Mathematics Grade 5: Volume 1 9798634175676

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Copyright © 2020, April 2, by Hassan A. Shoukr All rights reserved. This book or any portion thereof may not be reproduced or used in any manner whatsoever without the express written permission of the author except for the use of brief quotations in a book review. ASIN: B086S6CX7N ISBN: 979-8-63417-567-6 ISBN: 979-8-64036-739-3

‫ ﺑﺎﻟﮭﯿﺌﺔ اﻟﻌﺎﻣﺔ ﻟﻠﻜﺘﺎب‬13263 \ 2000 ‫رﻗﻢ اﻹﯾﺪاع‬ 3080‫ﺗﺮﺧﯿﺺ وزارة اﻟﺘﺮﺑﯿﺔ و اﻟﺘﻌﻠﯿﻢ ﺑﻤﺼﺮ رﻗﻢ‬

Dear parent, teacher, and student, The series of Mathematics Grade for primary stage is based on several principles: 1-

Mathematics is the queen of other sciences, i.e. all science has a certain amount of mathematics.

2-

Therefore, it must be accessible to all students and with different abilities, attitudes and inclinations...

Based on the previous principles, we did all of the following: 1) The mathematics grade series for primary

stage explains in details each topic of the book in simple, new and easy ways with solved examples followed by graduated exercises. Our strategy is analysis then construction, we analyze the mathematics topic into its initial blocks (small lessons); then we build every two blocks then every three and so on until the topic is finally built in a pyramid form from all its apparent aspects. 2) The mathematics grade series is built on using

the colors well to help the reader to understand quickly every part.

3) In addition to, the dictionary of new terms and

words in the course at the end of the book. We did it in English-Arabic 4) In the end of the book, there are five self-tests

as a bank of problems to measure the abilities of self-analysis and self-assessment. 5) After self-Test, there are ten Exam Style

Papers to ensure the capabilities of using the different skills in the books. Mathematics Grade 5 Volume 1 has the topics: place value; approximation or rounding; multiplication of numeral decimals; division of numeral decimals; Related numbers, length, weight, capacity, land, time, and its applications; related lines; geometric constructions; areas of triangle, cube, and cuboid(prism); volumes of cube and cuboid. We are pleased to know your opinion and observations about the book with our correspondence Hassan A. Shoukr

Contents Subject

Page

How much do you remember? --------------

Arithmetic

1 7

A-I Approximation -------------------------------------

8

♦ Place value ------------------------------------------

8

♦ Approximation to an integer --------------------

11

♦ Approximation to a decimal ---------------------

17

♦ Approximation to a measurement unit ---------

22

A-II Multiplication of Numeral Decimal: --------

31

♦ Decimal point ---------------------------------------

31

♦ Multiplying by 1-digit -----------------------------

34

♦ Multiplying by 2- digits ---------------------------

36

♦ Multiplying by 3- digits -------------------------

38

♦ Applications ---------------------------------------

41

A-III Division of Numeral Decimals: -------------

43

♦ Parts of division -------------------------------------

43

♦ Number of digits of the quotient ----------------

45

♦ Decimal point. --------------------------------------

46

♦ Dividing by 1- digit. -------------------------------

48

♦ Dividing by 2- digits -------------------------------

51

♦ Dividing by 3- digits. ------------------------------

53

♦ Applications. ----------------------------------------

56

A-IV Related Numbers: ---------------------------------♦

♦

♦

♦

♦

63

Measurement units of length ---------------------

63

• Addition -------------------------------------------

65

• Subtraction ---------------------------------------

67

• Applications --------------------------------------

69

Measurement units of weight --------------------

71

• Addition and Subtraction -----------------------

73

•

Applications -------------------------------------

76

Measurement units of capacity-------------------

77

• Addition and Subtraction -----------------------

79

• Applications --------------------------------------

82

Measurement units of land------------------------

84

• Addition and Subtraction -----------------------

87

• Applications --------------------------------------

90

Measurement units of time. ----------------------

92

• Addition -------------------------------------------

96

• Subtraction ----------------------------------------

99

• Applications --------------------------------------

101

Geometry G-I Related lines: -----------------------------------------

105 106

♦

Two intersecting lines. ---------------------------

106

♦

Two orthogonal lines. ---------------------------

108

♦

Two parallel lines ---------------------------------

111

G II Geometric Constructions -------------------♦

Constructing perpendiculars----------------------

♦

115 115

Applications ----------------------------------------- 120 • Construct a square --------------------------------

120

• Construct a rectangle ------------------------ ----

120

• Construct a height of a triangle. ----------------

120

Constructing parallels ----------------------------

123

G III Areas --------------------------------------------

127

♦ ♦

Triangle ----------------------------------------------

127

♦

Cube --------------------------------------------------

131

• Lateral area ---------------------------------------

132

• Total area ------------------------------------------

132

• Applications --------------------------------------

133

Cuboid -----------------------------------------------

135

• Lateral area --------------------------------------

135

• Total area ----------------------------------------

137

♦

• Applications -------------------------------------- 140

G IV Volumes --------------------------------------------

143

♦

Units -------------------------------------------------

143

♦

Cubes and Cuboids ------------------------------

145

♦

Applications ---------------------------------------- 146 •

Capacity ----------------------------------------•

149

Word problems --------------------------------- 149

S Self Tests and Exams---------------------------------♦

Self-Tests--------------------------------------------

♦

Exam Style Papers ------------------------------

Mathematical Terms. ---------------------------

153 153 185

201

How Much Do You Remember? Table Work 1×0 =

2×0 =

3×0 =

4×0 =

1×1 =

2×1 =

3×1 =

4×1 =

1×2 =

2×2 =

3×2 =

4×2 =

1×3 =

2×3 =

3×3 =

4×3 =

1×4 =

2×4 =

3×4 =

4×4 =

1×5 =

2×5 =

3×5 =

4×5 =

1×6 =

2×6 =

3×6 =

4×6 =

1×7 =

2×7 =

3×7 =

4×7 =

1×8 =

2×8 =

3×8 =

4×8 =

1×9 =

2×9 =

3×9 =

4×9 =

1 × 10 =

2 × 10 =

3 × 10 =

4 × 10 =

1 × 11 =

2 × 11 =

3 × 11 =

4 × 11 =

1 × 12 =

2 × 12 =

3 × 12 =

4 × 12 =

5×0 =

6×0 =

7×0 =

8×0 =

5×1 =

6×1 =

7×1 =

8×1 =

5×2 =

6×2 =

7×2 =

8×2 =

5×3 =

6×3 =

7×3 =

8×3 =

5×4 =

6×4 =

7×4 =

8×4 =

5×5 =

6×5 =

7×5 =

8×5 =

5×6 =

6×6 =

7×6 =

8×6 =

5×7 =

6×7 =

7×7 =

8×7 =

5×8 =

6×8 =

7×8 =

8×8 =

5×9 =

6×9 =

7×9 =

8×9 =

5 × 10 =

6 × 10 =

7 × 10 =

8 × 10 =

5 × 11 =

6 × 11 =

7 × 11 =

8 × 11 =

5 × 12 =

6 × 12 =

7 × 12 =

8 × 12 =

9×0 =

10 × 0 =

11 × 0 =

12 × 0 =

9×1 =

10 × 1 =

11 × 1 =

12 × 1 =

9×2 =

10 × 2 =

11 × 2 =

12 × 2 =

9×3 =

10 × 3 =

11 × 3 =

12 × 3 =

9×4 =

10 × 4 =

11 × 4 =

12 × 4 =

9×5 =

10 × 5 =

11 × 5 =

12 × 5 =

9×6 =

10 × 6 =

11 × 6 =

12 × 6 =

9×7 =

10 × 7 =

11 × 7 =

12 × 7 =

9×8 =

10 × 8 =

11 × 8 =

12 × 8 =

9×9 =

10 × 9 =

11 × 9 =

12 × 9 =

9 × 10 =

10 × 10 =

11 × 10 =

12 × 10 =

9 × 11 =

10 × 11 =

11 × 11 =

12 × 11 =

9 × 12 =

10 × 12 =

11 × 12 =

12 × 12 =

Easy Revision Tests Test 1

Test 2

Test 3

Test 4

1)6 + 5 =

8+3=

9+4

=

7+6=

2)13

15

17

=

14

6=

7=

8

9=

3)8 × 5 =

7×9=

4×8

=

5×7=

4)16 ÷ 4 =

25 ÷ 5 =

18 ÷ 3

=

27 ÷ 9 =

5)30 × 10 =

20 × 4 =

10 ×50 =

40 × 6 =

6)500 × 6 =

300 ÷ 30 =

200 ÷20 =

100÷100 =

7)35 × 0 =

342 ÷ 8 =

212 × 4 =

32300÷100 =

8)469 ÷7 =

576 × 5 =

576 × 5 =

575 × 5 =

9)786 ÷12 =

780 ÷10 =

0 ÷1384 =

1384 ÷8 =

10)4800 × 2 =

5001 × 3 =

561 ÷ 3 =

832 × 8 =

Test 6

Test 7

Test 5

Test 8

1)9 + 3 =

7+5=

8+4=

5+9=

2)16

12

14

11

7=

9=

6=

4=

3)6 × 9 =

9×5=

7×4=

3×8=

4)21 ÷ 3 =

36 ÷ 4 =

35 ÷ 7 =

40 ÷ 8 =

5)40 × 10 =

50 × 4 =

5 × 100 =

100 × 2 =

6)50 ÷ 10 =

400 ÷ 100=

100 ÷10 =

30 ÷ 30 =

7)1.80 + 3 =

2.46 + 8 =

4.20 3 =

5.23 4 =

8)3.30 2.05 =

2.3 1.36=

1.36 + 1.3 =

1.45 0.16 =

9)48 × 20 =

73 × 7 =

36 × 12 =

56 × 32 =

10)168 ÷ 12 =

165 ÷ 11 =

252 ÷ 14 =

260 ÷ 13 =

Test 10

Test 11

Test 12

Test 9 1)8 + 5 =

7+8=

9+6=

7+9=

2)13 7 =

16

15

14 8 =

3)3 × 9 =

8×7=

5 × 12 =

6×8=

4)36 ÷ 6 =

49 ÷ 7 =

56 ÷ 8 =

63 ÷ 7 =

5)20 × 10 =

3 × 100 =

40 × 100 =

100 × 20 =

6)60 ÷ 10 =

4000÷100=

200 ÷10 =

300 ÷ 300 =

7)42 × 25 =

123 × 17 =

7.4 + 2.2 =

8.2 + 0.8 =

8)720 ÷ 18 =

560 ÷ 16 =

11.07

1 1 − = 2 3

1 1 + = 3 4

1 ×9 = 3

2÷

1 2

9) −

1 = 4

1 2

10) × 8 = Test 13

9=

Test 14

6=

0.8 = 6.21 2.3 = 1 1 + = 3 5

1 = 4

3÷

Test 15

1 = 2

Test 16

1)12 + 9 =

14 + 7 =

19 + 4 =

17 + 6 =

2)23 16 =

25

37

34 19 =

3)4 × 12 =

9×8=

7×6=

5 × 11 =

4)81 ÷ 9 =

42 ÷ 6 =

96 ÷ 8 =

64 ÷ 8 =

5)30 × 100 =

5 × 1000 =

20 × 50 =

500 × 10 =

6)200 ÷ 100 =

500 ÷ 50=

6000 ÷10 =

4000 ÷ 400 =

7)68 × 43 =

70 × 15 =

0.61 + 3.0 =

1.17 + 2.7 =

8)693 ÷ 21 =

936 ÷ 52 =

12.87

17 =

18 =

3.3 = 9.36 0.39 =

1 = 2

2 1 + = 3 4

5 3 − = 8 5

3 3 − = 4 5

10) ÷ 3 =

5 ÷5 = 9

1 1 × = 4 5

1 1 × = 4 6

2 3

9) + 3 8

Test 17

Test 18

Test 19

Test 20

1)19 + 13 =

17 + 15 =

28 + 14 =

25 + 19 =

2)36

32

41

34

19 =

17 =

24 =

16 =

3)8 × 8 =

12 + 9 =

11× 11 =

7 × 12 =

4)72 ÷ 6 =

132 ÷ 12 =

72 ÷ 8 =

54 ÷ 9 =

5)20 × 30 =

40 × 50 =

200 × 50 =

500 × 20 =

6)5000 ÷500 =

4000 ÷2000 = 6000 ÷300 =

8000 ÷ 40 =

7)61 × 30 =

117 × 23 =

17.45 + 0.08 =

1.68 + 45 =

8)1107 ÷ 27 =

621 ÷ 23 =

21.40 =

23.35

5 3 − = 12 8

9 7 = − 15 20

1 1 1 + + = 2 3 9

2 1 5 + + = 3 4 8

4 3 ÷ = 5 4

1 1 1 ×1 = 2 3

3 1 1 ×1 = 5 9

9)

3 4

5 8

10) ÷ =

4.38

18 =

If you have the queen Mathematics you will be a king. H a s s a n A. S h o u k r

Approximation Place Value Well, dear pupil comes to learn how to determine the place value of a digit whether an integer or a decimal number. Look: dear pupil Thousands Hundreds

Tens

Units

Tenth

Hundredth Thousandth ...

... ... ... 3 9 8 0 . 1 4 6 ... ... ...

Note: The point . separates between two readings: 1) 2)

One with th lies on the R. H. S(*) of . or after . . Other with s . or before . .

or without s - lies on the L. H. S(**) of

Example 1: Complete using

after or

before

i) A tens is the 2nd digit the point.

ii) A tenth is the 1st digit the point. Solution

A tens is the 2nd digit before A tenth is the 1st digit after the point. the point. Tens has s , so we ve chosen before

*

R. H. S means Right hand side.

**

L. H. S means Left- hand side.

Tenth has th , so we ve chosen after

Exercise 1 : Complete using after or before : i)

A tenth is 1st digit

ii)

A hundredth is 2nd digit

iii)

A ten-thousands is 5th digit

iv)

A hundred-millions is 9th digit

v)

A thousandth is 4th digit

vi)

A hundred-thousands is 6th digit

the point . . the point . . the point . . the point . .

vii) A ten-millionth is 7th digit viii) A millions is 7th digit

the point . .

the point . .

the point . .

ix)

A tens is 2nd-digit

the point .

x)

A units is 1st digit

the point . .

xi)

A hundred-thousandth is 5th digit

xii) A thousandth is 3rd digit xiii) A millionth is 6th digit

the point . . the point . .

xv) A ten-thousandth is 4th digit xvi) A hundred-millionth is 8th digit xvii) A hundreds is 3rd digit

the point . .

the point . .

xiv) A ten-millions is 8th digit

♦

the point . .

the point . . the point . .

the point . .

How to determine the place value of a digit:

Example 2: Complete: i) The place value of 4 in 3153.3463 is Solution

4 is 2nd digit after . in 3153.3463. Thus 4 is hundredth digit

The place value of 4 in 3153.3463 is hundredth

Exercise 2: A- Complete: i)

The place value of 3 in 3148.1051 is

ii) The place value of 5 in 3148.1051 is iii) The place value of 0 in 3148.1051 is iv) The place value of 4 in 3154.3163 is v) The place value of 3 in 4136.1041 is vi) The place value of 1 in 136352.0305 is vii) The place value of 9 in 3190005.3 is viii) The place value of 6 in 138910.613 is ix) The place value of 7 in 3.5638197 is x) The place value of 8 in 2.3638521 is B- Determine the place value of the bold red digit in each of: xi) 4563.41386

xii) 461500.3621

xiii) 56.398410281

xiv)38631436.51

xv) 41.3063148612

xvi)3.1989763813

xvii) 123862.38195

xviii)9386319486.3

xix)1.4563281925

xx) 56282193.617

C- Complete: xxi)The tenth in 393.4538 is xxii) The hundreds in 956123.4805 is xxiii)The hundredth in 15638.1253 is xxiv)The millionth in 4.55109802 is xxv) The ten-thousandth in 0.005328 is xxvi)The ten-millionth in 0.003004 is

xxvii) The hundred-thousandth in 0.100428 is xxviii)The thousands in 438631 is xxix) The ten-millions in 12386481.36 is xxx) The unit in 3.456381 is xxxi) The millions in 41238619.313861 is xxxii) The hundred-thousands in 4381539.26205 is

Approximation to Integers Well, dear pupil comes now to know how to approximate to an integer that means approximation to units, tens, hundreds, thousands,

Defn: Mohamed wants to walk from Nagy to Ahmed. Nagy

Beginning

Mohamed

Ahmed

Half

End.

 Mohamed is nearer Nagy than Ahmed. So, we say, Mohamed is approximately at the beginning.

Beginning

Half

End.

Mohamed is nearer Ahmed than Nagy. So, we say,  Mohamed is approximately at the end

Beginning

Half

End.

 Also here Mohamed is nearer Ahmed than Nagy, where he walks half of the distance.

Rule From the previous, we get: i) The number is approximated to O added to the previous number, when it is less than half or < 5 , if we neglect the point. ii) The number is approximated to 1 added to the previous number, when it s greater than or equal half or ≥ 5 , if we neglect the point. Example 3: i)

Approximate 35638 to the nearest ten. Solution

The base number It s the tens 3

The right hand, it s the right hand to the ten 8

35638 30

31

32

33

34

35

36

37

38

39

40

You note that 38 > 35 or 8 > 5 . So, we add 1 to the base number ten - 3 to become 3 + 1 = 4 - and change all the digits after the new base number 4 into zeros

Thus; the required number is: 35640 It s called Or approximately equal . We can write: 35638 ≅ 35640 approximated to the nearest tens ii) Approximate 563819 to the nearest hundreds. Solution Please pupil to solve the problem. Try to answer the following questions: Don t forget all these steps on your mind.

1-

What s the base number? It s hundreds i.e 8

2-

What s the right hand to the base number? It s 1

3-

To what is 1 approximated 0 or 1 ? It s approximated to 0 because the right hand 1 < 5

4-

What s the new base number It s 8 + 0 = 8

5-

What s the required number? It s 563800 Or 563819 ≅ 563800 iii)

Don t forget: all digits after the new base number become zeros

Approximate 1565894 to the nearest ten-thousands Solution

Please

pupil. Try to answer by yourself the questions:

1-

What s the base number? It s

2-

What s the right hand to the base number? It s

3-

To what is approximated 0 or 1 ? It s approximated to because the right hand

Don t forget all these steps on your mind.

5.

4-

What s the new base number? It s + =

5-

What s the required number? It s

Or 1565894 ≅ iv)

to the nearest ten-thousands.

Approximate 56381538 to the nearest thousand.

Solution Please pupil, can you ask yourself the questions and write the answer direct? As the following: 1

56381538 ≅ 56382000 approximated to the nearest thousand. Note: Please dear pupil. Now do you know to ask yourself the questions, then write the final result direct without writing the method?. Exercise 3: i) Approximate 4563819 to the nearest ten. ii) Approximate 35861938 to the nearest ten- thousand. iii) Approximate 38631489 to the nearest million. iv) Approximate 138631285 to the nearest hundred. v) Approximate 3863819 to the nearest ten- million. vi) Approximate 23861927 to the nearest hundred thousand. vii) Approximate 153863195 to the nearest hundreds million. viii) Approximate 53851201 to the nearest thousand. ix) Approximate 38631928 to the nearest unit. Complete: x)

45387 ≅

to the nearest 10

xi)

5638193 ≅

to the nearest 1000

xii) 3861931 ≅

to the nearest hundred

xiii) 4032153 ≅

to the nearest hundred- thousand

xiv) 15638231 ≅

to the nearest million

xv) 43280321 ≅

to the nearest unit- million

xvi) 53829153 ≅

to the nearest hundred- millions

xvii) 938631400 ≅

to the nearest hundred- thousands

xviii)100008156 ≅

to the nearest ten- thousands

xix) 563819381 ≅

to the nearest hundred- millions

xx) 40000381 ≅

to the nearest ten- thousands

♦

Trick of

9 means tricky problems:

Example 4: i)

Approximate 49545

to the nearest thousand.

Solution Please problem.

pupil, attention where there is a device here in this

1- What s the base number? It s thousand i.e 9

Don t forget, all these steps on your mind

2- What s the right hand number to the base number? It s 5 3- To what is 5 approximated 0 or 1 ? It s approximated to 1 because the 5 =5 4- What s the new base number? It s 9 + 1 = 10

right

hand

Oh, there is a trick, where we don t instead 9 of 10, but we add 1 to left hand of the base number which is 4 to become 4 + 1 = 5 and all the digits after the new left hand number are zeros. 5It s 50000 Or

What s the required number? 1

5

4 9 5 4 5 ≅ 50000

to the nearest thousands

10 1

ii) Approximate 32963813 to the nearest hundred thousands. Solution Dear the following:

pupil, now we can solve direct with attention as

1 3

3 2 9 6 3813 ≅ 33000000 10 (to the nearest hundred thousands) 1

iii)

Approximate 5638993 to the nearest hundred. 1 9

Solution

563 8 9 9 3 ≅ 5639000

10

to the nearest hundred.

1

Exercise 4 : i)

Approximate 4596 to the nearest ten.

ii) Approximate 59838 to the nearest thousand. iii) Approximate 5983628 to the nearest hundred thousands. iv)

Approximate 1309923151 to the nearest million.

v)

Approximate 53992 to the nearest hundred.

vi)

Approximate 14795151 to the nearest ten- thousands.

vii) Approximate 498368001 to the nearest ten- millions. viii) Approximate 1569721 to the nearest unit- thousand. ix)

Approximate 53296328 to the nearest hundred- thousands.

x)

Approximate 9838153 to the nearest million.

Approximation to Decimals Well, dear pupil comes to learn how to approximate to a decimal that means approximation to tenth, hundredth, thousandth, Note: As the same previous method, you can approximate to a decimal, but you must pay attention: Unit

= Whole number

the same meaning

1 Tenth = 1-decimal place = 10

the same meaning

1 100 1 Thousandth = 3-decimal places = 1000

Hundredth = 2-decimal places =

Ten-thousandth = 4-decimal places =

the same meaning

the same meaning

1 the same meaning 10000

And so on

Example 5: i)

Approximate 353.486 to the nearest tenth. Solution We can apply the same previous method as the following: 1-

What s the base number? It s tenth i.e. 4

2-

What s the right hand to the base number? It s 8

3-

To what s 8 approximate 0 or 1 ? It s approximated to 1 because: The right hand 8 > 5

4-

What s the new base number? It s 4 + 1 = 5

5-

What s the required number? It s 353.500

Or

Don t forget: all the digits after the new base number are zeros.

1

3 5 3.4 8 6 ≅ 353.500 5

ii)

Don t forget, all these steps on your mind

to the nearest tenth

Approximate 4.38631 to the nearest 3-decimal places.

Solution 3-decimal places =

1 = thousandth, so we can write the 1000 Don t forget to ask your self, the questions to get the right answer

answer direct as the following: 0

4.3 2 6 3 1 ≅ 4.32600 6

iii)

to the nearest 3-decimals places

Approximate 0.5638 to the nearest whole number. Solution 1

0.5 6 3 8 ≅ 1.0000 ≅ 1 1

iv)

Zeros after the point has no meaning, but before the point is very important.

to the nearest whole number.

Approximate 36.48396 to the nearest ten-thousandth. Solution There is the device of 1

9 .

4

3 6. 4 8 3 9 6 ≅ 36.48400 ≅ 36.484 10

Exercise 5:

1

i) Approximate 0.0038.

to the nearest thousandth

ii) Approximate 363.485.

to the nearest unit

iii) Approximate 4.9836

to the nearest tenth.

iv) Approximate 53.856

to the nearest tens.

v) Approximate 486.43

to the nearest hundreds.

vi) Approximate 369.78153.

to the nearest tens- thousandth

vii) Approximate 2413.3281.

to the nearest thousands

viii)

to the nearest hundredth

ix)

Approximate 4696.481. Approximate 56.3895.

to the nearest 3-decimal places

x)

Approximate 0.368396

to the nearest

1 . 100000

1 100

Complete : xi)

4.3638 ≅

to the nearest

xii)

3698.413 ≅

to the nearest 10

xiii) 5638.386 ≅

to the nearest

1 10

xiv)

63984.96 ≅

to the nearest tenth

xv)

5638.38619 ≅

to the nearest thousandth

xvi)

5638.38619 ≅

to the nearest 1000

xvii)

2.563894 ≅

to the nearest

xviii) 5.99998 ≅

1 100000

to the nearest ten- thousandth

xix)

4.8 ≅

to the nearest whole number

xx)

5.36819 ≅

to the nearest 2- decimal places

♦

Approximation with operations (+, -):

Example 6: i)

436.486 + 28.12 =

≅

to the nearest

Solution

1 10

It s allowed using calculator .

Get the final result of addition firstly by your calculator as the following: 0

Thus 436.486 + 38.12 = 4 7 4. 6 06 6

≅ 4 7 4. 600 ii)

Approximate 483.13

19.563 to the nearest hundredth.

Solution By calculator: 1

483.13

It s allowed using calculator.

19.563 = 4 6 3 . 5 6 7 7

≅463.5 70 ≅463.5 7 Exercise 6: Find the result for each of the following then approximate it as shown: i)

3.53864 + 56.3158

ii)

5.3863 − 0.538

iii) 35.4 + 27

1 10 1 to the nearest 1000

to the nearest

to the nearest ten 1 100

iv)

9.362 − 3.2386

to the nearest

v) vi)

4.8629 + 5 4515 + 3.5631

to the nearest whole number to the nearest 100 1 100

vii) ( 34.35 + 3.362 ) − 2.41

to the nearest

viii) ( 5.3881 − 2.56 ) + 33.5

to the nearest unit to the nearest

1 10

ix)

( 5.386 + 3.5 ) − ( 4.38 + 0.008 )

x)

( 23.5 − 3.005 ) + ( 28.3 − 13.314 ) to the nearest 2- decimal places

Approximation to Measurement Units Well, dear pupil here in this chapter, we will show to approximate from a measurement unit to another. ♦

Measurement Units Of Length: Here is another useful arrangement of the table of metric units of length. *

Kilo-

10

**

10 1000

metre

10

deci-

10

centi-

10 1000

milli-

10

1000000

•

Approximation from unit to itself:

To approximate from unit to itself which means approximation to unit s digit or whole number as the previous. Example 7: i)

Approximate 356.538m

to the nearest metre.

Solution Here, we ll approximate from metre to metre. 1

3 5 6 .538 m ≅ 357 m. 7

Exercise 7: i) *

Approximate 43.385 m

This dash means hecto-, kilo- = 10 hectoThis dash means deca-, hecto- = 10 deca-

**

to the nearest metre.

ii)

Approximate 539.638cm

to the nearest metre.

iii)

Approximate 56.83 km

to the nearest kilometre.

iv)

Approximate 563.538dm

to the nearest decimetre.

v)

Approximate 3899.38 mm

to the nearest millimetres.

Complete: vi)

563.433 m

≅

m.

vii) 59.638 cm ≅

cm.

viii) 9386.5 dm ≅

dm.

ix)

56.938 mm ≅

mm.

x)

56.836 km ≅

km.

•

Approximation from unit to another one:

To approximate from first unit to second unit, follow the following steps: 1- Transform from the first unit to the second. 2- Approximate the result to the nearest unit or whole number. Example 8: i)

Approximate 5538.53 cm to the nearest metre. Solution 5538.53 cm = 5 5 3 8 . 5 3 ÷ 100 m = 55.3853 m 1

= 5 5. 3 8 5 3 m 5

≅ 5 5. 0000 m. ≅ 55m

Remember that changing from the small unit into the great unit is division.

b)

Approximate 6.6386 km to the nearest metre. Solution 6.6386 km

= 6 . 6 3 8 6 × 1000 m = 6638.6 m 1

= 6 6 3 8. 6

m

≅6639

m.

9

Exercise 8:

Remember that changing from the great unit into the small unit is multiplication.

i)

Approximate 5.3863 m

to the nearest centimetre.

ii)

Approximate 56386.563 m

to the nearest kilometre.

iii)

Approximate5638.5638dm

to the nearest metre.

iv)

Approximate 23998.5 cm

to the nearest metre.

v)

Approximate 55.3863 km

to the nearest metre.

vi)

Approximate 638.5329 m

to the nearest millimetre.

vii)

Approximate 53853.48 cm

to the nearest decimetre

viii)

Approximate 5539.548 dm

to the nearest kilometre.

ix)

Approximate 5.5638 km to

the nearest decimetre.

x)

Approximate 5385 cm

to the nearest metre.

Complete: xi)

5389.537 m ≅

km.

xii)

xiii)

53.8632 dm ≅

mm.

xiv)

53855 mm ≅

xv)

38659 m ≅

xvi)

9.3856381km ≅

km.

638.55638 cm ≅

mm

m. cm.

♦

Measurement Units of Weight: Well, dear pupil here is another useful arrangement of the table of metric units of weight. gram (*).

Kilo 10

10 1000

10

Example 9: i)

Approximate 356.486 gm

to the nearest gram.

Solution 0

3 5 6. 4 8 6 gm ≅ 356 gm.

Required to approximate from unit to itself which means approximation to the nearest unit

6

ii) Approximate 5639.9 gm

to the nearest kilogram.

Solution 5639.9 gm = 5639.9 ÷ 1000 kg. 1

= 5. 6 3 9 9 kg 6

Required, here to approximate from gram to kilogram which means you must transform to kilogram firstly, then approximate to the nearest unit.

≅ 6 kg. iii) Approximate 35.35968 kg

to the nearest gram.

Solution 35.3598 kg = 35.35968 × 1000 gm 1

6

= 3 5 3 5 9. 6 8 gm ≅ 3 5 3 60 gm 10

(*)

1 10 grams (gm) = 1 decagram.

10 decagrams (dag) = 1 hecto

10 hectograms (hg) = 1kilo.

Exercise 9: i)

Approximate 538.953 gm

to the nearest gram.

ii) Approximate 5638.538 gm

to the nearest kilogram.

iii) Approximate 5389.563 kg

to the nearest kilogram.

iv) Approximate 5.38693 kg

to the nearest gram.

v) Weigh yourself, then approximate your weight to the nearest kilogram. Complete: vi)

53.53986 gm ≅

gm.

vii)

5385.38 gm ≅

viii)

93.56893 kg ≅

kg.

ix)

936.53851 kg ≅

x)

535.38 gm + 5.35548 kg ≅

♦

kg. gm.

gm

Measurement Units of Coins:

Well, dear pupil units.

here is another useful arrangement of coinPound

Piaster

10

10 100

Example 10: i)

Approximate 538.563 P.T to the nearest piaster. Solution 1

5 3 8. 5 6 3 P.T ≅ 5 3 9 P.T 9

ii)

Approximate 53.5389 L.E

Here is approximation from unit to itself remember the method

to the nearest piaster.

Solution 53.5389 pound = 5 3 . 5 3 8 9 × 100 piasters = 5 3 5 3. 8 9

piasters

1

= 5 3 5 3. 8 9

piasters

≅5354

piasters.

4

iii)

Approximate 5369.85 piasters to the nearest pounds. Solution 5369.85 piasters = 5 3 6 9 .8 5 ÷ 100 pounds = 5 3. 6 9 8 5

pounds

1

= 5 3. 6 9 8 5

pounds

≅54

pounds.

4

Exercise 10: i)

Approximate 5386.93 piasters to the nearest piaster.

ii)

Approximate 36.3486 pounds to the nearest piaster.

iii)

Approximate 5353 piasters to the nearest pound.

iv)

Approximate 539.85 pounds to the nearest pound.

v)

Approximate 53599.3 piasters to the nearest pound. Complete:

vi)

L.E 563.386 ≅ L. E

viii) P.T 5389.96 ≅ L. E x)

L.E 0.056 ≅ P. T

vii) P.T 56.9854 ≅ P. T ix)

L.E 9.38696 ≅ P. T

xi)

P.T 2154.12 ≅ L.E

Measurement Units of Time:

♦

Well, dear pupil hour-units.

here is another useful arrangement of H

min

60

sec 60

3600

Example 11: i)

Approximate 33

1 H 4

to the nearest hour.

Solution By calculator:

Remember, from unit to itself.

0

1 4

33 H = 33. 25 H.

ii)

3

≅ 33 H Approximate 4 H, 55min Solution By calculator: 1

to the nearest hour. Remember, we ve divided by 60 because of 1H = 60 min

4H , 55min = 4. 9166H 5

≅ 5H iii)

Approximate 5min, 24 sec Solution By calculator:

to the nearest min.

0

5 min, 24sec = 5.4 min 5

≅ 5 min Exercise 11:

Remember, we ve divided by 60 because of 1min = 60 sec

i)

Approximate 53

3 min to the nearest minute. 4

ii)

Approximate 49

2 H to the nearest hour. 3

iii)

Approximate 56

2 sec to the nearest second. 4

iv)

Approximate 55 min, 43 sec to the nearest minute.

v)

Approximate 5 H, 35 min to the nearest hour.

vi)

Approximate 49 H, 31 min to the nearest hour. Complete:

vii)

35

1 H≅ 3

viii)

59

3 min ≅ 4

min =

ix)

32

2 min ≅ 3

min.

x)

45

2 sec ≅ 4

xi)

4 H, 25 min ≅

xii)

34 min, 51 sec ≅

H. H.

sec. H. min.

Multiplication of Numeral Decimals Decimal Point We can carry out the multiplication of numeral decimals as the following: for example: Don t forget: Walk with point to the left in division.

a)

1.2 × 2 =

12 2 12 2 12 24 ×2= × = = 24 ÷ 10 = 2.4 × = 10 1 10 1 10 10

Or: ×

b)

1.2

the point after 1-digit to the right.

2

the point after 0-digit to the right.

2.4

the point after 1-digit to the right.

1.32 × 3 =

+

132 396 132 3 132 3 ×3= = 3.96 × = × = 100 100 100 1 100 1

Or: ×

c)

1.32

the point after 2- digits to the right.

3

the point after 0- digit to the right.

3.96

the point after 2- digits to the right.

2.32 × 0.3 =

+

232 3 232 3 696 = = = 696 ÷ 1000 × × 100 10 100 10 1000

= 0.696

Or: 2.32

the point after 2- digits to the right.

0.3

the point after 1- digit to the right.

0.696

the point after 3- digits to the right.

×

+

Rule For the previous examples, we find that: At multiplying two numeral decimals looks like the normal multiplication but in case of numeral decimals, we put the point in the result after the sum of the decimal places in the two numeral decimals to the right. Firstly, write the missing.

Example 12: Compete without using calculator: a)

If 328 × 4 = 1312 Then 3.28 × 4 = Solution 3.28 × 4 = 13.12 In 3.28 the point after 2-digits to the right and in 4 the point after 0-digit to the right, so that in the result, the point after 2 + 0 = 2-digits to the right.

b)

If 428 × 5 = 2115

Then 4.25 × 0.5 =

Solution 4.23 × 0.5 = 2.115 In 4.23 the point after 2-digits to the right and in 0.5 the point after 1-digit to the right, so that in the result, the point after 2 + 1 = 3-digits to the right.

c)

If 536 × 16 = 8576

Then 5.36 × 0.16 =

Solution Please pupil. Can you write the result direct as the following? 2

+ 2 = 4 5.36 ×0.16 =0.8572 Exercise 12: Complete without using calculator: i)

If 363 × 2 = 726

ii) If 456 × 16 = 7296 iii) If 36

× 23 = 828

Then 36.3 × 2

=

Then 4.56 × 1.6

=

Then 0.6.3 × 2.3 =

iv) If 515 × 5 = 2575

Then 0.515 × 0.5

v) If 623 × 4 = 2492

Then 62.3 × 0.04 =

vi) If 568 × 3 = 1704

Then 5.68 × 0.003 =

vii) If 1238 × 6

Then 1.238 × 0.6 =

= 7428

=

viii)If 5005 × 81 = 405405

Then 0.05005 × 8 =

ix) If 312 × 423 = 131976

Then 3.12 × 4.23 =

x) If 438 × 568 = 248784

Then 4.38 × 0.0568 =

Multiplying by 1-digit Well, dear pupil comes to learn how to get the product of two numeral decimals, one of them of 1-digit number. Example 13: i)

Find the result: 3.56 × 0.3

It is allowed using calculator.

Solution By calculator: Thus, 3.56 × 0.3 = 1.068 Exercise 13: i)

Find the result for each of the following: 43.5 ii) 53.8 × 6 × 0.7

iii)

3.56 × 9

iv)

2.53 × 0.8

v)

0.438 × 7

vi)

0.489 × 0.3

vii)

1.386 × 0.5

viii)

0.0486 × 0.04

ix)

2.409 × 0.002 •

x)

0.4863 × 0.05

Multiplication with Approximation:

Example 14: Find the result, then approximate the result to the nearest unit: i)

3.56 × 0.4 Solution By calculator: 0

Thus, 3.56 × 0.4 = 1. 424 1

≅ 1. 000 = 1 Exercise 14: Find the result for each, then approximate it as shown opposite: i)

53.4 × 0.2

to the nearest tenth

ii)

5.89 × 7

to the nearest unit.

iii)

0.563 × 9

to the nearest

1 . 10

iv)

0.486 × 0.6

to the nearest

1 . 100

v)

2.46 × 0.08

to the nearest

1 . 1000

vi)

0.806 × 0.9

to the nearest one decimal places.

vii)

0.005 × 0.0031

to the nearest 3-d.p.

viii)

(0.38 × 5) + (2.36 × 0.9) to the nearest 2-d.p.

ix)

(5.386 × 0.4) + 23.48

to the nearest

x)

123.638 − (4.38 × 9)

to the nearest whole number.

1 . 100

Multiplying by 2 digits Well, dear pupil comes to learn how to get the product of two numerals, one of them of 2-digit number. Example 15: i)

Find the result: 3.28 × 2.3 Solution By calculator: Thus, 3.28 × 2.3 = 7.544

Exercise 15: Find the result for each of the following: i) ×

13.2 12

ii)

3.28 × 13

iii)

1.53 2.1

iv)

5.38 × 0.31

0.328 × 3.1

vi)

0.196 × 0.28

× v) vii)

viii)

ix)

0.056 × 0.12

x)

xi)

1.381 × 0.016

xii)

•

0.563 × 0.051 0.3815 × 0.0011

Multiplication With Approximation:

Example 16: Find the result, then approximate it to the nearest

i)

3.15 × 0.15

1 : 10

Solution By calculator: 1

Thus, 3.15 × 0.15 = 0.4 7 25 5

≅ 0.5

Exercise 16: Find the result for each of the following , then approximate it as shown: i)

13.6 × 11

to the nearest unit.

ii)

5.38 × 19

to the nearest 1-decimal place.

iii)

13.28 × 2.5

to the nearest

iv)

15.32 × 0.15

to the nearest 3-decimal places

v)

1.386 × 2.3

to the nearest

vi)

2.386 × 35

to the nearest tenth.

vii)

0.0538 × 5.1

to the nearest hundredth.

viii)

( 3.48 × 1.1 ) + 31.56 to the nearest unit.

ix)

538.3 − (1.386 × 0.5 to the nearest hundredth.

1 . 100 1 . 1000

Multiplying bY 3-digits Well, dear pupil comes to learn how to get the product of two numerals, one of them of 3-digit number. Example 17: Find the result for each: i) 3.38 × 43.5 Solution By calculator: Thus, 3.38 × 43.5 = 147.03

Exercise 17: Find the result for each of the following: i)

1.38 × 236

ii)

13.6 × 238

iii)

1.38 × 23.2

iv)

5.63 × 2.82

v)

11.36 × 1.06

vi)

vii)

0.286 × 0.203

viii)

0.106 × 0.0153

ix)

1.362 × 0.101

x)

0.2386 × 1.32

•

×

0.386 1.36

Multiplication with Approximation:

Example 18: Find the result, then approximate it to the nearest i)

35.1 × 2.33 Solution By calculator: 0

Thus, 35.1 × 2.33 = 81.7 8 3 8

≅ 81.78

1 : 100

Exercise 18: Find the result of each of the following the, approximate as shown: i)

2.38 × 13.2

to the nearest tenth.

ii)

44.9 × 1.36

to the nearest hundredth.

iii)

5.39 × 2.06

to the nearest 3-decimal places.

iv)

0.386 × 52.3

to the nearest

1 . 10

v)

0.456 × 1.58

to the nearest

1 . 1000

vi)

1.365 × 0.381

to the nearest 4- decimal places.

vii) 386.5 × 0.0531

to the nearest unit.

viii) ( 1.38 × 0.315 ) + 46.38

to the nearest thousandth.

ix)

363.41 − (4.39 × 0.281 )

x)

(4.38 × 1.39) + (0.361 × 1.98)

to the nearest

1 . 100

To the nearest tenth.

Applications on Multiplication Well, dear pupil comes to learn how to solve the word problems on multiplication of numerals. Example 19: i)

Hany covers 41.3 meters per minute on his bicycle. What s the distance can he cover in 5.3 minutes? Solution By calculator: To find the distance covered in 5.3 minute, we find the product: 41.3 × 5.3.

Thus, 41.3 × 5.3 = 218.89 He covers in 5.3 min 218.89 m. ii)

A box contain 318 tins of oil for L.E 3.21 each. What s the price of the box of oil approximate L.E? Solution By calculator, you ll get: 1

To find the price of the box of oil, we find the product: 318 × 3.21

318 × 3.21 = 1 0 1 0. 7 8 pounds 1

≅1021

pounds.

The price of the box of oil approximated to L. E = 1021 pounds. Exercise 19: i)

Samy covers 43.8 km per hour by his car. What s the distance can he cover in 3.28 hours?

ii)

Nagy bought 35.3 kg of banana for 2.35 pounds each. What s the price of banana approximated to L. E?

iii)

Ahmed went to the market and he has got 15.3 pounds. He bought 11.5 kg of potatoes for 0.35 pounds each. What s the remainder with Ahmed?

iv)

Waleed has 315.3 dollars for 1.35 pounds each. What s he has in pounds?

v)

If you walk 143.3 meters per hours. Then what are you can walk in 5.3 hours?

vi)

The dimensions of a rectangle are 3.15 m and 0.38 m. What s the surface area of this rectangle?

vii)

If the side length of a piece of land looks like a square whose side length 3.38 km. Then what s the surface area of land approximated to the nearest kilometer?

viii)

Mohammed bought a T.V with L. E 3181.351. What s the price of 12 sets approximated to L. E?

ix)

Mona s length is 2.36 m. What s the total length of 11 girls has the same length of Mona?

x)

13.51 m 5.18 m 1.36 m

0.35 m

What s the surface area of colored part?

Division of Numeral Decimals Parts of Division What s your name? It s Divided by or over.

37

÷

What s your name? It s the Dividend.

What s your name? It s the Quotient.

5

=

7

What s your name? It s the Divisor.

What s your name? It s the remainder.

+ r2 What s your name? It s equals or is equal to.

Can be read as: 37 is divided by 5 equals 7 and remainder 2 Example 20: Complete: a) If 2876 ÷ 23 = 125 + r1. Then the dividend is , the divisor is , the quotient is and the remainder is Solution

Don t forget the name of each one and its place.

If 2876 ÷ 23 = 125 + r1. Then the dividend is 2876, the divisor is 23, the quotient is 125 and the remainder is 1

Exercise 20: Complete: i)

If 351 ÷ 5 = 70 + r1. Then the dividend is , the divisor is , the quotient is and the remainder is

ii)

If 2829 ÷ 23 = 123. Then the divisor is , the quotient is , the dividend is and the remainder is

iii)

If 308.75 ÷ 2.5 = 123.5. Then the quotient is , the dividend is , the divisor is and the remainder is

iv)

If 55.35 ÷ = 12.3. Then the quotient is , the dividend is , the divisor is 45 and the remainder is

v)

If 2.496 ÷ 0.8 = . Then the quotient is 3.12, the dividend is , the divisor is and the remainder is

vi)

If ÷ 0.23 = 0.005. Then the quotient is , the dividend is 0.00115, the divisor is and the remainder is

vii) 2533 is divided by 3, equals 844 and remainder 1. Thus can you write the problem? viii) 5930 is divided by 13, equals 456 and remainder 2. Thus can you write the problem? ix)

If the divisor is 25, the quotient is 125 and the dividend is 3125. Then can you write the problem?

x)

If the remainder is 3, the dividend is 1163, the quotient is 232 and the divisor is 5. Then can you write the problem?

Number of Digits of The Quotient Well, dear pupil comes to learn how to determine the number of digits of quotient without carrying out the division? Example 21: Determine the number of digits of quotient of: i) 256 ÷ 3? Solution 356 > 3 356 > 30 add zero 356 > 300 add 2nd zero 356 < 3 000 add 3rd zero.

Note that you ll add zeros until the divisor becomes greater than the dividend. Then the number of digits of the quotient is the number of added zeros.

Thus, the number of digits of the quotient is 3- digits. ii)

5238 ÷ 36? Solution 5238 5238 5238 5238

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