x = 165° x = 33° C 10. Option (3) is correct. Given: 2a = 38° D ⇒

a = 38 = 19° 2

Since, AC || BD, ∠ACB = ∠CBD = 19°

A

2a

a b B E

304

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

In ∆ABC, ∠CAB + ∠ABC + ∠BCA = 180° ⇒ 38° + b + 19° = 180° ⇒ b + 57° = 180° ⇒ b = 123° Now, ∠ABC + 5∠CBD = 123° + 5(19°) ⇒ ∠ABC + 5∠CBD = 218° 11. Option (3) is correct. Given: AD is bisector of ∠BAC ∴ ∠CAE = ∠EAB = a (say)

In ∆EDG, ∠DEG ∴ ∠DEG + ∠EDG + ∠EGD ⇒ ∠EGD ⇒ ∠EGD Now,

a a

b b B

A

Also, BD is bisector of ∠CBF ∴ ∠EBD = ∠DBF = b (say) In ∆ABC, ∠CBF is the exterior angle at B. ∴ 34° + 2a = 2b ⇒ b = 17 + a ….(i) In ∆ABC, ∠ABC + ∠BAC + ∠ACB = 180° ⇒ ∠ABC + 2a + 34° = 180° ⇒ ∠ABC = 146° − 2a ….(ii) In ∆ABD, ∠ABD + ∠BAD + ∠ADB = 180° ⇒ (146° − 2a + b) + a + ∠ADB = 180° ⇒ 146° − 2a + a + 17° + a + ∠ADB = 180° (from (i)) ⇒ ∠ADB = 180° − 163° ⇒ ∠ADB = 17°

C

….(i)

D z

34° E a a

bb

A

B

F

Now, in ∆ABD, ∠DBF is the exterior angle. ∴ b =a + z From (i) and (ii), z = 17°

….(ii)

12. Option (4) is correct. Given: ABCDE is a regular pentagon. Since each angle in a regular pentagon is 108° J

C E

D

AQ is the bisector of angle A ∠BAQ =

Q

C

90° = 45° 2

In ∆ABQ, ⇒∠BAQ + ∠ABQ + ∠BQA = 180° ⇒ 45° + 60° + ∠BQA = 180° ⇒ ∠BQA = 180° − 105° ⇒ ∠BQA = 75° ∴ ∠BQA = ∠PQA = 75° 15. Option (4) is correct. Given: Centre at (1, −2) and radius = 4 cm Equation of a circle whose centre at (h, k) and radius ‘a’ is (x − h)2 + (y − k)2 = a2 2 2 ( x − 1) 2 + ( y + 2 ) = (4) ⇒

x 2 − 2 x + 1 + y 2 + 4 y + 4 = 16

⇒

x 2 + y 2 − 2 x + 4 y = 11

Triangle

Points to Remember :

H

G

∠JAE = 180°

⇒

∠BAJ + ∠BAE = 180°

⇒

∠BAJ + 108° = 180°

⇒

30° P

A triangle is a closed two-dimensional figure with three straight sides.

B

Since,

60° B

 Revision Notes

I

A F

A

Topic-2

∠ABC = 108°

60° = 30° 2

∴ Measure of angle PAQ is 30° 14. Option (4) is correct. Given: AB ⊥ BC In ∆ABC, ∠A + ∠B + ∠C = 180° ⇒ ∠A + 60° + 30° = 180° ⇒ ∠A = 90°

⇒

Shortcut Method :

∴

a =

⇒

F

In ∆ABC, ∠CBF is the exterior angle. ∴ 2b = 2a + 34 ⇒ b =a + 17

396° = 66°. 6

13. Option (2) is correct. By given data, we have In ∆APC, ∠APC + ∠PCA + ∠PAC = 180° ⇒ 90° + 30° + 2a = 180° ⇒ 2a = 180° − 120° = 60°

?

34° E

108° + 72° + 3 ( 72 ) ∠ABC + 2 ∠EGD + ∠BAT = 6 6

=

D

C

= ∠EDG = 72° = 180° = 180° − 144° = 36°

∠BAJ = 72°

(linear pairs)

• A triangle has three sides, three vertices and three angles. • The sum of the three interior angles of a triangle is always 180°. • The sum of the length of two sides of a triangle is always greater than the length of the third side. • A triangle with vertices P, Q and R is denoted as ∆PQR. • The area of a triangle is equal to half of the product of its base and height. • Difference of two sides of a triangle is always smaller than the third side of that triangle.

305

QUANTITATIVE APTITUDE: GEOMETRY

• Altitude is also called the height of the triangle. • The three altitudes of a triangle are concurrent.

P

• The point of concurrence is called orthocentre of the triangle.

Equilateral triangle : Q

R

Triangles are grouped in two different ways : (i) With regard to their sides. (a) A triangle is called equilateral, when all its sides are equal. i.e., AB = BC = CA Also, ∠A = ∠B = ∠C (b) Isosceles triangle is a triangle whose two sides are equal. AB = BC

A

• Acute triangle: The orthocentre lies inside the triangle.

D C

B

• Obtuse triangle: The orthocentre lies outside the triangle.

B

C A

• Right triangle: The orthocentre lies at the vertex containing the right angle.

Median :

Also, ∠B = ∠C (c) A triangle is called scalene, when no two sides are equal. AB ≠ BC ≠ CA

A

The orthocentre appears to be directly in the centre of a triangle, creating three triangles of equal area. In the following ∆ABC, O is the orthocentre. Area ∆AOB = Area ∆BOC = Area ∆COA

A

A median of a triangle is the straight line segment joining a vertex to the midpoint of the opposite side.

B

Also, ∠A ≠ ∠B ≠ ∠C (ii) With regards to angles. A (a) A triangle is called a right angled triangle, when one of its angle is 90°. • ∆ABC is right angled triangle, ∠B = 90°, AC is called hypotenuse. • Right angled triangle satisfies pythogoras theorem. B i.e., AC2 = AB2 + BC2

C

Centroid : The medians are concurrent. The point of concurrence is called the centroid of the triangle.

F

B

G

D

E

C

• Centroid is always lies inside the triangles. • Centroid divides the median in the ratio 2 : 1. • In ∆ABC, G is centroid such that AG : GD = 2 : 1 • Median divides area of triangle in two equal parts. • Distance of the centroid from the vertex in an equilateral triangle

C

(b) A triangle is called an obtuse angled triangle when one of its angle is obtuse. ∠B is obtuse (> 90°) A

is

a 3 , where a is the side of the equilateral triangle. 3

Perpendicular bisectors of the sides : A line passing through the midpoint of a side and perpendicular to it is called the perpendicular bisector of the side. The point where three perpendicular bisectors of the sides meet is called circumcentre of the triangle. • ‘O’ is the circumcentre of the triangle ABC and it is equidistance from all the three vertices. A

B

C O

(c) A triangle is called an acute angled triangle when all the three angles are acute ∠A, ∠B and ∠C are acute angles. B

A

B

C

• The distance between circumcentre and one of the vertices of the triangle is called circumradius. A C

Altitudes:

O

An altitude of a triangle is a line drawn from any vertex perpendicular to the opposite side. In ∆ABC, AD is an altitude. A

B

• The circle drawn taking ‘O’ as the centre passes through all the three vertices. The circle is called circumcircle of the triangle. • Circumradius, R =

B

D

C

C

abc 4 × area of the triangle

where a, b and c are sides of triangle.

306

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Angular bisectors of a triangle :

• Angles, ∠BIC = 90° + ∠AIC = 90° +

1 ∠A 2

B

C

1 ∠B 2

R

• We denote similar triangle as ∆ABC ~ ∆PQR

Congruency of triangles : B

xx

C

Area of ∆ABC Semi perimeter of ∆ABC

Two triangles are congruent if (i) Their corresponding sides are equal. (ii) Their corresponding angles are equal. AB = PQ, AC = PR, BC = QR & ∠A = ∠P, ∠B = ∠Q & ∠C = ∠R

P

A

Inradius of a right angle triangle r=

Q

AB AC BC i.e., = = PQ PR QR

A

1 ∠AIB = 90° + ∠C 2

Inradius :

P

A

An angular bisector in a triangle is a line which bisects an angle of the triangle and extends to the opposite side • Angular bisectors are concurrent. The point of the concurrency is called the incentre of the triangle.

Base + Perpendicular − Hypotenuse 2 A

B

C

Q

R

We write two triangles are congruent using the symbol “ ≅ ” i.e., ∆ABC ≅ ∆PQR

Points to Remember :

I B

C

Similar triangle :

The two triangles ∆ABC and ∆PQR are said to be similar triangles if any two corresponding angles are equal. In ∆ABC and ∆PQR, ∠B = ∠Q and ∠C = ∠R Two angles are equal, then definitely third angles are also equal. • Corresponding sides of both the triangles are in proportional to each other.

• If two triangles are similar, then the ratio of the area of both the triangles is proportional to the square of the ratio of their corresponding sides. • The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it. • The perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the original triangle and also to each other. • If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angle 0.

Objective Type Questions 1. In the given figure, DE || BC, the value of x is: (CUET 2023) A x2

15

E

D

5

3 B

C

(1) 12 (2) 11 (3) 10 (4) 9 2. Find the coordinates of centroid of ∆ABC if the mid point of BC is D(2, 4) and vertex A is (2, −3). (CUET 2023) (1) (1, 0)

2  (2)  0 , −  5 

(3) (2, 5)

6. The areas of two similar triangles are 324 cm2 and 225 cm2. If the altitude of the smaller triangle is 10 cm, then the altitude of the bigger triangle, in centimetres, is (1) 12 (2) 16 (3) 14 (4) 18 7. If in any triangle, the angles are in the ratio of 1 : 2 : 1, then what will be the ratio of its sides ? (1) 1 : 2 : 3 (2) 2 : 1 : 2 (3) l : 2 : 1 (4) 1 : 2 : 1 8. In the given figure, BD is perpendicular to AC, then what will be the measure of ∠AEB ? A 35°

 5 (4)  2 ,   3

3. The sides of a triangle are in the ratio 1 : 1 : 1 . If the semi3 4 5 perimeter of the triangle is 47 cm, then what is the length of the longest side? (CUET 2023) (1) 20 cm (2) 40 cm (3) 44 cm (4) 30 cm 4. Find the area of an equilateral triangle each of whose sides measures 4 cm: (CUET 2023) (3) 3 3 cm 2 (4) 3 3 m 2 (1) 4 3 cm 2 (2) 4 3 m 2 5. In ∆PQR, S and T are mid points of PQ and PR, respectively. If ∠QPR = 75° and ∠PRQ = 40°, then ∠TSQ is (1) 135° (2) 120° (3) 105° (4) 115°

D

B

45°

E

C

(1) 80° (2) 100° (3) 60° (4) 45° 9. In ∆XYZ, if G is the centroid and XL is the median with length 18 cm, then the length of XG is (1) 14 cm (2) 16 cm (3) 12 cm (4) 10 cm 10. A triangle is NOT said to be a right-angled triangle if its sides measure (1) 5 cm, 12 cm and 13 cm (2) 5 cm, 7 cm and 9 cm (3) 6 cm, 8 cm and 10 cm (4) 3 cm, 4 cm and 5 cm

307

QUANTITATIVE APTITUDE: GEOMETRY

11. What is the area of a triangle whose sides measure 5 cm, 6 cm and 7 cm ? (2) 12.8484 cm2 (1) 10.9797 cm2 2 (3) 16.4545 cm (4) 14.6969 cm2 12. In ∆ABC, AB = AC, and ∠BAC is 50°. Then ∠ABC and ∠BCA are, respectively (1) 50° and 55° (2) 65° and 65° (3) 70° and 75° (4) 55° and 55° 13. The side MN of ∆LMN is produced to X. If ∠LNX =117° and ∠M=

1 ∠ L, then ∠ L is 2

(1) 77°

47 =

(3) 76°

47x = 47 × 2 x =2 Length of longest side = 20 × 2 = 40 cm 4. Option (1) is correct. Given that side of equilateral triangle = 4 cm 3 × side2 using, area of equilateral triangle = 4 ⇒ ⇒

(4) 75°

AB BD 14. In ∆ABC, D is a point on BC. If = , ∠B = 75° and ∠C AC DC

= 45°, then ∠BAD is equal to (1) 50° (2) 30° (3) 60° (4) 45° 15. The centroid of an equilateral triangle ∆PQR is L. If PQ = 18 cm, then the length of PL is (2) 5 3 cm

(3) 6 3 cm

5. Option (4) is correct. Given: ∠QPR = 75°, ∠PRQ = 40° P

75° S

(4) 4 3 cm

1. (2)

2. (4)

4. (1)

9. (3)

10. (2) 11. (4) 12. (2) 13. (2) 14. (2) 15. (3)

5. (4)

6. (1)

7. (4)

8. (1)

15

D

E

3

5 C

B

By using Thale’s theorem, AD AE = AB AC ⇒

324 x =    10  225

⇒

⇒ 20 x − 40 = 15 + 15x

x2 =

⇒

⇒ 5x = 55 ⇒ x = 11 2. Option (4) is correct. Using property of centroid that it divides AD into a ratio of 2 : 1. So, coordinates of centroid A (2, -3)  2 × 2 + 1 × 2 2 × 4 + 1 ( −3 )  = ,  2+1  2+1  6 5 = ,  3 3 B

D (2, 4)

C

1 1 1 : : 3 4 5

= 20 : 15 : 12 Let the sides of triangle = 20x, 15x and 12x a+b+c Using, semi-perimeter = 2

2

2

324 × 100 = 144 225

⇒ x = 144 = 12 ∴ The altitude of the bigger triangle is 12 cm. 7. Option (4) is correct. Given: Angles are in the ratio 1 : 2 : 1 ∴ x + 2x + x = 180° ⇒ 4x = 180° ⇒ x = 45° Now, by sin rule a b c = = sin 45° sin 90° sin 45° a c b = =  1   1  1     2 2

⇒

3. Option (2) is correct. Given that, The ratio of sides of triangle =

R

 side1  Area of ∆ 1 =  Area of ∆ 2  side 2 

x−2 15 = 3 + x − 2 15 + 5

 5 = 2,   3

40°

In ∆PQR, ∠QPR + ∠PRQ + ∠PQR = 180° ⇒ 75° + 40° + ∠PQR = 180° ⇒ ∠PQR = 180° − 115° ∠PQR = 65° Since, S and T are mid-points. (given) So, ST | | QR (line joining the mid points of opposite sides in a triangle all parallel and half of the third side) We have, ∠TSQ + ∠SQR = 180° ⇒ ∠TSQ = 180° − 65° = 115° 6. Option (1) is correct. We have, if two triangles are similar, then the ratio of the area of both the triangles is proportional to the square of the ratio of their corresponding sides, altitudes.

A

x2

T

Q

Answers with Explanations 1. Option (2) is correct.

3 × 16 4

= 4 3 cm 2

Answer Key 3. (2)

20 x + 15x + 12 x 2

=

(2) 78°

(1) 3 3 cm

⇒

⇒

b=

2a and b =

2c 1

∴

a:b:c =

i.e.,

a:b:c = 1: 2 : 1

2

: 1:

1 2

308

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

13. Option (2) is correct.

8. Option (1) is correct. Given: In ∆ABC, BD ⊥ AC. In ∆DBC, ∠ΒDC = 90° and ∠DBC = 45° Now, ∠DBC + ∠BDC + ∠DCB = 180° ⇒ 45° + 90° + ∠DCB = 180° ⇒ ∠DCB = 45° In ∆AEC, ∠EAC + ∠ACE + ∠AEC = 180° ⇒ 35° + 45° + ∠AEC = 180° ⇒ ∠AEC = 180° − 80° ⇒ ∠AEC = 100° Since, ∠BEC = 180° ⇒ ∠AEB + ∠AEC = 180° ⇒ ∠AEB = 180° − 100° = 80° 9. Option (3) is correct. Given: G is centroid, XL = 18 cm Centroid divides the median in the ratio 2 : 1.

1 ∠L, ∠LNX = 117° 2

Given: In ∆LMN, LM =

L 2a

117°

a M

N

a =

⇒

G

Given: Z

Let XG = 2x and GL = x Then, 2x + x = 18 ⇒ 3x = 18 ∴ x = 6 cm Now, XG = 2x = 2 × 6 = 12 cm 10. Option (2) is correct. If the sides 5, 7 and 9 are the sides of the right angled triangle, then it doesn’t satisfy Pythagoras theorem. i.e., 92 = 52 + 72 ⇒ 81 = 25 + 49 ⇒ 81 ≠ 74 ∴ 5, 7 and 9 are not the sides of the right angle triangle 11. Option (4) is correct. Let a = 5, b = 6, c = 7 cm Then,

S = =

AB BD = AC DC

∴ AD is the bisector of ∠A. Now, in ∆ABC ∠ABC + ∠BCA + ∠BAC ⇒ 75° + 45° + ∠BAC ⇒ ∠BAC ⇒ ∠BAC Also, we have ∠BAD + ∠DAC ⇒ ∠BAD + ∠BAD

= 180° = 180° = 180° − 120° = 60° ….(i) = ∠BAC = 60° (∴ AD is the angle bisector) 2∠BAD = 60° ∠BAD = 30°

⇒ ∴ 15. Option (3) is correct. Given: PQR is an equilateral triangle. We have, PL = QL = x Also PQ = 18 cm P

a+b+c 2

S ( S − a )( S − b )( S − c )

A =

9 ( 9 − 5 )( 9 − 6 )( 9 − 7 )

A =

9×4×3×2

x

18

Q

x =

130° = 65° 2

∴ ∠ABC = x = 65° and ∠BCA = x = 65°

R

18

a 3 18 × 3 PL = = 3 3

Now,

PL = 6 3 cm

⇒

(Distance of the centroid from vertex in equilateral triangle is =

= 14.6969 cm2

⇒

18

x

A = 3×2 6 = 6 6 12. Option (2) is correct. Given: In ∆ABC, AB = AC and ∠BAC = 50° ∴ ∆ABC is an isosceles triangle. ∴ ∠ABC = ∠BCA = x Now, in ∆ABC, ∠ABC + ∠BCA + ∠BAC = 180° ⇒ x + x + 50° = 180° ⇒ 2x = 180° − 50° ⇒ 2x = 130°

(given)

L

18 5+6+7 = = 9 cm 2 2

Area (A) =

117° = 39° 3

∠L = 2a = 2 × 39 = 78°

∴ 14. Option (2) is correct.

L

X

In ∆LMN, ∠MLN + ∠LMN = 117° (Exterior angle property) ⇒ a + 2a = 117° ⇒ 3a = 117°

X

Y

1 ∠L = a 2

Let ∠L be 2a, then, ∠M =

Topic-3 ….(i)

a 3 ) 3

Circle

 Revision Notes Circle : Set of all points in a plane which are at a fixed distance from a fixed point is called a circle. In the following circle ‘O’ is centre, ‘r’ is radius, and ‘d’ is diameter

d

O r

309

QUANTITATIVE APTITUDE: GEOMETRY

Tangent to a circle : A line which touches the circle at exactly one point is called tangent to the circle.

T

P

D

B A

O

Q

C

E

1 ∠A = (∠DOE − ∠BOC) 2

O In the above circle, PQ is a tangent to the circle at point T.

• When two chords intersect inside a circle, then AE ⋅ EB = CE ⋅ ED

• For every point on the circle, there is a unique tangent passing through it.

D

A

Points to Remember : E

• Tangent is always perpendicular to the radius of the circle at the point of tangency.

A

O

B

C

• If two secants are drawn to a circle from one exterior points then

90° T

AB × AD = AC × AE

D

B

B

A

• The two tangents drawn from an external point to a circle are equal in length. PA = PB (Tangents)

O C

E

• If one secant and one tangent are drawn to a circle from one exterior point, then AB2 = AC × AD

A O

B

A

P

C

O

B In the quadrilateral OAPB,

D

Cyclic Quadrilateral :

∠AOB + ∠BPA = 180° • The two direct common tangents drawn to two circles are equal in length. W X P O

O'

A quadrilateral circumscribed in a circle is called a cyclic quadrilateral. In the given cyclic quadrilateral ABCD,

∠A + ∠C = 180° and ∠B + ∠D = 180°

Z

A

Y

B

In the given figure,WX = YZ

O

• The length of a direct common tangent to two circle is

(

d 2 − r1 − r2

)

2

, where d is the distance between the centres of

D

the circles, and r1 and r2 are the radii of the given circles.

• The point of intersection of the direct common tangents and the centre of the circles are collinear.

Secant : A secant is a line that intersects a circle in exactly two points

C

Also, ∠A + ∠B + ∠C + ∠D = 360° • Radius of a cyclic quadrilateral Let a, b, c and d be the successive sides then R=

1 4

( ab + cd )( ac + bd )( ad + bc ) , where S is the semi-perimeter ( S − a)( S − b )( S − d )

of cyclic quadrilateral Points to Remember : • When two secants intersect outside a circle, then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

• Area of a cyclic quadrilateral A =

( S − a)( S − b )( S − c )( S − d )

• If there’s a quadrilateral which is inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.

310

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

• Let p and q be the diagonals, then pq = ac + bd • If E is the point of intersection of the two diagonals, then AE × EC = BE × ED • The exterior angle formed if any one side of the cyclic quadrilateral produced is equal to the interior angle opposite to it. Points to Remember : Angle in a circle : • Inscribed angles subtended by the same arc are equal.

x

x

x

A

• The angle subtended by an arc at the circumference of a circle is half of the angle subtended by the same arc at the centre of the circle.

x

B

O • Angles subtended by the diameter is 90°.

2x

Objective Type Questions 1. In the following figure, if ∠ABC = 95° ∠FED = 115° (not to scale). Then angle ∠APC is equal to : E

If ∠CDE = 85°; ∠DCF = 94°, then the value of ∠ABF + ∠EAB is : (1) 182°

F D

P

C

A B

(1) 120° (2) 150° (3) 135° (4) 155° 2. If in the following figure (not to the scale), ∠ACB = 135° and the radius of the circle is 2 2 cm, then the length of the chord AB is : D

A

A

(4) 168°

(1) 10 cm

(2) 18 cm

(3) 16 cm

(4) 12 cm

7. In a circle. PQ and RS are two diameters that are perpendicular to each other. Find the length of the chord PR. (1)

PQ 2

(2)

(3) 2 PQ

2 PQ

(4)

PQ 2

8. A circle is centred at O. Two tangents AP and AQ are drawn from an external point A. If ∠POQ = 118°, then ∠PAQ is equal to : (2) 62°

(3) 72°

(4) 98°

9. In the given figure O is the centre of the circle. If PRQ = 64°, then what is the measure of angle OPQ ? R

C

(1) 3 2 cm (2) 4 2 cm (3) 4 cm (4) 6 cm 3. To a circle with centre at O, two tangents AP and AQ are drawn from an external point A, If the ∠PAQ = 80°, then the ∠POQ is : (1) 90° (2) 80° (3) 70° (4) 100° 4. Two tangents AP and AQ are drawn to a circle with centre O from an external point A, where P and Q are points on the circle. If AP= 12 cm and ∠PAQ = 60°, then the length of chord PQ is : (1) 12 cm (2) 10 cm (3) 24 cm (4) 16 cm 5. In the given figure, chords AD and BC in the circle are extended to E and F, respectively.

(3) 179°

6. Two parallel chords are drawn in a circle of diameter 20 cm. The length of one chord is 16 cm and the distance between the two chords is 12 cm. The length of the other chord is :

(1) 112° B

(2) 194°

64° O P

(1) 26°

Q

(2) 32°

(3) 24°

(4) 36°

10. In the figure : C

G

B

150° A

B

D

C

‘G’ is the centre of the circle. Find the ∠ACB when ∠AGB = 150°

E

F

(1) 50°

(2) 60°

(3) 65°

(4) 75°

311

QUANTITATIVE APTITUDE: GEOMETRY

11. In the given figure : X

x cm Y 4 cm Z

6 cm

T

XYZ is a secant and ZT is a tangent. What is the value of x. (1) 9 cm (2) 5 cm (3) 8 cm (4) 7 cm 12. In the given figure, if OQ = QR, then the value of m is :

In ∆AFP, ∠AFP + ∠FPA + ∠PAF ⇒ 85° + ∠FPA + 65° ⇒ ∠FPA We know that, ∠FPC ⇒ ∠FPA + ∠APC ⇒ ∠APC ⇒ ∠APC 2. Option (3) is correct.

= 180° = 180° = 30° = 180° (Linear pair angles) = 180° = 180° − 30° = 150° D

P O2 2

Q S

m°

n°

O

B

R

A 135° C

Given: (1) 3n° (2) n° (3) 2n° (4) 4n° 13. In the given figure, PQRS is a cyclic quadrilateral. What is the measure of the ∠PQR, if PQ is parallel to SR ? S

P

R

110° Q

(1) 70° (2) 110° (3) 80° (4) 100° 14. If two concentric circles are of radii 13 cm and 12 cm, respectively, then the length of the chord of the larger circle which touches the smaller circle is : (1) 35 cm (2) 10 cm (3) 15 cm (4) 25 cm 15. A 5 cm long perpendicular is drawn from the centre of a circle to a 24 cm long chord. Find the diameter of the circle. (1) 32 cm (2) 13 cm (3) 30 cm (4) 26 cm Answer Key 1. (2)

2. (3)

3. (4)

4. (1)

5. (3)

6. (3)

7. (4)

∠ACB = 135°

Radius = 2 2 cm ∠ACB + ∠ADB = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°) ⇒ 135° + ∠ADB = 180° ⇒ ∠ADB = 45° O is the centre of the circle ∠AOB = 2∠ADB ⇒ ∠AOB = 2 × 45° = 90° OA = OB = 2 2 = Radius



∴ ∆AOB is right angled triangle. OB2 + OA2 = AB2 2

8+8

AB =

∴

16 = 4 cm

3. Option (4) is correct. We have by properties of tangent P

8. (2) O ?

Answers with Explanations

A

80° Q

∠POQ + ∠PAQ = 180° ⇒ ∠POQ = 180° − 80° ∴ ∠POQ = 100° 4. Option (1) is correct.

E

F P

P

D C

A

[Sum of the opposite angles of a cyclic quadrilateral is 180°] = 180° = 180° − 95° = 85° = 180° = 65°

12 c m 60°

O

B

Join A to F,

In quadrilateral ABCF, ∠ABC + ∠AFC ⇒ ∠AFC ⇒ ∠AFC In quadrilateral AFED, ∠FAD + ∠FED ⇒ ∠FAD = 180° − 115°

2

=

9. (1) 10. (4) 11. (2) 12. (1) 13. (2) 14. (2) 15. (4)

1. Option (2) is correct.

(2 2 ) + (2 2 )

AB =

⇒

A

Q

We have,

PA = QA (Tangents from same external point is equal) ∠APQ = ∠AQP = x

∴ In ∆APQ; ∠PAQ + ∠APQ + ∠AQP = 180° 60° + 2x = 180° ⇒ 2x = 120° ⇒ x = 60° ∴ All the angles are 60° ∴ The ∆PAQ is equilateral triangle ∴ PQ = 12 cm

312

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

7. Option (4) is correct. We have,

5. Option (3) is correct. Given, ∠CDE = 85°, ∠DCF = 94°

R

B

A

P

D

C

E

F

S

We have, ∠ADE = 180° i.e., ∠ADC + ∠CDE = 180° ⇒ ∠ADC = 180° − 85°

(Linear pair)

⇒ ⇒

PR2 = PO2 + OR2 PR2 = 2PO2 PR = 2 PO PQ = 2PO

∴

∠ADC = 95°

We know that,

Also,

∠BCF = 180°

∴

PR =

⇒

PR =

∠DCF + ∠BCD = 180° ⇒

∠BCD = 180° − 94°

∴

∠BCD = 86°

Now,

∠DAB + ∠ΒCD = 180° ∠DAB = 180° − 86° = 94° …(i)

= ∠EAB Also, ⇒

∠CBA = 180° − 95° = 85°

⇒

∠CBA = ∠ABF = 85°

From (i) and (ii), ∠ABF + ∠FAB = 85° + 94° = 179° Shortcut Method : From the properties of quadrilateral we know that: ∠EDC = ∠ABF = 85° ∠DCF = ∠EAB = 94° ∴ ∠ABF + ∠EAB = 85° + 94° = 179°

C

D

Given: diameter = 20 cm ∴ Radius, r = OB = 10 cm Also, given distance between two chords is 12 cm. Let PO be x then OQ = 12 − x In ∆POB, ⇒ ⇒ ⇒ ⇒ In ∆OQD,

PQ

PB2 + PO2 = OB2 82 + x2 = 102 x2 = 100 − 64 x2 = 36 x = 6 cm OD2 = OQ2 + QD2 100 = (12 − 6)2 + (

⇒

1 CD2 = 64 4

⇒

CD =

2

O 118°

….(ii)

A

Q

⇒ 118° + ∠PAQ = 180° ∴ ∠PAQ = 62° 9. Option (1) is correct. Given: PRQ = 64° R

64° O P

Q

(d = 2r)

PQ 2× 2

P

6. Option (3) is correct. Let AB and CD be two chords 16 cm B P x cm 10 O (12 – x)

Q

OP = OQ (Radius of the circle) ∆POQ is an isosceles triangle ∴ ∠OPQ = ∠OQP = x ∴ ∠PRQ = 64° ∴ ∠OQP = 2 × 64° = 128° (Angle subtended by an arc at the circumference of the circle is half of the angle subtended by the same arc at the centre.) Now, in ∆OPQ, ∠POQ + ∠OPQ + ∠OQP = 180° ⇒ 128° + x + x = 180° ⇒ 2x = 180° − 128° ⇒ x = 26° ∴ ∠OPQ = x = 26° 10. Option (4) is correct. We have result : Angle subtended by an arc at the circumference of the circle is half of the angle subtended by the same arc at the centre. C

1 CD)2 2

64 × 4 = 16 cm

( PO = OR = r) ….(i)

8. Option (2) is correct. Given: ∠POQ = 118° We have, ∠POQ + ∠PAQ = 180°

∠ADC + ∠CBA = 180°

A

Q

O

G 150° A

∴

B

∠AGB = 2∠ACB

313

QUANTITATIVE APTITUDE: GEOMETRY

14. Option (2) is correct. Let AB be the chord which touches the smaller circle at D. Then, from the given data OD = 12 cm and OB = 13 cm Draw OD ⊥ AB. We have, AD = DB [Perpendicular from the centre on the chord bisects the chord.]

150° = ∠ACB 2

⇒

∴ ∠ACB = 75° 11. Option (2) is correct. Given: xyz is a secant and ZT is the tangent We have, (ZT)2 = YZ × ZX X

x cm

D

A

Y

12

4 cm 6 cm

T

(6)2 = 4(4 + x) 36 = 16 + 4x 4x = 20

⇒ ⇒

x =

⇒

In ∆OBD, ⇒

20 = 5 cm 4

O

DB = 169 − 144 = 5 cm AB = DB + AD = DB + DB = 10 cm

Let chord be AB = 24 cm and perpendicular be OD = 5 cm

Q S

122 + DB2 = 132

15. Option (4) is correct.

P

m°

OD2 + DB2 = OB2

⇒ Now,

12. Option (1) is correct. Given: OQ = QR ∴ ∠QOR = ∠QRO = n°

A n°

13

O

Z

B

5

R

∠OQP = ∠QOR + ∠QRO [By the property of exterior angle] =2n° Now, OP = OQ = Radius ∴ ∠OPQ = ∠OQP = 2n° [Angles opposite to equal sides] In ∆OPR, ∠POS = ∠OPR + ∠ORP [By the property of exterior angle] m = 2n° + n° =3n° 13. Option (2) is correct. Given: PQ | SR and PQRS is cyclic quadrilateral Then, ∠PQR = ∠SPQ ∴ ∠PQR = 110°

D 12

B

r

O

In ∆OQR,

DB =

1 AB = 12 cm 2

[Perpendicular from the centre on the chord bisects the chord.] ⇒

OB2 = OD2 + DB2

⇒

OB2 = 52 + 122

⇒

OB = =

25 + 144 169 = 13 cm

Radius = 13 cm, ∴ diameter = 2r = 2 × 13 = 26 cm

0

10

20

30

40

2012

Year

30%

20%

2014

500 400 300 200 100 0 2015

0

50

100

150

200

250

A

B

Year

2016

2017

0

1000

2000

3000

4000

A

B

C

Data Interpretation

Sub-divided Bar Graph

Pie-Chart

Multiple Bar Graph h

2013

Simple Bar Graph h

Bar Graphs

25%

15% 10%

Graphs

Line Graph

Representing data Systematically in Rows and Columns

Tables

Simple Table: In a simple table, only one attribute (quality) or specialty of the data is presented

First Level

Second Level

Trace the Mind Map

=

Third Level



Value of the sector × 360° Total value Value of sector Angle of the sector = × Total value 360°

Angle of asector

D

Complex Table: More than one attribute or characteristics of the data are presented.

314 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Chapter

Data Interpretation

12

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

Tables

12 1

Bar and Line Graphs

14

Pie-Chart

15

Types of Data Interpretation • Tables • Bar and Line Graphs • Pie-Chart

Tables

Topic-1

 Revision Notes

Data Interpretation: Tables :

Scan to know more about

Table is often used to present a set of numerical this topic data. It helps the person to make comparisons and draw quick conclusions. It provides the reader greater objectivity in the data. Tabular presentation makes complicated information Data easier to understand. Its another advantage is that one can see all the information at a glance. Interpretation (Tabular Data) Tabular presentation usually consists of a table title followed by columns and rows containing data. While looking at the table, carefully read the table title and headings/nomenclature of the columns and the rows. The table title gives a general idea of the type and objective of the data presented. The column and row nomenclatures indicate the specific kind of information contained in them respectively. We present below an example of tabular presentation of annual expenditure of 5 families during the last 4 years. Annual expenditure of five families (in ` Thousands) Year →

MERITS OF TABULATION (a) Tabulation is the next stage after collection and compilation of the data. (b) It simplifies the data. (c) It gives a general idea of trend and pattern within the data. (d) It provides a gateway for further statistical analysis. (e) In tabulation comparable data are kept close, so that a comparable study of these data becomes easy. (f) It makes the data suitable for further diagramatic and graphic representation. (g) It saves time and space, as maximum information is expressed in a small space without repetition. TYPES OF TABLES Statistical tables are formed on the basis of purpose, originality and construction. Keeping in view the present pattern of questions asked in competitive exams of today, we will limit ourselves to the study of tabulation on the basis of construction. This type of tabulation can be divided into two categories, namely: (i) Simple Tables (ii) Complex Tables (i) Simple Tables: In a simple table, only one attribute (quality) or speciality of the data is presented. (ii) Complex Tables: In a complex table, more than one attribute or characteristic of the data are presented. The complex tables are of three types: (a) Two-way Tables (b) Three-way Tables (c) Manifold Tables

I

30

45

50

70

II

45

50

55

80

Example: Classification of 100 students based on the marks obtained by them in Physics and Chemistry in an examination. Marks out of 50 40 and 30 and 20 and 10 and 0 and Subject above above above above above

III

50

55

60

65

Physics

9

32

80

92

100

IV

25

35

40

60

Chemistry

4

21

66

81

100

V

55

45

65

85

Average

7

27

73

87

100

Families ↓

2018

2019

2020

2021

MEANING OF TABULATION Tabulation is one of the most important devices for the presentation of the data in a condensed and comprehensive form. A statistical table is the logical listing of related quantitative data in vertical columns and horizontal rows of numbers with sufficient explanatory and qualifying words, phrases and statements in the form of titles.

The number of students scoring less than 40% marks in aggregate is :  40  Sol. We have 40% of 50 =  × 50  = 20  100  Required number = Number of students scoring less than 20 marks in aggregate. = 100 − Number of students scoring 20 and above marks in aggregate = 100 − 73 = 27

316

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions Directions (Q.1 – 4) : The table shows the production of different types of cars (in thousands) Years 2012 2013 2014 2015 2016 Cars A 30 35 48 45 56 B 42 48 40 38 56 C 48 36 38 35 44 D 51 24 30 46 54 E 20 42 40 35 43 1. If the data related to the production of cars of type E is represented by a pie chart, then the central angle of the sector representing the data of production of cars in 2013 will be (1) 102° (2) 84° (3) 70° (4) 80° 2. What is the ratio of the total production of cars of type A in 2014 and type C in 2013 taken together to the total production of cars of type B in 2016 and type E in 2015 taken together? (1) 12 : 13 (2) 11 : 12 (3) 10 : 11 (4) 12 : 11 3. The total production of type B cars in 2012, 2014 and 2015 taken together is approximately what per cent more than the total production of type A cars in 2013 and 2016 taken together? (1) 31.9 % (2) 33.2 % (3) 36.3 % (4) 34.4 % 4. The number of years, in which the produc-tion of cars of type B is less than the average production of type D cars over the years, is : (1) 4 (2) 1 (3) 3 (4) 2 Directions (Q. 5 – 8) : The table shows the production of different types of cars (in thousands). Years 2013 2014 2015 2016 2017 Cars A 35 40 48 50 36 B 39 45 54 60 72 C 52 25 32 54 45 D 50 42 45 46 47 E 36 46 42 48 55 5. If the data regarding the production of cars of type B is represented by a pie-chart, then the angle of the sector representing the production of cars in 2016 will be (1) 80° (2) 96° (3) 60° (4) 72° 6. The total production of cars of type B in 2013, 2014, 2015 and 2017 taken together is what per cent less than the total production of all types of cars in 2017? (Correct to one decimal place) (1) 18.2 % (2) 18.4 % (3) 15.8 % (4) 17.6 % 7. The ratio of the total production of cars of type C and E taken together in 2013 to the total production of cars of type D in 2014 and 2016 and type E in 2017 taken together is : (1) 8 : 13 (2) 5 : 8 (3) 13 : 32 (4) 8 : 11 8. The production of cars of type A in 2015 and of type C in 2013 taken together is approximately what per cent of the total production of cars of type D in five years? (1) 40.2 % (2) 42.4 % (3) 43.5 % (4) 42.8 % Directions (Q. 9 – 12) : The table shows the production of different types of cars (in thousands). Years

Cars A B C D E

2014

2015

2016

2017

2018

64 48 33 25 40

56 54 42 45 48

57 63 48 40 52

63 64 57 55 61

70 72 64 35 60

9. The ratio of the total production of type A cars in 2015 and type B cars in 2014 taken together to the total production of type C cars in 2017 and type E cars in 2018 taken together is: (1) 16 : 19 (2) 4 : 5 (3) 8 : 9 (4) 34 : 39 10. If the data related to the production of type D cars is represented by a pie chart, then the central angle of the sector representing production of cars in 2015 will be (1) 72° (2) 63° (3) 81° (4) 99° 11. The total production of type D cars during 2015 to 2017 is what per cent less than the total production of type E cars during 2014, 2015, 2016 and 2018 taken together? (1) 35 % (2) 32 % (3) 28 % (4) 30 % 12. The total production of type C cars in 2015 and type E cars in 2018 taken together is what per cent of the total production of cars in 2014 and 2017 taken together ? (1) 22 %

(2) 25 %

(3) 20 %

(4) 27 %

Answer Key 1. (2)

2. (1)

3. (1)

4. (4)

5. (1)

9. (3)

10. (3) 11. (4) 12. (3)

6. (4)

7. (1)

8. (3)

Answers with Explanations 1. Option (2) is correct. Production of cars of type E Total value = 20 + 42 + 40 + 35 + 43 = 180 We have, angle of a sector = Angle of a sector (2013) =

Value of the sector × 360° Total value

42 × 360° = 84° 180

2. Option (1) is correct. Production of type A in 2014 + Production of type C in 2013 = 48 + 36 = 84 Production of type B in 2016 + Production of type E in 2015 = 56 + 35 = 91 The ratio is 84 : 91 ⇒ 12 : 13 ∴ The required ratio is 12 : 13. 3. Option (1) is correct. The total production of type B cars in 2012, 2014 and 2015 taken together = 42 + 40 + 38 = 120 The total production of type A cars in 2013 and 2016 taken together = 35 + 56 = 91 ∴

Required % =

4. Option (4) is correct.

(120 − 91) × 100 91

=31.868

Average production of type D : = =

 31.9 %

51 + 24 + 30 + 46 + 54 5 205 = 41 5

In 2014 and 2015, the production of type B car is less than 41. ∴ The required answer is 2 5. Option (1) is correct. Production of type B cars, Total value = 39 + 45 + 54 + 60 + 72 = 270 Angle of a sector (2016) =

60 × 360° = 80° 270

6. Option (4) is correct. Sum of type B cars in 2013, 2014, 2015 and 2017 = 39 + 45 + 54 + 72 = 210 Sum of all types of car in 2017 = 36 + 72 + 45 + 47 + 55 = 255 ( 255 − 210 ) × 100 ∴ Required % = 255

317

QUANTITATIVE APTITUDE: DATA INTERPRETATION

=

45 × 100 255

= 17.647

 17.6 %

7. Option (1) is correct. Production of type C cars in 2013 + Produc-tion of type E cars in 2013 = 52 + 36 = 88 Production of type D cars in 2014 + Production of type D cars in 2016 + Production of type E cars in 2017 = 42 + 46 + 55 = 143 Now, the ratio = 88 : 143 = 8 : 13 ∴ The required ratio is 8 : 13. 8. Option (3) is correct. Production of type A cars in 2015 + (Production of type C cars in 2013 = 48 + 52 = 100 Total production of type D cars in five years = 50 + 42 + 45 + 46 + 47 = 230 100 ∴ Required percentage = × 100 230 =43.478

 43.5%

9. Option (3) is correct. Production of type A cars in 2015 + Production of type B cars in 2014 = 56 + 48 = 104 Production of type C cars in 2017 + Production type E cars in 2018) = 57 + 60 = 117 104 8 ∴ Required ratio = = Or 8 : 9 117 9 10. Option (3) is correct. Production of type D cars Total value = 25 + 45 + 40 + 55 + 35 = 200 Value of the sector × 360° We have, angle of sector = Total value = Angle of sector (2015) 45 = × 360° = 81° 200 11. Option (4) is correct. Total production of type D cars from 2015 to 2017 = 45 + 40 + 55 = 140 Total production of type E cars in 2014, 2015, 2016 and 2018 = 40 + 48 + 52 + 60 = 200 60 × 100 = 30%. % less = 200 12. Option (3) is correct. Production of type C cars in 2015 + Production of type E cars in 2018 = 42 + 60 = 102 Total production of cars in 2014 and 2017 is = 64 + 48 + 33 + 25 + 40 + 63 + 64 + 57 + 55 + 61 = 510 102 × 100 = 20% ∴ Required percentage = 510

Topic-2

Graphs

 Revision Notes Graph : In Mathematics a graph can be defined as a pictorial representation or a diagram that represents data or values in an organised manner. Here, we will see two types, of graphs. (i) Bar graphs (ii) Line graphs (i) Bar graphs : A bar graph can be defined as a chart or a graphical representation of data, quantities or numbers using bars or strips.

Bar graphs are used to compare and contrast numbers, frequencies or other measures of distinct categories of data. Types of Bar Diagram The various types of bar diagrams which are most commonly used are mentioned below: 1. Simple Bar Diagram 2. Sub-divided Bar Diagram 3. Multiple Bar Diagram 4. Percentage Bar Diagram 1. Simple Bar Diagram Simple bar diagram is the simplest and the easiest of the bar diagrams. It is used to represent only one dependent variable. Example : Production of fertilisers by a company(in 10,000 tonnes) over the year. 100 90 80 70 60 50 40 30 20 10 0

90 75

70

60 45

40 25

2000

2001

2002

2003 Year

2004

2005

2006

2. Sub-divided Bar Diagram A simple bar diagram can represent only one characteristic at a time. For example, the total number of students studying in a University for the last 10 years can easily be expressed by simple bar diagram, but it cannot show the faculty wise distribution of students. This limitation of bar diagram is overcome by subdivided bar diagrams. Sub-divided bar graphs : These are used to represent the breakdown of a total into its component bars. A bar is divided into different segments, each segment represents a given component. Different shades, colours, designs, etc., are used to distinguish the various components. Example : Number of students in different years in college, studying Zoology, Botany, Chemistry and Physics subjects. 140 120

27

100 80 60

35

26 25 30

20 28

35

35

40 20

45

40

2011

2012

35

22

Physics

20

Chemistry

20

Botany Zoology

40

0 2013 Year

2014

3. Multiple Bar Diagram When a combination of inter-related variables is to be presented graphically, multiple bar diagrams are used. These are extended forms of simple bar diagrams. Here, many aspects of the given data are presented simultaneously and as such are very useful for direct comparison between two or more phenomena by representing them with separate bars of different shades or colours.

318

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Multiple bar diagram : Multiple bar diagrams are used to represent various sets of interlinked data. Import and Export

Example : Number of vehicles manufactured by companies over the years 160

250000

140 120

200000

148

139

128

120

100

107

100

80

150000 100000

Import

60

Export

40 20

50000

0 2016

0 2015

2016

2017 Year

2017

2018

2019

2020

2021

Year

2018

(ii) Line graph : Line graphs are used to show how quantity changes. The line goes up, indicating an increase in quantity, and when the line goes down, it signifies a decrease. If the line is horizontal, the quantity remains constant.

4. Percentage Bar Diagram Sub-divided bar diagram presented graphically on percentage basis is termed percentage bar diagram. They are specially useful for the diagrammatic representation of the relative changes in the data. A percentage bar diagram is used to highlight the relative importance of the different component parts to the whole.

Objective Type Questions Direction (Q. 1 – 4) : From the graph given below, answer the following question. (CUET 2022)

How many states have at most 30% or less non-electrified villages? (1) 3 (2) 4 (3) 1 (4) 2 Directions (Q. 5 – 8) : The given bar graph shows the number of student, in Class 4, Class 5, Class 6 and Class 7 who got first class, second class and failed in the final examinations. 50

45 42

45 38

40 35

30

30

25

25 20 15

15 12

10 7

10 5

8 5

3

0 Class-4

Class-5 First Class

1. The total number of people owing books more than ‘40’ is (1) 60 (2) 31 (3) 14 (4) 30 2. The number of people owing books more than 20 but less than 60 is (1) 14 (2) 16 (3) 30 (4) 40 3. The total number of people surveyed is (1) 16 (2) 56 (3) 54 (4) 55 4. The given bar graph represents the percentage of non-electrified villages in six states A, B, C, D, E and F. Study the graph and answer the question that follows. Percentage of non-electrified villages 70 60 60 50

Percentage

50 40 40 30 30 20 20 10 10 0 A

B

C

D State

E

F

Class-6 Second Class

Class-7

Failed

5. What is the pass percentage of Class 4? (1) 94.66% (2) 91.23% (3) 95% (4) 90.45% 6 Which class has the highest number of students who passed? (1) Class 4 (2) Class 5 (3) Class 6 (4) Class 7 7. Which class has the least number of students who passed? (1) Class 4 (2) Class 6 (3) Class 7 (4) Class 5 8. What is the pass percentage of Class 6? (1) 87.83% (2) 90.12% (3) 86.66% (4) 89.96% Directions (Q. 9 – 11): Study the following line graph and answer the questions.

319

QUANTITATIVE APTITUDE: DATA INTERPRETATION

9. What was the difference between the average exports of all three companies in 1994 and 1995 ? (1) `5 crores (2) `20 crores (3) `10 crores (4) `30 crores 10. In which year was the difference between the exports of companies Y and Z maximum? (1) 1994 (2) 1997 (3) 1995 (4) 1996 11. In which year was the export of all three companies maximum? (1) 1996 (2) 1998 (3) 1995 (4) 1999 Directions (Q. 12 – 15): The line graph shows the number of vacancies for executives in a certain company. Study the diagram and answer the following questions. Number of Vacancies 120

100

90

100 80

40

65

55

60 35

40

30

∴

2011

2012

2013

2014

2015

2016

12. In which year was the number of vacancies lesser than that of the previous year? (1) 2015 (2) 2014 (3) 2016 (4) 2012 13. What was the difference in the number of vacancies between the years 2016 and 2014? (1) 60 (2) 50 (3) 70 (4) 40 14. The number of vacancies in 2014 was lesser than that in 2016 by ________. (1) 1.167 (2) 0.7 (3) 0.6 (4) 2.333 15. The salary of an executive in the company is ` 30,000, what was the increase in the expense (in ` lakhs) due to salaries that had to be paid when posts were filled for the vacancies in the year 2017? (2) 6

(3) 9

(4) 12

Answer Key 2. (3)

3. (4)

4. (1)

5. (3)

6. (1)

7. (2)

42 + 10 × 100 = 86.66% 60

9. Option (3) is correct. Average exports of all three companies:

2017

Year

(1) 3

60

6. Option (1) is correct. The highest number of students who passed, that means less failed. [Because each class have 60 strength] ∴ From the graph we have, class−4 has the highest number of students who passed. 7. Option (2) is correct. The least number of students passed means, number of failed students is more. ∴ From the given Bar graph, we can say class−6 has the least number of students who passed. 8. Option (3) is correct. Total strength of the class 6 = 42 + 10 + 8 = 60 Pass percentage =

0

( 45 + 12 ) × 100

= 57 × 100 = 95% 60

20

1. (2)

Pass percentage =

8. (3)

9. (3) 10. (3) 11. (4) 12. (2) 13. (3) 14. (2) 15. (4)

Answers with Explanations 1. Option (2) is correct. The total number of people owing books more than 40 = 14 + 10 + 7 = 31 2. Option (3) is correct. The number of people owing books more than 20 but less than 60 = 16 + 14 = 30 3. Option (4) is correct. The total number of people surveyed = 16 + 14 + 10 + 8 +7 = 55. 4. Option (1) is correct. State A has 10% non-electrified villages State C has 20% non-electrified villages State E has 30% non-electrified villages ∴ 3 states have at most 30% or less non-electrified villages. 5. Option (3) is correct. From the given data Class − 4 strength = 45 + 12 + 3 = 60

 40 + 60 + 90   crores 3  

Year 1994 = `  =`

190 crores 3

 40 + 60 + 120   crores 3  

Year 1995 = `  =` Required difference =

220 crores 3

30 220 190 =` crores = `10 crores − 3 3 3

10. Option (3) is correct. Difference between exports by companies Y and Z: Year 1994 = `(90 – 40) crores = `50 crores Year 1995= `(120 – 60) crores = `60 crores Year 1996= `(90 – 60) crores = `30 crores Year 1997= `(100 – 80) crores = `20 crores Year 1998 = `(100 – 50) crores = `50 crores 11. Option (4) is correct. Total exports by all three companies: Year 1996, `(60 + 70 + 90) crores = `220 crores Year 1998, `(50 + 80 + 100) crores = 230 Year 1995, `(40 + 60 + 120) crores = `220 crores Year 1999 = `(100 + 120 + 140) crores = `360 crores 12. Option (2) is correct. In year 2014, the number of vacancies was lesser than that of previous year. 13. Option (3) is correct. Difference in the number of vacancies between the years 2016 and 2014 = 100 – 30 = 70 14. Option (2) is correct. Number of vacancies in the year 2016 = 100 Number of vacancies in the year 2014 = 30 Decrease in number = 100 – 30 = 70 ∴

Required decrease =

70 = 0.7 100

15. Option (4) is correct. Number of vacancies in the year 2017 = 40 ∴ Required increase in expenditure =`(40 × 30,000) =`12 lakhs

320 Topic-3

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

(iii) Since the total angle at the centre of the circle is 360°, the total magnitude of the various components is taken to be equal to 360°. In other words, 360° is taken as 100% and vice versa.

Pie Chart

 Revision Notes

A pie-chart is a pictorial representation of the numerical data by non-intersecting adjacent sectors of the circle such that area of each sector is proportional to the magnitude of the data represented by the sector. Just as sub-divided and percentage bars are used to represent the total magnitude and its different components, the circle representing the total may be divided into different segments representing certain proportion or percentage of the different components/parts to the total. Such a sub-divided circle diagram is called pie-chart because the entire graph looks like a pie and the components resemble slices cut from a pie. Given below are the important points regarding Pie Chart (i) Converting values to degree

(iv) Since 1 per cent of the total value is equal to

percentage of the component parts can be converted to degrees by multiplying each of the them by 3.6. Example : Various expenditures incurred in publishing a book. Royalty 15%

Transportation cost 10%

Values of the sector × 360° Total value

Angle of a sector =

Printing 20%

Promotion cost 10%

(ii) Converting degree to value

Paper cost 15%

Binding 30%

Angle of the sector × Total value 360°

Value of any sector =

(iii) Value is represented in percentage, then value of any sector Per cent of the sector × Total value = 100°

(iv)If the values of charts are in percentage, then these should sum upto 100% and if it is in degrees then the sum should be 360°. Some Important Points: (i) Different sectors of a pie-chart represent various component parts. (ii) Each of the component value is expressed either as a percentage or fractional ratio of the respective total or as sectoral angle of the respective total.

360 = 3.6°, the 100

1. What is the central angle of the sector corresponding to the expenditure incurred on royalty? Sol.

Central angle (Royalty) =

15 × 360° = 54° 100

2. If for an edition of the book, the cost of paper is `56,250, then find the promotion cost for this edition. Sol. Let the promotion cost for this edition be `P Then 15 : 10 = 56,250 : P 56, 250 × 10 = `37,500 ⇒ P= 15

Objective Type Questions 1. The given pie chart shows the percentage of students enrolled into the colleges A, B, C, D, E and F in a city, and the table shows the ratio of boys to girls in the college. STUDENTS ENROLLED A

B

C

D

E

F

11%

19%

16% 18% 22% 14%

College

B:G

A

9:4

B

5:9

C

3:4

D

7:2

E

l:4

F

3:2

Based on this information, if the total number of students is 9800, then the number of girls in the college B is : (1) 560 (2) 280 (3) 1008 (4) 504 Direction (Q. 2 – 5) : The given pie chart shows the marks obtained in an examination by a student (in degrees). Observe the pie chart and answer the question that follows. Mark obtained in examination Physical Education 50°

Operator P E 23%

Operator Q

A 25%

E 21%

A 16%

B 14%

Maths 80°

English 60°

Biology 55°

2. If the total marks are 720, then the marks obtained in Maths are : (1) 80 (2) 140 (3) 120 (4) 160 3. If the total marks are 720, then the marks obtained in English is what percentage of the marks obtained in Maths? (1) 60% (2) 55% (3) 50% (4) 75% 4. If total marks are 720, then the marks obtained in Chemistry, Biology and Maths together is what percentage of the total marks? (1) 30% (2) 40% (3) 50% (4) 60% 5. If the total marks are 720, then the difference between the total marks obtained in Physics, Maths and Physical education together and the total marks obtained in Chemistry, Biology and English together out of the total marks is : (1) 90 (2) 110 (3) 100 (4) 80 6. The percentage of customers of two network operators P and Q across the cities A, B, C, D, E is shown in the given pie charts.

Physics 70° Chemistry 45°

D 19%

B 18% C 15%

D 26% C 23%

Based on the information in the pie charts, if the customers of operator P are 3.6 lakhs, and the customers of operator Q are 4.2 lakhs, then in the city C, the positive difference between the customers of the operators is : (1) 44,500 (2) 42,600 (3) 48,000 (4) 52,000

321

QUANTITATIVE APTITUDE: DATA INTERPRETATION

7. Study the given pie-chart and table carefully and answer the question that follows. The percentage-wise distribution of lecturers of five different subjects in a university is shown in the pie-chart. The total number of lecturers is 500. Ratio of male to female lecturers : Lecturers

Physics 22% Mathematics 30%

Chemistry 20% Zoology 13% Botany 15%

Male : Females

Mathematics

7:3

Physics

2:3

Chemistry

4:1

Botany

3:5

Zoology

2:5

Ratio of male to female lecturers : Lecturers Male : Females Mathematics 7:3 Physics 2:3 Chemistry 4:1 Botany 3:5 Zoology 2:5 What is the difference in the number of female lecturers in Chemistry and Mathematics? (1) 30 (2) 22 (3) 20 (4) 25 Directions (Q. 11 – 12) : The given pie-chart represents the percentage of students enrolled in five different sports. The total number of students is 2800. Cricket 23%

Find the number of male lecturers in Physics (1) 42 (2) 46 (3) 44 (4) 40 8. The following pie-chart shows the percentage-wise distribution of the number of students in five different schools P, Q, R, S and T. Total number of students in all five schools together is 8400. Q 18%

P 26%

R 14% T 20%

S 22%

The number of students in school T is what percentage of the total number of students in schools Q and S together.? (1) 45% (2) 50% (3) 40% (4) 55% 9. The following chart shows the marks (in degrees) scored by a student in different subjects — English, Hindi, History, Economics and Political Science — in an examination. Total marks obtained in the examination are 600. Observe the chart and answer the question. Political Science 65

Tennis 14%

Hindi 55

History 70

What is the difference between marks scored in History and marks scored in Hindi? (1) 40 (2) 30 (3) 25 (4) 15 10. Study the given pie-chart and table carefully and answer the question that follows. The percentage-wise distribution of lecturers of five different subjects in a university is shown in the pie-chart. The total number of lecturers is 500.

11. If 24 students playing Cricket are shifted to Kabaddi, then find the new ratio of the number of students in Cricket to those in Kabaddi. (1) 30 : 17 (2) 30 : 13 (3) 31 : 18 (4) 31 : 16 12. What is the average of students enrolled in Hockey and Tennis together? (1) 460 (2) 540 (3) 580 (4) 560 Direction (Q. 13 – 14) : Study the following pie-chart and answer the given questions. The pie-chart shows budget expenditure of a company in the year 2018 (percentage distribution) on different heads A, B, C, D and E. E 10% D 15%

B 8%

13. If `165 crore were spent in the year 2018 on A, what would have been the total expenditure for that year (in `crores)? (1) `320 (2) `400 (3) `350 (4) `300 14. The central angle of the sector representing expenditure on head D is : (1) 45° (2) 52° (3) 56° (4) 54° 15. The given pie chart shows the percentage of students enrolled for the courses A, B, C, D and E in a university and the table shows the percentage of students that passed out of the enrolled students in that course. % STUDENTS ENROLLED A

B

C

D

E

18%

26%

24%

12% 20%

Chemistry 20% Zoology 13% Botany 15%

A 55%

C 12%

Physics 22% Mathematics 30%

Hockey 26%

Kabaddi 12%

English 90

Economics 80

Football 25%

Courses

% passed out

A

76

B

82

C

80

D

90

E

75

If the total number of students is 60,000, then the total number of students who did not pass in the courses A and C are : (1) 7,852 (2) 4,992 (3) 8,254 (4) 7,628

322

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Answer Key 1. (3)

2. (4)

3. (4)

4. (3)

5. (4)

6. (2)

7. (3)

8. (2)

9. (3) 10. (4) 11. (3) 12. (4) 13. (4) 14. (4) 15. (2)

Answers with Explanations 1. Option (3) is correct. Given: Total number of students is 9800. ∴ The number of students in college B = value of the sector =

Total angle of Chemistry, Biology and English = 45° + 55° + 60° = 160° Total marks obtained in Chemistry, Biology and English =

∴ Difference in marks = 400 − 320 = 80 Shortcut Method : Required difference

( 70 + 80 + 50 ) − ( 45 + 55 + 60 )  =  × 720 = 80

16 × 9800 = 1568 100

360

In College B, the ratio of boys to girls = 5 : 9 ∴ 5x + 9x = 1568 ⇒ 14x = 1568

6. Option (2) is correct. Customers of P in city C =

x = 1568 = 112

⇒

Shortcut Method : No. of girls in college B =

16 9 × × 9800 100 14

= 1008

2. Option (4) is correct. Given: Total marks = 720 Value of the sector (Maths) =

Customers of Q in city C =

80 ο × 720 = 160 360 ο

ο Marks obtained in English = 60 ο × 720 = 120

= 96,600 Difference = 96,600 − 54,000 = 42,600 7. Option (3) is correct. Given: Total number of lecturers = 500.

We have, 2x + 3x ⇒ 5x ⇒ x Number of male lecturers is

22 2 100 5

× × 500 = 44 =

160

60 × 100 = 75% 80

8. Option (2) is correct. Given: Total number of students in all five schools together = 8400. Number of students in school T = 20 × 8400 = 1680 100

4. Option (3) is correct.

ο

Marks in Chemistry = 45 ο × 720 = 90

Number of students in school Q and S

360

=

ο

Marks in Biology = 55 ο × 720 = 110 360 ο

Marks in Maths = 80 ο × 720 = 160 360

Sum of the marks of Chemistry, Biology and Maths = 90 + 110 + 160 = 360 360 Required percentage = × 100 = 50% 720

Shortcut Method : Required percentage =

= 110 = 110 = 22 2x = 2 × 22 = 44

The number of male lecturers in Physics

Required percentage = 120 × 100 = 75%

Shortcut Method :

22 × 500 = 110 100

Shortcut Method :

360

∴

23 × 4, 20, 000 100

∴ Number of Physics lecturers =

∴ Marks obtained in Maths are 160. 3. Option (4) is correct. Marks obtained in Maths = 160

Required % =

15 × 3, 60, 000 100

= 54,000

14

∴ Number of girls in the college = 9x = 9 × 112 = 1008

∴

160 × 720 = 320 360

( 45 + 55 + 80 ) 360

× 100 = 50%

5. Option (4) is correct. Given: Total marks = 720. Total angle of Physics, Maths and Physical Education = 70° + 80° + 50° = 200° Now, total marks obtained in Physics, Maths and Physical 200 Education = × 720 = 400 360

∴

(18 + 22 ) × 8400 100

= 3,360

Required percentage = 1680 × 100 = 50% 3360

Shortcut Method : Required percentage =

20

(18 + 22 )

× 100 = 50%

9. Option (3) is correct. Given: Total marks obtained in examination 600. Marks scored in History = Marks scored in Hindi =

70 ο × 600 = 116.66 360 ο

55ο × 600 = 91.66 360 ο

Difference = 116.66− 91.66 = 25 Shortcut Method : Required difference =

( 70° − 55° ) 360°

× 600 = 25

323

QUANTITATIVE APTITUDE: DATA INTERPRETATION

10. Option (4) is correct. Given, total number of lectures = 500

Number of Chemistry lecturer = 20 × 500 = 100 100

We have, 4x + x = 100 ⇒ 5x = 100 ⇒ x = 20 Chemistry female lecturer = 20. Number of Mathematics lecturer =

30 × 500 = 150 100

We have, 7 : 3 ⇒ 7x + 3x = 150 ⇒ 10x = 150 ⇒ x = 15 Mathematics female lecturers = 3 × 15 = 45 ∴ Difference = 45 − 20 = 25 Required difference 30 3 20 1 × × 500 − × × 500 = 45 – 20 = 25 100 10 100 5

11. Option (3) is correct. Given: Total number of students = 2800 Number of students who enrolled in Cricket = 23 × 2800 = 644

Required average =

= 12 × 2800 = 336 100 24 students who enrolled in Cricket shifted to Kabaddi ∴

( 26 + 14 ) 2 × 100

Now, total money = 100 × 165 = 300 55

∴ Total expenditure for the year 2018 is `300 crores. 14. Option (4) is correct. 15 × 360 = 54° 100

15. Option (2) is correct. Given: Total number of students = 60,000 Number of students who enrolled for the course A =

C =

= 31 : 18. 12. Option (4) is correct. Given: Total number of students = 2800 Number of students who enrolled in Hockey

26 = × 2800 = 728 100

Number of students who enrolled in Tennis

= 14 × 2800 = 392 100

20 × 60, 000 = 12,000 100

Number of students who didn’t pass in course A =

24 × 10, 800 = 2,592 100

Number of students who didn’t pass in course C =

Kabaddi players = 336 + 24 = 360 = C : K = 620 : 360

18 × 60, 000 = 10,800 100

Number of students who enrolled for the course

Cricket players = 644 − 24 = 620

The new ratio to Cricket of Kabaddi is

× 2800 = 560

13. Option (4) is correct. Given: 55% of the money spent on A in 2018. That is nothing but `165 crore.

100

Number of students who enrolled in Kabaddi

1120 728 + 392 = = 560 2 2

Shortcut Method :

We have, angle of the sector =

Shortcut Method : =

Average =

20 × 12, 000 = 2,400 100

∴ Number of students who didn’t pass in courses A and C = 2,592 + 2,400 = 4,992 Shortcut Method : The total number of students failed in course A and C 100 − 80 ) = 18 × (100 − 76 ) × 60, 000 + 20 × ( × 60, 000 100

100

100

= 18 × 24 × 6 + 20 × 20 × 6 = 2,592 + 2,400 = 4,992

100

1 0

cos A

tan A

3

3 2 1

1 2

1

2

1

2

1

180 • Angle in Degrees = Angle in Radians =  180 • 1 radians = 

0

sin A

of gle ssion n A pre de B

A

O

O

le o A ng

A

Second Level

Trace the Mind Map

  • cosec    = sec  2 

    90    2 

Third Level



  • sec    = cosec  2  

  • cot    = tan  2 

First Level

on va ti f e le

B

  • tan    = cot  2 

  • cos    = sin  2 

  • sin    = cos  2 

Trigonometric Functions

180 = C

Degree

Height and Distance Elevation and Depression Angles

• sin2  + cos2  = 1 • 1 + tan2  = sec2  • 1 + cot2  = cosec2  sin  • tan   cos 

• Domain • Range • Period

 180  RC =     

Radian

Fundamental relation 1 = cosec  • sin  1 = sec  • cos  1 = cot  • tan 

Pythagoras theorem Consider the triangle given below • On Y-axis [90  θ, 270  θ] where a is the perpendicular sin  cos, sec  cosec b is the base, c is the and tan  cot c hypotenuse. a • On X-axis 180  , 360  θ]  2 2 2 b sin  sin, cos  cos c =a +b similarly, for other trigonometric ratios.

Trigonometry

 (not defined)

0

1

 2

90°

Trigonometric Identities

3

1 2

3 2

Trigonometry Ratios Table 30 45 60 Angles (In Degrees) 0    Angles (In Radians) 0 6 4 3

324 Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Angle of depression Angle of the point on the object (below horizontal level) viewed by the observer which is formed by line of sight (with horizontal level.)

angle of elevation

tower

Angle of elevation Angle formed by line of sight with horizontal level

Line of sight Line drawn from eyes of the observer to the point on the object.

tower

angle of depression

0° 0 1 0

θ sinθ cosθ tanθ

Height and Distance

bodies

• Determining distance between celestral

• Calculate distance of the shape from sea

towers

• Measuring heights of buildings and

en u ot H 3 2 1 2

45° 1 2 1 2

1

30° 1 2

3 2 1 3

3

not defined

0

1

90°

height base tan θ =

60°

base hypotenuse cos θ =

height hypotenuse sin θ =

B

height

A

base

se C

θ

yp

Applications:

QUANTITATIVE APTITUDE: TRIGONOMETRY

325

Chapter

13

Trigonometry

Chapter Analysis Concept Name

2021

2023

Additional Questions

Degree and Radian Measures and Trigonometric Ratios

3

12

Standard Identities of Trigonometry and Height and Distance

2

13

Types of Trigonometry

Trigonometric Ratios

Let ∆ABC be a right angled triangle, with 90° at a vertex A, i.e., ∠A = 90° Let ∠ABC = θ, [0° < θ < 90°] or

• Degree and Radian Measures and Trigonometric Ratios • Standard Identities of Trigonometry and Height and Distance

Topic-1

2022

Degree and Radian Measure and Trigonometric Ratios

C

 Revision Notes

The word ‘trigonometry’ is derived from the Greek words ‘trigonon’ (‘triangle’) and ‘metron’ (‘to measure’). Angle : Angle is a measure of rotation of a given ray about its initial point in a plane. The original ray is called ‘initial side’ and the final position of the ray is called the ‘terminal side’ of the angle. The point of rotation is called vertex. • Angle has a positive measure if it is measured in the anticlockwise direction and a negative measure if it is measured in the clockwise direction. ide

B

lS

ina

O

m Ter angle Initia l Side

Degree and Radian Measures

A

Scan to know more about this topic

Trigonometry

1 2 r θ square units 2

Example : Convert degree to radian : 11π C π = (i) 330° = 330° × 6 180°

Example : Convert radian to degree : (i)

5π C 5π 180 = × = 300° 3 3 π

sin θ =

AC perpendicular = BC hypotenuse

cos θ =

base hypotenuse

=

AB BC

perpendicular = AC base AB sin θ tan θ = cos θ

A

Also,

Using above 3, we are defining 3 more ratios

th

Area of the sector OAB, A =

B

tan θ =

of revolution, the angle is said to measure one degree, written as 1°. • A degree is divided into 60 equal parts, each part is called as minute. • A minute is divided into 60 equal parts, each part is called as second. 1° = 60′ [minute, denoted by ′] 1′ = 60″ [second, denoted by ″] Radian : A radian is an angle whose corresponding arc in a circle is equal to the radius of the circle. Denoted by 1C, one radian • Radian is a constant angle and πC = 180° r • Length of an arc of a circle, l = rθ c O ‘r’ is radius of the circle and θ is the angle. r •

θ

base

Then we call, ‘AB’ as base side and ‘AC’ is called as perpendicular side. The side perpendicular to 90°, ‘BC’ is hypotenuse of ∆ABC. We define standard trigonometric ratios sine, cosine and tangent are as follows.

Degree : If a rotation from the initial side to terminal side is  1    360 

hypotenuse

perpendicular

cosec θ =

1 sin θ

F

1 cos θ 1 cos θ cot θ = ; cot θ = tan θ sin θ

sec θ =

Example : In ∆DEF, sin θ = cosec θ = B r A

5

4

D

4 perpendicular = 5 hypotenuse

1 5 hypotenuse = = sin θ 4 perpendicular

cos θ =

3 base = 5 hypotenuse

sec θ =

1 hypotenuse 5 = = cos θ 3 base

tan θ =

4 perpendicular = 3 base

cot θ =

1 base 3 = = tan θ perpendicular 4

3



E

327

QUANTITATIVE APTITUDE: TRIGONOMETRY

Standard Angles of Trigonometric Ratios 0°(0)

π 30°   6

π 45°   4

π 60°   3

π 90°   2

180°( π)

sin θ

0

1 2

1

3 2

1

0

cos θ

1

3 2

1

1 2

0

−1

tan θ

0

ND

0

cosec θ

ND

1

ND

sec θ

1

ND

−1

cot θ

ND

0

ND

2

1 3

2 2 3

3

2

1

3 2

2

3

2

2

1

1

3

Domain and Range of Trigonometric Functions Function, f(x) sin x cos x

Domain, D (–∞, +∞) (–∞, +∞)

π + nπ 2 π + nπ All real numbers except 2

tan x

(–∞, ∞)

All real numbers except

sec x cosec x cot x

(–∞, –1] ∪ [1, ∞)

All real numbers except nπ All real numbers except nπ

Allied Angles The angles of the form, −θ, 90° ± θ, 180° ± θ, 270° ± θ and 360° ± θ are called allied angles • If the trigonometric ratios are expressed in terms of either 180° ± θ or 360° ± θ, the functions remain the same. Example: sin(180° ± θ) will give ‘sin’ angle only. • If the trigonometric ratios are expressed in terms of either 90° ± θ or 270° ± θ, the functions change sin θ  cos θ, tan θ

•

Range, R [–1, 1] [–1, 1]

 cot θ, sec θ  cosec θ Example: sin (90° − θ) = cos θ, sec(90° − θ) = cosec θ, etc. The sign of trigonometric ratios depends on the quadrants in which the angles lie. In the first quadrant, all ratio are positive. In second quadrant

(–∞, –1] ∪ [1, ∞) (–∞, ∞) sin, cosec all are negative. In third quadrant, tan, cot are positive and the rest of all are negative. In fourth quadrant, cos, sec are positive and the rest of all are negative. i.e., All students take coffee In I quadrant → All +ve S A In II quadrant → sin and cosec +ve T C In III quadrant → tan and cot +ve In IV quadrant → cos and sec +ve 3 2

Example :sin 120° = sin(90° + 30°) = cos 30° = cos 210° = cos(180° + 30°) = −cos 30° = −

3 2

Objective Type Questions 1. In ∆ABC with AB = 5 cm, BC = 12 cm, AC = 13 cm and ∠B = 90°, which of the following is/are not correct? (CUET 2023) 12 13 (a) tan C = (b) cosec A = 13 12 (c) sin B =

5 13

12 (e) cosC = 13

(d) tan A =

12 15

Choose the correct answer from the options given below: (1) (a), (b) and (c) only (2) (a) and (b) only (3) (a), (c) and (d) only (4) (b), (c) and (d) only

4 5

2. If in ∆ABC ∠A + ∠B = 90° and sin B = ,then find the value (CUET 2023)

of cosA. (1) 1

3 (2) 5

4 (3) 5

(4) 0

328

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

3. If sinθ + cosecθ = 2, then what is the value of sin2θ + cosec2 θ? (1) 4

(2) 8

(CUET 2023) (4) 16

(3) 2

4. In a triangle ABC, ∠ABC = 75° and ∠ACB = π° . The circular 4 measure of ∠BAC is p p p 5p radian (2) radian (3) radian (4) radian 2 3 6 12 5p . 5. The circular measure of an angle of an isosceles triangle is 9 Circular measure of one of the other angles must be (1)

(1) 5 p 18

2p (3) 9

(2) 5 p 9

(4) 4 p

(2) 120°

9

(3) 108°

(4) 180°

7. If the sum of two angles is 135° and their difference is then the circular measure of the greater angle is (1)

2p 3

(2)

3p 5

(3)

5p 12

(4)

p , 12

p 3

π and sec2θ + tan2θ = 7, then θ is 2 p p 5p p (1) radian (2) radian (3) radian (4) radian 3 6 12 5

8. If 0 ≤ θ ≤

θ

 2θ 

9. If tan   tan   = 1 , what is the value (in degrees) of θ? 2  5  (1) 45° (2) 90° (3) 100° 10. The minimum value of 2 sin2 θ + 3 cos2 θ is (1) 0 (2) 3 (3) 2 11. If cosec 39° = x, the value of –

1 sin 2 51° sec 2 39°

(1)

(2)

x2 − 1

(4) 120° (4) 1

1 + sin2 39° + tan2 51° cosec 2 51°

(1)

(2)

3

1 − x2

(3) x2 – 1

(4) 1 – x2

3 2

1 (3) 2

tan 57° + cot 37° is equal to tan 33° + cot 53°

(1) tan 33° cot 57° (2) tan 57° cot 37° (3) tan 33° cot 53° (4) tan 53° cot 37° 15. If 2 (cos2θ – sin2θ) = 1, 0 is a positive acute angle, then the value of θ is (1) 60°

(2) 30°

(3) 45°

(4) 22

1 2



2. (3)

3. (3)

5 12 13

A

12

sin B = 1, sin C =

5 13

tan A =

12 5

B

2. Option (3) is correct. Given that: ∠A + ∠B We know that in a triangle, ∠A + ∠B + ∠C So, 90° + ∠C or ∠C

C

= 90° = 180° = 180° = 180 − 90° = 90°

Also given that 4 sin B = from right angle triangle, 5 sin B =

B

AC AB AC AB

∴ cosA = =

90

4 5

C

A

3. Option (3) is correct. Given: sinθ + cosecθ = 2 Squaring both sides, (sinθ + cosecθ)2 = 22 ⇒ sin2θ + cosec2θ + 2sinθ.cosecθ = 4 ⇒ sin2θ + cosec2θ = 4 − 2 (Since, sinθ.cosecθ = 1)

⇒ sin2θ + cosec2θ = 2

4. (2)

5. (3)

∠ABC = 75° and ∠ACB = p 4 [∴ 180° = π radian]

6. (3)

7. (3)

9. (3) 10. (3) 11. (3) 12. (2) 13. (4) 14. (2) 15. (2)

8. (2)

A

∴ ∠ABC = 75 × 9 180

= 5p 12

75°

π 5π ∴∠BAC = π − − 4 12 12 π − 3π − 5π = 12 4π = 12 π = radian 3

5. Option (3) is correct. The sum of two remaining angles, =π−

Answer Key 1. (3)

cosec A =

(4) 1

13. If ∠A and ∠B are complementary to each other, then the value of sec2 A + sec2 B – sec2 A. sec2 B is (1) 1 (2) –1 (3) 2 (4) 0 14. The expression

tan C =

4. Option (2) is correct.

is

tan θ + cot θ 12. If = 2 (0 ≤ θ ≤ 90°), then the value of sin θ is tan θ − cot θ 2

1. Option (3) is correct. By using right angle triangle property:

12 cos C = 13

6.  3π  radian is equal to  5  (1) 100°

Answers with Explanations

=

5π 9

4π 9

∴ Each angle =

1 4 π 2π × = 2 9 9

B

π

4

C

329

QUANTITATIVE APTITUDE: TRIGONOMETRY

6. Option (3) is correct.

= 2 (sin2 θ + cos2 θ) + cos2 θ = 2 + cos2 θ (We know that, sin2 θ + cos2 θ = 1) ⇒ Minimum value of cos θ = –1 [ cos 0° = 1, cos 90° = 0] But cos2 θ ≥ 0, ∴ The required minimum value is = 2 + 0 = 2 11. Option (3) is correct.

We know that π radian = 180°

180 3π 3π ∴ radian = × 5 π 5

= 108° 7. Option (3) is correct. Given, two angles are A and B, where A > B ∴ A + B = 135°

1 + sin 2 39° + tan 2 51° cos ec 2 51° −

 135 × π  =  radian  180 

1 sin 2 51° sec 2 39°

= sin 2 51° + sin 2 39° + tan 2 ( 90° − 39° )

 3π  or A + B =   radian  4  π A−B = 12

...(i) ...(ii)

−

1 sin 2 ( 90° − 39° ) sec 2 39°

= cos2 39° + sin 2 39° + cot 2 39° −

On adding equations (i) and (ii), we get 3π π 9 π + π + = 4 12 12 10 π 5π = = 12 6 5π ∴A = 6×2 2A =

=

= 1 + cot2 39° – 1 = cosec2 39° – 1 = x2 – 1 12. Option (2) is correct.

8. Option (2) is correct. sec2 θ + tan2 θ or 1 + tan2 θ + tan2 θ or 2 tan2 θ or tan2 θ

θ = 60°

∴ 180° = π radian

π π ∴ 60° = × 60 = radian 180 3

9. Option (3) is correct.

2θ θ tan × tanθ = 12θ =1 2 tan ×5tan 2 5 1 θ or tan = θ 1 2θ or 2 tan = or tan 2θ 2 tan 5 5 2θ 5 θ 2θ   or tan = tan  90° −  2 5   = cot

or

θ 2θ = 90° − 2 5

tan θ + cot θ + tan θ − cot θ 2 + 1 = tan θ + cot θ − tan θ + cot θ 2 − 1

or

2 tan θ 3 = 2 cot θ 1

or

sin θ sin θ =3 cos θ cos θ

or

sin2 θ = 3 cos2 θ

or

sin2 θ = 3 (1 – sin2 θ)

or

sin2 θ = 3 – 3 sin2 θ

or

4 sin2 θ = 3

or

sin2 θ =

or

sin θ =

1 1 1 + − cos2 A sin 2 A sin 2 A cos2 A

=

or

5θ + 4 θ = 90° 10

sin 2 A + cos2 A − 1 sin 2 A ⋅ cos2 A

=

or

9θ = 900

1−1 =0 sin 2 A cos2 A

900 = 100° 9

10. Option (3) is correct. 2 sin2 θ + 3 cos2 θ = 2 sin2 θ + 2 cos2 θ + cos2 θ

3 2

=

θ 2θ + = 90° 2 5

θ=

3 4

13. Option (4) is correct. Given that, A + B = 90° or B = 90° – A So, sec2 A + sec2 B – sec2 A sec2 B = sec2 A + cosec2 A – sec2 A cosec2 A [ sec(90° – A) = cosec A]

or

or

[ cosec 39° = x]

Using componendo and dividendo rule, we get =7 =7 =7–1=6 =3

tan θ = 3 = tan 60°

or

or

[We know that sin (90° – θ) = cos θ tan (90° – θ) = cot θ]

Given that, tan θ + cot θ 2 = tan θ − cot θ 1

5π 12

or

1 cos2 39° sec 2 39°

14. Option (2) is correct. Given that,

tan 57° + cot 37° cot 33° + tan 53° = tan 33° + cot 53° tan 33° + cot 53°

[ tan (90° – θ) = cot θ, cot (90° – θ) = tan θ]

330

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Height and Distance

1 + tan 53° tan 33° = 1 tan 33° + tan 53° 1 + tan 53° tan 33° tan 53° × tan 33° tan 53° + 1 tan 33°

= tan 53° cot 33° = cot 37° tan 57° 15. Option (2) is correct. Given that, 2 (cos2 θ – sin2 θ) = 1

or

2 sin2 θ =

or

sin2 θ =

or

sin θ = ±

or

Topic-2

of ngle

θ = 30°

( θ is +ve angle)

A

2 tan θ 1 − tan 2 θ

tan 2θ =

• •

sin 3θ = 3 sin θ − 4 sin3 θ cos 3θ = 4 cos3 θ − 3 cos θ

•

tan 3θ =

Horizontal line

of ne Li P

Example: If the height of a tower is 2 3 m and the length of its shadow is 2 metres. Find the angle of elevation of the top of the tower. Sol. Let AB be the tower and CA be its shadow as shown is the figure. AB 2 3 = = CA 2

Then,

tan θ =

⇒ ⇒

tan θ = 3 tan θ = tan 60°

3

B

2 3m



C

Town A

2m

θ = 60°

⇒

∴ The angle of elevation is 60°. Example: The angle of depression of a stone from the top of a tower on the ground is 45°. If the stone is away from the building at a distance of 120° metres. Find the height of the tower. Sol. Let C be a stone and AB is a tower with height ‘h’ metres. In ∆ABC, A

45°

Tower (h)

y = −2 is not possible because −2 does not belong to range of sin x. ∴ y = −1 i.e., sin x = −1 ⇒ sin x = −sin 90° ⇒ sin x = sin(180° + 90°) ⇒ sin x = sin 270° ∴ x = 270°

ssi pre

f de

le o

Ang

3 tan θ − tan 3 θ 1 − 3 tan 2 θ

• sin4 θ + cos4 θ = 1 − 2 sin2 θ cos2 θ • sin6 θ + cos6 θ = 1 − 3 sin2 θ cos2 θ Example : Find x, if 2 sin2 x + 6 sin x + 4 = 0 Sol. Put sin x = y, We get, 2y2 + 6y + 4 = 0 ⇒ 2y2 + 4y + 2y + 4 = 0 ⇒ 2y(y + 2) + 2(y + 2) = 0 ⇒ (y + 2) (2y + 2) = 0 ⇒ y = −2 or 2y = −2 ⇒ y = −1

O

on

 Revision Notes Mainly, we are using 3 standard identities to solve problems. 1. sin2 θ + cos2 θ = 1 2. 1 + tan2 θ = sec2 θ 3. 1 + cot2 θ = cosec2 θ In addition to these, there are some more identities. • sin 2θ = 2 sin θ cos θ • cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ

A

Horizontal line

Here, ∠POA is angle of elevation. (ii) Angle of Depression : Suppose a man standing at a point ‘O’ looks down at an object P placed below the level of his eyes, then the angle which the line of sight makes with the horizontal line is called the ‘angle of depression’. Here, ∠AOP is angle of depression.

Standard Identities of Trigonometry and Height and Distance

•

A

O

1 = sin 30° 2

ation

elev

ht

(1 – 2 sin2 θ) =

ht

or

ig

(1 – sin2 θ – sin2 θ) =

of s

or

1 2 1 2 1 2 1 2 1 4

e

cos2 θ – sin2 θ =

Li n

or

P

sig

=

In this concept, we will solve the problems on height and distance of different object by using trigonometric properties. There are mainly two concepts that are explained below : (i) Angle of Elevation : Suppose a man standing at a point ‘O’, looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal line is called the ‘angle of elevation’.

90° B

D Angle of depression

45° Stone C 120 m

331

QUANTITATIVE APTITUDE: TRIGONOMETRY

tan 45° =

In ∆ACD,

h 120

A h

⇒ ⇒

h = 120 metres

30 m

B

h 1 = 120

Example: From a point on the ground, the angle of elevation of the bottom and top of a tower fixed at the top of a 30 m height building are 45° and 60°, respectively. Find the height of the tower. Sol. Let BC be the height of the building and h be the height of the tower. Now, in ∆BCD,

tan 45° = CD =

⇒

30 CD

60°

C

30 + h tan 60° = CD

CD =

⇒ From (i) and (ii),

30 =

30 30 = = 30 metres tan 45° 1

…(i)

45°

D

30 + h 30 + h = tan 60° 3

…(ii)

30 + h

⇒

3 30 3 = 30 + h

⇒

h = 30

(

)

3 −1 m

Hence, the height of the tower is 30 ( 3 – 1) m.

Objective Type Questions 1. From a point P on the ground the angle of elevation of the top of 10 m high building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from the point P is 45°. Length of the flagstaff is: (Take 3 = 1.732 ) (CUET 2023) (1) 7 m (2) 7.24 m (3) 7.56 m (4) 7.32 m 2. A man standing on the bank of a river observes that the angle subtended by a tree standing on the opposite bank is 60° on his side of bank. When he moved away 24m from the bank, he finds the angle to be 30°. Find the breadth of the river: (CUET 2023) (1) 12 (2) 18 (3) 26 (4) 20 3. If sec2 x − 3 sec x + 2 = 0, then the value of x (0° < x < 90°) is (1) 45° (2) 15° (3) 60° (4) 30° 4. Find x, if 2 sin2 x − 1 = 0. π (1) 4

π (2) 2

(3) 0

(4) π

5. If a sin A + b cos A = c, then a cos A − b sin A is equal to : (1)

a2 − b 2 − c 2

(2)

a2 + b 2 − c 2

(3)

a2 + b 2 + c 2

(4)

a2 − b 2 + c 2

6. If sin θ − cos θ = (1)

41 29

1 , find the value of sin θ + cos θ. 29

(2)

42 29

(3)

22 29

(4)

30 29

7. The elimination of θ from x cos θ − y sin θ = 2 and x sin θ + y cos θ = 4 will give : (2) x2 + y2 = 20 (1) 3x2 − y2 = 20 (3) 3x2 + y2 = 20 (4) x2 − y2 = 20 8. If sin θ = (1)

2p 1 + p2

p2 − 1 , then cos θ is equal to : p2 + 1

(2)

p p −1 2

(3)

p 1 + p2

(4)

2p p2 − 1

9. The angle of elevation of a flying drone from a point on the ground is 60°. After flying for 5 seconds the angle of elevation drops to 30°. If the drone is flying horizontally at a constant height of 1, 000 3m , the distance travelled by the drone is: (1) 2,000 m (2) 1,000 m (3) 3,000 m (4) 4,000 m

10. From the top of house A in a street, the angle of elevation and depression of the top and foot of another house B on the opposite side of the street are 60° and 45°, respectively. If the height of house A is 36 m, then what is the height of house B? (nearest to an integer) (1) 93 m (2) 94 m (3) 98 m (4) 91 m 11. A pole of length 7 m is fixed vertically on the top of a tower, the angle of elevation of the top of the pole observed from a point on the ground is 60° and the angle of depression of the same point on the ground from the top of the tower is 45°. The height (in m) of the tower is : (2) 7(2 3 − 1) 7 (4) ( 3 + 2) 2

(1) 7 3 (3)

7 ( 3 − 1) 2

12. From the top of a hill 240 m high the angle of depression of the top and of the bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is: (1) 120( 3 − 1)

(2) 80( 3 − 1)

(3) 120(2 − 3 )

(4) 80(2 − 3 )

13. A 22 m long ladder (whose foot is on the ground) leans against a wall making an angle of 60° with the wall. What is the height (in m) of the point where the ladder touches the wall from the ground? (1) 11

(2) 11 3

(3) 11 2

(4)

22 2 3

14. As observed from the top of the lighthouse, 45 m high above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. The distance travelled by the ship during the period of observation is : (Correct to one decimal) (1) 36.9 m (2) 24.8 m (3) 33.4 m (4) 32.9 m Answer Key 1. (4)

2. (1)

3. (3)

4. (1)

5. (2)

6. (1)

9. (1)

10. (3) 11. (3) 12. (4) 13. (1) 14. (4)

7. (2)

8. (1)

332

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Answers with Explanations 1. Option (4) is correct. Let the length of flag = l In the ∆ DBC, 10 tan 30° = BC ⇒

1 3

=

A D

10 BC

45

⇒ BC = 10 3

30

C

Now in ∆ABC, 10 + l tan 45° = BC

x = π

⇒

10 m B

4

5. Option (2) is correct. Given, a sin A + b cos A = c Squaring on both the sides (a sin A + b cos A)2 = c2 ⇒ a2 sin2 A + b2 cos2 A + 2ab sin A cos A = c2 ⇒ a2[1 − cos2A] + b2[1 − sin2A] + 2ab sinA cosA = c2 ⇒ a2 − a2 cos2A + b2 − b2 sin2A + 2ab sin A cos A = c2 ⇒ a2 + b2 − c2 = a2 cos2 A + b2 sin2 A − 2ab sin A cos A ⇒ a2 + b2 − c2 = (a cos A − b sin A)2 ∴(a cos A − b sin A) = 6. Option (1) is correct. Given:

⇒ 10 3 = 10 + l

24m

⇒ sin2 θ + cos2 θ − 2 sin θ cos θ = 1

841

A

2 sin θ cos θ = 1 − 1

⇒

841

Now, consider (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ

60° x

C

(sin θ + cos θ)2 = 1 + 1 − 1

y y ⇒ x= x 3

...(1)

In ΔABD, tan 30° =

y x + 24

...(2)

By equation (1) and (2) 1

x 3 ⇒ x = 12 = 3 x + 24 So, breadth of the river = 12 m

3. Option (3) is correct. Given: sec2 x − 3 sec x + 2 = 0 Put sec x = y. Then, y2 − 3y + 2 = 0 2 ⇒ y − 2y − y + 2 = 0 ⇒ y(y − 2) − 1(y − 2) = 0 ⇒ (y − 2)(y − 1) = 0 ⇒ y = 2 or y = 1 ⇒ sec x = 2 or sec x = 1 (not possible, ∴ sec x = 2 ⇒ sec x = sec 60° ∴ x = 60° 4. Option (1) is correct. Given: 2 sin2 x − 1 = 0 ⇒ 2 sin2 x = 1

⇒ ⇒

sin x = sin

π 4

= 1681 841

sin θ + cos θ =

∴

1681 = 41 29 841

7. Option (2) is correct. Given: x cos θ − y sin θ = 2 …(i) x sin θ + y cos θ = 4 …(ii) Multiply equation (i) by cos θ and multiply equation (ii) by sin θ x sin 2 θ + y sin θ cos = θ 4 sin θ

(

)

2 x cos2 θ + sin= θ 2 cos θ + 4 sin θ

⇒ x = 2 cos θ + 4 sin θ …(iii) Now, multiply equation (i) by sin θ and multiply equation (ii) by cos θ x cos θ sin θ − y sin2 θ = 2 sin θ x cos θ sin θ + y cos2 θ = 4 cos θ

( −) [  sec 60° = 2]

2 1 = 2

(sin θ + cos

θ)2

x cos2 θ − y sin θ cos = θ 2 cos θ

sin2 x = 1 sin x =

841

⇒

B

In ΔABC, tan 60° =

⇒

2

  29

⇒ D

1 29

(sin θ − cos θ)2 =  1 

y 30°

sin θ − cos θ =

Squaring on both the sides,

l = 10 (1.732 – 1) = 7.32 m 2. Option (1) is correct. Let the length of tree is y metre. And breadth of river is x metre. ⇒

⇒

a2 + b 2 − c 2

1 [taking (+) value] 2

( −)

( −)

− y = 2 sin θ − 4 cos θ y = 4 cos θ − 2 sin θ Now, x2 + y2 = 4 cos2 θ + 16 sin2 θ + 16 cos2 θ + 4 sin2 θ + 16 cos θ sin θ − 16 cos θ sin θ ⇒ x2 + y2 = 4(cos2 θ + sin2 θ) + 16(sin2 θ + cos2 θ) ⇒ x2 + y2 = 4 + 16 ⇒ x2 + y2 = 20 8. Option (1) is correct. Given:

sin θ =

p2 − 1 p2 + 1

333

QUANTITATIVE APTITUDE: TRIGONOMETRY

A p + 1 2

p –1

3x = 7 + x

⇒

θ

B BC =

7+x = tan 60° x

In ∆ACD,

2

AC 2 − AB 2 =

(p

2

) (

)

2

+ 1 − p2 − 1

⇒ BC =

p4 + 1 + 2p2 − p4 − 1 + 2p2

⇒ BC =

4p 2 = 2p

∴ cos θ =

2p p2 + 1

x =

⇒

C 2

7 ( 3 − 1)

=

7 ( 3 − 1) m 2

12. Option (4) is correct.

9. Option (1) is correct.

E

A Height = 240 m = AB

60° 30° D Also, Now, in ∆ACB, ⇒ ⇒ Now, in ∆BED, ⇒

C

B

ED = 1, 000 3m ED = AC tan 60° = AC

(Given)

⇒ ⇒ Then, height of

tan 60° =

⇒

BC = 80 3 m

⇒

BC

AE ED AE 1 = 80 3 3

Again, in ∆AED,

3 = 1, 000 3 BC

tan 30° =

1,000 m = BC

⇒

BD

∴

AE = 80 m

Now

DC = EB = AB – AE

tan 30° = ED 1 = 1, 000 3 3 BD

⇒ BD = 3,000 m ∴ DC (distance covered) by = 3,000 – 1,000 = 2,000 m 10. Option (3) is correct. Height = 36 m AC In ∆ABC, tan 45° = BA 60° 45° ⇒ AB = 36 m In ∆CXY,

AB BC 240 3 = BC

Now, in ∆ABC,

XY tan 60° = CX XY 3 = 36

B = BY = 36 + 36 3 = 98.28 m = 98 m

Y

45°

DC – BC = 160 − 80 3

∴

= 80(2 − 3 ) m

X

13. Option (1) is correct. NOTE : Angle is made with wall. So,

36

36

XY = 36 3 m

= 240 – 80 = 160 m

∠ACB = 90o – 60o = 30o Given:

AC = 22 m

A

B

36

60° 2 1

11. Option (3) is correct. Let height of the tower = x m

30° B

3

C

But, according to T-ratio values in ∆ABC,

AB 1

BC 3

AC 2

×11 11 m Required height = 11 m

×11 22

334

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

14. Option (4) is correct.

In ∆ABD,

A 30° 45°

∴ In ∆ABC,

45 m

B Let

⇒

30°

45° D

C

AB = 45 m (Given) CD = x

⇒ ⇒

tan 45° =

AB BD

BD = 45 m tan 30° =

AB BC

1 45 = 3 45 + x 45 + x = 45 3 x = 45( 3 − 1)

x = 32.85 ≈ 32.9 m [ 3 = 1.73]



nC

nP r

r



r

nPr

n nr

nC

n n 0 + C1 +  Cn = 2

n

nC

r=

nC n−r

n nC  r r nr

n = Total no. of items r = No. of items taken for arrangement

Combinations

Sum Rule

Probability

First Level

Second Level

Trace the Mind Map

Third Level



P(A ∪ B) =P(A) × P(B)

Independent Events

Favourable Outcomes Total Outcomes

Mutually Exclusive Events P(A ∩ B)=0 P(A ∪ B)=P(A)+P(B)

= P(A) + P(B) – P(A ∩ B)

P=

Mutually Exclusive Events = m + n

General P(A ∪ B)

Permutations, Combinations and Probability

Permutations

Independent Events No. of ways = m × n

Product Rule

QUANTITATIVE APTITUDE: PERMUTATIONS & COMBINATIONS & PROBABILITY

335

Chapter

14

Permutations and Combinations & Probability

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

4

7

Permutations and Combinations & Probability •

The number of ways in which (m + n) things can be divided into two groups containing m and n things, respectively ( m + n)! . =

Rule of Product: If there are ‘m’ ways to do something and there are ‘n’ ways to do another, then the total number of Scan to know more about ways of doing both things is ‘m × n’.

•

If the number of things are equal, say m = n, total ways of grouping = 2m !

Rule 

Probability:

Types of Permutations and Combinations & Probability  Revision Notes

m! n!

of Sum: If there are ‘m’ ways to do something and there are ‘n’ ways to do another and we cannot do both at the same time, then there are ‘m + n’ ways to choose one of the actions.

Permutations and Combinations •

n

•

n

•

n! Pr = ( n − r )!

this topic

1. Permutations and Combinations Scan to know more about this topic

n! Cr = ( n − r )! r ! n

Pr r! n Cr = n Cn −r n

Cr =

•

n

C0 + n C1 + n C 2 + n C3 + ... + n Cn = 2 n



Number of ways of distributing ‘n’ identical things among ‘r’ persons such that each person may get any no. of things

•

=

•

n+r −1

2 !( m !)2

•

Probability of an event



=

2. Permutations and Combinations

Cr − 1

If out of n things, p are exactly alike of one kind, q exactly alike of second kind and r exactly alike of third kind and the rest are different, then the number of permutations of n things taken all at a time = n ! p!q!r !

•

Total number of ways in which a selection can be made by taking some or all out of (p + q + r +....) items where p are of one type, q are of second type and r are of another type, and so on = {(p + 1) (q + 1) (r + 1) ...} - 1 • The number of different relative arrangements for n different things arranged on a circle = (n - 1)!

Number of favourable outcomes Number of all possible outcomes

• The probability of E not occurring, denoted by P (not E), is given by

P (not E) or P ( E ) = 1 - P(E)

•

Odds in favour =

•

Scan to know more about this topic

Probability

Number of favourable cases Number of unfavourable cases Number of unfavourable cases Odds against = Number of favourable cases

•

If two events are said to be mutuality exclusive, then if one happens, the other cannot happen and vice versa. In other words, the events have no simultaneous occurrence. In general P(A or B)= P(A) + P(B) - P(A Ç B) if A, B are mutually exclusive, then P(A Ç B) = 0 If A, B are independent, then P(A Ç B) = P(A) × P(B) • Addition law of probability: If E and F are two mutually exclusive events, then the probability that either event E or event F will occur in a single trial is given by: P(E or F) = P(E) + P(F) If the events are not mutually exclusive, then P(E or F) = P(E) + P(F) - P(E and F together). • Multiplication law of probability: If the events E and F are independent, then P(E and F) = P(E) × P(F).

Objective Type Questions 1. A game consists of tossing a coin 3 times. Hanif wins if all the tosses give the same result. What is the probability that he loses the game? (CUET 2023) 2 3 1 1 (1) (2) (3) (4) 3 4 2 3 2. If P (E) = 0.05, what is the probability of ‘ not E’? (CUET 2023) (1) 0.05 (2) 0.45 (3) 0.95 (4) 0.55 3. How many 3-digit even numbers can be formed from the digits (0 – 9) if repetitions of digits are allowed? (CUET 2023) (1) 350 (2) 400 (3) 450 (4) 1000 4. Tickets numbered from 1 to 20 are mixed and a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3? (CUET 2023)

3 (2) 3 (3) 2 (4) 1 (1) 10 5 2 20

5. In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is : (1) 200 (2) 216 (3) 235 (4) 256 6. Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S ? (1) 228 (2) 216 (3) 294 (4) 192

337

QUANTITATIVE APTITUDE: PERMUTATIONS & COMBINATIONS & PROBABILITY

7. There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done ? (1) 144 (2) 180 (3) 192 (4) 360 Directions (Q. 8): Let S be the set of all pairs (i, j) where 1 ≤ i ≤ j < n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1,2) and (2, 3) are also friends, but (1,4) and (2, 3) are enemies. 8. For general n, how many enemies will each member of S have ? (2) 1 ( n 2 - 3n - 2 ) 2 (4) 1 ( n 2 − 5n + 6 ) 2

(1) n - 3 (3) 2n - 7

Directions (Q. 9 and 10): The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

Y X E

9. Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is : (1) 60 (2) 75 (3) 45 (4) 90 10. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is : (1) 1170 (2) 630 (3) 792 (4) 1200 11. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed ? (1) 499 (2) 500 (3) 375 (4) 376 Answer Key 2. (3)

3. (3)

Total 3 digit numbers ending with zero = 9 × 10 × 1 = 90 Total 3 digit numbers ending with even digit, {3, 4, 6, 8} = 9 × 10 × 4 = 360 So, total even number of 3 digit that can be formed by given digits = 90 + 360 = 450 4. Option (1) is correct. Multiple of 3 in between 1 and 20 = 3, 6, 9, 12, 15, 18 So, probability that the ticket drawn bears multiple of 3 = =

4. (1)

5. (1)

6. (2)

7. (1)

8. (4)

9. (4) 10. (1) 11. (4)

Answers with Explanations 1. Option (2) is correct. Given that, coin is tossed 3 times and Hanif will win if all the tosses give same result. Cases in Hanif’s favour = [H, H, H],(T, T, T) Total cases = 8 (8 - 2) 6 3 So, required probability = = = 8 8 4 2. Option (3) is correct. Given that: P(E) = 0.05 then probability of not E will be: P ( E ) = 1 − 0.05 = 0.95

6 20

3 10

5. Option (1) is correct. Let there be g girls and b boys. Number of games between two girls = gC2 Number of games between two boys = bC2 ∴ ⇒ ⇒ ⇒ Also,

F

1. (2)

3. Option (3) is correct. Given that the digits are available from 0 to 9.

g

( g - 1) = 45 2

g2 - g - 90 = 0 (g - 10)(g + 9) = 0 g = 10 b

( b - 1) = 190 2

⇒ b2 - b - 380 = 0 ⇒ (b + 19)(b - 20) = 0 ⇒ b = 20 ∴ Total number of games = (g + b)C2 = (10 + 20)C2 = 30C2 = 435 ∴ Number of games in which one player is a boy and the other is a girl = 435 - 45 - 190 = 200 6. Option (2) is correct. Let O and E represent odd and even digits, respectively. ∴ S can have digits of the form O _ O _ E or O _ E _ O or E _O_O Case 1: O _ O _ E The first digit can be chosen in 3 ways out of 1, 3 and 5. The third can be chosen in 2 ways. The fifth digit can be chosen in 2 ways after which the second and fourth digits can be chosen in 2 ways. ∴ There are 3 × 2 × 2 × 3 = 24 ways in which this number can be written. 12 out of these ways will have 2 in the rightmost position and 12 will have 4 in the rightmost position. ∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72 Case 2: O _ E _ O This number can again be written in 24 ways. 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position. Thus, the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72 Case 3: E _ O _ O This number can also be written in 24 ways. As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position. ∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72 ∴ The sum of the digits in the rightmost position of the numbers is S = 72 + 72 + 72 = 216.

338

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

7. Option (1) is correct. Task 2 can be assigned in 2 ways (either to person 3 or person 4). Task 1 can then be assigned in 3 ways (persons 3 or 4, 5 and 6) The remaining 4 tasks can be assigned to the remaining 4 persons in 4! = 24 ways ∴ The assignment can be done in 24 × 2 × 3 = 144 ways. 8. Option (4) is correct. Enemies of every pair are the pairs formed with all members other than the two in the member itself. ∴ If there are n elements then each member has (n - 2)

C2 =

n 2 − 5n + 6 . 2

9. Option (4) is correct. For the shortest route. Neelam follows the following path: A→E→F→B

From B to C, Neelam follows the route: Case I: B → X → C Case II: B → Y → C Case I: B → X → C No. of ways to reach from B to X: ( 5 + 1)! = 6 5 ! × 1!

No. of ways to reach from X to C : 2 So, total number of paths are 6 × 2 = 12 ways. Case II: B → Y → C: There is just one way. Therefore, from B to C, there are 6 × 2 + 1 = 13 ways ∴ Total number of ways of reaching from A to C, via B = 90 × 13 = 1170. 11. Option (4) is correct.

No. of ways to reach from A to E: = ( 2 + 2 )! = 6

In other words, we need to find the total number of 4-digit numbers not more than 4000 using the digits 0, 1, 2, 3 and 4.

No. of ways to reach from E to F: 1

The digits at the hundreds place can be selected in 5 ways.

2! × 2!

No. of ways to reach from F to B: = ( 4 + 2 )! = 15 4! × 2!

Hence, total number of possible shortest paths = 6 × 1 × 15 = 90 10. Option (1) is correct. Neelam has to reach C via B. From A to B. the number of paths are 90, as found in question.

The digit at the thousands place can be selected in 3 ways. The digits at the tens place can be selected in 5 ways. The digits at the units place can be selected in 5 ways. Therefore, the total number of 4-digit numbers less than 4000 is equal to 3 × 5 × 5 × 5 = 375. Therefore, the total number of 4-digit numbers not more than 4000 is equal to 375 + 1 = 376.

Chapter

15

Set Theory

Chapter Analysis Concept Name

2021

2022

2023

Additional Questions

Set Theory

 Revision Notes

6 Union of Sets

A set is nothing but a collection of distinct objects. The objects could be numbers, letters, etc.

Types of Sets: •

•

Mathematically, A Ç B = {x : x ∈ A and x ∈ B}. For example, A = {1, 3} and B = {3, 4, 5}, then A Ç B = {3} Difference of Sets: If A and B are two sets then their difference (A - B) is a set containing elements that are only present in A

A

AB

Intersection of Sets

A

B

U

A

B

Shaded Region = A'

Venn Diagram for 3 Sets Let A, B and C be three sets and U be the universal set then the Venn diagram for 3 sets can be constructed as following: 1 A

and B - A = {x : x ∈ B and x ∈ A}. For example, A = {1, 3} and B = {3, 4, 5}, then A - B = {1}, and B - A = {4, 5} A Venn diagram is a diagram that shows all possible logical relations between a finite A B collection of sets. They can be used to depict two sets, three sets, 4 sets or more. A two set Venn diagram is U given alongside This diagram is called Venn Diagram where there are two sets A and B and U is the universal set. ∪ → Union ∈ → Belongs to

AÇB

Complement of A Set

and not in B. Mathematically, A - B = {x : x ∈ A and x ∈ B}

Venn Diagrams:

B

{

Empty Set or Null Set: A empty set is the one which does not contain any element. The set is denoted by the symbol and n0 = 0, as the set is empty. • Subset of A Set: A set B is subset of set S, if the set B is contained in the set, i.e., all the elements of set B are present in set S. For example, consider B = {1, 3} and S = {1, 3, 4, 5} Then, both 1 and 3 are present in the set S. Hence, B is subset of S. Note: A null set is subset of every set. • Equal Sets : Two sets are called to be equal if they contain same elements. • Universal Set : A universal set is the set containing all the objects, including itself of which all other sets are subsets. It is generally denoted by letter ‘U’. A perfect example for universal set can be set of Real Numbers and the set of Natural numbers, Integers, Rational Numbers, etc., are the subsets of it. • Power Set : A power set is the set of all the subsets of a set say S, it contains both null set and S itself. • Union of Sets : In set theory, union of collection of sets is a set containing every element from those sets, i.e., if there are two sets A and B then their Union contain elements which are in A, in B, or both A and B. It is denoted by ‘’. If we have two set A and B then, their union is denoted as A ∪ B and is called A union B. Mathematically, A ∪ B = {x : x ∈ A or x ∈ B}. For example, A = {1, 3} and B = {3, 4, 5}, then A Ç B = {1, 3, 4, 5} • Intersection of Sets: Intersection of collection of sets containing elements that are common to all the set, i.e., if A and B are two sets then their intersection is the set containing elements that are commonly present on both of them. It is denoted by symbol ‘Ç’.

4

5 7

2 B

6

C 3

Some basic formula for Venn Diagram:

•• n(A Ç B) = n(A) + n(B) - n(A Ç B) •• n(A È B È C) = n(A) + n(B) + n(C) - n(A Ç B) - n(B Ç C) - n(B ∩ C) + n(A ∩ B ∩ C)

340

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test)

Objective Type Questions 1. Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose ? (1) 5 (2) 10 (3) 9 (4) 15 2. A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarised information regarding readership in 3 months is given below: Only September: 18; September but not August: 23; September and July: 8; September: 28; July: 48; July and August: 10; None of the three months: 24. What is the number of surveyed people who have read exactly two consecutive issues (out of the three) ? (1) 7 (2) 9 (3) 12 (4) 14 3. Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is : (1) 52 (2) 42 (3) 36 (4) 64 4. If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be : (1) 23 (2) 26 (3) 96 (4) 93 5. A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is : (1) 38 (2) 32 (3) 45 (4) 43 6. Students in a college have to choose at least two subjects from Chemistry, Mathematics and Physics. The number of students choosing all three subjects is 18, choosing Mathematics as one of their subjects is 23 and choosing Physics as one of their subjects is 25. The smallest possible number of students who could choose Chemistry as one of their subjects is: (1) 20 (2) 19 (3) 22 (4) 21

F2 F3 F1 F1 E3

II round calls :

F1

F1, F2, F3 F1, F2, F3, F4 F2, F3, E2 F2, F3, E1, E2, E3) all know F1, F2, F3, E1, E2, E3) all know

F2 9

E1

F3

7

8

2. (2)

3. (1)

4. (4)

5. (4)

6. (1)

Answers with Explanations 1. Option (3) is correct. Frenchmen : F1, F2, F3 Englishmen: E1, E2, E3 Let E1 knows French I round of calls:

1 F1

3

2 F2

6

th

In the 6 call, E1 knows all the secrets. Similarly, after 9th call, everybody knows all the secrets. 2. Option (2) is correct.

–

–

100 - 24 = 76 had read at least one issue. So, if x people read all the three issues, then (8 - x) people read only the September and July issues. 23 people read the September issue but not the August issue. ∴ 18 + 8 - x = 23 ⇒x=3 As 28 people read the September issue, [28 - (8 - 3) - 3 - 18] = 2 people read only the August and September issues. As 10 people read the July and August issues, 10 - 3 = 7 people read only the July and August issues. ∴ The number of people who have read exactly two consecutive issues = 7 + 2 = 9. 3. Option (1) is correct. Let the number of students who studying only H be h, only E be e, only H and P but not E be x, only E and P but not H be y H

E

Answer Key 1. (3)

20

h

4

F3

Persons secrets know after I-round F1 F1, F2

5 E2

E3

e

10 x

y

O

E1

E2 or E3

P

Given only P = 0 All three = 10; Studying only H and E but not P = 20 Given number of students studying H = Number of students studying E = h + x + 20 + 10 = e + y + 20 + 10 h + x = e + y total number of students = 74

341

QUANTITATIVE APTITUDE: SET THEORY

h + x + 20 + 10 + e + y = 74 h + x + e + y = 44 h + x + h + x = 44 h + x = 22 Therefore, the number of students studying H = h + x + 20 + 10 = 22 + 20 + 10 = 52.

Therefore,

F (144)

26 63

4. Option (4) is correct. Let the number of students who like both pizza and burger be ‘m’. The number of students who like neither of them be n. P

B

105 - m

m

134 - m

n

From Venn diagram 105 - m + m + 134 - m + n = 200 Þ m - n = 39 ∴ The possible values of (m, n) are (39, 0) (40, 1)…….(105, 66) ∴ The number of students who like only burger is lies in the range [134 - 105, 134 - 39] = [29, 95] ∴ From options, the possible answer 93. 5. Option (4) is correct. Let the number of members playing all three games be x. Given, that all the members play at least one of these three games, hence the union of these three sets = 256. Therefore, 256 = 144 + 123 + 132) - (58 + 25 + 63) + x or x = 3. Fitting the numbers in the Venn diagram, we get

58 60

55 3 22

47 C (132)

43 T (123)

25

Hence, the number of members playing only tennis = 43. 6. Option (1) is correct. Given, only P= 0 only M = 0 only C = 0 P(25)

M(23)

x

O

O

18 y

z O C

Let only P&M = x only P&C = y only C&M = z for minimising Chemistry students, we have to maximise only P & M which is 5. But those who choose P as one of their subject. So value of y = 2 and zmin = 0 Total number of students who choose Chemistry as one of their subject = 18 + 2 = 20

Measures of Central Tendency

Chapter

16

Chapter Analysis Concept Name

2021

2022

Measures of Central Tendency

Types of Measures of Central Tendency

2023

Additional Questions

1

11

Mode: Mode is the highest bar in a bar chart or histogram.

 Revision Notes

Mode

25

Measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. The mean, median and mode are all valid measures of central tendency. Mean: The mean is equal to the sum of all values in the data set divided by the number of values in the data set.

20 15 10

x + x 2 + x3 + .........x12 Mean = 1 n

5

Median: The median is middle score for a set of data that has been arranged in order of magnitude. i.e., the median of 2, 3, 4, 6, 9, 10, 11 is 6.

0

[1,5]

[5,9]

[9,13]

[13,17]

[17,21]

[21,25]

Objective Type Questions 1. For the data 11, 15, 13, 12, 10, 8, 11, 7, 15, 11, 13, 7, 11, which of the following is true? (CUET 2023) (1) Mean = Median = Mode (2) Mean > Median > Mode (3) Mean > Mode = Median (4) Mean < Median < Mode 2. What is the median of all possible factors of 120? (1) 10 (2) 11 (3) 12 (4) 13∙5 3. Consider the following table in respect of students of 4 schools who appeared in a test: School

Number of students

Average marks in the test

I

60

60

II

50

80

III

50

40

IV 50 x If the average marks of the students of all four schools are 58, then how many students appeared from School–IV ? (1) 38 (2) 40 (3) 42 (4) 44 4. The data of different natural numbers 4, 7, 10, 14, 2x + 3, 2x + 5, 22, 23, 30, 50 are in ascending order. How many possible values are there for the median of the data for various values of x? (1) Only one value (2) Only two values (3) Only three values (4) Five values 5. If M is the mean of the first 100 even natural numbers, then what is the mean of the first 100 odd natural numbers ? (1) M – 1 (2) M (3) M + 1 (4) M+2 6. Mean marks of 50 students were found to be 78.4. But later it was detected that 95 was misread as 59 and 25 was misread as 52. What is the difference between correct mean and incorrect mean ? (1) 0.04 (2) 0.08 (3) 0.12 (4) 0.18

7. What is the median of the following data ? 2, 3, –1, 2, 6, 8, 9 (1) 2 (2) 3 (3) 4 (4) 5 8. What is the arithmetic mean of the first ten composite numbers ? (1) 8.5 (2) 9.5 (3) 10.2 (4) 11.2 9. What is the median of the data 3, 5, 9, 4, 6, 11, 18 ? (1) 6 (2) 6.5 (3) 7 (4) 7.5 10. Which one of the following pairs is correctly matched ? (1) Median – Graphical location (2) Mean – Graphical location (3) Geometric mean – Ogive (4) Mode – Ogive 11. The mean marks obtained by 300 students in a subject are 60. The mean of top 100 students was found to be 80 and the mean of last 100 students was found to be 50. The mean marks of the remaining 100 students are (1) 70 (2) 65 (3) 60 (4) 50 12. Consider the following distribution :



Class

Frequency

0 – 20

17

20 – 40

28

40 – 60

32

60 – 80

f

80 – 100

19

If the mean of the above distribution is 50, what is the value of f? (1) 24 (2) 34 (3) 56 (4) 96

343

QUANTITATIVE APTITUDE: MEASURES OF CENTRAL TENDENCY

Hint: If n is even, then the median is given by the mean of

Answer Key 1. (3)

2. (2)

3. (2)

4. (4)

5. (1)

6. (4)

7. (2)

8. (4)

9. (1) 10. (1) 11. (4) 12. (1)

Answers with Explanations

Sum of numbers 144 = 13 13

Median of numbers = 11 (Mid number of given data) Mode of numbers = 11 (Most frequent number in given data) So, mean > mode = Median 2. Option (2) is correct. Let N = 120 Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 Total number of factors are 16, which is even.

 Tn + 1 , n is odd  2  Median =   Tn + Tn +1  2 2 , n is even  2

+1

 

⇒ 2 + 4 + 6 + . . . + 200 = M 100

100

⇒ 2

( 200 + 2 ) 100

=M

⇒ M = 101 Now, mean of first 100 odd natural numbers = 1 + 3 + . . . + 199 100 100 1 ( + 199 ) = 2 100

= 100 = M – 1

Hints:

∑x

i

n

6. Option (4) is correct. Given: Incorrect mean marks of 50 student = 78.4 And 95 misread as 59 and 25 misread as 52 ∴ Incorrect total marks = (78.4) 50 = 3920 Now, correct total marks = 3920 – 59 + 95 – 52 + 25 = 3929

= T9 = 12

So, median = 1 (10 + 12) = 11

So,

2

3. Option (2) is correct. School

 2   

2. Sum of n terms of AP = n (First + last term ) 2

2

2

th

observation and  n  + 1 observation.

1. Mean of x1, x2, x3 . . . xn is given by

Here, T= T= 10 n 8

Tn

th

5. Option (1) is correct. Given: Mean of first 100 even natural numbers = M

1. Option (3) is correct. Given numbers = 11, 15, 13, 12, 10, 8, 11, 7, 15, 11, 13, 7, 11 Ascending order of given numbers = 7, 7, 8, 10, 11, 11, 11, 11, 12, 13, 13, 15, 15 = Mean of numbers

n 2  

correct mean marks = 3929 = 78.58 50

∴ Difference between correct mean and incorrect mean

I

60

60

II

50

80

III

50

40

IV

X

50

⇒ Avg. marks of all students = 58 ⇒ (60)(60) + (50)(80) + (50)(40) + 50 (x) 60 + 50 + 50 + x

= 78.58 – 78.4 = 0.18

Avg marks in the test

Number of students

Hint: Mean of x1, x2, . . . ,xn given by =

n

7. Option (2) is correct. Let us first arrange the values in ascending order –1, 2, 2, 3, 6, 8 and 9 Total terms are 7 Median =  n + 1 

∴

= 58

⇒ 3600 + 4000 + 2000 + 50x = 58 (160) + 58x ⇒ 8x = 320 ⇒ x = 40



2



∴ Median = T5 + T6

th



=  7 + 1 

Hint: Recall the formula of average and solved further. 4. Option (4) is correct. Given: 4, 7, 10, 14, 2x + 3, 2x + 5, 22, 23, 30, 50 are in ascending order and all are natural numbers. Total number = 10 2 2x + 3 + 2x + 5 ⇒ Median = = 2x + 4 2

å xi .

2 

th

observation observation

= 4th observation observation = 3 8. Option (4) is correct. First ten composite numbers are 4, 6, 8, 9 ,10, 12, 14, 15, 16, 18 4th

Mean = 4+6+8+9+10+12+14+15+16+18 10

…(i)

⇒ Numbers in ascending order ∴ 2x + 3 > 14 amd 2x + 5 < 22 ⇒ x > 5.5 and x < 8.5   For x = 6, 6.5, 7, 7.5, 8, 2x+3 and 2x+5 are natural numabers. ∴ Five values of median is possible

= 112 = 11.2 10

9. Option (1) is correct. Let us arrange the given data in ascending order. 3, 4, 5, 6, 9, 11 and 18 Here, the number of observations is odd.

344

Oswaal CUET (UG) Chapter-wise and Topic-wise (General Test) th

 7 + 1   term Median =  2 

= 4th term Therefore 4th term is the median ∴ Median = 6 10. Option (1) is correct. Median is linked with graphical location. From ogive, we find the median. 11. Option (4) is correct. Total marks obtained by 300 students = 300 × 60 = 18000 Total marks obtained by top 100 students = 100 × 80 = 8000 Total marks obtained by last 100 students = 100 × 50 = 5000 Total marks obtained by the remaining 100 students = 18000 – 8000 – 5000 = 5000 So, mean marks of the remaining 100 students = 5000 = 50 100

12. Option (1) is correct. Class

ClassMark (xi)

Frequency (fi)

fixi

0 – 20

10

17

170

20 – 40

30

28

840

40 – 60

50

32

1600

60 – 80

70

f

70f

80 – 100

90

19

1710

90 + f

4320 + 70f

Mean = Σi xi Σfi

⇒ ⇒ ⇒ ⇒ ⇒

50 = 4320 + 70 f 96 + f

4800 + 50f = 4320 + 70f 70f – 50f = 4800 – 4320 20f = 480 480 f = = 24 20