Manhatton GMAT Strategy Guide: Equations, Inequalities, and VIC's 0979017521, 9780979017520, 9780979017551, 9780979017575

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Equations, Inequalities,

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EQUATIONS, INEQUALITIES, & VIC'S GMAT

Preparation Guide

This essential guide covers algebra

various forms

in all its

(and disguises) on the GMAT. Master fundamental

techniques and nuanced strategies to help you solve for

unknown

variables of every type.

The Manhattan GMAT Prep Advantage: Attain True Content Mastery:

Learn

Sophisticated Strategies For Top Scores

GMAT Algebra:

Special Features:

Appendix includes categorized lists of real algebraic problems that have appeared on past

•

Key Algebraic Concepts

•

Basic Equations

•

In-depth Strategies

•

Exponential Equations

•

Essential Rules

•

Formulas

GMAT exams

(published

•

Detailed Examples

•

Functions

separately by

GMAC*).

•

Skill-building

•

Inequalities

•

Comprehensive Explanations

•

VIC Problems

Problem Sets

•

•

Bonus chapter on attacking data sufficiency algebra problems.

8 Strategy Books Available from Manhattan GMAT Prep: Math Guides: Number Properties Fractions, Decimals. & Percents

Equations, Inequalities,

Word Translations

Reading Comprehension

•GMAT and GMAC

Geometry

Verbal Guides: Critical Reasoning

are registered trademarks of the Graduate

Management Admission Council which

& VICs Sentence Correction

neither sponsors nor endorses this product.

EQUATIONS, INEQUALITIES,

& VIC's Math Preparation Guide This essential guide covers algebra in (and disguises) on the

GMAT.

all its

various forms

Master fundamental tech-

niques and nuanced strategies to help you solve for

unknown

variables of every type.

Equations, Inequalities, and VIC's 10-digit International Standard

13-digit International Standard

Copyright

©

2007

MG

GMAT

Preparation Guide, 2007 Edition

Book Number: 0-9790175-2-1 Book Number: 978-0-9790175-2-0

Prep, Inc.

ALL RIGHTS RESERVED. No part of this work may be reproduced or used in any form or by any means— graphic, electronic, or mechanical, including photocopying, recording, taping, lisher,

MG

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8

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Fractions, Decimals,

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Equations, Inequalities,

Word

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isbn: 978-0-9790 175-4-4

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ManfiattanGM/KY'PreD the

1.

BASIC EQUATIONS In Action

2.

EXPONENTIAL EQUATIONS

27 35

QUADRATIC EQUATIONS

39

Problems

Solutions

47 49

FORMULAS

53

Solutions

FUNCTIONS

67

Problems

73

Solutions

75

INEQUALITIES

77

In Action

7.

Problems

61 63

In Action

6.

33

Problems

Solutions

In Action

5.

21

23

In Action

4.

Problems

Solutions

85 87

VIC'S

91

In Action

97 99

Problems

Solutions

8.

STRATEGIES FOR DATA SUFFICIENCY Sample Data

9.

11

Solutions

In Action

3.

Problems

Sufficiency Rephrasing

OFFICIAL GUIDE Problem Solving Data Sufficiency

List List

new standard

PROBLEM SETS

105 111

119 122 123

TABLE OF CONTENTS

Chapter

1

EQUATIONS, INEQUALITIES, & VIC's

BASIC

EQUATIONS

In This Chapter

.

.

.

-Variable Equations

•

Solving

•

Simultaneous Equations: Solving by Substitution

•

Simultaneous Equations: Solving by Combination

•

Simultaneous Equations: 3 Equations

•

Mismatch Problems

•

Combo

•

Absolute Value Equations

I

Problems: Manipulations

BASIC EQUATIONS STRATEGY

Chapter

1

BASIC EQUATIONS Algebra

Basic

is.

therefore, an essential

GMAT equations

different types of

1)

An

GMAT. Your

one of the major math topics tested on the

is

equations

are those that

BASIC

equation with

component

ability to solve

to success.

DO NOT involve exponents. There are GMAT expects you to solve:

several

equations that the

variable

1

2) Simultaneous equations with 2 or 3 variables 3)

Mismatch Equations

4)

Combos

To solve tions,

5) Equations with absolute value

Several of the preceding basic equation types probably look familiar to you. Others—

Mismatch Equations and Combos— are unique

counter to some of the rules you

that

whatever you do to one side,

particularly

basic equa-

remember

may have

attuned to the particular subtleties of

GMAT

you must also do to the other side.

favorites that run

learned in high-school algebra. Being

GMAT equations can be the difference between

an

average score and an excellent one.

Solving 1-Variable Equations Equations with one variable should be familiar to you from previous encounters with algebra.

In order to solve

of the equation. In doing

-variable equations, simply isolate the variable

1

so.

make

sure

you perform

on one side

identical operations to both sides

of the equation. Here are three examples:

3.x

+

5

= 26

Subtract 5 from both sides.

Divide both sides by

3.

x=l

w = Iw 0= 16m1

1

— 1

= I6w =

w

+

3

1 1

Subtract

w

Add

both sides.

1

to

from both

Divide both sides by

sides.

16.

16

Subtract 3 from both sides.

^=2

Multiply both sides by

9.

9

p=

18

ManfiattanGMKT'Prep the

new standard

l

I

.

Chapter

1

BASIC EQUATIONS STRATEGY Simultaneous Equations; Solving by Substitution the (AIM asks \ou to solve a system of equations with more than one You might be given two equations with two variables, or three equations with variables. In either case, there are two primary ways of solving simultaneous

Sometimes \

ariable.

three

equations: by substitution, or by combination.

Solve the following for

x and y.

x+y = 9 2x = 5y + 4

l

k

es er

substitution

when-

1

Solve the

first

equation for

x.

one variable can be

x +y = 9 x = 9-y

easiK expressed in

terms of the other

2.

Substitute this solution into the second equation wherever x appears.

2x = 5y + 4 2(9

3.

-y) = Sy + 4

Solve the second equation for >>.

2(9 18

-y) = 5y + 4 5y + 4

-2y = 14

= 7y

2=y 4.

Substitute your solution for>' into the

x+y = 9 x+2 = 9 x= 7

!ManfJattanGMfiiY'Preo 14

the

new standard

first

equation in order to solve for

x.

>

BASIC EQUATIONS STRATEGY

Chapter

1

Simultaneous Equations: Solving by Combination Alternatively,

you can solve simultaneous equations by combination. to eliminate one of the variables.

In this

method,

add or subtract the two equations Solve the following for

x and y.

x+y = 9 2x = 5y + 4 1.

Line up the terms of the equations.

Use combination whenever

x+y = 9

it

is

easy to manip-

ulate the equations so

2x - 5 v = 4

that the coefficients for

one variable are the 2. If

you plan

to

add the equations, multiply one or both of the equations so

coefficient of a variable in one equation in the other equation.

If

you plan

tions so that the coefficient

is

the

OPPOSITE

in

one equation

is

SAME

or

OPPOSITE.

of that variable's coefficient

to subtract them, multiply

of a variable

that the

one or both of the equa-

the

SAME

as that variable's

coefficient in the other equation.

-2(x

+v=

9)

2x - 5 v = 4 3.

Add

—

-2x-2y=

-11

Note that the x

2x - 5y = 4

now

coefficients are

opposites.

the equations to eliminate one of the variables.

-2x-2v=-18 + 2y-5v= 4 -ly = -14 4.

Solve the resulting equation for the

unknown

variable.

-7v = -14

5.

Substitute into one of the original equations to solve for the second variable.

x+v= 9 x +2 = 9

x= l

ManfiattanGM/KT'Prep the

new standard

15

.

Chapter

1

BASIC EQUATIONS STRATEGY Simultaneous Equations: 3 Equations The procedure for solving a system of three equations with 3 variables is exactly same You can use substitution or combination. This example uses both: Solve the following for h\

x+

Solve three simultane-

1.

The

M'

=

+

m-

13

-

2h'

first

and

i

v

= 3a -

2v

.v,

the

2

= x +y

equation

is

already solved for>\

ous equations step-bystep

Keep

y=x + w

careful track

of your work

to

careless errors.

avoid

2.

Substitute for v in the second and third equations.

Substitute for v in the second equation:

Substitute for>' in the third equation:

+ w) + w = 3x - 2 2x + 2w + w = 3x-2

2{x

13 13

-x + 3w=-2 3.

Multiply the

first

- 2w = x + (x + w) — 2w = 2x + w

3w+2jc=13

of the resulting two-variable equations by (—

1)

and combine them

with addition.

x - 3w + 2x + 3w

2 13

3jc= 15

4.

Use your

Therefore, x

5

solution for x to determine solutions for the other

3w+ 2jc= 3v^ 10= 3w = w—

y=x+ w y = 5+ y=6

13

13

1

3 \

ManfiattanGM fiJ'Preo n

=

the

new standard

two

variables.

BASIC EQUATIONS STRATEGY

Chapter

1

Mismatch Problems Consider the following

you

which you might have learned in a basic algebra course: If you need 2 equations. If you are trying variables, you need 3 equations, etc. The GMAT loves to trick rule,

are trying to solve for 2 different variables,

to solve for 3 different

you by taking advantage of your

MISMATCH

faith in this easily

problems, which are particularly

of the

test,

to the

number of given

are those in

common on

rule.

the Data Sufficiency portion

which the number of unknown variables does

Do

equations.

school algebra, because

MISMATCH problems

misapplied

it is

NOT correspond

not try to apply that old rule you learned in high-

not applicable for

many

GMAT equation problems.

must be solved on a case-by-case

All

basis. Don't assume

Solve for

a:

given the following two equations:

that the

number of equations must be equal to the

3* (1)

2y It is

+

=3

number of variables.

(2)y + z =

5

2z

tempting to say that these two equations are not sufficient to solve for

are 3 variables and only 2 equations.

However, note

you

only asks you to solve for

to solve for all three variables.

First,

solve the

first

for x, in terms of y

x,

does

x, since there

NOT ask

which IS

possible:

Then, notice that the second equation gives

equation

and

It

that the question

us a value for y

z:

+ z, which we can

substitute

into the first equation in order to solve for x.

3x

x = 2(y + x = 2(5)

2y + 2z 3x = 3(2v + 2z)

x=

x = 2y + 2z x = 2(y+z)

Now

consider an example in which 2 equations with 2

sufficient to solve a

Solve for

x given

(1)

It is

z)

10

unknowns

are actually

NOT

problem:

the following two equations:

x + 12>=15

(2)

j + 4j = 5

tempting to say that these 2 equations are surely sufficient to solve for x, since there and only 2 variables. However, notice that both equations actually

are 2 equations

express the same relationship between x and

v.

That

is,

they are the same equation!

you divide the first equation by 3, you will obtain the second equation. Alternatively, if you multiply the second equation by 3, you will obtain the first equation. If

Thus,

we have

only one equation with 2 variables.

We

do

NOT have

sufficient informa-

tion to solve for x.

ManfiattanGM AT'Prep the

new standard

it

Chapter

BASIC EQUATIONS STRATEGY

1

Combo The

Problems: Manipulations GMAT often asks you to solve for a combination of variables,

problems.

or example, a question might ask, what

1

In these cases, since

generally try to

NOT try

you are not asked

Only

called

the value of x

COMBO

+ y?

one specific variable, you should

to solve for

to solve for the individual variables right

manipulate the given equation(s) so that the

the equation.

is

COMBO

away. Instead, you should is

isolated

try to solve for the individual variables .after

on one

side of

you've exhausted

all

other avenues.

There are 3 easy manipulations that are the key To soh

combo,

G for a variable

You can use

the

acronym

to solving

most

COMBO

problems.

MUD to remember them.

isolate the

combo on one the equation.

side

of

M:

Multiply or divide the whole equation by a certain number.

U:

Unsquare or square both

D:

Distribute and factor.

sides.

Here are two examples, each of which uses one or more of the manipulations above:

If x

=

7

~

J what ,

x =

is

2x + yl

1-y—

Here, getting rid of the denominator by multiplying both sides of the equation

-y Zx+y = l

2x = l

If

V2f +

r

=

\ It It 6/

5,

what

is

is

the key to isolating the

combo

3r + 6f? Here, getting rid of the square root by squaring both sides

-

of the equation

+ r = 25 + 3r - 75

the

is

the first step. Then, multiplying the

whole equation by

ManltattanGM AT"Prep 18

by 2

on one side of the equation.

new standard

3

forms the combo

in question.

BASIC EQUATIONS STRATEGY

Chapter

1

Absolute Value Equations Equations that involve absolute value generally have tant to consider this rule

The following

value.

when

thinking about

three-step

TWO SOLUTIONS.

GMAT questions that

method should be used

It is

impor-

involve absolute

for solving equations that

include a variable inside an absolute value sign.

Solve for j, given that \y\

1.

+

7

=

15.

Isolate the expression within the absolute value brackets.

LkI

+ 7=15 lvl

=

8

Absolute value equations

2.

Remove

case, case,

the absolute value brackets and solve the equation for 2 cases.

assume the variable assume the variable

is

positive.

inside the absolute value brackets

is

negative.

CASE

y=S 3.

Check

In case

1,

to see if

we

tion (since 8

each solution

set the is

Thus, both solutions are valid. condition. For example, if

y

is

then

positive,

we it

for

negative cases.

v < v=-8

2:

>

0.

The

y=

solution,

8 meets this condi-

greater than 0).

we set the following (since —8 is less than 0).

y = — 2,

must be solved

both the positive and

valid.

following condition: v

In case 2, tion

is

first

For the second

inside the absolute value brackets

CASE \:y>0

For the

we

could discard

condition:

y




set a condition that v

valid if

that

it

this condi-

violated the pre-set

and came

this solution (since, in effect,

must be -2, a statement

y = -8 meets

to a solution that

we would

be saying that

if

does not make logical sense).

Equations that involve a variable inside the absolute value brackets generally have two solutions: one for when the expression inside the brackets is positive, and one for when the expression inside the brackets

when

the solution

is

is

negative.

The exception

to this general rule is

zero.

ManfiattanGMATPrep the

new standard

14

Chapter

1

BASIC EQUATIONS STRATEGY Absolute Value Equations with Variable Expressions \ppl\ the

same

3-step

method when

there

is

a variable expression within the absolute

\alue brackets

Solve for h\ given that

12 +

|

m-

-4

|

=30.

Again, isolate the expression within the absolute value brackets and consider both cases.

1.

Don't forget

to deter-

:

\w — 4

CASE

1

mine the \ahdit\ of

u-4=

each of your

w-

to absolute

\

solution.-*

=18

1

:

w-

CASE 2: w - 4 u--4= -18

4 >

18




solution,

w=

(since 22

22,

-4=

is

18,

valid because

which

is

it

meets the preset condition

that

greater than 0.)

any relevant preset conditions.

The second solution, w = — 14, is valid because it meets w - 4 < (since — 14 - 4 = — 18, which is less than 0.)

fyianfiattanGM AT'Prep 20

the

new standard

the preset condition that

..

IN

Problem

ACTION

1.

Chapter

1

Set

For problems #1-6, solve for .

BASIC EQUATIONS PROBLEM SET

all

unknowns.

3x - 6 = x- 6, 5

2.

3.

4.

x+2

5

4+x

9

22

-|v+

14

=20

|

|6+x| =2jc+1

5.

y = 2x + 9and7x + 3y = -51

6.

3x + v + 7z = 15 and x + 2v + (-4z) =

For problems #7-10, determine whether tions.

7.

(Do

is

1

and 5x +

2_y

+

8z

possible to solve for

jc

=

19

using the given equa-

not solve.)

^- = land -^=14 4

6a 8.

it

-

3x + 2a = 8 and 9x + 6a = 24

26+x = 8and

+ 86 + 2* = 4

9.

3a +

10.

4a + lb + 9x = 17 and 3a +

12a

2>b

+ 3x =

3

and 9a +

—+4=9 4

9

For problems #11-15, solve for the specified expression.

x+ v '

1 1

Given

that

=

1

7,

what

is

x + y?

3

12.

Given

=

that

21,

what is—? b

b 13.

Given

that lO.v

14.

Given

that 5jc

5

Given

that

1

—

+ lOy = x

+y +

81,

what

+ 9y + 4 = 2x + 3y + 31,

— = 4 and x + y = 3

,

is

x +y?

what

what

is

is

x + 2y?

z?

ManfiattanGM AT'Prey the

new standard

21

I

1

IN

ACTION ANSWER KEY 1.

Chapter

BASIC EQUATIONS SOLUTIONS

1

jc=12: 3x - 6

=

,

x- 6

5

- 6 = 5(x - 6) 3x - 6 = 5x - 30 24 = 2x 3jc

Solve by multiplying both sides by 5 to eliminate the denominator. Then, distribute

and

isolate the variable

on the

left side.

\2=x

*'•* x+2

5

4+x

9

= 5(4+;c) = 20 + 5* 4x = 2

+

2)

9x+

18

9(x

x=

Cross-multiply to eliminate the denominators.

Then, distribute and solve.

— 1

2

3.

y ={-16, -12}: First, isolate the

22

-\y +

\y+

Case

14

|

brackets.

=2

v+

1:

20

14

expression

14

>

Case

j=-12

is

x=

5:

Solve

this

y + 14

Case

less than

1:

When

positive and one in

which

it

in is

which the negative.


>+ 14 = 2

4.

2:

expression within the absolute value

Then, solve for two cases, one

is

positive and one in

is

not valid, since

it

which

it

violates the condition that

—6.

6

+x

is

positive (meaning

= 2x + + x = 2x +

6 +x 6

|

5=x

1

1

x > -6)

Case

2:

When

6

+x

is

negative (meaning x

6+x= 6+x=

~(2x+ -2x -

< -6)

1) 1

3x=-7 7

not valid!

ManhattanGMPiT'Preo the

new standard

2.*

Chapter v

IN

BASIC EQUATIONS SOLUTIONS

1

-6;

ACTION ANSWER KEY

\= -3: •

7.v

b

7.v

3(2*

-

9)

-

I3v

v

6.

= 22a

-

^ =

x= -l:y = 4:z 3.v

«

Solve

this

system by substitution. Substitute the

value given for v

27= -51 13v = -78 x = -6

in the first

ond equation. Then,

equation into the sec-

distribute,

combine

and solve. Once you get a value for

back

like terms,

x, substitute

into the first equation to obtain the value

it

of y.

2(-6) + 9 = -3

=

+ v + Iz =

(2)[3x

= -51

27 = -51

Si

2:

1

5

and x + 2v + (-4z) = -1 and 5x + 2y + 8z = 19 6x + 2y+ 14z = 30 + -x - 2 v + 4z = 1 + 18z = 31 5x

+ v + 7z= 15] + 2y-4z) = -l]

(-!)[(*

First,

combine equations

by addition. Multiply them each by a

and gle

(1)

(2)

number

as

shown

sin-

to

eliminate one variable.

(-!)[*

+ 2y + (-4z)] = -l

-x - 2y + 4z = 1 5s + 2v + 8z= 19 + 12z = 20 4x

Then, combine equations (2) and (3). In this case, you can eliminate a variable by altering only one of the equations.

(4)[5x+ 18z = 31] (-5)[4jc+ 12z

->

= 20]

-»

20x + 72z = 124 + -20x-60z = -100 12z = 24 z

=

-3+y+\4= 11 +y=

4x + 24 = 20

4x= -4

x=-l

2-variable equations to isolate the variable z.

2

3x+y + 7z = 15 3(-l) + v + 7(2)=15

4x+ 12z = 20 4x+ 12(2) = 20

Then, combine the resulting

Finally, plug the

known

value for z into a 2-variable

15

equation to get the value of

15

x.

y=4

Then, plug the

known

val-

ues of z and x into any of the original equations to get the

value of y.

ManfiattanGM/KT'Prev u

the

new standard

IN

ACTION ANSWER KEY

basic equations solutions

Chapter

1

YES:

This problem contains 3 variables and 2 equations. However, this is not enough to conclude you cannot solve for*. You must check to see if you can solve by isolating a combination of variables, as shown below: 7.

that

vx = _ land ,

6a

— Ta

.

.

= 14

4

2 x = (6Ta) and Ta = 56

x = (6

We

8.

X

56)

2

can find a value for

x.

NO:

that

This problem contains 2 variables and 2 equations. However, this is not enough to conclude you can solve for*. If one equation is merely a multiple of the other one, then you will not have

unique solution for

x.

In this case, the second equation

merely

is

3 times the first.

a

Therefore, they

cannot be combined to find the value of x.

9.

YES:

This problem contains 3 variables and 2 equations. However, this is not enough to conclude you cannot solve forx. You must check to see if you can solve by eliminating all the variables but as shown below:

that x,

(4)[3a

+ 2b + x =

S]

->

\2a

+ Sb + 4x = 32

- 12a + Sb + 2x = 4 2* = 28

x=

10.

that are,

14

YES:

This problem contains 3 variables and 3 equations. However, this is not enough to conclude you can solve for x. You must check to make sure that they are 3 unique equations. In fact, they as you can see by multiplying each equation to establish a common coefficient in the x term:

(3)[3a

+ 3b + 3x =

(81)[9a

+

x

9a + 9b + 9x = 9

3]

— + — = 9]

-»

729a + 20.25Z> + 9x = 9

9 4a + 7b + 9x= 17

you could do so by systematically combining these equations to isolate indiHowever, since you are only asked to decide whether or not you could solve for v. it

If required to solve for x,

vidual variables. is

sufficient to establish that these are 3 distinct equations.

ManhattanG MAT Prep the

new standard

25

1

Chapter i

12.

BASIC EQUATIONS SOLUTIONS

1

.v

+ r-51:

-

=

v

•

v

-

l

- 17

51

v

20:

b a + b

+

= 21

1

=21

= 20

13.

*+.>'

=

9:

10x+ lOv = *+>> + 9jc

81

+ 9y = 81

x+y = 9

14x + 2>=9: 5x + 9y + 4 = 2x + 3y 3x + 6y = 27 3(x

I5.z=

+

3

+ 2y) = 27 x + 2y = 9

12:

zx

+ zv

=4

x+y =

3

z(x+y) = 36 3z = 36

z=12

fytanfiattanGM/KT'Prep 2C

the

new standard

IN

ACTION ANSWER KEY

Chapter 2 EQUATIONS, INEQUALITIES, & VIC's

EXPONENTIAL EQUATIONS

In This Chapter

.

•

Even Exponent Equations: 2 Solutions

•

Odd

•

Same Base

•

Eliminating Roots: Square Both Sides

Exponents: or

1

Solution

Same Exponent

EXPONENTIAL EQUATIONS STRATEGY

Chapter

2

EXPONENTIAL EQUATIONS The

GMAT tests more than your knowledge of basic

equations. In fact, the

complicates equations by including exponents or roots with the

unknown

GMAT

variables.

Exponential equations take various forms. Here are some examples:

There are two keys

1)

Know

covered

the

to

RULES

in the

=

21

achieving success with exponential equations:

for exponents

Number

heart. In particular,

Vx+15

f + 3=x

3 x = -125

and

roots.

You

Properties Strategy Pack.

you should review

will recall that these rules

It is

the rules for

essential to

know

were

these rules by

Even exponents are

combining exponential expressions.

dangerous! They hide the sign of the base.

2)

Remember

that

EVEN EXPONENTS

DANGEROUS

are

because they hide the sign

of the base. In general, equations with even exponents have 2 solutions.

Even Exponent Equations: 2 Solutions Recall from the rules of exponents that even exponents are dangerous in the hands of the

GMAT test writers.

Why

are they dangerous?

Even exponents hide

the sign of the base.

As

a result,

equations that involve variables with even exponents can have both a positive and a negative solution.

Exponential equations with even exponents must be carefully analyzed, because they

some examples:

often have 2 solutions. Here are

= 9

jt

Here, x has two solutions: 3 and -3.

2 a - 5 = 12

By a

Note

2

that not all equations with

2 .x

+

3

=

3

adding 5

=

17.

to

By

BE ON THE LOOKOUT

where k

is

any positive

^+9= 2

x =

integer.

we

can rewrite

this

equation as

V 17 and -V 17.

even exponents have 2 solutions. For example: subtracting 3 from both sides,

equation as x~

Also,

both sides,

Thus, a has two solutions:

for

=

0, so

we

can rewrite

x has only one solution:

even exponential equations

These equations have

in the

this

0.

form x2

+k=

,

NO REAL solutions:

Squaring can never produce a negative number!

-9

ManfiattanGMPJPrep the

new standard

29

Chapter

2

EXPONENTIAL EQUATIONS STRATEGY

(\U

Exponentsi

1

Solution

involve only odd exponents or cube roots have only

quations thai

I

Here,

125

v

has only

(— 5)(— 5)(— 5)

24 3

solution: -5.

(3)(3)(3)(3)(3)

solution:

You can

will not

see that

work with

positive 5.

solution: 3. You can see that = 243. This wULnot work with negative

Here, v has only

y

1

= -125. This

1

1

3.

[fan equation includes some variables with odd exponents and some variables with

even exponents, Rewrite exponential

treat

it

as dangerous, as

make

nents in an equation

it

it

is

likely to

have 2 solutions.

Any even expo-

dangerous.

equations mi lhe> ha\e either the the >aine

vime have or exponent

Same Base or Same Exponent In

problems

that involve exponential expressions

imperative to

REWRITE

appears on both sides of the the bases or the exponents

.

BOTH

sides of the equation,

it

is

same base or the same exponent exponential equation. Once you do this, you can eliminate

and rewrite the remainder as an equation.

Solve the following equation for w: (4

1

on

the bases so that either the

hO = 32|H>-1 ) K

Rewrite the bases so that the same base appears on both sides of the equation. Right now, the

left

side has a base of 4 and the right side has a base of 32.

Notice that both 4 and 32 can be expressed as powers of 2.

We 2.

can rewrite 4 as 2

3

' = 32" 5 )") = (2

2 M 3

((2

)

and we can rewrite 32 as 2

2

.

1

3

=

5

(2

r

1

Eliminate the identical bases, rewrite the exponents as an equation, and solve.

=

5vv

w=

-5

6vv

—

5

^ManfJattanGM/KT'Preo N

5

Simplify the equation using the rules of exponents.

((2 )")

4.

,

Plug the rewritten bases into the original equation.

(4")

3.

2

the

new standard

EXPONENTIAL EQUATIONS STRATEGY

Chapter

2

Eliminating Roots: Square Both Sides The most

effective

way

problems

to solve

bols (variable square roots)

is

to square

Solve the following equation for

Vs- -12 s

-

-12

that involve variables underneath radical

sym-

both sides of the equation.

5:

Vs -

12

=

7

==

7

Squaring both sides of the equation eliminates the radical

==

49

symbol and allows us

to solve for 5

more

easily.

5 = 61

When

Given that

V36 -

8

= Vl2-6, what

is

6?

eliminating radi-

cals,

you may some-

times need to square

V36-8 = Vl236-8= 12-6 46-8=12

both sides of the equa-

Squaring both sides of the equation eliminates both radical

symbols and allows us

to solve for

6 more

tion

more than once.

easily.

46 = 20 6

=

5

After you've solved for the variable, check that the solution works in the original equation.

Squaring both sides can actually introduce an extraneous solution.

Remember (or zero)

number

also that the written square root

symbol only works over positive numbers zero). The square root of a negative

and only yields positive numbers (or

is

not defined in the real

Vx~ =

means x = 25

5

At the same

time, there are

For equations that involve cube Solve t he

fol lowing

vy-% -27 = v -8 -3 =

-19 =

number system.

two numbers whose square

roots, solve

by cubing both

is

25: 5 and -5.

sides of the equation:

equation for y: -3 = ^s/y-S

Cubing both

sides of the equation eliminates the radical.

v

Man fiattanGM AT Prep the

new standard

ii

x

IN ACTION

Problem

EXPONENTIAL EQUATIONS PROBLEM SET

Given

that

vt +

2.

Given

that

Vm

3.

Given

that

V V2x + 23

4.

If

is

=

4

what

6,

6.

Given

that

4

V2w — =

5

what

11,

what

,

is

is

w?

x?

which of the following could be

4 (B)u 3 + u + u 5

2x+y= — 3

2

and

=

= — 125, and d1 =

= 81, x + z + (P.

.r

Given

k and

that

is /?

(C)u

9

a negative

(D)w"

13

number?

+ w 13

(E) u

3

-

v21 \7

I

Simplify:

8.

+

-u 6

5.

If x

=

8

a positive integer,

(A)w 7

7.

2

Set

1.

m

Chapter

v

4,

+

3,

what

solve for x and

is

of

the smallest possible value

4

m

v.

are positive integers and that (x )(x

8 )

=

k m (x )

and

m = k+

1,

solve for k and m.

9.

Given

that (3*)

10.

Given

that

4

=

27,

what

x + y = 13 and

kl

is

vy

For problems #11-13, given that x

is

=x—

1,

solve for the positive values of x and

an integer greater than

1,

v.

determine whether each of

the following expressions can be an integer.

n

+x^

4

+x~2

3

+x°+x 5

11.

x

12.

x

13.

x~

For problems #14 and #15, use the following information: At a shoe sale, prices are

14.

If a bill,

15.

store's fairly drastic

defined as the square root often dollars more than the original dollar price.

customer buys a pair of discounted shoes and gets change for a twenty-dollar what is the highest possible original cost of the shoes (in whole dollars)?

A salesman

sells a

shoehorn for the sale price of 80 dollars, even though

item was not supposed to have been part of the storewide store

owner

tries to

original price.

buy

it

How much

sale.

When

this

the shoe

back, the crafty purchaser holds out for twice the

of a profit does the crafty buyer make?

ManfiattanGM AT'Prep the

new standard

33

1

IN

ACTION ANSWER KEY 1.

t

+

8 = -6 8

==

36

=

28

--

t

m

x=

-

+ 4 = 2m \5 = m

to eliminate the radical.

Square both sides

to eliminate the radicals.

Then, solve for

/.

1

11

Square both sides;

2x = 4

4.

(E): If u

5.

x 126

is

this eliminates the larger radical sign

on

the left side of the equation.

Then, subtract 23 from both

sides to isolate the variable.

Square both sides again

to

eliminate the radical. Finally, divide both sides by 2 to find

x= 2

the value of x.

a positive integer, w

8

>

w

3

Therefore, u

.

J

-

w

8


(unction rulex

each term of

it>

in

place

Sn = n +

N here

The

3

mh

term of this sequence

example, the

a function

in the

term

mpia

first

term

sequence

in this

in this

2

is

+

3

defined by the rule w +

is

sequence

is

=

first

The

5.

1

+

=

3

For

3.

The second

4.

ten terms

of the

sequence are as follows:

4,5,6,7,8,9,

Qn =

n~

The

+4

/7th

term of this sequence

example, the term

10, 11, 12, 13

first

term

sequence

in this

in this

2

is

is

defined by the rule n~ +

sequence

+4 =

The

8.

1+4 =

is

first

5.

4.

For

The second

ten terms of the

sequence are as follows: 5,8, 13,20,29,40,53,68,85, 104

Defining Rules for Sequences The important thing

to

remember about sequence problems

is

that

YOU MUST be

given

RULE in order to find a particular number in a sequence. It is tempting (BUT WRONG!) to try to solve sequence problems without the RULE. For example:

the

If

It

57 =

and 58 =

5

looks like this

However,

RULE

this

what

is

( 1 )

59 ?

a sequence that counts

deduction

is

NOT

5, 6, 8, 11, 15.

by

1

's,

so

is

tempting to say that S9

The sequence might be

Without the

rule, there is

(1) If

you

some

trial

and

error.

no way

are told that the difference

(2) If

is

be

However,

at least

it

sure.

one term, you can

between successive terms x,

7.

is

where k and x are

to follow:

always the

real

numbers

equal to the difference between successive terms.

you are

terms b,

is

to

However, there are some guidelines

same, the rule will take the form of kn +

and k

we

=

were not given the

5, 6, 7, 8, 9...

information about the form of the sequence and (2)

find the rule using

told that the difference

always the same, the rule

and c are

real

numbers.

ManfiattanGM AT'Prep 58

it

mathematically valid, because

for this particular sequence.

might also be

Given

is

6,

the

new standard

between the difference between successive will take the

2 form of an + bn +

c,

where

a,

FORMULAS STRATEGY The

difference

terms

>

20

The

is

the same.

rule for this

sequence

24

is

kn

The difference

> > >

between successive

between the difference between successive terms

+ x, 38

where k-A.

28

>

Chapter 4

is

the same.

The

rule for this sequence is

an + bn +

c.

13

51

Consider the two sequences above. Many

assume

In the first sequence,

sive term

is

form An +

that

We

are told that the difference

means

the constant. This

x.

we

between each succes-

sequence must be

that the equation for this

in the

can find the value of x by using any of the terms in the sequence. For

Set

We

first

term

4(1)+*= x=

in the

sequence (n

where

between successive

12, since 4(2)

Therefore, An

has a value of 16.

1)

12

can confirm that x

x=

x,

16

=

by using another term

12

For example, the second term that

+

k equals the difference

terms.

example:

The

sequences are of

the form kn

+

12

+

12

=

in the

20.

the rule for the

is

in the sequence.

sequence (n = 2) has a value of 20. This verifies

first

sequence above.

We

can use

this

formula to

find any term in the sequence.

The second sequence is a bit more complicated. Assume that we are told that the difference between the difference of each successive term is constant. Since the difference between the difference between successive terms is the same, we can set up a system of 2 equations for this sequence using the formula an + bn + c as follows: The

first

term

in the

The second term The

We

third term in the

=

1 1

mula It is

.

has a value of 18:

+ b(l) + c = 18. + b(2) + c = 27. + 6(3) + c - 38.

a(l) a{2)

sequence (n = 3) has a value of 38:

we

can solve

this

a(3)

system of equations for the three unknowns: a = 1,6 =

Therefore, the formula for this sequence

to find

any term

in the

rr

is

+ 6n +

11.

We

can use

BOTH

(1) information about the

ence between each successive term

form kn +

6,

and

this for-

sequence.

important to remember that you can derive a rule for a sequence only

provided with

the

1)

sequence (n = 2) has a value of 27:

have three different equations with three unknowns. Using substitution or another

method, c

sequence (n =

in the

in the

form of the sequence

sequence

x") and (2) some consecutive terms

is

if

you are

(e.g. "the differ-

constant" or "the sequence takes

in the

sequence.

ManhattanG MAT Prep the

new standard

:>'»

Chapter

FORMULAS STRATEGY

4

Sequence Problems: Alternate Method I

cm

simple additive sequences,

the next term,

each number

If

in

which the same Dumber

is

added

to

each term

to yield

you can use the following alternative method:

the sixth

in a

number

is

sequence 32,

what

is

is

mure than the previous number, and number?

three

the 100th

Instead of finding the rule tor this sequence, consider the following reasoning:

From

the sixth to th*

2s2. there 1 1

a

problem \cvms

require too

is

one hundredth term, there are 94 "jumps" of

3.

Since 94

X

3

=

an increase of 2.i>

to solve

it

Sequences and Patterns Some sequences ple,

If

S„ = 3", what

is

the units digit of 565 ? ,65

Clearly,

you cannot be expected

assume

that there is a pattern in the

3 3 3 3 3 3 3

3

to multiply out 3

!

=3 =9 = 27

Note the pattern of the units

= = = =

tiple

4 5

6 7

8

81

243

729 2,187

[repeating]...

of 4,

is

Also note

to

1.

in the pattern.

1

6,561

ManfJattanGMATPrep 60

digits in the

the

new standard

Therefore, you must

powers of 3:

in the pattern.

9, 7,

1, 3,

of SIV when n

a mul-

is

You can use

the multiples of 4 as

Since 65

more than 64

closest multiple of 4), the units digit of

always follows

GMAT.

that the units digit

always equal

"anchor points"

on the

powers of three.

2 3

For exam-

are easier to look at in terms of patterns, rather than rules.

consider the following:

is

S65

1

will

be

3,

(the

which

.

ACTION

IN

Problem 1

FORMULAS PROBLEM SET

Set

Given

that

Given

that

Given

that

A* B = 4A

\f y

u

-B, what

=

the value of (3 * 2) * 3?

is

what

*—,

x+z

is

A 2 + B 2 + 2AB, what

is

A+

For problem #4, use the following information: x e.g. 3

=> 7 = 3 + 4 + 5 + 6 +

4.

What

5.

Life expectancy

the value of

is

monthly

is

( 1

00

=>

1

is

( 1

25

=>

and

is

1

,

G

G = GMAT

1)

+

(x

+

2)

...

+ y.

50)?

where S = shoe score.

is

expressed by the formula tb

4

If

.

b

is

is

B=

average

GMAT score

is

is

his shoe size?

— where w

is

the width of

is

—

the grade of rubber

spring factor for an insole that

If Elvin's

size,

what

expectancy

his life

for spring factor in a shoe insole

maximum

Cost

7.

-

50)

and

bill,

the insole in centimeters and x the

9?

if

=> y - x + (x +

defined by the formula

electric bill in dollars,

The formula

B,

7.

twice his monthly electric

6.

Chapter 4

50,

,

on a scale of

3 centimeters

1

to 9.

What

is

wide?

doubled, by what factor has the cost

increased?

(A) 2

8.

(B) 6

(D) 16

(C) 8

(E)

If the scale

model of a cube sculpture

what

volume of the model,

is

the

V*

is .5

if the

cm

per every

volume of the

1

m of the real

real sculpture

is

sculpture,

64

m3 ?

For problems #9-10, use the following information: The "competitive edge" of a baseball

—W

3

team

is

defined by the formula

sents the

9.

where

W represents the number of wins, and L repre-

number of losses.

This year, the

GMAT All-Stars

losses that they

10.

r-,

had

last year.

tripled the

By what

number of wins and halved

the

number of

factor did their "competitive edge" increase?

The "competitive edge" of the LSAT Juveniles was 256 times as high in year A as it was in year B. If the team won one quarter as many games in year B as it did in year A, what was the percentage increase from year A to year B in the number of games it lost?

ManfiattanG MAT Prep the

new standard

61

Chapter

IN

FORMULAS PROBLEM SET

4 11.

If the radius

of

> circle

circle to the area

is

tripled.

\\

of the whole new

bat

is

the ratio of the area of half the original

circle.'

For problems --12-13. use the following sequence: A„ = 3 -

12.

What

is

ACTION

8/1.

'

I

What is! -

14.

It"

each number

number 15.

If

Sn =

is

4" + 5"'

sequence

in a

46. what

'

+

is

3,

is

three

what

is

the

the previous number,

the units digit of 5 100 ?

ManhattanGM/KT'Prep 62

more than

the rule for this sequence?

new standard

and the eighth

IN

ACTION ANSWER KEY 1.

3*2:

37: First, simplify

2. 2:

Plug the numbers

corresponding 5 variable F

4(3)

-2=12-2=10.

formula

—=

x+z

3.A + B={3,-3}: A 2 + B 2 + 2AB = 9 2 (A + B) = 9 OR A + B = -3 A +B=3

4.

Then, solve 10 *

formula, matching up the

in the grid into the in the

=

4+5

ponents,

the

we

numbers from 125

First, set the

expression

by solving 1

3+4 1

+

to 150.

Since

we

in

= 40 -

3

=

37.

each section with the

2.

formula equal to

9.

Then, factor the both sides, taking

sum of all

the

numbers from 100 to 150, and the between these two com-

are finding the difference

sum of all

the

numbers from 100 between

(1

to 124.

You can

=> 5) - (3 =>

think of

5).

5 5

+2

There are 25 numbers from 100 to 124 (124 - 100 +

sum of these numbers,

find the

number

A 2 + B 2 + 2AB. Unsquare

a simpler problem: find the difference

+2 + 3+4 +

-

4(10) - 3

both the positive and negative roots into account.

are essentially finding just the

this logically

=

3:

9

2,800: This problem contains two components: the

sum of all

Chapter 4

formulas solutions

1).

Remember

multiply by the average term: (100

to

add one before you're done! To

+ 124) + 2=112. 25 X 112 =

2,800.

5.

Size 50:

2SB

Substitute IB for G in the formula. Note that the term IB appears in both the numerator and denominator, so

50

5=50

6. 6:

Determine the Let s s

=

—w

2

they cancel out.

maximum

spring factor by setting

x — 9.

spring factor

+x 3

s

7.

=

(D): Pick is 2,

by

(3)2

+

9

numbers the cost

to see

a factor of

what happens

When

is 16/.

,

b

is

to the cost

doubled

when b is doubled. If the original value of b new cost value is 256r. The cost has increased

to 4, the

or 16.

16

hat

we

negative,

the range of possible values for a?

above equation, it seems that we should simply divide both However, because we don't know whether y is positive or

the

of the inequalit\ b\

sides

is

y.

NOT ALLOWED to divide

are

both sides of the equation by

positive, then the solution to the inequality is.v

we

are di\ iding an inequality

solution x

When >ou

must

< 3.

However,

y

if

thus, the sign flips

is

negative,

and yields the

3.

multipK or

or \anable.

tlip the

Do not multiply or divide inequalities by unknowns (unless you know the sign of the unknown). What you should do instead is move all the terms to one side of the inequality and factor:

This example leads us to the following rule:

diMiie b> a negative

number

>

by a negative number;

y.

\ou

Men of the

inequality

xv

- 3v
-*

more eonstrained:

and second inequalities so

Simplify the inequalities, so that the inequalities

9

Ka

15_-

the inequality symbols point in the

same

direction.

Then,

line

up

and combine.




the

d=6

3d

numbers we have

-2 = 6-2=4.

each answer choice.

We plug

in 2 for a, 3 for b,

4 for

c,

search of our target number: 4.

d, in

Incorrect

(A) 72

72

72

72

cd

(4)(6)

24

(B)

=

144

144

144

cd

(4)(6)

24

Incorrect

3

Incorrect

(C)

a(\44-lcd) _ 2(144 -

(2)(4)(6))

_

(2)(96)

=

(

cda 144a

-led _

cd

(144)(2)-(2)(4)(6) = (4)(6)

^

CORRECT

48

(4)(6)(2)

j40-= t

n

Incorrect

24

Just as with other VIC's, be sure to test every

answer choice

to

make

sure that only one

works.

ManfiattanGM AT'Prep the

new standard

IS

.

ACTION

IN

Problem

Chapter

VIC PROBLEM SET

7

Set

Solve each problem with a tracking chart.

1

If x, y,

and z are consecutive

Maria will be x years old

(A)x-3

T pages

(A)*» v

(C)x +

old was she 9 years ago?

(E)x + 6

(D)x-21

3

how many

per minute,

(C)il

,B)i?2.

T

How

in 12 years.

(B)x-20

If Cecil reads

which of the following must be an integer?

integers,

hours will

it

him

take

A town's oldest inhabitant If the oldest inhabitant is

is

500 pages?

(E)^I

(D) 30007

500

6

607/

to read

x years older than the sum of the ages of the Yow triplets. old, how old will one of the triplets be in 20

now J years

years? ,

M

(A)

5.

K

is

7-50

(L)

(B)KG

7.

J years. how many

Mr. and Mrs. Wiley have a child every they have a child 2 years from now,

(A)^p-+1 If b

= ,

rtjvJ+x-20 (h)

(D)

is

Their oldest child

is

now T years

If

old.

children will they have in total 9

(Qy+y

(B)JT+l then what

_7-x + 60

y + x-50

G is an integer. Which of the following cannot be odd? (C)K-G (D)2K+G (E)3(/C-G)

an even number, and

(A)K+G 6.

3(7+20)

,„, (B)

(E)^y^

(D)77-l

a?

4

wm 8.

If 77

3 _ = x

(A)x2

9.

If

5^-2

- x + 30

x -5

P N =— — T M

,

then

MT

is

andx

j*+6

(D)

«±i

**5, then //is equivalent to

(E)36+6 which of the following.'

2 3 (D) x - Sx - 3x

3 (C)x -25

2

(E).r

+ .r+10

equivalent to which of the following?

NP (B)^p

The average of A, B, and g-C

(A)^

(Q

2 3 (B)x + 3x + 3x

-x-6

(A)NP 10.

mJ ^6

MN

NP (D)^

(C)^C is D. What

(B,^£

3

(C)

is

— PT 2

(E)

the average of A and

°-^

C

3

(D)

5?

°^; J * C

(E)

9

3to/ia»aftGMATPrep the

new standard

97

— Chapter

IN

VIC PROBLEM SET

7 11.

[fa

It

(A)

—

is

whal percent of 67

1

(
-|r

(l))

^-

wins 1/ dollars in a billiards tournament and

what

1

./

(E)c + 20 dollars at each of seven county

her average prize (in dollars)?

is

2£±v

v3.

then a

(B)200&

Iessic

tairs.

1

.

\i:oc

I

12

20bt

ACTION

(B)i£lZ

\

If-

=

(A)Uv-- llv+

19)

ra

»*v a>)^t£

which of the following

9,

is

(E)M

+

^

equivalent to LI

9 (B) x-

(C)

+

1

Ijc

-

(*--7Xx

-28

-4)

9

,

tracking Chait, the

tvenge Of A and B

It

n.

numbers shown is

\

lest

5

m

variable

number

A

3

B

4

(

8

D

5

the

each answer

choice to find the one thai yields the target Dumber 3.5.

H

\

ACTION ANSWER KEY 1

Avoid selecting consecu-

(

Incorrect ti\e integers to represent

A, B, and C.

3D-C

s 3(5)^8 = 7 =35

3D

-B

A

+

C

_ 3(5)

-3 -4 +

8

+

fi

C

_ 3(5) +

Incorrect

1

a = 2Qbc 120

true, as

2 = 60, a

-s-

one

the

3+4

+ 8 _ 30

for all the vari-

ables in the problem.

Incorrect

shown

is

and c

make

that

the equation

in the tracking chart to the right.

6000% of b.

that yields the target

B 2000c = 2000 x i

c

3

20

20

3

Since

Test each answer choice to find

number 6000. Incorrect

CORRECT

= 6000

(C)

Incorrect

(D)

Incorrect

2000 (E) c

2000

+ 20 =

12. (A): If

20 dollars

3

+ 20 = 23

Incorrect

Tessie wins 10 dollars in a billiards tournament and at

each of seven county

=

18.75.

fairs,

her average prize

You might recognize

this as

variable

is

choice A.

However, you can also solve the problem by testing each answer choice. +J 7-/ = .8.75 = 11.25 (A) CORRECT ;

M

(D)™

8

A/ + 7

number

M

10

J

20

Incorrect

8

./

= 30

Incorrect

7

A/ + 7J _ 150

(E)M+y

= 90 7

Incorrect

ManfiattanGM/KY'Prep 102

have

to pick different

numbers

(A) 20c = 20 x 3 = 60 i

you do

will

Incorrect

(B): Select values for a, b,

.

want

3

(E)9

1

If

D

same value. Remember, you always

_16_

3

3D + A +

and

the

3

(D)

this/fi

2

2

2

CORRECT

the

new standard

Incorrect

.

IN

ACTION ANSWER KEY

x 2 - llx + 28

L= =

3,

—4

L=

then

.

Test each answer choice to find the one that yields the target

4 (A)I(x2 -llx+19) = -^-(9-33 +

(B)

x2 + l\x- 28 (C)

9

9 (jc

+

14. (A):

Use

=

20 -—

+

4)

_

Incorrect

14

—

is

each run 100

/

Incorrect

109

to variables A, B, C,

and

D so that Kate's

greater than Amelia's rate

the equation

Kate: 100

CORRECT

_ 70

(10)(7)

Assign values

d=rt\o

—4

9

81+28

+ 28

number

Incorrect

+ 33-28

9

7)(jc

9x2

rate

19)

(x-7)(x-4) _ (-4X-1) _ 4

(D)

7

-"* + 28 =9

*2 13. (C):

If x

Chapter

VIC SOLUTIONS

calculate

how

fast

Kate and Amelia will

feet.

= 5t = 20s

Amelia: 100 t

= 4/ = 25

variable

number

A B

10

C

12

D

3

2

s

Kate will beat Amelia by 5 seconds. Test each answer choice

to find the

one

that yields the target

number of 5.

(A)

(B)

(C)

(D)

100(AD-BC)_ 100(30-24)

AC

_

\00BC-\00DA _ 2400 - 3000

AC 20 - 36

100

100

100

Incorrect

120

AB -CD

AD-CB

CORRECT

120

=

30-24 =

-.16

Incorrect

.06

Incorrect

100

ManfiattanGMAiT'Prep the

new standard

103

Chapter

IN

VIC SOLUTIONS

7 \l:

1

irst.

Pick numbers

assign

DUmben

that translate

to represent A. ).

and

/.

variable

easih into percent*.

KHofZ- 50% of 100 -50 10% of 50

/ Decreasing

10O(10)(50)(100)-(10)(2500)(100)

1.000.000

(B)

A?-)' _

XZ-Y

(10)(100)

-50 _

100

number

=

2.5

2.5

CORRECT

9

Incorrect

.095

Incorrect

10.000

XYZ-2Y

_ (10)(50)(100) -2(50) _

100

100

XYZ-2Y

= (10)(50)(100)-2(50) =

10.000

10.000

Incorrect

QQ

Incorrect

^

m

ManfiattanGM AT'Preo 104

/

1.000.000

_ (10)(100)-50

10.000

E)

50

100

100 (C)

10

}

2.5.

Ji answer ehoiee to find the one that \ields the target

lOOATZ-AI-Z _

n u in her

A

5

Y%, or h> 50%, yields

this result b>

ACTION ANSWER KEY

the

new standard

1

Chapter 8 EQUATIONS, INEQUALITIES, &

VIC's

STRATEGIES FOR DATA SUFFICIENCY

In This Chapter

Rephrasing:

Mil)

.

.

.

Manipulations

Sample Rephrasings

for

Challenging Problems

DATA SUFFICIENCY STRATEGY

Chapter 8

MUD Manipulations

Rephrasing:

Data sufficiency problems that involve algebraic equations and inequalities can usually be solved through algebraic manipulations, such as the ones strategy guide. ers,

you

will

some

In

need

to

cases,

you

will

we have covered

in this

need to manipulate the original question;

in oth-

manipulate the statements. Sometimes, you will need to manipu-

late both.

Remember,

the major manipulations include:

Multiplication and division by

some number

Unsquaring and squaring Distributing and factoring

You can also consider combination and substitution as manipulations for combining two or more equations. Likewise, combining inequalities is also a manipulation. When you

Remember

the three

MUD manipulations.

use manipulations to rephrase, you will often uncover very simple questions that have

been disguised by the

GMAT writers to

look complicated.

lsp>q? (1) (2)

-3p < -3q p - r >q-

r

ALONE is sufficient, but statement (2) alone is not sufficient. alone is not sufficient. ALONE is sufficient, but statement BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. EACH statement ALONE is sufficient. Statements and (2) together are NOT sufficient.

A

Statement

B

( 1 )

Statement (2)

C

D E

( 1 )

( 1 )

You can

rephrase both statements by performing simple manipulations.

Rephrase statement (1) by dividing both sides by —3. The result isp > must switch the direction of an inequality when multiplying or dividing

q, since it

by

you

a negative

number.

Rephrase statement

Now

this

question

(2)

is

by adding

simply:

r to both sides.

Is

The

result

is

p

>

q.

p>q?

(D/>> H (2)p>q Clearly, either statement

is

sufficient to

answer the question, since the rephrased stateto this data sufficiency problem is (D):

ments exactly match the question. The answer

EACH

statement

ALONE

is sufficient.

ManfiattanGM/KT' Prep the

new standard

107

DATA SUFFICIENCY STRATEGY

Chapter 8

Sometimes, rephrasing but

ma\

Consider

As you cated.

Mon

ncc

j

oomph

\anable cxpre-.-

in j

data >uftkien-

itself!

infonnation that you need to answer the question.

to the

example:

this

\\ hat

When \ou

statement ma) not uncover the answer to the question

i

you closer

gel

the \alue of r +

is

(1)

rs

(2)

s

—

—

ut = 8 +

rr

//?

- us

= 6

t

mind the variable or variable combo you are seeking how you manipulate equations and inequalities. In order we will need to isolate r + uon one variable combo r *

rephrase, always keep in

This can help guide

isolate

determine the \alue of the

to to

//.

side of an equation.

c> problem, look, tor

MJM

to

manipulate

it

Manipulate statement

mon

by moving

1 )

(

all

com-

the variables to one side and factoring out

terms.

— rt + r{s - /)

= 8 /) = 8

ut

-

u(s

-i-

r

-

us

rs

S

+u v

-

t

We

have manipulated statement (1) so that r + u is isolated. Although we still do not have a value for r + u, the information uncovered by this manipulation becomes important

On r

+

once

we

look

statement

at

its

own, statement

u.

It

simply

tells

(2)

is

insufficient because

us that s

However, when we look

at

(2).

-

t

=

it

tells

us nothing about the value of

6.

both statements together,

we

can plug the value of 5

vided by statement (2) into our rephrased statement (1) to get a value for r +

r

s

-

t

to this data sufficiency

sufficient, but

NEITHER

statement

ManfiattanG MAT Prep 108

pro-

8

+u=

The answer

t

u.

the

new standard

BOTH

problem

is

(C):

ALONE

is

sufficient.

statements

TOGETHER

are

DATA SUFFICIENCY STRATEGY

Chapter 8

Consider the following problem which involves manipulating the statements and then numbers.

testing

If

ab =

8, is

a greater than b?

(1)

-3b > -18

(2)

lb

>

8

First rephrase statement (1)

you divide both

If

by manipulating

sides of the inequality in statement (1)

forget to flip the direction of the inequality

You can

test

whether a

is

the inequality

numbers

to see

when

—36 > — 18.

by -3, you get b