GMAT Math Review - gmatcat.com Introduction to GMAT Math After a 5-minute break, you are assigned to the Quantitative section, which consists of 37 multiple-choice questions

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GMAT Math Review Table of Contents

Chapter 1: Arithmetic..................................................................................................6 1. Integers & Number Theory ...................................................................................... 6 2. Fractions................................................................................................................... 83. Decimals ................................................................................................................ 10 4. Percent.....................................................................................................................115. Ratio and Proportion .............................................................................................. 12 6. Exponents & Roots ................................................................................................ 14 7. Sets ......................................................................................................................... 168. Permutation and Combination ............................................................................... 18 9. Sequences............................................................................................................... 2010. Probability............................................................................................................ 2211. Standard Deviation............................................................................................... 27

Chapter 2: Algebra.....................................................................................................28 1. Simple Equation..................................................................................................... 28 2. Simultaneous Equations......................................................................................... 30 3. Quadratic Equation ................................................................................................ 33 4. Defined Functions.................................................................................................. 35 5. Inequalities ............................................................................................................. 366. Factoring ................................................................................................................ 38

Chapter 3: Geometry .................................................................................................40 1. Lines and Angles.................................................................................................... 40 2. Triangles................................................................................................................. 463. Quadrilaterals ......................................................................................................... 554. Circles .................................................................................................................... 60 5. Solids...................................................................................................................... 636. Coordinate Geometry............................................................................................. 65

Chapter 4: Word Problem.........................................................................................72 1. Interest, Discount and Profit .................................................................................. 72 2. Rate & Time........................................................................................................... 74 3. Work....................................................................................................................... 764. Averages and Medians ........................................................................................... 77 5. Mixture................................................................................................................... 78

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6. Age Problem........................................................................................................... 79 7. Doubling ................................................................................................................ 81 8. Sales Commission.................................................................................................. 82 9. Decision Tree ......................................................................................................... 83 10. Data Interpretation ............................................................................................... 85

Chapter 5 Data Sufficiency .......................................................................................88 Introduction................................................................................................................ 88 Section 1 Four Test-taking Strategies ........................................................................ 89 Section 2 Data Sufficiency Trick Questions .............................................................. 91

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Introduction to GMAT Math

After a 5-minute break, you are assigned to the Quantitative section, which consists of 37 multiple-choice questions. About two-thirds of them are problem-solving questions. A sample problem-solving question looks like this:

Example: If m is an odd integer, which one of the following is an even integer?

(A) 45 +m

(B) 2m

(C) 12 +m

(D) 2−m

(E) )1( +mm

The other one-thirds is the Data Sufficiency question. Unlike the problem-solving questions, the data sufficiency doesn’t require you to work out a solution to the presented question. Instead, it expects you to determine whether one or two conditions are sufficient to solve the presented questions.

The GMAT math section doesn’t test your specific knowledge in mathematics. Rather, it tests your problem solving ability. Therefore, calculus or another advanced math topic is never covered on GMAT math. The high school math knowledge is sufficient to answer a typical GMAT math question. Picking up the correct answer, however, is not as easy as its knowledge base. As we said, only the basic math concepts are chosen so that everyone taking GMAT is put on a fair arena. To make the test effective, the test writer should design questions that some answers correctly while other answers incorrectly. The best solution is to create complex questions or questions with numerous tricks.

How to use GMAT Math Review

We divide this course into five chapters, the first three of which are the comprehensive review of basic math concepts, including arithmetic, algebra, and geometry, the fourth is exclusively to draw on the difficult word problems, and the final is to cover the other part of the quantitative section--data sufficiency. Even you do not meet any problem in the math concepts, we encourage you to pay attention to the example questions in each section, since each example represents the most common question you will encounter in the real test.

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General Math Strategies: Substitution

Substitution, also known as plug-in, is one of the most effective strategies for solving complicated math questions. It is a method in which we plug numbers that fit the question’s parameters into the answer choices, and then determine which one is the best answer. In most cases, you have to plug in two different numbers before you eliminate four choices.

The substitution method is also very helpful for double checking your answers. Since you have nearly two minutes for each question, you won’t have sufficient time to do formal formula or calculating. The plug-in is a perfect solution. The numbers you choose for substitution should have the properties given in the problems. For example, if the question asks for an even integer, you should insert integer, and even integer.

Example: If m is an odd integer, which one of the following is an even integer?

(A) 45 +m

(B) 2m

(C) 12 +m

(D) 2−m

(E) )1( +mm

Solution

m is an odd integer, so we choose an odd integer for m, say 3, and plug it into each answer choice.

Choice A: 1943545 =+×=+m , which is an odd integer, therefore, A is incorrect.

Choice B: 23

2=

m . 1.5 is not an integer, let alone even integer.

Choice C: 713212 =+×=+m , which is not an even integer, either.

Choice D: 1232 =−=−m , which is an odd integer. Eliminate this choice.

Choice E: 12)13(3)1( =+×=+mm . 12 is an even integer. Therefore, E is the correct answer.

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Therefore, the correct answer is D.

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Chapter 1: Arithmetic

1. Integers & Number Theory

An integer is a whole number, such as -3, -2, -1, 0, 1, 2, 3. Zero is an integer.

An even integer is any number that is divisible by 2, such as -4, -2, 0, 2, 4. Otherwise, it is an odd integer, such as -3, -1, 1, 3.

Rules of adding and multiplying odd/even integers

even + even = even

odd + odd = even

even + odd = odd

even x even = even

odd x odd = odd

even x odd = even

A factor, also called as divisor, is an integer that divides another number resulting in an integer. For example, 2 is a factor of 10 since 5210 ×= . If an integer is not divisible by another integer, there exists remainder. For example, when the number 10 is divided by 3, the quotient is 3, and the remainder is 1, since 13310 +×= .

A prime number is a positive integer that has exactly two different positive divisors, 1 and itself. For example, 2, 3, 5, and 7 are prime numbers. 8 is not a prime number, since 8 has four different positive divisors, 1, 2, 4, and 8. The number 1 is not a prime number.

The numbers -1, 0, 1, 2 are consecutive integers.

Consecutive even numbers: 2, 4, 6, 8, 10

Consecutive odd numbers: 1, 3, 5, 7, 9

A positive number is a number greater than zero, such as 1, 2, 3.5, 4.5

A negative number is a number less than zero, such as -4.5, -3.5, -2, -1.

Rules of multiplying positive/negative number

Positive × Positive = Positive

Positive × Negative = Negative

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Negative × Negative = Positive

Example #1(Low-level)

If x is a positive integer, then x(x - 1)(x - 2) is always

(A) an positive odd (B) an positive even (C) divisible by 2 (D) a negative (E) a positive

Solution

If x is a positive integer, then the three numbers of x, (x - 1) and (x - 2) are consecutive integers. Therefore, at least one of them is even and so is their sum. So, the correct answer is C.

Example #2(Middle-level)

If m is a positive integer and k - 2 = 3m, which of the following can be a value of k?

(A) 81 (B) 27 (C) 12 (D) 9 (E) 5

Solution

As each of the choices is substituted for k, the sum k - 2 can be examined to determine whether or not it is a power of 3. The sums corresponding to the answer choices are 79, 25, 10, 7, and 3, respectively. Note that 3 = 31, 9 = 32, 27 = 33, and 81 = 34, only 3 is a power of 3. So the correct answer is E.

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2. Fractions

The expression 52 is a fraction, where 2 is the numerator, and 5 is the denominator. The

denominator can be any number, but 0.

The greatest common divisor (GCD) divides both numerator and denominator of a fraction, resulting in a fraction with lowest terms. For example, the GCD of 6 and 15 is 3 since the fraction

156 can be reduced to lowest terms

52 by dividing both 6 and 15 by 3.

Here, 156 and

52 are called equivalent fraction, since they represent the numbers with same

values.

The least common multiple (LCM) is an integer that can convert all fractions to ones with the same

denominator while with the least possible terms. For example, the LCM of 21 and

41 is 4 since

21 can be written as

42 , which has the same denominator as that of

41 . While the integer of 8

can also write both of 21 and

41 with the same denominator 8 (

84

21= ;

82

41= ), it is not the LCM,

because 8 is in higher term than 4.

Adding and Subtracting Fractions

The LCM is very helpful in adding and subtracting fractions with different denominators. For

example, when we add 125 and

165 , we can find the LCM 48. Then the original two fractions can

be written as:

4820

125=

4815

165= Therefore,

4835

4815

4820

165

125

=+=+

We can use the same method to subtract two fractions.

Multiplying and Dividing Fractions

To multiply fractions, simply multiply all numerators to form one numerator and all denominators to form one denominator. For example,

9

28030

875523

85

72

53

=××××

=××

Then, find the LGD 10, and divide it from both numerator and denominator. The final fraction

becomes283 .

To divide fractions, say 72

53÷ , invert the divisor (its reciprocal,

27 ) and multiply the two fractions,

i.e.,1021

27

53

72

53

=×=÷ .

The fraction 1021 can also be written as

1012 , a number that consists of an integer and a fraction,

formally known as mixed number.


If the product ba ⋅ is positive, which of the following must be true?

(A) a > 0(B) b > 0

(C) ba > 0

(D) a – b > 0(E) a + b > 0

Solution

If ba ⋅ is positive, ba is also a positive. Therefore, the correct answer is C.


If x, y, and z are positive numbers and x + y = z, which of the following must be greater than 0?

(A) z

yx −

(B) y

zx −

(C) x

zy −

(D) z

xy −

(E) x

yz −

Solution

Since x = z – y and x is a positive number, z – y > 0. Therefore, 0>−x

yz . E is the correct answer.

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3. Decimals

Decimal and percent (which we will consider in the following section) are the two derivatives from

fraction. Decimal is the horizontal expression of fraction. For example, 215.0 = . A decimal contains

a decimal point, the position of which determines the place value of the digits. For examples, 123.456

Digits before decimal point

1: hundreds

2: tens

3: units

Digits after decimal point

4: tenths

5: hundredths

6: thousandths

Scientific notation is an expression where a decimal is written as the product of a number with only one digit to the left of the decimal point and a power of 10. For example, 123.456 can be expressed as 1.23456x10 .

Example (Low-level)

=00002.200007.7

(A) 3.0005 (B) 3.50015 (C) 3.50001 (D) 3.500015 (E) 3.5

Solution

5.327

)00001.1(2)00001.1(7

00002.200007.7

=== , so E is the correct answer.

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4. Percent

A percent can be represented as a fraction with a denominator of 100, or as a decimal. For

example, 22% is a percent, which is equivalent to the fraction 10022 or decimal 0.22. To change a

decimal number to a percent, we simply multiply by 100 or move the decimal two places to the right.


The price of a share of stock S was 439 when the stock market was opened. The price became

217 when market was closed. Which of the following is closest to the percent decrease in the

price of stock S?

(A) 19%(B) 23%(C) 24%(D) 30%(E) 31%

Solution

The percent decrease equals %08.23399

393039

439

215

439

439

217

439

==−

=−

=−

, which is

closest to 23%. Therefore, the correct answer is B.


An increase of 30 percent on a stock followed by an increase of 20 percent amounts to

(A) the same as an increase of 20 percent followed by an increase of 30 percent(B) less than one 25 percent increase(C) the same as one 25 percent increase(D) less than an increase of 20 percent followed by an increase of 30 percent(E) the same as one 50 percent increase

Solution

A is the correct answer. If p is the original price, then the 30 percent increase in price results in a price of 1.3p. The next 20 percent increase in price results in a price of 1.3(1.2p), or 1.56p. Thus, the price increased by 1.56p – p = 0.56p. Only A has the same percent of increase.

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5. Ratio and Proportion

A ratio is a fraction, such as 53 . A ratio can be expressed in several ways. For example, the ratio

of 53 may be written as 3:5, 3 to 5, or

53 .

A proportion is an expression that two ratios are equal, such as 106

53= . In most cases, one the

four number is unknown, such as 105

3 x= . To solve this proportion, just cross multiply. For

example:

1053 x=

The unknown x is then found by cross multiplying:

3105 ×=x

Hence, 6=x


The ratios of profit generated by business units A, B, and C in 2000 is 2:3:5, respectively. If A generates profit of $40 million that year, what was the profit of generated by unit C in the same year?

(A) $20 million (B) $30 million (C) $50 million (D) $60 million (E) $100 million

Solution

Based on the ratio 2:3:5, the total profit t was divided as follows:

t102

was given to A, t103

was given to B, and t105

was given to C. Since t102

= $40 millions,

t = 200$)40(2

10= and 100$200$

105

105

=×=t . So the correct answer choice is E.

Example #2(high-level)

The ratio, by weight, of coffee to mate to water in a certain cup of coffee is 6:2:15. The coffee is altered so that the ratio of coffee to mate is halved while the ratio of coffee to water is doubled. If the altered coffee will contain 16 grams of mate, how many grams of water does it contain after alteration?

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(A) 12 (B) 15 (C) 30 (D) 60 (E) 100

Solution

The current ratio of coffee to mate is 6:2, when halved, the ratio becomes 3:2 or 6:4 or 12:8. The current ratio of coffee to water is 6:15, when doubled, the ratio is 12:15. The ratio of cofee to mate to water, when combined, becomes 12:8:15. Based on the ratio 12:8:15, the total grams t was divided as follows:

t3810

was coffee, t388

was mate, and t3815

was water. Since t388

= 16, t = 76)16(838

=

and then 30763815

3815

=×=t . So the correct answer choice is C.

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6. Exponents & Roots

Exponent is used to express long products of the same number. For example, when a number a

is to be used n times as a factor in a product, aaaaa ××××× ... , it can be written as na , where na is called a power, b is called a base, n is called an exponent, and there are n factors of a .

n a is the nth root of a , where a is called the base, n is called the index, and is called

the radical. The most commonly used roots are square root and cube root. Every positive number

a has two square roots, one positive and the other negative. 2 a denotes the positive root, while

2 a− denotes the negative root. For cube root, every number a has exactly one cube root, such

as 3 a .

Rules:

(1) )())(( baba xxx +=

(2) )( bab

ax

xx −=

(3) aaa xyyx )())(( =

(4) a

aa

yx

yx

=⎟⎟⎠

⎞⎜⎜⎝

⎛

(5) abba xx =)(

(6) a

a

xx 1

=−

(7) 10 =x

(8) b aba

xx =

Example (Low-level)

If 272x-3 = 3-x+5, then x =

(A) –2

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(B) –1 (C) 0 (D) 1 (E) 2

Solution

E is the correct answer. Since 272x+3 = (33)2x+3 = 26x-9, it follows, by equating exponents, that 6x - 9 = -x + 5, or x = 2.

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7. Sets

A set is a collection of various numbers, such as { }201 ，，− , where -1, 0, and 2 are called the

elements of the set.

The union of 1s and 2s is the set of all elements that are in 1s or in 2s or in both. It is

expressed as 21 ss ∪ . The intersection of 1s and 2s is the set of all elements that are both

in 1s and 2s , expressed as 21 ss ∩ .

The Venn diagram is often used to describe the relationship between two or more sets. For

example, the relations of 1s and 2s which share some (but not all) common elements can be

diagrammed as below.

Example (Middle-level)

If 80 percent of a class answered the first question on an exam correctly, 40 percent answered the second question on the test correctly, and 30 percent answered both correctly, what percent answered neither of the questions correctly?

(A) 10%(B) 20%(C) 30%(D) 40%(E) 80%

Solution

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Using the diagram above, we have deduced some new facts:

Percent of students who answered only first question correctly: 50% Percent of students who answered only second question correctly: 10% Percent of students who answered both questions correctly: 30%

Therefore, the percent of students who answered neither of the questions correctly is

100% - 50% - 10% - 30% = 10%. The correct answer is A.

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8. Permutation and Combination

Permutation is the ordering of various elements, such as abc, bac, cba, …. The number of ways of

ordering n different elements can be expressed as nnP or !n , where

123)...2)(1(! ⋅⋅−−== nnnnP nn . For example, 120123455

5 =××××=P

Combination is the selection of m objects from n elements. The number of ways of selecting

m objects from n elements can be written as mnC . For example,

62

121234

)12(121234

)!24(!2!42

4 =××

=××××××

=−

=C

The value of mnC is given by

)!(!!

mnmnC m

n −= .

Rule: mnn

mn CC −=


How many three-element subsets of {-2, -1, 0, 1, 2} are there that contain the pair of elements -2 and 2?

(A) One (B) Two (C) Three (D) Four (E) Five

Solution

There are a total of three pairs of three-element subsets that contain -2 and 2: {-2, -1, 2}, {-2, 0, 2}, and {-2, 1, 2}. Therefore, the correct answer is C.


If 10 persons meet at a meeting and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 36 (B) 45 (C) 90 (D) 100 (E) 28,800

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Solution

The total number of handshakes is 9 + 8 + 7 + 6 + 5 + 4+ 3 + 2 + 1 = 45. Therefore, the correct answer is B.

Example #3(High-level)

In how many arrangements can a director seat 3 actors and 3 actresses in a row of 6 seats if the actresses are to have the first, third, and fifth seats?

(A) 3 (B) 6 (C) 9 (D) 36 (E) 720

Solution

The number of arrangements to seat 3 actresses is 612333 =××=P , and the number of

arrangements to seat 3 actors is 612333 =××=P . Therefore, the number of arrangements

under the condition that 3 actresses are to have the first, third, and fifth seats is 6 x 6 = 36. The correct answer is D.

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9. Sequences

A sequence is an ordered list of numbers, such as 1, 3, 5, 7, 9, … where 1 is the first term, 3 is the second term, and 5 is the third term. The ellipsis symbol indicates that the sequence will continue. For the above sequence, the next term would be 11, then 13, and so on. This sequence is called arithmetic progression, in which the different between any two consecutive terms is constant. The sum of the first n terms of an arithmetic sequence can be expressed as the following formula:

[ ]2

)1()1(22 11

dnnnadnansn−

+=−+= , where 1a is the first term, and d is the common

difference.

Geometric progression is another type of sequence where the ratio of any two consecutive terms is constant. For example, -1, 3, -9, 27, -81, …The sum of the first n terms of a geometric sequence is

qqas

n

n −−

=1

)1( , where a is the first term, and q is the common ratio.


How many multiples of 3 are there between 6 and 42, exclusive?

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14

Solution

To solve this problem, we can list each number that is divisible by 3 between 6 and 42. They are 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. There are a total of 11 multiples. We can also solve this problem by using geometric progression. Let 3 be the first term, and the common ratio is 3. Therefore, 42 is the 14th term, or there are a total of 14 terms in this progression. By excluding 3, 6, and 12, there are 11 terms. Therefore, the correct answer is B.


If a sequence of 9 consecutive even integers with increasing values has 12 as its 6th term, what is the sum of the terms of the sequence?

(A) 120 (B) 102 (C) 98 (D) 90 (E) 80

Solution

Since the sixth term is 12, the fifth is 10, the fourth is 8, the third is 6, the second is 4, and the first is 2. We got the progression:

21

2, 4, 6, 8, 10, 12, 14, 16, 18

The sum is 9072182

2)19(9292

)1(1 =+=

−+×=

−+=

dnnnasn . The correct answer is D.


In an increasing sequence of 10 consecutive even integers, the sum of the first 5 numbers is 120. What is the sum of the last 5 integers in the sequence?

(A) 140 (B) 150 (C) 160 (D) 170 (E) 180

Solution

We let 1a be the first term, then the complete sequence is as below:

1a , 21 +a , 41 +a , 61 +a , 81 +a , 101 +a , 121 +a , 141 +a , 161 +a , 181 +a or

1a , 21 +a , 41 +a , 61 +a , 81 +a , 101 +a , 2)10( 1 ++a , 4)10( 1 ++a , 6)10( 1 ++a ,

8)10( 1 ++a

Since the sum of the first 5 numbers is 120, the sum of the last numbers is 120 + 5 x 10 = 170. Therefore, the correct answer is D.

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10. Probability

Probability questions are becoming increasingly common in GMAT CAT. Probability represents the most difficult questions. Therefore high scorers will be more likely to encounter them.

A. Simple Probability

A simple probability represents a situation where exactly one event occurs. It is written as:

P (A) = (the number of outcomes in A) / (the total number of possible outcomes)

For example, what is the probability that a card drawn randomly from a deck of cards will be an ace? Since of the 52 cards in the deck, 4 are aces, the probability is 4/52. The same principle can be applied to the problem of determining the probability of obtaining different sums from throwing a pair of dice. There are a total of 36 possible outcomes when a pair of dice is thrown.

Dice 1 Dice 2 Sum Dice 1 Dice 2 Sum Dice 1 Dice 2 Sum

1 1 2 2 1 3 3 1 4

1 2 3 2 2 4 3 2 5

1 3 4 2 3 5 3 3 6

1 4 5 2 4 6 3 4 7

1 5 6 2 5 7 3 5 8

1 6 7 2 6 8 3 6 9

Dice 1 Dice 2 Dice 2 Dice 1 Dice 2 Dice 2 Dice 1 Dice 2 Dice 2

4 1 5 5 1 6 6 1 7

4 2 6 5 2 7 6 2 8

4 3 7 5 3 8 6 3 9

4 4 8 5 4 9 6 4 10

4 5 9 5 5 10 6 5 11

4 6 10 5 6 11 6 6 12

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To calculate the probability that the sum of the two dice will equal 6, calculate the number of outcomes that sum to 6 and divide by the total number of outcomes (36). Since five of the outcomes have a total of 6 (1,5; 2,4; 3,3; 4,2; 5,1), the probability of the two dice adding up to 6 is 5/36. Similarly, the probability of obtaining a sum of 10 is computed by dividing the number of outcomes that the sum is 10 by the total number of outcomes (36). The probability is therefore 3/36=1/12.

Example (Low-level)

A committee is composed of x women and y men. If 6 women and 5 men are added to the committee, and if one person is selected at random from the enlarged committee, then the probability that a woman is selected can be represented by

(A) yx

(B) yx

x+

(C) 56

++

yx

(D) 6

6++

+yx

x

(E) 11

6++

+yx

x

Solution

Since each member of the enlarged committee represents a possible outcome, there are

11)56( ++=+++ yxyx possible outcomes. Therefore, the probability is 11

6++

+yx

x or E.

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B. Conditional probability

A conditional probability is the probability of an event given that another event has occurred. For example, what is the probability that the total of two dice will be greater than 7 given that the first die is a 4? This can be computed by considering only outcomes for which the first die is a 4. Then, determine the proportion of these outcomes that total more than 7. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 4, and of these, there are three that total more than 7 (4,4; 4,5; 4,6). The probability of a total greater than 7 given that the first die is 4 is therefore 3/6 = 1/2.


Balls numbered consecutively from 1 through 250 are placed in a box. What is the probability that a ball selected randomly will have an even number given that the number has a hundreds digit of 1?

(A) 52

(B) 21

(C) 53

(D) 25099

(E) 249100

Solution

There are 100 outcomes that the ball selected will have a number with a hundreds digit of 1: 100, 101, …, 199. Since of these 100 numbers, 50 are even, the probability is 50/100=1/2. The correct answer is B.

C. Probability of A and B

If A and B are Independent A and B are two events. If A and B are independent, then the probability that events A and B both occur is: p(A and B) = p(A) x p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. What is the probability that a fair coin will come up with heads twice in a row? Two events must occur: a head on the first toss and a head on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 x 1/2 = 1/4.


Andy, Bob, and Cherry each try independently to solve a problem. If their individual probabilities

for success are 31 ,

21 , and

53 , respectively, what is the probability that Andy and Bob, but not

Cherry, will solve the problem ?

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(A) 811

(B) 87

(C) 649

(D) 645

(E) 643

Solution

The probability that Both Andy and Bob will solve the problem is 61

21

31

=× . Since the probability of

success for Cherry is 53 , the probability of failure for Cherry is

52

531 =− . Therefore, the

probability that Andy and Bob, but not Cherry will solve the problem is 151

52

61

=× .

If A and B are Not Independent If A and B are not independent, then the probability of A and B is p(A and B) = p(A) x p(B|A) where p(B|A) is the conditional probability of B given A. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17 and the probability of A and B is: 1/13 x 1/17 = 1/221.


A jar contains only m red balls and n green balls. One ball is drawn randomly from the jar and is not replaced. A second ball is then drawn randomly from the jar. What is the probability that the first ball drawn is red and the second ball drawn is green?

(A) ⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

+ nmn

nmm

(B) ⎟⎠⎞

⎜⎝⎛

+−

⎟⎠⎞

⎜⎝⎛

+−

nmm

nmm 11

(C) ⎟⎠⎞

⎜⎝⎛

−+⎟⎠⎞

⎜⎝⎛

+ 1nmn

nmm

(D) ⎟⎠⎞

⎜⎝⎛

−+−

⎟⎠⎞

⎜⎝⎛

+ 11

nmm

nmm

(E) nm

mn+

Solution

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The probability that the first ball drawn will be red is nm

m+

. Since the first ball is not replaced, the

number of remaining balls is 1−+ nm and the probability that the second ball will be green is

1−+ nmn . Therefore, the probability that the first ball drawn is red and the second ball drawn is

green is ⎟⎠⎞

⎜⎝⎛

−+⎟⎠⎞

⎜⎝⎛

+ 1nmn

nmm or C is the correct answer.

D. Probability of A or BIf events A and B are mutually exclusive, then the probability of A or B is simply:p(A or B) = p(A) + p(B).What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible to getboth a 1 and a 6, these two events are mutually exclusive.

Therefore, p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3


In a box, there are 60 red marbles, 50 green marbles, 30 black marbles, and 20 white marbles. What is the probability that the marble will be either red or white?

(A) 61

(B) 51

(C) 41

(D) 31

(E) 21

Solution

There are a total of 60 + 50 + 30 + 20 = 160 marbles. The probability that marble will be red is

83

16060

= and the probability that marble will be white is81

16020

= . Therefore, the probability that

the marble will be either red or white is 21

84

81

83

==+ . The correct answer is E.

27

11. Standard Deviation

The standard deviation, a concept in Normal Distribution, is a statistic that tells you how tightly all the various examples are clustered around the mean in a set of data. Before we address, let’s first address the concept of variance. The variance is a measure of how spread out a distribution is. It is computed as the average squared deviation of each number from its mean. For example, for the numbers 0, 3, and 6, the mean is 3 and the variance is:

63

9093

)36()33()30( 2222 =

++=

−+−+−=σ

The formula for the standard deviation is very simple: it is the square root of the variance. Here, it is the positive value of σ -- 6 .

Standard deviation is the most commonly used measure of spread. Why? On the above formula, we find that the closer each number toward to the mean, the smaller the standard deviation is. In other word, the standard deviation will tell you how diverse the numbers are.

In business, the standard deviation is often used by investors to measure the risk of a stock. The basic idea is that the standard deviation is a measure of volatility: the more a stock's returns vary from the stock's average return, the more volatile the stock. Consider the following two stock s and their respective returns over the last three months. Both stocks end up increasing in value from $12 to $15. However, Stock A's monthly returns range from -1% to 2% while Stock B's range from -12% to 15%. Here, the standard deviation of the three month’s returns for Stock A is smaller thanthat for Stock B. Therefore, Stock B is considered more risk than Stock A.

It seems a little difficult to understand the full meanings of standard deviation. The question in this subject, however, is much easier. Let’s look at an example:

Example (Low-level)

The mean and standard deviation of a certain normal distribution are 18.5 and 2.5. What value is exactly 3 standard deviations bigger than the mean?

(A) 21.5(B) 22(C) 23.5(D) 25(E) 26

Solution

Since the mean is 18.5 and one standard is 2.5, 3 standard deviations bigger than the man is 18.5+3x2.5=18.5+7.5=26. The correct answer is E.

28

Chapter 2: Algebra

In this section, we mainly talk about simultaneous equations, quadratic equations and defined functions. At the end of this chapter, we will let you know how to use factoring to simply an algebra expression.

1. Simple Equation

An equation that contains only one unknown is called simple equation. The following algebraic expression is a simple equation.

xx −=− 513


If 463

21

=+

x

then x =

(A) – 2

(B) 4823

−

(C) 61

(D) - 6(E) 3

Solution

Since 463

21

=+

x

, 81

421

63 ==+x

. Thus, 823

824

813

816

−=−=−=x

, or 4823

8236

−=−

=x .

B is the correct answer.


If n is an integer such that 24n = (21

) n-10, then n =

(A) 2 (B) 3 (C) 4(D) 5 (E) 6

Solution

29

Since 24n = (21

) n-10, 24n = ( 2 ) –(n-10) or 4n = – (n – 10) = – n + 10. Therefore, 5n = 10 or n = 2. A is

the correct answer.

30

2. Simultaneous Equations

An equation that contains two unknown is called simultaneous equation. The following algebraic expressions are two simultaneous equations.

1172 =+ yx

75 =+ yx

Here, the two equations have the same solutions. There are major two methods to solve two linear equations in two unknowns: Substitution and Adding/Subtracting.

A. The substitution method:

a. Express one of the unknowns in terms of the other using either of the two equations.

b. Substitute the expression into the other equation to obtain an equation which contains only one unknown.

c. Solve the simple equation and get value of the unknown.

d. Substitute the value into either of the original equations to find the value of the other unknown.

Example

(1) 1172 =+ yx

(2) 75 =+ yx

Solution (using the substitution method)

Using the substitution method, the second equation gives

yx 57 −=

Substitute this into the first equation and solve for y :

1172 =+ yx

117)57(2 =+− yy

1171014 =+− yy

11314 =− y

31

33 =y

1=y

Substitute 1=y into the second equation and solve for x :

75 =+ yx

215757 =×−=−= yx

Therefore, the solution is 2=x , 1=y .

B. The adding or subtracting method:

a. Multiply one equation by a properly chosen number to make the coefficients of one of the unknowns the same in both equations.

b. Either add or subtract the equations to eliminate one of the unknowns.

c. Solve the resulting equation for the remaining unknown.

d. Substitute the value of the known unknown into either of the two original equations and solve for the other unknown.

Example

(1) 1172 =+ yx

(2) 75 =+ yx

Solution (using the adding/subtracting method)

We first multiply equation (2) by the number of 2 to get

(2) 14102 =+ yx

Subtract the two equations

(1) 1172 =+ yx

(2) 14102 =+ yx

The yields the new equation: 33 =y .

Therefore, 1=y

32

Substitute 1=y back into the second equation and solve for x :

75 =+ yx

715 =×+x

257 =−=x

Therefore, the solution is 2=x , 1=y .

33

3. Quadratic Equation

The following algebraic expression is a quadratic equation.

02 =++ cbxax , where a , b , and c are constants.

If either b or c is zero, the equation is relatively easy to solve. If not, we should consider factoring or quadratic formula. If a quadratic equation is not easily factored, then its roots can always be found using the quadratic formula:

02 =++ cbxax ( 0≠a ) where a, b, and c are constants.

aacbbx

242 −±−

= , or a

acbbx2

42

1−+−

= and a

acbbx2

42

2−−−

=

Now, we try to use this formula to solve a sample quadratic equation:

0432 =−+ xx

Here, 1=a , 3=by , and 4−=c

Then, 2

)4(49212

)4(14332

4 22

1−×−+−

=×

−××−+−=

−+−=

aacbbx

23

252

2252

24492

=+−

=+−

=×++−

=

so, 23

1 =x

And, 2

)4(49212

)4(14322

4 22

2−×−−−

=×

−××−−−=

−−−=

aacbbx

27

27

252

2252

24492

−=−

=−−

=−−

=×+−−

=

so, 27

2 −=x


If 0)73)(2( =−−x

x and x ≠ 2, then x =

(A) 7(B) 3

34

(C) 37

(D) 2

(E) 21

Solution

Since 0)73)(2( =−−x

x , it follows that either 02 =−x or 073 =−x

. That is, either 2=x or

37

=x . But 2≠x is given, so 37

=x . Answer: C


Which of the following equations has a root in common with x2 – 5x + 4 = 0?

(A) x2 + 4x – 1 = 0 (B) x2 + x + 1 =0 (C) 2x2 – 5x – 6 =0 (D) x2 – 1 =0 (E) x2 + x – 1 =0

Solution

Since x2 – 5x + 4 = (x - 4)(x - 1), the roots of x2 – 5x + 4 = 0 are 4 and 1. when these two values are substituted in each of the five choices to determine whether or not they satisfy the equation, only in D does a value satisfy the equation. So, D is the correct answer.

35

4. Defined Functions

Besides the above three equations, the test writers may create new functions, known as defined functions. In these problems, you are given a symbol and a mathematical expression or description that defines the symbol. Defined functions are common on the GMAT. However, once you get used to them, defined function can be some of the easiest problems on the test.


For any positive integer n, n >1, the “length” of n is the number of positive primes (not necessarily distinct) whose product is n. For example, the length of 10 is 2 since 10 = (2)(5). Which of the following integers has length 3?

(A) 4 (B) 6 (C) 50 (D) 60 (E) 100

Solution

Since 50 = (2)(5)(5), the length of 50 is 3. Therefore, the correct answer is C.


In an Einstein theory, the relationship between energy and mass is expressed in the equation 2mcE = , where E, in Joule, represents energy, m, in kilogram, represents mass, and c is a

constant. If the energy for a mass with 9 kilograms is 17101.8 × Joule, what is the energy, in Joule, for a mass with 5 kilograms?

(A) 17104.5×

(B) 17105×

(C) 16104.5×

(D) 15104.5× (E) 4.5

Solution

In order to compute 2mcE = when m = 5, the value of the constant c must be determined.

Since E = 17101.8 × Joule when m = 9 kilograms, substituting these values into the formula

yields 216 91081 c=× , or c = 8103× . Therefore, when m = 5 kilograms, the energy, in Joule is 1728 105.4)103(5 ×=××=E . A is the correct answer.

36

5. Inequalities

An inequality is a statement that compares two numbers or expressions. The following symbols are used in inequalities:

> greater than, for example, 2>x

< less than, for example, 52 <y

≥ greater than or equal to, for example, 2≥x

≤ less than or equal to, for example, 52 ≤y

≠ not equal to, for example, 105 ≠y


Which of the following inequalities is equivalent to 6x – 7 < 11?

(A) x > -18 (B) x > -3 (C) x > 3 (D) x < 3 (E) x < -3

Solution

From 6x – 7 < 11, it follows that 6x < 18 or x <3. Therefore, the correct answer is D.


If 32

<a and 05 =− ax , which of the following must be true?

(A) 51

>x

(B) 152

>x

(C) 32

<x

(D) 51

<x

(E) 4<x

Solution

It follows from x – 5a = 0 that a = x51

. So 32

<a implies 32

51

<x , or 3

10<x , which means x <

4 (this choice). None of the other choices must be true (although 51

>x and 152

>x could be

37

true.). So, E is the correct answer.

38

6. Factoring

To factor an algebraic expression is to rewrite it as a product of two or more expressions, called factors. Factoring is a very useful method to solve quadratic equation. Also, some of the GMAT questions may ask you to compare two algebraic expressions. Before you pick up the answer, you may have to first simplify one of the two expressions. The most commonly used factoring rules are the following three:

(1) )( yxaayax ±=±

(2) ))((22 yxyxyx −+=−

(3) 222 )(2 yxyxyx ±=+±

Sometimes, factoring can save you much work of calculating. Look at the following example:


If 22 2 yxyx −= , then, in terms of x, y =

(A) x (B) 2x

(C) 2x

(D) 0 (E) –2x

Solution

Since the equation 22 2 yxyx −= equals 02 22 =+− yxyx or 0)( 2 =− yx , in terms of x, y = x . The correct answer is A.


If x = 99, then =−

−−4

432

xxx

(A) 973 (B) 660 (C) 100 (D) 99 (E) 72

Solution

If you just substitute x = 99 to the above expression, it may take you several minutes to work out

39

the solution. However, if we first factor the numerator, you would find it much easier.

Since 14

)1)(4(4

432+=

−+−

=−

−− xx

xxx

xx, x = 99 is given, therefore, the original expression

equals 100. C is the correct answer.

40

Chapter 3: Geometry

1. Lines and Angles

The following figure is a line, which can be referred to as line AB. The part of the line from A to B is called a line segment.

If two lines intersect at an angles with the measure of 90 are perpendicular, denoted by 21 ll ⊥ .

Two lines are parallel if the two lines that in the same plane do not intersect, denoted by 21 || ll .

If a third line intersect two parallel lines, then the angle measures are related as indicated as below:

41

4321 ∠=∠=∠=∠ 8765 ∠=∠=∠=∠

If two lines intersect, the opposite angles are called vertical angles and have the same measure.

BODAOC ∠=∠

AODCOB ∠=∠

42

In the above figure, AOC∠ and BOD∠ are vertical angles, and BOC∠ and AOD∠ are vertical angles.

There are special kinds of angles:

A) An acute angle has measure less than 90

B) An obtuse angle has measure greater than 90

C) A right Angle has measure 90

D) Complementary Angles are two angles that sum to 90 .

E) Supplementary Angles are two angles that sum to 180 (or a straight line).

43

44


In the figure above, line 1 is parallel to line 2. If ∠ 1 = 110 , then ∠ 2 =

(A) 10 (B) 20 (C) 30 (D) 40 (E) 50

Solution

Since line 1 is parallel to line 2, ∠ 3=∠ 1. Also, ∠ 3 and ∠ 4 are vertical angles, so ∠ 4=∠ 3. Since ∠ 1 = 110 , ∠ 4 = 110 . Therefore, ∠ 2 = 110 -90 =20 or the correct answer is B.


45

In the figure above, AB = AC and OB = OC. If ∠ BCO = X , ∠ ABO = 10+X , and ∠ CAO = 60 - X , then what is the value of X?

(A) 30(B) 25(C) 20(D) 10(E) 5

Solution

Since OB = OC, ∠ BCO = ∠CBO = X . So, ∠ ABC = ∠CBO + ∠ ABO = X + 10 +X = 10 + 2X . Since AB = AC, ∠ ACB=∠ ABC=(10 + 2X) . Also, ∠ BAC = 2∠ CAO = 2(60 - X )=(120-2 X) . For a triangle, ∠ ACB + ∠ ABC +∠ BAC =180 or (120-2X) + 2(10 + 2X) = 180 .Therefore, X = 20 or the correct answer is C.

46

2. Triangles

A triangle has three sides and three angles.

Properties of triangles

1) The sum of its three angles is 180 .

2) The sum of the lengths of any two of the sides is greater than the length of the third side.

The altitude of a triangle is the segment drawn from a vertex perpendicular to the side opposite that vertex. The opposite side is called the base.

(the length of the altitude) x (the length of the base)The area of a triangle =

2

CD x AB Here, the area of ABC∆ = 2

47

AD x CB The area of ABC∆ is also equal to

2

A triangle that has two equal sides is an isosceles triangle. It has the following properties:

1) Two sides have equal lengths.

2) The angles opposite the equal sides are also equal.

3) BD = DC

An equilateral triangle have all three sides equal. It has the following properties:

1) All three sides are equal.

2) Each of its three angles is 60 .

48

3) BD = DC = 1/2(BC) = 1/2 (AB) = 1/2 (AC)

A right triangle is a triangle that has a 90 angle. It has the following properties:

1) 222 BCACAB += (Pythagorean Theorem), where AB is its hypotenuse and AC and BC are its

legs.

2) The legs are always shorter than the hypotenuse.

3) 2ACxCBABCS =∆

4) ooo 90=+ yx

49

50

Most Commonly Used Triangles

There are 3 types of right triangles that show up often on the GMAT: 3 - 4 - 5 triangle, 30 - 60 - 90 triangle, and 45 - 45 - 90 triangle.

1) 3 - 4 - 5 triangle: 3 - 4 - 5 are the lowest terms that satisfy Pythagorean Theorem (5 = 3 + 4 ). Many right angles can be expressed as 3n-4n-5n, where n is a variable.

......

2) 30 - 60 - 90 triangle: The ratio of its short leg to its hypotenuse is 1 : 2.

2ABBC =

51

3) 45 - 45 - 90 triangle: The 45 - 45 - 90 triangle has equal legs, therefore, it is also an

isosceles triangle. The length of the hypotenuse is 2 times of that of one legs.

2ABBCAC ==

52


Which of the following groups of numbers could be the lengths of the sides of a triangle?

(A) 2, 3, and 5

(B) 1, 31 , and 2

(C) 32 ,

41 , and

53

(D) 4, 5, and 12 (E) 2, 5, and 7

Solution

The sum of the lengths of any two of the sides is greater than the length of the third side. The

correct answer is C, where 32 +

41 >

53 ,

32 +

53 >

41 , and

53 +

41 >

32 .


Student A and Student B live 13 miles apart. On one day, they drove to meet at a town that is directly east of Student A’s home and directly south of Student B’s home, the total number of miles they drove were 4 miles more than the number of miles either A or B drove to meet at one of their homes. If Student A lives closer to the town than does Student B, what is the distance, in miles, between Student B’s home and the town?

(A) 4 (B) 5 (C) 7 (D) 9 (E) 12

Solution

53

We transform the set-up into the above figure. Based on the original conditions, we conclude:

5−+= BCACAB

222 ACBCAB +=

Since AB = 13, we get the following equations:

(1) 413 −+= BCAC or BCAC −= 17

(2) 22213 ACBC +=

Substitute BCAC −= 17 into the second equation and solve for BC :

222 )17(13 BCBC −+=

We get 121 =BC and 52 =BC .

Student A lives closer to the town than does Student B, the distance between Student B’s home and the town is 12 miles. The correct answer is E.


A stick 56 feet long is cut into three pieces. The three pieces then form a right triangle with an area of 168 square feet. If the hypotenuse is 16 feet longer than one of the other two sides, then what is the length, in feet, of the hypotenuse?

(A) 7(B) 12(C) 24(D) 25(E) 26

Solution

To solve this problem, we first let a, b, and c be the length of each side, and c for hypotenuse. Then, we get the following three equations:

(1) 222 cba =+

(2) 56=++ cba

(3) 168=ab

Our mission is to work out the values of a, b, and c. We transform equation (2) to

(4) cba −=+ 56

Then, multiple 2 in each side of equation (3) and get

(5) 3362 =ab

54

We merge equation (5) into equation (1), and get

3362 222 +=++ cabba or 336)( 22 +=+ cba

Then, substitute cba −=+ 56 into the above equation:

336)56( 22 +=− cc or 336112562 =− c

Therefore, 25328112

336562=−=

−=c . The correct answer is D.

55

3. Quadrilaterals

A quadrilateral is a polygon with four sides. A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

A parallelogram has the following properties:

AD || BC, AB || CD

AB =CD, AD = BC

AE = EC, DE = EB

S (area) = BF x DC

A rectangle is a parallelogram with right angles. A square is a rectangle with all sides of equal length.

Rectangle

o90=∠=∠=∠=∠ DCBA

AD = BC, AB = CD

S = AB x AD = BC x DC

P = 2 x (AD + CD)

56

Square o90=∠=∠=∠=∠ DCBA

DACDBCAB ===

2222 DACDBCABS ==== DACDBCABP ×=×=×=×= 4444

o45=∠DAC

A trapezoid is a quadrilateral with two sides that are parallel. The area of trapezoid ABCD is equal to

(AB + CD) x AE S(ABCD) =

2


What is the area of the trapezoid above?

(A) 12 (B) 13 (C) 14 (D) 21 (E) 28

Solution

57

The area of above trapezoid is 1322

)76(=×

+. So, B is the correct answer.

58


In the figure above, O is the center of half circle and has a radius of 3, what is the area of the region enclosed by the figure above?

(A) π20 (B) π618 +

(C) π2918 +

(D) π2916 +

(E) π316 +

Solution

The figure above represents a half circle with a radius of 3 and a rectangle with a width of 3 and a

length of 6. The area of the whole figure is ππ29183

2163 2 +=+× . Therefore, the correct

answer is C.


A drawer wants to draw a circle inside a square that is 6 on a side. If the circle has the greatest possible area, what is the radius of the circle?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Solution

The circle that has the greatest possible area must inscribe the square as shown in the following figure.

59

Since the side of square in length equals the diameter of the circle, the radius of the circle is 3 in length. The correct answer is C.

60

4. Circles

The following figure is a circle. Segment CD is a chord. Segment AB that passes through the center (O) of the circle is a diameter. Segment OG is a radius. A line that just touches a circle is called a tangent. In the following figure, EF is a tangent.

1) rd 2=

2) o90=∠OGF

3) rncecircumfere π2=

4) 2rarea π=

5) 14.3

722

≈=π

An inscribed angle has its vertex on the circle itself, and its measure is 1/2 of the measure of the arc it intercepts:

ACBAOB ∠=∠21

Triangle ABC is a right triangle, since

ACBAOB ∠=∠21

o180=∠AOB .

61

62


O is the center of circle shown above. If o30=∠ABC , then what is the ratio of the area of ∆ABC to the area of circle O?

(A) 43

to π

(B) 23

to π

(C) 3 to π (D) 3 to 1 (E) 5 to π

Solution

We can conclude from the above figure that ∆ABC is a right triangle. Since o30=∠ABC ,

ABAC21

= and ABABABABACABBC23

43)

21( 22222 ==−=−= . Then, the

area of ∆ABC is 283

23

21

21

21 ABABABBCACS =××=⋅= . The area of the circle O is

4)

2(

222 ABABrS ×===πππ . Therefore, the ratio of the area of ∆ABC to the area of circle O

is4

:8

3 22 ABAB π or

23

to π . B is the correct answer.

63

5. Solids

There are two major solids that you will encounter on the GMAT test. They are the rectangular solid and circular cylinder. A rectangular solid in which all edges are of equal length is a cube. The volume of a rectangular solid is the product of its three sides: Volume = Length x Breadth x Height. The volume of a circular cylinder is the products of the area of the base and the height: Volume = Area of Base x Height.

V = Lbh V = r h

surface area = 2 x (AB x BC) x (BC x BF) x (DC x CG) surface area = 2 rh + 2 r


An empty swimming pool is 40 meters long, 15 meters wide and 2 meters deep. If water is being filled at the rate of 3 cubic meters per minutes, what is the time it takes to fill the swimming pool?

(A) 8 hours(B) 6 hours and 40 minutes(C) 5 hours(D) 4 hours and 20 minutes(E) 4 hours

Solution

The volume of the swimming pool is (40)(15)(2) = 1,200 cubic meter. Since the pool is filled at the rate of 3 cubic meters per minute, it takes 1,200/3 = 400 minutes or 6 hours and 40 minutes to fill the box. Therefore, B is the correct answer.


64

In a supermarket, Liquid X is sold in two kinds of right cylindrical container, one with radius of 2 centimeters and height 5 centimeters, the other with radius 5 centimeters and height 10 centimeters. If the milk in smaller container is sold for $9.0, and that in the larger is sold for $75.0, then what is the ratio of the cost per cubic centimeters in smaller container to that in larger container?

(A) 5 to 3 (B) 5 to 4 (C) 3 to 2 (D) 2 to 1 (E) 3 to 1

Solution

The cost per cubic centimeters in smaller container is ππ 20

952

92 =×

and that in larger

container is πππ 10

3250

75105

752 ==×

.Therefore, the ratio is ππ 10

3:20

9 or 3:2. C is the

correct answer.

65

6. Coordinate Geometry

The following figure is coordinate plane. The horizontal line is the x-axis and the perpendicular vertical line is y-axis. The two axes divide the plane into four areas, quadrant I, quadrant II, quadrant III, and quadrant IV. Any point in the plane has an x-coordinate and a y-coordinate. For example, point P is identified by an ordered pair (2,4). P(x, y) can represent any point in the plane, where x represents x-coordinate, and y represents y-coordinate.

Distance Formula

The distance between two points in the plane can be measured by the Pythagorean Theorem. Look at the following coordinate plane, you will find two points, A and B. to yield the distance between A and B, just diagram two segment, one perpendicular to x-axis, and the other perpendicular to y-axis. The two segments intersects at point C. There generates a right triangle. We notice that to measure the segment AB is in fact to measure the hypotenuse of the triangle ABC.

Therefore, [ ] [ ] 132521636)3(1)2(4 2222 ==+=−−+−−=+= ACBCAB

66

When the two points A and B are not defined, we can represent them with A (x, y) and B (a, b). There comes out the general distance formula:

2222 )()()()( axbyxaybAB −+−=−+−=

67

Slope Formula

The GMAT question also often asks you to measure the slope of a line in the coordinate plane. The slope is defined as the ratio of the difference in the y-coordinates to the difference in the x-coordinates.

68

Dividing the vertical change by the horizontal change results in slope formula.

xayb

m−−

=

Here, m represents the slope.

When the slope of a line is defined, the third, fourth and fifth point in the line is simultaneously defined by the following linear equation:

bmxy +=

Here, m represents the slope and the constant term b is the y-intercept. The x-intercept is when y = 0 in the above equation.

y-intercept: b x-intercept: m

bx −=

69

As far as two points are defined, the above linear equation was defined. In other words, the slope m, y-intercept and x-intercept are defined.

Example:

In the following coordinate plane, two points in coordinate plane are given, A and B. What is the linear equation for the line?

70

Solution:

To solve this problem, plug the digits in the line into the slope intercept equation.

xaybm

−−

=

y = -1, b = 3

x = -2, a = 2

144

)2(2)1(3

==−−−−

=−−

=xaybm

The slope is 1. That is m=1. Therefore, the original formula is bxy += .

Plug either point A into the new formula, and get

b+−=− 21

Thus, b = 1

Finally, the linear equation is defined as:

71

1+= xy

And the y-intercept is 1; x-intercept is -1.

72

Chapter 4: Word Problem

When simple concept is used to create math question in a complicated manner, it becomes a problem, especially for international students. This chapter is dedicated to identify the most commonly found word problems and then develop specific techniques to solve each of the “problems”.

The following represents a typical procedure to solve word problems:

Step 1: Choose a variable (e.g. x) to represent the unknown quantity, or the figure that you are asked to get.

Step 2: Express the question as an algebraic expression (simple equation) using the variable defined at Step 1.

Step 3: Solve the equation in Step 2

Step 4: Eliminate answers that are outside of the ballpark, and pick up the correct answer.

1. Interest, Discount and Profit

With simple interest, the interest is computed by the following formula:

Interest = Principal x Interest Rate x Time

The interest rate can be annual or monthly interest rate.

Discount is the percent reduced on the original price of a product. Profit is equal to selling price minus cost.

Profit = selling price - cost


If Jennifer invested $65,000 at 12 percent annual interest which is compounded monthly, then the total value of the investment, in dollars, at the end of 7 months would be

(A) 65,000(1.01)7

(B) 65,000(1.12)7

(C) 65,000(1.07)7

(D) 65,000+65,000(0.01)7

(E) 65,000 + 65,000(0.08) 7

Solution

Since the annual interest rate is 12 percent, the monthly rate is 1 percent. Then, at the end of first

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month, the total value will be 65,000(1.01)1 and that at the end of 7 months will be 65,000(1.01)7. Therefore, the correct answer is A.


George made a down payment of $10,000 and borrowed the balance on an apartment that cost $85,000. The balance with interest was paid in 60 monthly payments of $1,600 each. What is the total amount of interest paid by George?

(A) $11,000 (B) $21,000 (C) $22,000 (D) $23,000 (E) $24,000

Solution

The total amount of money that George paid is $10,000 + 60 x $1,600 = $106,000. Therefore, the amount of interest paid is $106,000 - $85,000 = $21,000. The correct answer is B.

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2. Rate & Time

The distance that an object moves is equal to the product of the average speed at which it moves and the amount of time it takes to move that distance.

Distance = Rate x Time or D = R x T


If a printer prints 8 copies per minutes, then, at the same rate, how many copies does it print in 3 hours?

(A) 480 (B) 600 (C) 960 (D) 1,440 (E) 2,400

Solution

The production rate of each printer is 8 x 60 = 480 copies per hour. In 3 hours, it will print 480 x 3 = 1,440. Therefore, the correct answer choice is D.


If Mr. Clinton had driven 2 hours longer on a certain day and at an average rate of 10 miles per hour faster, he would have driven 120 more miles on that day than he actually did. If he had driven 4 hours longer and at an average rate of 20 miles per hour faster on that day, then how many more miles would he have driven than he actually did?

(A) 140 (B) 180 (C) 240 (D) 260 (E) 280

Solution

Let r be the current speed and t be the current number of hours Mr. Clinton drives. Then we get the following equation:

(r + 10) (t + 2) – rt = 120, or

rt + 2r + 10t + 20 –rt = 120

2r + 10t = 100

r + 5t = 50

The number of miles he would drive on that day if he had driven 4 hours longer and at a rate of 20 miles hour faster is:

(r + 20) (t + 4)

Then, the extra number of miles is

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(r + 20) (t + 4) – rt = rt + 4r + 20t + 80 – rt = 4r + 20t + 80 = 4(r + 5t) + 80

Since r + 5t = 50, then 4(r + 5t) + 80 = 200 + 80 = 280.

Therefore, E is the correct answer.

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3. Work

For a single person or machine, the work problem is quite similar to the problem of Rate and Time, and can be expressed as the following formula:

Work = Rate x Time or W = R x T

When two persons or machines work together, the basic formula is:

CBA111

=+ , where A and B are the amount of time it takes A and B, respectively, to complete a

project when working alone, and C is the amount of time it takes A and B to do the project when working together.


Used alone, one pipe fills an empty swimming pool in 12 hours and the second pipe fills the same pool in x hours. If both pipes are used together, it will take 3 hours to fill the pool. What is the value of x?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Since the first pipe fills the tank in 12 hour, it fills 121

of the bank in one hour. The second pipe fills

the tank in x hours, so it fills x1

of the tank in one hour. Together they fill x1

121+ of the tank in

one hour. At this rate, if 3 is the number of hours needed to fill the tank, then 1)1121(3 =+

x or

311

121

=+x

. So, 4=x . The correct answer is D.

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4. Averages and Medians

The average is the sum of a list of numbers divided by the total number of numbers in the list. The median is the middle number in an ordered list of numbers in ascending or descending sequence. The median is determined by position, and is a different concept from the average.


What is the average of the numbers 3, 4, 5, 5, and 8?

(A) 8 (B) 7 (C) 5 (D) 4.8 (E) 4.5

Solution

The sum of these five numbers is 3 + 4 + 5 + 5 + 8 = 25. The average is 5525

= . Therefore, the

correct answer is C.


The average of 8 numbers is 7.5. If one of them is deleted, the average of the remaining 7 numbers becomes 8.5. What is the deleted number?

(A) 0 (B) 0.1 (C) 0.2 (D) 0.4 (E) 0.5

Solution

The sum of the 8 numbers is 8(7.5) = 60; the sum of the 7 remaining numbers is 7(8.5) = 59.5. Thus, the deleted number must be 60 – 59.5 = 0.5. E is the correct answer.

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5. Mixture

In a mixture problem, you are required to determine the characteristics of the resulting mixture when substances with different characteristics are combined. The key to these problems is that the combined total of the concentrations in the two parts must be the same as the whole mixture.


Thirty kilograms of solution A contain 80 percent water and 20 percent liquid M. If 50 percent of water evaporates from this solution, what percent of this new solution is liquid M?

(A) 10%(B) 12%(C) 13%

(D) %3133

(E) 40%

Solution

The amount of liquid M in the solution is 0.2×30 or 6 kilograms and the amount of water is 24 kilograms. After the evaporation of the water, the total amount of water is 24/2 = 12 kilograms. The

percent of liquid M in the new solution is thus %3133

31

1266

==+

. Therefore, D is the correct

answer.

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6. Age Problem

Generally, we solve this problem by letting x be a person's current age and then the person's age a years ago will be ax − and the person's age a years in future will be ax + .

Example #1 (Low-level)

If Jennifer was 18 years old 6 years ago, how old was she m years ago?

(A) 24 + m(B) m - 24(C) 14 – m(D) 24 – m(E) 12 + m

Solution

Since Jennifer was 18 years old 6 years ago, her age now is 18 + 6 = 24. m years ago, Jennifer was m years younger, so her age then was 24 – m. The correct answer is D.

Example #2 (High-level)

Five years ago, Al was 5 years older than Sandra, and Sandra was twice as old as Teresa. Which of the following is not true?

(A) Al is 5 years older than Sandra(B) Teresa is twice as old as Sandra(C) Al is older than Teresa.(D) In 2 years, Sandra will be 5 years younger than Al.(E) In 3 years, Teresa will be younger than Al.

Solution

To pick up one choice that is not true, we first work out the current ages of these three people. Let x be Al’s current age, then Al’s age five years ago was x – 5, Sandra’s age was x – 10, and

Teresa’s age was 21

(x – 10). Therefore, Sandra’s current age is x – 10 + 5 = x – 5, and Teresa’s

current age is 21

(x – 10) + 5 =21

x. That is:

Al: x

Sandra: x – 5

Teresa: 21

x

Now, let’s look at the five choices one by one. Al is 5 years older than Sandra, therefore, A is true.

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For B, Is Teresa twice as old as Sandra? No. Are C, D, and E true? Yes. Therefore, B is the best answer is B.

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7. Doubling

For this kind of question, the key is to find the base number, time period, and final number. A formula to figure out doubling question would be:

2nx(base number) = final number, where n is the number of time period.

Let’s look at an example.

Example (Low-level)

The population of a bacteria doubles every 3 hours. How many hours will it take for the population to grow from 300 to 19,200 bacteria?

(A) 12(B) 16(C) 18(D) 20(E) 24

Solution

Since after each 3-hour period, the backteria polupation will be 600, 1,200, 2,400, 4,800, 9,600, and then 19,200, the total numbers of 3-hour period taken are 6. Therefore, it will take 6 x 3 hours or 18 hours for the population to grow from 300 to 19,200.

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8. Sales Commission

The following formula is useful to solve sales commission problem.

Total Earnings/Income = Base salary + Sales Figure x Commission Rate.

A typical commission question looks like this:

A special award to a contributing employee includes a 5 percent salary increase plus a $600 bonus. If this equals a 10 percent increase in salary, then what is the employee’s salary, in dollars, before the special award?

(A) 600 (B) 1,200 (C) 6,000 (D) 12,000 (E) 18,000

To solve this problem, we first use x as the employee’s original salary, then the increase in the employee’s incomes is $600 + 5%x. Since the increase is 10 percent of x, it follows that $600 +

5%x = 10%x, that is 5%x = $600, therefore x = .000,12$05.0

600$= B is the correct answer.

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9. Decision Tree

Decision tree is a kind of graphical representation of decisions involved in the choice of statistical procedures. Here, we use this definition to describe a type of math question. A typical decision tree question will be as following:

In a certain college, 80 percent of the freshmen (first-year students) lived on campus, and 60 percent of those who live on campus live in dormitory. If 1,200 first year students live in campus dormitory, how many first year students were there in this college?

(A) 2,500 (B) 2,400 (C) 2,000 (D) 1,500 (E) 1,200

There are two ways to solve this question. The one is to express this question in a simple equation. But here, I’d like to introduce a diagramming method, also known as decision tree. Step 1 Divide the first-year students into two groups: one who live on campus (80%) and the other one who live off campus (20%). Step 2 Further divide the subgroup living on campus into the one who live in dormitory (60%) and the other one who live in apartment (40%). Step 3 Since the population for sub-subgroup (who live in campus dormitory) is 1,200, then the

subgroup would be 000,2%60

200,1= .

Step 4 Then the total graduating class would be 500,2%80

000,2=

84

Therefore, A is the correct answer.

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10. Data Interpretation

Sometimes, you are asked to interpret data provided in a table, graphic or charts. This type of question rarely involves significant calculating. Rather, in most cases, you simply interpret the data.

• Before you begin calculating, make sure you understand what you are being asked to do. Notethat some question does not ask you to do calculation.

• Check the proper columns, dates, or figures to determine what information is needed to solvethe problem. Generally, only some not all information is needed to solve each question.

• Be sure that your answer is in thousands, millions, or whatever the question calls for. Do not beconfused with the units.

Example #1(Low-level) Application Fees

The table above shows application fees in 1985 to 1986 application season and in 2003 to 2004 season for School A, B, C, D, and E. For which school shown would the application rate has the greatest percent increase over the two seasons?

(A) School A(B) School B(C) School C(D) School D(E) School E

Solution

The percent increase is %100%10010

1020=×

− for School A, %300%100

151560

=×−

for

School B, %50%10020

2030=×

− for School C, %140%100

5050120

=×−

for School D, and

%200%10030

3090=×

− for School E. Therefore, the correct answer is B.

School A School B School C School D School E

1985-1986 $10 $15 $20 $50 $30

2003-2004 $20 $60 $30 $120 $90

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According to the graph above, what percent of the students for the graduating class seek jobs in Finance?

(A) 16 %(B) 15 %(C) 14 %(D) 13%(E) 12%

Solution

Since the six figures sum up to %100 , 10032)5(151313 =++++++ XX or 11=X .

Therefore, the percent in Finance is %16)%5( =+X . The best answer is A.


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0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

1985 1990 1995 2000

College A College B College C

The table above shows the averaged GPA of admitted students for College A, B, and C from 1985, 1990, 1995 to 2000. Which of the following is closest to the increase in the averaged combined GPA for College A, B, and C from 1985 to 2000?

(A) 0.6 (B) 0.5 (C) 0.4 (D) 0.3 (E) 0.2

Solution

The average GPA for College A, B, and C in 1985 is around 3.1 and that in 2000 is around 3.6. Therefore, 0.5 (3.6-3.1) is closest to the increase from 1985 to 2000. The correct answer is B.

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Chapter 5 Data Sufficiency

Introduction

Unlike the problem solving questions, Data Sufficiency questions do not require you to solve a problem. Rather, the data sufficiency questions require you to find whether the statements provide enough information to solve the question. Many test-takers have difficulty with the Data Sufficiency questions because they never encounter such questions before the GMAT. As a matter of fact, when you become familiar with these types of questions, you will find it easier to solve than the standard problem solving questions.

A sample GMAT Data Sufficiency questions look like this:

What is the value of x ?

(1) 2=y

(2) 3−= yx

The five answer choices after the question do not give five optional values for you to choose. Instead, they ask for the sufficiency of the above statement.

Five answer choices:

(A) if statement (1) alone is sufficient to answer the question, but statement (2) alone is not;

(B) if statement (2) alone is sufficient to answer the question, but statement (1) alone is not;

(C) if statements (1) and (2) taken together are sufficient to answer the question, even though neither statement alone is sufficient;

(D) If either statement alone is sufficient to answer the question;

(E) If statements (1) and (2) taken together are not sufficient to answer the question.

Note:

1. Diagrams accompanying problems agree with information given in the questions, but may not agree with additional information given in statements (1) and (2).

2. All numbers used are real numbers.

Directions:

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In each of the problems, a question is followed by two statements containing certain data. You are to determine whether the data provided by the statements are sufficient to answer the question. Choose the correct answer based up on the statement's data, your knowledge of mathematics, and your familiarity with everyday facts (such as number of minutes in an hour or cents in a dollar).

Section 1 Four Test-taking Strategies

1. Solve the questions in term of sufficiency

Don’t waste your valuable time in solving a problem. Instead, you are required to determine whether information provided in Statement 1 and/or Statement 2 is sufficient to solve the problem. If you are trying to figure out "what is the exact value," or "is this question true or false?", then you are falling into the trap set by the test writers. Keep clear in mind throughout the math section that you are to determine whether each statement provides enough information to solve the question.

2. Memorize the five answer choices

Don’t waste your time in reading the directions or answer choices on the test day. The directions and answer choices never change. Memorize the five answer choices before the test day.

(A) if statement (1) alone is sufficient to answer the question, but statement (2) alone is not;

(B) if statement (2) alone is sufficient to answer the question, but statement (1) alone is not;

(C) if statements (1) and (2) taken together are sufficient to answer the question, even though neither statement alone is sufficient;

(D) If either statement alone is sufficient to answer the question;

(E) If statements (1) and (2) taken together are not sufficient to answer the question.

3. Use Process of Elimination

GMAT Data Sufficiency section is a good place to use POE. Unlike in other sections, in the Data Sufficiency section, you will always be able to eliminate at least two choices. The benefits of eliminating some choices are obvious. When you can eliminate one or more answer choice, the odds of getting right answer is significantly increasing.

Statement Sufficiency Choices Eliminated Odds of getting right answers

(1) alone is sufficient B, C, E 50%

(1) alone is not sufficient A, D 33%

(2) alone is sufficient A, C, E 50%

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(2) alone is not sufficient B, D 33%

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Section 2 Data Sufficiency Trick Questions

As I said at the beginning of the GMAT Math Review, simple math concepts are applied to create GMAT math questions. But the real math questions are not as easy as their original math knowledge because the test writers always create questions which are filled of tricks. If you are careful enough, you are probably fooled by them.

However, the tricks used in Data Sufficiency questions are few and can be identified one by one in practicing GMAT questions. In the following passages, we will introduce the two major tricks that frequently appear in math test. Before we start, let me first summarize the general strategies to sidestep any trick questions.

• Never question the validity or soundness of the original question.

• Consider each statement individually.

• Mechanically progress through the two statements.

• Never solve the problem, only determine the sufficiency.

If you are able to stick to the above four approaches, you will not be fooled by the test writers.

1. Time-Wasting Trick

The test writers make these questions to make you waste time so that you are unable to finish the test on time. The test writers intentionally load tricks on Data Sufficiency questions. The unskilled test-takers are often fooled by these traps. A sample trick question looks like this:

Is 0132 >+− xx ?

1) x is equal to 3

2) x is a negative

Some test-takers are anxious about the outcome. That is whether the expression is greater than zero. When they read the first statement x = 3, they simply plug the value of 3 in order to determine the final result. You may say that you would not be fooled here. However, the Data Sufficiency is neither tested in a separate section nor in consecutive sequence. Rather, it randomly appears with the problem solving question. You are not informed in advance whether the next question is a problem solving question or a data sufficiency one. If you encounter a problem solving question right before the above question, you are very likely to be fooled at this one.

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2. The “C for A” Trick

In this trick, the correct answer should be A, but many unskilled test-takers may choose C. In other words, combining Statement 1 and 2 can solve the problem, but Statement 1 alone can solve it. Here, the trap is the choice of C. A sample “C for A” trick question looks like this:

What is the value of yx 2+ ?

(1) 3763 =+ yx

(2) 2=y

Here, the question asks for the value of yx 2+ . After a first glance, you may consider it by

combining the two statements into simultaneous equations. In this way, you can definitely get values of both x and y , and then plug the values into the expression of yx 2+ to solve the

question. That’s to say Statement 1 and Statement 2 together can solve the problem.

However, after a closer look at Statement 1, you will find it provides enough information to answer the original question: the value of yx 2+ . By factoring, the equation of 3763 =+ yx can be

turned into: 37)2(3 =+ yx . That is 3

372 =+ yx . Therefore, the figure of 3

37 is the value of

yx 2+ . You may find it a little amazing. But actually, Statement 1 is sufficient to solve the problem.

Therefore, the correct answer should be A, not C.

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Practice SECTION 1 30 Minutes 25 Questions

1. If the revenue generated by Company X was $12 million in 1998, what was the revenuegenerated by Company X in 1996?

(1) The revenue generated by Company X increased by 12 per year from 1996 to 1998.(2) The revenue generated by Company X in 1998 was 1.31 times its revenues in 1996.

2. What is the original price of a computer?

By what percent has the price of an overcoat been reduced?

(1) The computer was sold at $420.(2) The price of the computer on sale was $30 less than its original price.

3. What is the length of a rectangular?

(1) The area of the rectangular is 397 square feet.(2) The perimeter of the rectangular is 62 feet.

4. What is the value of x23

?

(1) 64 =−x(2) 0>x

5. Is Bill older than Al?

(1) Ted is not as old as Al.(2) Bill is older than Ted.

6. What is the perimeter of the parallelogram above?

(1) AD > DC(2) AB + BC = 12

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7. Is a2 equal to ab?

(1) 0≠a (2) ab =

8. Is 3.2 the averaged GPA in the class with 60 students?

(1) Exactly 29 students got GPA above 3.2 and the other 29 got GPA below 3.2. (2) The other two students got GPA of 3.2 respectively.

9. What was Athlete M’s average running speed in meter per second during a 100-meter sprint?

(1) It took Athlete M 7 seconds in the sprint. (2) The maximum speed was 13 meter per second and the minimum speed was 6 meters per

second during this sprint.

10. Is ∆ABC is a right triangle?

(1) o90=∠+∠ BA

(2) o90=∠C

11. Is n an integer greater than 4?

(1) 3n is a positive integer.

(2) 3n

is a positive integer.

12. What is the area of a right triangle with a hypotenuse of 5 feet?

(1) The length of one side is 3. (2) The length of the other side is 4.

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13. What is the ratio of the area of a square to the area of a circle?

(1) The perimeter of the square is twice the circumference of the circle.(2) The length of each side equals the radius of the circle.

14. Is a – b > c – 2?

(1) a > b and c < 2(2) a – 1< b + c – 1.

15. How old is Peter today?

(1) Two years ago, the age of Ted was twice that of Peter.(2) Ted is 23 years today.

16. What is the area of the triangle above?

(1) AD = 3.(2) BD = 5.

17. What is the ratio of the area of theΔABC to the area of the Circle O?

(1) The area of Circle O is π36 .

(2) The area of ΔABC is2

15.

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18. Isba

an integer?

(1) b is a prime number . (2) a is the product of 2 and a prime number.

19. What is the value of k?

(1) 0232 =+− kk

(2) 11

131

<+

<k

20. Is quadrilateral ABCD a square?

(1) AB = BC = CD = DA

(2) o90=∠A

21. If Y = 362 – X and X is a positive integer greater than 10 and less than 20, what is the value of X?

(1) Y is a prime number. (2) Y is not divisible by 3.

22. A trader sold a product at a price that would make 20 percent profit. What is the original cost at which the trader purchased the product?

(1) The original cost is $12.05. (2) The selling price is $15.30.

23. If k is an integer between 100 and 110, exclusively, what is the value of k?

(1) k is divisible by 3. (2) k is divisible by 7.

24. Is x2 – y2 an integer?

(1) x2 + y2 is an integer. (2) x2 > y2.

25. If each of the four sides of a rectangular is reduced by 3 feet, what is the area after changes were made?

(1) The area of the rectangular before changes were made is 28. (2) The area of the rectangular has been reduced by 30 percent.

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1. Who drives faster, Driver A or Driver B?

(1) Drive A drives at a speed of 90 miles per hour. (2) The difference between their driving speeds is 10 miles per hour.

2. How many months can John use Company X’s email account?

(1) Company X charges $5.4 per month. (2) John has total budget of $108 for Company X’s email account.

3. Line AB, Line CD, and Line EF are three lines in a plane. Does Line AB parallel to Line EF?

(1) Line AB parallels to Line CD. (2) Line CD is perpendicular to Line EF.

4. Dently College accepted 32 percent of its applicants in 2003. How many students attend Dently College in 2003?

(1) Exactly 18 percent of applicants who was accepted did not attend Dently. (2) There were total of 3,200 applicants to Dently College.

5. What is the value of k?

(1) 3k > 0 (2) k < 2

6. Inside a square is a circle. If the circle has the greatest possible area, what is the difference between the area of the circle and the area of the square?

(1) The length of each side of the square is 4. (2) The area of the circle is π4 .

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7. What is the average of the three numbers a, b, and c?

(1) The average of a and b is 5. (2) The average of a and c is 4.

8. Jennifer purchased Brand X pencil at $0.30 each and Brand Y pencil at $0.25. If Jennifer purchased 12 Brand X pencils, how many Brand Y pencils did Jennifer purchase?

(1) Jennifer purchased a total of 30 pencils. (2) For the 30 pencils, Jennifer had paid $8.50.

9. What fraction of Teresa’s salary was from commission last year?

(1) Teresa’s salary has increased by 5.6% since last year. (2) Last year Teresa received a commission of $6,000 and a salary of $23,000.

10. For number x, y, and z, 21

=−z

xy. What is the value of

yzy 2−

?

(1) xy 2=

(2)3xz =

11. How old is Stephen now?

(1) Stephen is twice old as he was eight years ago. (2) Stephen was 8 years old eight years ago.

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12. What is the value of B?

(1) The average of A, B, and C is 3. (2) The average of A and C is 2.

13. If x and y are integers, is the number yx 23 − divisible by 3?

(1) y is divisible by 6.

(2) 2

3 yx − is divisible by 3.

14. What is the volume of a circular cylinder?

(1) The surface of the circular cylinder is 20. (2) The radius of each base is 2.

15. If xz≠0, what is the value of ?2

)2(3

2

zxxyzzyx

−

−

(1) x = - 3 (2) z = 5

16. What is the area of a rectangular?

(1) The length of a diagonal is 5. (2) The length of the short side is 3.

17. If an integer x is divisible by 3, is x divisible by 2?

(1) x is divisible by 6 (2) x is an even.

18. E is a point on side AC of ΔABC. Is ΔABC is an isosceles triangle?

(1) BE┴AC and AE = EC. (2) The area of triangular region ABE is equal to the area of triangular region EBC

19. If a is an integer, what is the value of a?

(1) a < b (2) 1 < b < 2

20. If the three vertical points of triangle ABC are on the Circle O, is triangle ABC a right triangle?

(1) The length of BC equals the length of a diameter of Circle O.

(2) 222 ABACBC =−

21. Which is greater, 2x

or 2x

?

(1) 22 yx > (2) x and y are both integers

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22. Is 22 yx > ?

(1) yx < . (2) x and y are both positive numbers.

23. If x is in Set S, then 2x is also in Set S. Is 16 in Set S?

(1) 2 is in S. (2) 256 is in S.

24. If the perimeter of triangular region ABC is 33, what is the area of ABC?

(1) The length of one side of ABC is 11. (2) ABC is a equilateral triangle.

25. If a and b are consecutive odd integers, and a > b, what is the value of b?

(1) a = 1. (2) ab < 0.

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1. What is the length of a rectangular?

(1) The sum of the length and the width of the rectangular is 29.(2) The length of the rectangular is 4 more than the width of the rectangular.

2. If Peter earned a commission that is 20 percent of his salary, what is Peter’s salary?

(1) Peter’s salary is $8,000 more than his commission.(2) The commission rate is 5 percent of his total $150,000.

3. What was Ted’s average grade for all of his courses?

(1) His grade in math was 92, and his grade in business was 79.(2) His grade in social science was 67.

4. If x = 1, what is the value of y?

(1) x – z = 3y(2) x – 3y = 1

5. A retailer wishes to transport his goods of 10 tons in weight from Town A to Town B that is 120miles in distance. If the retailer can choose to hire a truck company that charges by ton or by mile,which way costs less?

(1) The truck company charges $50 per ton.(2) The truck company charges $6.0 per mile.

6. If x + y is a prime, what is the value of x + y?

(1) x > y > 0(2) x and y are both prime.

7. What is the area of ΔABC?

(1) o60=∠A(2) BC = 3

8. Andrew has total of $160 bills, some of which are in $20 denomination, and the other are in $10denomination. How many 20-dollar bills does Andrew have?

(1) There are total of 9 bills.(2) The number of 20-dollar bills is 5 more than the number of 10-dollar bills he has.

9. City B is somewhere between City A and City C. It takes 40 minutes for a person to drive fromCity A to City C, a distance of 120 miles. What is the distance between City A and City B?

(1) The road from A to B takes 12 minutes.(2) The speed during distance of AB is 180 miles per hour.

10. What is the perimeter of an equilateral triangle?

(1) The length of each side is 7(2) The area of the triangle is 9.

11. Is k an even integer?

102

(1) k is the square of an integer.(2) k is the cube of an integer.

12. If Bill got 5 more admission offers than Al, how many offers did Bill get?

(1) If Bill got 2 more offers, the number of offers Bill got will be twice that of Al.(2) The number of offers Bill got is more than 5.

103

13. In the first round, a certain applicant was accepted exactly by 60 percent of schools applied to.What is the combined acceptance rate for the first and second rounds?

(1) In the second round, the applicant applied to 5 schools.(2) The applicant was accepted by all schools applied in the second round.

14. If a and b are integers, is a + b + 1 divisible by 3?

(1) When a is divided by 3, the remainder is 1.(2) When b is divided by 3, the remainder is 1.

15. Is x2 > 1?

(1) |x| < 1

(2) |x| <||

1x

16. Student A, Student B, and Student C are to share the cost of an apartment equally, what is thecost per student?

(1) If there had been 1 more student and all 3 had shared the cost equally, the cost per memberwould have been $120 less.

(2) There are total of 4 bedrooms.

17. Rectangle ABCD is inscribed in a circle as shown above. What is the circumference of thecircle?

(1) The area of the rectangular is π100 , a number that is twice the area of the rectangular.(2) The length of the radius of the circle is 7.

104

18. What is the area of triangle ABC?

(1) o60=∠A(2) BC =3.

19. Is a > b?

(1) 22 ba =(2) ab > 0

20. A circle is inscribed in a square. What is the area of the circle?

(1) The length of a side of the square is twice that of a radius of the circle.(2) The perimeter of the square is 3.

21.If x can be any number from 0 to 9, what is the value of 1.x 5?

(1) 1.x 5 is divisible by 5.(2) x is a prime number.

22. Last year, Andrew got commission and salary of $24,567. What was Andrew’s salary last year?

(1) Last year, the amount of Andrew’s commission is 25 percent of that of Andrew’s salary.(2) If the commission rate were reduced by 2 percent, Andrew’s total earning (commission and

salary) will be $ 2,860 less.

105

23. Is a greater than b?

(1) a-b+3<0(2) ab>0

24. Is quadrilateral ABCD a rectangle?

(1) The measure of ∠ABC is 90o

(2) The measure of ∠CDAR is 90o

25. If y is an integer, is 222 xy − an integer?

(1) 222 xy + is the square of an integer.

(2) 02 22 =+ xyAnswer

Section 1: DCCAE BBEAD EDDDC DDECC EDCEA

Section 2: ECCCE DEDBB DDDDA CAACD EAABD

Section 3: CDEBC BEDAD EAECD AAECB CDAEB

The End of GMAT Math Review

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GMAT Math Review - gmatcat.com Introduction to GMAT Math After a 5-minute break, you are assigned to the Quantitative section, which consists of 37 multiple-choice questions