We are extremely happy that you have chosen to study with ... · Sentence Correction/Critical Reasoning/Reading Comprehension Total Time: Approximately 3 hours, 50 minutes Content

Dear Veritas Prep student:

We are extremely happy that you have chosen to study with the 2012 version of the Veritas Prep lesson books, which we firmly believe will provide you with the best GMAT preparation available anywhere. As with the first release of anything new, however, these books do contain a handful of typos.

The attached pages include reprints of each page on which your books contain a typo. These corrected pages will allow you to complete your books, by simply printing the file and inserting the replacement pages according to their book and page numbers. To use this file, please take note of the book-by-book dividers within, as all of the Arithmetic fixes will come after the Arithmetic cover page, the Algebra fixes will come after the Algebra cover page, etc.

We thank you for your patience with these errors and replacements and hope that they don’t get in the way of the true value of the books that you’re using. We’re confident that these lessons represent your best opportunity to maximize your GMAT score, and we’re honored to be a part of your study program.

Happy studying!

The Veritas Prep Academic Team

Foundationsof GMAT Logic

Foundations | 9

GMAT OVERVIEW

Format of the GMAT

Sections Questions Allotted Time

Interactive Tutorial 0

Analytical Writing Assessment 1 30 minutes

Integrated Reasoning 12 30 minutes

Optional break 8 minutes

Quantitative 37 75 minutesProblem Solving/Data Sufficiency

Optional break 8 minutes

Verbal 41 75 minutesSentence Correction/Critical Reasoning/Reading Comprehension

Total Time: Approximately 3 hours, 50 minutes

Content on the GMAT

Analytical Writing Assessment The Analytical Writing Assessment (AWA) is the first section of the GMAT, and provides schools with a writing sample with which to gauge your communication ability. In the 12th lesson of this course you will learn strategies to master the AWA efficiently, saving mental stamina for the remainder of the test while posting a high score.

GMAT OVERVIEW

Format of the GMAT

Arithmetic

Arithmetic | 39

SKILLBUILDER

SKILLB

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SKILLBUILDER

Multiplication and Division Decimal Drills

1. Which is greater: 102.4 ÷ 25.6, or 1.25•3.25?

2. Whichisgreater:2.25÷0.15,or7.501•2.22?

3. Whatis25.6•0.16?

4. What is 8.1 ÷ 2.7?

5. What is 24.36 ÷ 1.2?

6. What is 222.2 ÷ 1.1?

7. Whichisgreater:100,or24.445•3.98125?

8. Whichisgreater:2.25•0.4,or(0.3)2?

9. Which is greater: (0.3)4 or (0.9)2?

10. Which is greater: 1.234 ÷ 0.01234, or (1.234)3?

Arithmetic | 55

SKILLBUILDER

SKILLB

UILD

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SKILLBUILDER

Adding and Subtracting Fractions Drill

1. What is the sum of 14 +

13 +

16 ?

2. What is the result of 29 +

14 –

16

+ 1

18?

3. What is the result of 25 –

110 +

34

– 3

20?

Lowest Common Denominator

When adding and subtracting fractions you can simplify calculations and save time by finding the lowest common denominator. To do so, recognize that the process of finding the lowest common denominator (LCD) is the same as finding the least common multiple of the denominators.

As an exercise, find the lowest common denominator of the fractions:

What is 16 +

18 +

19

+ 1

10 ?

Arithmetic | 56

Adding and Subtracting Fractions Drill Solutions

1. 34

2. 1336

3. 9

10

LCD Solution

The least common multiple of 6, 8, 9, and 10 is:

6=2•3

8=2•2•2

9=3•3

10=2•5

LCM=2•2•2•3•3•5=360

So the problem can be rewritten as:

60

360 + 45

360 + 40

360 + 36

360 = 181360

Arithmetic | 65

SKILLBUILDER

SKILLB

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Factors/Multiples/Divisibility Drill

Please do the prime factorization for each of the following integers:

1. 56

2. 46

3. 36

4. 42

5. 24

6. 81

7. 32

8. 74

9. 124

10. 343

11. 707

12. 10,000

13. What are all the positive factors of 56?



Arithmetic | 68

Greatest Common Factor

The greatest common factor (GCF) is the largest number that divides evenly into each of a given set of numbers.

Calculating the GCF is similar to calculating LCM. The first step is to express each number as the product of its prime factors. With the GCF, however, your job is to find exactly what is common between them. For example:

What is the greatest common factor of 10 and 20?

10: 2 • 5

20: 2 • 2 • 5

Thepairingof2•5iscommontoboth,so10isthegreatestcommonfactor.

What is the greatest common factor of 12 and 18?

12: 2 • 2 • 3

18: 2 • 3 • 3

Eachof12and18hasacommon2•3,so6isthegreatestcommonfactor.

N O T E : 2 and 3 are each “common factors,” but for the greatest common factor, you need to find all of the matching prime factors.

What is the greatest common factor of 36, 72, and 162?

36:2•2 • 3 • 3

72:2•2•2 • 3 • 3

162: 2 • 3 • 3 •3•3

Thegreatestcommonfactoris2•3•3=18.

Arithmetic | 82

2. Percentages must be taken “of” a value.

Percentages don’t exist on their own; they must represent a fraction of an existing value. Perhaps the single greatest mistake that examinees make with percent questions is that they often take the percentage of the incorrect value. When taking a percentage, ask yourself the question “of what?” so that you organize your thoughts appropriately. Consider this example:

Example

What is 25 increased by 20%?

25 + 20

100 …

The answer clearly isn’t 25.2, so what comes next?

The 20% must be a fraction of a value; in this case, it’s a fraction of itself.

25 increases by 20% (of itself)

25 + 20

100 (25) =

25 + 15 (25) = 25 + 5 = 30

Percent Increases and Decreases

Increases by + Decreases by –

Examples

10 increases by 20% 20 decreases by 20%

10 + 20

100 (10) 20 – 20

100 (20)

10 + 200100 20 –

400100

10 + 2 = 12 20 – 4 = 16

Arithmetic | 97

LESSON

Division

In the previous subtraction and multiplication examples, you have seen how the authors of the GMAT can take a simple operation that you have performed nearly all your life and make it confusing. Strategically, it’s often helpful to perform that same operation with small numbers in a “parallel problem” so that you can isolate steps and relationships in those operations.

Traditionally, students have struggled the most with this abstract/reverse application of standard operations when division has been involved. Consider this problem:

3. When m is divided by n, the remainder is 14. If mn

= 65.4 , what is the value of n?

(A) 14

(B) 27

(C) 35

(D) 42

(E) 45

SECTION 1: ARITHMETIC CALCULATIONS

Division1

A r i t h m e t i c | 111

LESSON

Divisibility Drill

Is 21,865,014 divisible by:

3?

6?

9?

Reduce 321219 :

Solutions to Divisibility Drill: Yes, Yes, Yes,

10773

SECTION 2: FACTORS, MULTIPLES, AND THE NUMBER LINE

Divisibility Rules2

Arithmetic | 128

Review from Skillbuilder: LCM and GCF

Least common multiple: the smallest non-zero number that is a multiple of two or more numbers

Greatest common factor: the largest number that divides evenly into each of a given set of numbers

Finding the LCM or GCF starts with the prime factorization of the numbers involved, as you will see in the drills below.

Example

Find the LCM of 56 and 20 .

Primefactorizationof56=2•2•2•7,whichcanbewrittenas23 •7.

Primefactorizationof20=2•2•5,whichcanbewrittenas22•5.

To determine the least common multiple (as with lowest common denominators, where you are finding the lowest common multiple of the various denominators), you must pick out each unique prime integer (in this example: 2, 5, 7) from the prime factorization of the numbers involved and represent each of those numbers with the highest exponent that appears in any one of the numbers involved (in other words, greatest number of times they appear). The least common multiple in this example is 23•5•7=280.

Example

Find the GCF of 110 and 66 .

Primefactorizationof110=2•5•11

Primefactorizationof66=2•3•11

To determine the greatest common factor, you must pick out what is exactly in common from the sets of prime factors. Here 110 and 66 share only one 2 and one 11, so the greatestcommonfactoris2•11=22.

Arithmetic | 175

HOMEWORK

HO

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OR

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HOMEWORK

Calculations

27. Which of the following is greater than 34

?

(A) 58

(B) 3750

(C) 1724

(D) 79

(E) 711

Arithmetic | 192

44. (.0032) (.028) (100)

(.008) (.07) (50)

(A) .16

(B) .32

(C) 3.2

(D) 32

(E) 64

Arithmetic | 202

54. For positive integers a and b, ab

= 0.6 . Which of the following CANNOT be the value of a?

(A) 42

(B) 63

(C) 105

(D) 137

(E) 174

Arithmetic | 209

HOMEWORK

HO

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OR

K

61. If x is y percent of z, is y greater than 100?

(1) zx

= 0.8

(2) x decreased by 20% is equal to z.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked

(C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient

(D) EACH statement ALONE is sufficient to answer the question asked

(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed

Arithmetic | 221

HOMEWORK

HO

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OR

K

73. Which of the following values is greatest?

(A) 11% of 89

(B) 22% of 78

(C) 33% of 67

(D) 44% of 56

(E) 55% of 45

Arithmetic | 222

Ratio

74. An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of real orange juice does it contain?

(A) 2

100

(B) 2

68

(C) 2

32

(D) 32

100

(E) 64

100

Arithmetic | 242

96. If x and y are integers and 450x = 120y, which of the following must be an integer?

(1) xy60

(2) 15x4y

(3) 4x

15y

(A) I only

(B) II only

(C) I and III only

(D) I and II only

(E) I, II, and III

Arithmetic | 244

98. What is the value of integer x?

(1) The lowest common multiple of x and 16 is 48.

(2) The greatest common factor of x and 16 is 4.






Arithmetic | 247

HOMEWORK

HO

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OR

K

101. For nonnegative integers x and y, 30x = 49y. Which of the following must be true?

(1) x > y

(2) y

10 is an integer.

(3) √—

x7

is an integer.

(A) I only

(B) II only

(C) III only

(D) I and II

(E) II and III

Arithmetic | 248

102. If a, b, and c are positive integers and a6 +

b5 =

c30

, is c divisible by 5?

(1) b is divisible by 5.

(2) a is even.






Arithmetic | 269

Answer key

answer key

Homeworklessons

1. B

2. D

3. C

4. A

5. B

6. C

7. D

8. E

9. E

10. E

11. B

12. C

13. C

14. B

15. C

16. B

17. E

18. B

19. E

20. C

21. B

22. B

23. E

24. E

25. D

26. D

27. D

28. C

29. A

30. C

31. D

32. C

33. C

34. D

35. D

36. B

37. D

38. D

39. A

40. B

41. E

42. C

43. B

44. B

45. B

46. B

47. B

48. D

49. C

50. A

51. D

52. A

53. A

54. D

55. B

56. A

57. C

58. A

59. E

60. E

61. D

62. C

63. C

64. B

65. A

66. D

67. D

68. A

69. E

70. C

71. D

72. C

73. E

74. E

75. D

76. D

77. B

78. E

79. C

80. A

81. D

82. D

83. C

84. A

85. B

86. D

87. B

88. D

89. C

90. C

91. B

92. E

93. D

94. D

95. E

96. D

97. C

98. C

99. D

100. A

101. B

102. A

103. D

104. C

105. E

106. B

107. B

108. E

109. E

110. C

111. D

112. E

113. B

114. E

115. B

116. C

117. C

118. E

119. C

120. D

Algebra

Algebra | 17

SKILLBUILDER

SKILLB

UILD

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Multiply by 1 Drills

1. What is

4x3

5x9

?

2. What is (5x

12 +

3x4)

x6

?

3. What is

y + 3y2

14

?

4. Simplify: n + 3n2 –

7n3 .

5. What is

x + 5x6

14

?

Algebra | 18

Solutions to Multiply by 1 Drills

1. Multiply the top and bottom by 9: (4x3

5x9

) ( 99 ) = 12x

5x =

125

2. Multiply the top and bottom by 12: (5x

12 + 3x4 )

x6

(1212) = 5x + 9x

2x = 14x2x = 7

3. Multiply the top and bottom by 4:

y + 3y2

14

= 4y + 6y

1 = 10y

4. Multiply all terms by 66

: 6n + 6(3n)

2 – 6(7n)

3

6 = 6n + 9n – 14n

6 = n6

5. Multiply the top and bottom by 12:

x + 5x6

14

(1212)

= 12x + 10x3

= 22x3

Algebra | 47

SKILLBUILDER

SKILLB

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Roots

While exponents ask you to multiply a value by itself a number of times, roots do exactly the opposite. Roots, instead, ask: What number multiplied by itself (a certain number of times) will produce this value? The most common root, the square root, asks: What number, squared, will produce this value? Accordingly, the square root of 16 asks for a number that, when squared, will produce 16. That number is 4, as you see in the first example below. Even though all positive numbers technically have two square roots (a positive one and a negative one) the symbol √

—– represents only the “principal”—that

is, the positive—square root. Furthermore, the word "principal" is often omitted, so that even when we say just “square root,” we are referring to the positive square root only.

Examples

If x2 = 16, then x = 4 OR –4 because 42 = 16 AND (–4)2 = 16 BUT

√—–16 = 4 only

√

——425 = 2

5 because ( 2

5 )2=

425

√—––––––16•36 = 42 •62because(4•6)2 = 42•62=16•36

√—–x12 = x6 because (x6)2 = x6 •2 = x12

(√—– x )2 = x by definition. The process of finding the square root and then squaring it is analogous to adding a value to x and then turning right back around and subtracting that value again, or to dividing x by some value and then immediately multiplying the result back by that value. Just as addition and subtraction “undo” each other (i.e., cancel each other’s effect) and just as multiplication and division do as well, so, too, do taking square roots and squaring precisely counteract each other. Note here that only positive values of x have real square roots, since it is impossible to multiply any real numberbyitselftoyieldanegativenumber(negative•negative=positive;positive•positive = positive).

√—–x2 = |x|, since regardless of whether x itself is positive or negative, squaring it will yield a positive number, and then the √—–

will instruct us to take specifically the positive square root of that number.

N O T E : While there are many roots, which we will cover in the coming pages, the standard is the principal square root, and the radical sign, √

—–, with no other notations,

represents precisely that.

√—––––––

Algebra | 50

Roots Drills

1. Simplify √—–2 + √

—–8 + √—–32

2. What is √—4

9 ?

3. What is √—–––√—–16 ?

4. Simplify √—–27 + √

—–75 + √—–48

5. What is 3√—–82 ?

6. What is √—–8 + √

—–18√—–2 + √

—–32?

7. If √—–x + √

—–50 = √—– –16x + √

—–8 , what is x?

8. What is √—36

225 ?

9. Simplify √—– –343 + √

—– –112 + √—– –63

10. What is √—– –363 + √

—––300√—–75 + √

—–12?

Algebra | 56

Solutions to Inequalities Drills

1. 12 + 3x > 15 + x

x > 32

2. 0 > x and – 1x > 17

– 1

17 < x < 0

3. 13 – 4x < 5 + 2x

x > 43

4. 12 + y > 4 – x and 12 – 3y > 9 x > –8 – y ; y < 1

5. 14 – x < y < 2x – 1

x > 14 – y ; x > y + 1

2

Algebra | 57

SKILLBUILDER

SKILLB

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Absolute Value

Absolute value, denoted by the symbol |n|, gives a quantity’s “distance from zero.”

-3 -2 -1 0 1 2 3

On the number line, 3 and –3 are equally far from 0 on each side. Therefore, they have the same absolute value.

When using absolute value in algebraic terms, it is important to note that there are two possible “input” values behind every absolute value: If a variable has an absolute value of 5, for example, the variable could be either 5 or –5, because each of those values is five units away from zero:

|x| = 5 x could be 5 or –5, because |–5| = 5 and |5| = 5

Accordingly, any equation in which an expression within absolute value bars is equated to some numerical value needs actually to be seen as two possible equations with no absolute value bars: one that equates the quantity that was within the absolute value bars to the numerical value given, and another that equates the quantity that was within the absolute value bars to the opposite of the numerical value given.

|x + 5| = 10 x + 5 = 10 OR x + 5 = –10

−5 − 5x = 5 OR

−5 − 5x = −15

|2x – 3| = 7 2x – 3 = 7 OR 2x – 3 = –7

+3 +32x = 10

+3 +32x = −4

x = 5 OR x = −2

Algebra | 115

LESSON

The Quadratic Formula: You Won’t Use it!

For quadratics in the form ax2 + bx + c = 0, you can find the solutions employing the quadratic formula:

x = – b ± √—–––––––

b2 – 4ac

2a

But recognize this: That formula is a long one to memorize if you don’t already have it committed to memory based on your prior math experience. It’s a bit tedious to employ. And the GMAT isn’t apt to require memorization for memorization’s sake. On the GMAT, two options should present themselves long before you ever have to employ the quadratic formula:

1. You can factor the equation using the method from the previous page.

2. You can plug in answer choices as potential values of x, probably much quicker than you could calculate using the quadratic formula.

Simply put, if you already have the quadratic formula memorized and in the “active” part of your brain, there’s a small chance that it could come in handy. If not, look for the opportunity to either factor of back-solve, and don’t worry about memorizing this formula.

SECTION 3: QUADRATICS

The Quadratic Formula: You Won’t Use it!3

Algebra | 135

LESSON

SECTION 5: INEQUALITIES

Inequalities with Absolute Value5

Inequalities with Absolute Value

While there are two approaches to these types of inequality problems, the most reliable is to create two separate inequalities that consider the two possible scenarios given by the absolute value sign (positive/zero and negative). Consider the following two examples:

|x| < 5

In this case, the absolute value of x is less than 5, and there are two possible scenarios: Either x is positive/zero or negative. If x is positive or zero, then the following case is true: x < 5. However, if x is negative, then the following inequality is true: -x < 5 or, after manipulating by –1, x > –5.

Taking them together, we know that x < 5 and x > –5. This can also be written as –5 < x < 5.

Here is the visual representation of that on the number line:

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6

|x| > 5

In this case the absolute value of x is greater than 5, and again there are two possible scenarios:

Either x is positive or negative. If x is positive, then the following case is true: x > 5. However, if x is negative, then the following inequality is true: –x > 5 or, after manipulating by –1, x < –5. Taking them together, we know that x > 5 or x < –5, which cannot be written as one statement. Here is the visual representation of that on the number line.

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6

Summarized in simplistic terms, absolute values with inequalities can be considered in the following manner:

|x| < y means −y < x < y

|x| > y means x > y or x < –y

Algebra | 189

HOMEWORK

HO

MEW

OR

K

56. If5•x√—––125 =

1

51x

, then x =

(A) –4

(B) –1

√—2

(C) 0

(D) 1√—2

(E) 1

Algebra | 226

93. In the sequence 1, 2, 2, …, an, …, an = an-1•an-2. The value of a13 is how many times the value of a11?

(A) 2

(B) 23

(C) 232

(D) 264

(E) 289

Algebra | 227

HOMEWORK

HO

MEW

OR

K

94. The infinite sequence a1, a2, …, an, … is such that a1 = 7, a2 = 8, a3 = 10, and an = an – 3 + 7 for values of n > 3. What is the remainder when an is divided by 7?

(1) n is a multiple of 3.

(2) n is an even number.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Geometry

Geometry | 52

Solutions

1. What is the surface area of a cube with volume 8 cm3?

24 cm2

2. What is the volume of a cylinder with radius 3 and height 8?

72π

3. What is the surface area of a cylinder with radius 3 and height 8?

66π

4. What is the greatest distance between any two points on a cylinder with radius 4 and height 8?

8 √__2

Geometry | 109

LESSON

SECTION 6: 3-DIMENSIONAL SHAPES

The only other major geometry concept area that is tested on the GMAT is 3-D shapes. While that may seem like an entirely new category, the GMAT only explicitly tests 3-D shapes that are based on everything you just learned from the previous questions. The only difference is that one dimension is added. You are responsible, for example, for a cube:

But if a question asks about a cone or sphere, for example, any necessary formulas (e.g., volume) will be specifically provided to you. 3-D shape problems require you to think spatially and to leverage what you know about the 2-D shapes that comprise them. Over the next few pages you will see multiple examples of how the GMAT tests 3-D shapes.

ESSE NTIAL PRO PE R TI ES O F COM MO N 3- D SHAPES

A cube: a square with a depth component.• Volume = (side)3

• Surface area = 6(side)2

A cylinder: a circle with a height component.• Volume=areaofacircle•heightofthecylinder• Surface area = area of two circles + area of the

rectangle formed by the height of the cylinder and the circumference of the circle.

A rectangular box: a rectangle with a height component.• Volume=length•width•height• Surface area = Add the areas of all six sides.

SECTION 6: DIMENSIONAL SHAPES

Essential Properties of Common 3-D Shapes6

Geometry | 124

LE AR N ING BY DO ING Geometry Is Often “Surprisingly Sufficient”

To start on this problem, it bears mentioning that, as geometry on the GMAT is really the art of leveraging known assets, you should be extremely leery of answer choice E in a Data Sufficiency context. Answer choice E doesn’t mean “I can’t solve this problem,” but rather “this problem cannot be solved”—not by Stephen Hawking, Will Hunting, or Chad Troutwine himself.

By the same token, you don’t really have to solve the problem to know that a statement (or both statements) is sufficient; if you know that, given the proper amount of time, patience, and facility with algebra, one could solve the problem; that proves sufficiency well enough. On Geometry-based Data Sufficiency problems, you often won’t actually do the math, but just be sure you can reach the finish line.

To attack this problem efficiently, first consider the given information. The fact that the rectangle is perfectly inscribed within a circle means that the line CB is the diameter of the circle (which is a key component of finding the arc length), and that multiplying anytwoadjacentsidesoftherectangle(forexample,AC•AB)willgiveyoutheareaof 40. You should also know that the diameter line, CB, is the hypotenuse of a right triangle with sides of AC and AB, and of a right triangle with sides of BD and CD. So the Pythagorean Theorem is in play.

Statement (1) provides the sum of the measures of angles m and n. But also important is the fact that, because triangles CAB and CDB are congruent, angle m equals angle n. (You could also prove this by noting that lines CD and AB are parallel and intersected by the same line CB.) So if m + n = 60 and m = n, then m = 30 and n = 30. Because both triangles are right triangles, then angles j and k must each equal 60. And knowing that you’re dealing with 30-60-90 triangles, you can determine that line AC (across from the 30-degree angle m) and line AB (across from the 60-degree angle k) have the ratio x: x√—

3 .SoAB•AC=40(theareaoftherectangle)andAB=AC•√—3 . This should be

enough for you; you know that you will be able to determine the lengths of either of those sides, and then complete the ratio x : x√—

3 :2x ratio of 30-60-90 triangles to find the diameter CB, with which you can find the circumference of the circle. Knowing the circumference, you can then use the inscribed angle j to find the proportion of arc CD to the overall circumference. Because this is a Data Sufficiency question, the actual numbers do not matter! Just that you can find a direct path to finding a solution if you did want one is enough.

Geometry | 137

HOMEWORK

HO

MEW

OR

K

28. A pizza with diameter of 12 inches is split into eight equally sized pieces. Four non-adjacent pieces are removed. What is the perimeter AOBCODEOFGOHA of the pizza now, including the inside edges of the slices?

(A) 48π + 48

(B) 24π + 48

(C) 24π + 24

(D) 6π + 48

(E) 6π + 24

Data Sufficiency

Data Sufficiency | 108

B C

A D

25. In parallelogram ABCD above, what is the measure of angle ADC?

(1) The measure of angle ADC is greater than 90 degrees.

(2) The measure of angle BCD is 70 degrees.







HOMEWORK

HO

MEW

OR

K

32. Is x > y?

(1) x = y + 2

(2) x2

= y – 1







HOMEWORK

HO

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Q R

P T S

78. In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ are parallel?

(1) Length PQ = Length PR

(2) Line TR bisects angle PRS.







HOMEWORK

HO

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88. What is the value of x + y?

(1) 4x2 – 4y2

2 ( x + y ) = 2x – 2y

(2) 3x + 2y = 24

(F) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked

(G) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked

(H) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient

(I) EACH statement ALONE is sufficient to answer the question asked

(J) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed


ANSWER KEY

ANSWER KEYANSWER KEY

1. B

2. C

3. D

4. A

5. E

6. E

7. C

8. A

9. C

10. B.

11. E

12. C

13. D

14. A

15. A

16. A

17. D

18. C

19. E

20. B

21. E

22. B

23. E

24. C

25. B

26. B

27. D

28. E

29. D

30. D

31. C

32. A

33. A

34. E

35. A

36. A

37. B

38. E

39. C

40. E

41. C

42. D

43. C

44. C

45. D

46. D

47. D

48. E

49. B

50. C

51. E

52. D

53. A

54. C

55. B

56. A

57. D

58. E

59. B

60. A

61. B

62. E

63. A

64. D

65. B

66. D

67. E

68. E

69. A

70. C

71. B

72. D

73. E

74. A

75. E

76. D

77. B

78. B

79. A

80. D

81. E

82. C

83. D

84. A

85. E

86. D

87. C

88. E

89. E

90. E

91. C

92. E

93. D

94. A

95. D

96. C

97. B

98. A

99. A

100. B

101. A

102. A

103. E

104. B

105. A

106. B

107. A

108. B

109. A

110. D

111. C

112. C

113. E

114. C

115. D

116. A

117. A

118. E

119. D

120. E

121. A

122. D

123. A

124. D

125. A

HOMEWORKLESSONS

Advanced Verbal Strategy

Advanced Verbal Strategy | 49

LESSON

SKILLS ME ET STR ATEGY Meaning Matters Most

This problem demonstrates clearly that logical meaning matters most in GMAT Sentence Correction questions. The authors of the GMAT know that you have worked to become adept at identifying parallel structures, rooting out singular-versus-plural verbs, noticing verb tenses, checking for logical modifiers, etc. So they write questions in which the no-brainer, quick-check methods actually require you to think—in which the meaning of the “eyeball test” winner is actually illogical. Always consider the meaning changes with Decision Points and you will be less likely to fall for some of these traps.

SKILLS ME ET STR ATEGY Every Decision Point Matters

Another big strategy takeaway from this problem is that you must assess every Decision Point. Often people assess only the fact that answer choice C contains the verb and answer choice E does not, and they miss entirely the choice with the word “other.” Testmakers are clever at hiding differences between answer choices and getting you to focus on the wrong Decision Point. Remember to always find and assess every single Decision Point!

SECTION 2: ADVANCED SENTENCE CORRECTION

Misdirection: Selling the orrect Answer2

Statistics and Combinatorics

Statistics & Combinatorics | 78

LE AR N ING BY DO ING Be Careful with Multiplication/Division in Sets

On a problem like this, it would be tempting to shortcut it as follows:

The sum of the set before the change is 6.8 x 5 (Sum = average X number of terms) = 34.

The sum of the set after one term is multiplied by 3 is 9.2 x 5 (Sum = average X number of terms) = 46.

Since the net gain after the change is 12, it seems logical to ask: What number multiplied by 3 will give me 12? The answer, of course, is 4, but that is the sucker choice on this problem. Why? Because the original number is still in the set. The correct question to ask yourself is this:

What number when multiplied by 3 creates a net change of 12 after the original number is subtracted? The answer to that question is 6. 6 x 3 = 18 – 6 = desired increase of 12. If you worked it out algebraically, this is clear:

Before change: a + b + c + d + e5

= 6.8 so a + b + c + d + e = 34

After one number is multiplied by 3: 3a + b + c + d + e5

= 9.2 so 3a + b + c + d + e = 46

Subtracting the first equation from the second, you see that 2a = 12 and a = 6. Answer choice E is correct.


LESSON

Permutations with Restrictions

Permutation problems get very difficult very quickly when restrictions are introduced. What does a restriction mean? It is some condition that limits somehow the total number of possibilities. Consider the question stem from a few example problems so you understand what these will look like:

1. Three couples go to the movies and sit in six consecutive seats. If the couples must sit together, in how many ways can they be seated?

2. There are three girls and three boys who need to be arranged in six consecutive seats. If the boys and girls must be alternated, in how many ways can they be seated?

3. Steve goes to the movies with four friends but refuses to sit next to one of them. In how many ways can they be arranged?

Now try a full problem:

14. John and four friends go to a Lakers game. In how many ways can they be seated in five consecutive seats, if John has to sit between any two of his friends?

(A) 144

(B) 120

(C) 96

(D) 72

(E) 48

SECTION 2: COMBINATORICS

Permutations with Restrictions2


55. Which of the following is equal to the average (arithmetic mean) of (n − 4)2 and (n + 4)2?

(A) n2

(B) n2 + 4

(C) n2 + 16

(D) n2 + 4n

(E) n2 + 16n


63. At a certain laboratory, chemical substances are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using seven colors?

(A) 21

(B) 35

(C) 105

(D) 135

(E) 210

WordProblems

Word Problems | 32

Solutions

1. To solve, set up your Matrix Box, pick a number of 100, and insert the information, being very careful about the wording:

WHITE BLUE TOTAL

COLONIAL 10 (25% of 40) 30 40

RANCH 30 30 60

TOTAL 40 60 100

There are 60 total ranch-style homes, and 30 of those are blue, so the answer is 50%.

2. To solve, pick a number that is divisible by 3, 4, and 5, and establish that as the total number of widgets produced last month. (The LCM of 3, 4, and 5 is 60, so use that.) Then fill in, starting with the totals for each group. Then insert 24 in sold and curved, using the fact that two-thirds of 45 were curved. Finally, answer the proper question, which asks what percent of the widgets not sold (15) were curved (12). 12/15 = 80%

CURVED NOT CURVED TOTAL

SOLD 24 21 45

NOT SOLD 12 3 15

TOTAL 36 24 60

Word Problems | 58

Solutions

1. Since Alexa won with 40 votes, it is fairly easy to conclude that the most number of votes Bill could have received is 39. Figuring out the minimum is much more complicated: If you know Alexa received 40, then you need to figure out the least number of votes that Bill could have received in 2nd and still account for 60 votes with the other three candidates. To find that number quickly, divide 60 by 4 and know that, if they could each have the same number of votes, the answer would be 15. Since they cannot, you want to arrange the candidates around 15 so that it remains the average but each candidate has a different number. Imagine, for instance, if Bill had 16, Charlie 15, Dan 14, and Ernie 13. That combination would not quite work, as the sum would be 58 (average is below 15). The smallest number Bill could have is 17 (the others could have 16, 15, and 12).

2. To figure out the most number of votes Charlie could receive, you want to minimize Dan and Ernie, and then split the remaining votes between Bill and Charlie. You know that Alexa has 40, and imagine Ernie and Dan had 2 and 1, respectively. That leaves 57 votes to be split as evenly as possible: 29 for Bill in second and 28 for Charlie in third. The answer is 28. To minimize Charlie, you should maximize Alexa and Bill at 40 and 39. That leaves 21 votes to be split among the remaining three candidates: 8 for Charlie, 7 for Dan, and 6 for Ernie. The answer is 8.

3. To maximize Dan, you’ll want to minimize the others. Alexa is fixed at 40, and you can minimize Ernie by putting him at 1. That leaves 59 votes to split between Bill, Charlie, and Dan, with B > C > D. Your goal, then? As even a distribution as possible, so you should look for a number below 59 that is divisible by 3. 57 = 3 * 19, but then you cannot give each of the three candidates a different total (there are only two votes left from the 59). So choose 54: That allows for each candidate to receive 18, and now you simply need to allocate the other 5 votes so that the candidates have different vote totals (either D = 18, B = 20, and C = 21; or D = 18, B = 19, and C = 23). To minimize Dan in fourth place is a bit easier: Simply give him 2 and Ernie 1—leaving Alexa with 40, Bill with 29, and Charlie with 28. The answer is 2.

Word Problems | 59

SKILLBUILDER

SKILLB

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4. To maximize Ernie, the thought process is similar – you want as even a distribution of B, C, D, and E as possible. Since Alexa is fixed at 40, then you have 60 votes left, and the most even distribution would be 15 each. But each value needs to be different, so you should look to allocate the votes around 15. B = 16, C = 15, D = 14 and E = 13 spreads the values out around the 15 average that you want and leaves you with 2 votes remaining. This isn’t enough to bump Ernie to 14 (you can’t do that without creating a tie among the other three), but you can give those votes to Bill and Charlie to keep them “ahead” and leave Ernie with a max of 13 votes. And, as you’ve already seen, it’s certainly possible for Ernie to receive just one vote, so his minimum total is 1.

5. If Alexa received 40 votes and Bill received 25 votes, that leaves 35 votes for third, fourth, and fifth places. If you divide 35 by 3, you see that the smallest third place could be is 13 votes for Charlie, which would leave 12 votes for Dan, and 10 votes for Ernie. Charlie could not have 12 votes as the most Dan could then have is 11, followed by 10 for Ernie. That only sums to 33 votes, which is not enough to reach the required 35. Yes, Charlie must have received at least 13 votes.

6. If Alexa received 40 votes and Charlie received 12 votes, that accounts for 52 votes. To consider possibilities for Dan, you must realize that at most Bill received 39 votes in second place. Adding 39 to 52 accounts for 91 total votes. Therefore, Dan and Ernie together must account for 9 votes minimum. This can only be done if Dan in fourth place has at least 5 votes, so yes, he must have received at least 5 votes.

Word Problems | 82

6. If a taxi driver charges x cents for the first quarter-mile of a trip and x5 cents

for each additional quarter-mile, what is the charge, in cents, for a trip whose distance in miles is the whole number y?

(A) x + xy

125

(B) 4x + 4xy

5

(C) 4x + xy

500

(D) 4x + xy

5

(E) xy25

Word Problems | 83

LESSON

LE AR N ING BY DO ING How to Pick Numbers

Setting up the algebra on this problem is difficult for most students, but number picking is relatively straightforward. To number pick efficiently, pick the smallest and easiest numbers given the set-up of the problem; in this example, make x be 5 cents (the smallest number divisible by 5) and y be 1 mile. Now use reasoning to determine how much the trip would cost. The first ¼ mile would be 5 cents and the three remaining ¼ miles would each cost 1 cent apiece. When x = 5 and y = 1, the total cost for the trip is 8 cents. Now plug those numbers into each answer choice to see which one gives 8 cents. The only one that gives you that is answer choice B (and the only other answer choice that is even close is answer choice D, which gives you 5 cents).

The difficulty with the algebra in this problem revolves around how to represent the number of remaining ¼ miles in terms of y. Here is the correct approach algebraically:

• Total cost of trip = Cost for first ¼ mile + Cost for remaining ¼ miles

• Cost for the first ¼ mile is clearly = x

• Cost for remaining ¼ miles = (price per remaining ¼ miles)(# of remaining ¼ miles)

• The price per remaining ¼ miles is given as x5 , but the # of remaining ¼ miles is

tricky:

• In a y-mile trip, the number of ¼ miles traveled = 4y. However, since you have already been charged for the first ¼ mile, the number of remaining ¼ miles = 4y – 1.

Plugging this all back in you see that:

Total Cost = x + ( x5

) (4y – 1)

Remove the parentheses: x + 4xy – x

5

Combine like terms: 5x5

+ 4xy – x

5 = 4xy + 4x

5

While you should be able to do the algebra (as you might have to on another problem), it is hard to deny that number picking is much easier on this difficult problem.

SECTION 2: THE PROBLEM SOLVING TOOLKIT

Number Picking2

Word Problems | 209

HOMEWORK

HO

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88. How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?

(A) 72

(B) 5

(C) 203

(D) 8

(E) 394

Analytical Writing Assessment& Integrated Reasoning

AWA/IR | 21

SKILLBUILDER

SKILLB

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Venn Diagrams

A bit of a departure from “graphs,” Venn Diagrams are nonetheless “graphics” and in the pool of testable graphics for the IR section. Venn Diagrams exist as tools to organize overlapping sets of information. Consider an example:

In the diagram below, each x represents one MBA applicant from a consulting firm.

Note a few things about the Venn Diagram:

1. To the left of the overlapping circles is a group of five data points, representing the “neither” category that fits in neither set.

2. The overlapping area between the circles (six x’s) represents the “both” group—in this case people who both “used an admissions consultant” and “were admitted to first-choice school.”

3. In the partial circles (three x’s in the left; six in the right), those data points represent the “only” categories. There are three applicants who “only” used an admissions consultant (but were not admitted to their top-choice schools) and six who “only” were admitted (but did not use an admissions consultant).

AWA/IR | 64

Let’s take a look at a Table Analysis problem:

Before looking at the questions, note a few things about this table:

1. There is quite a bit of data. Reading the table without looking at the questions first is a lost cause; you simply cannot process all of this in three minutes. Be question-driven!

2. The values are quite specific (to the tens digit for eight-figure numbers). You do not want to calculate these numbers—even with the on-screen calculator—unless absolutely necessary. Use estimates and logical determinations of when you need to truly calculate, as we will discuss.

3. The “Sort by” button at the top of the screen will be dynamic on a computer; you can change the organization of the table. Currently the table is sorted by Brand. But notice that the % Change totals (for both Unit and Dollar Sales) are all over the place; if a question asks about those, you can re-sort the column to better organize your search for the pertinent information.

AWA/IR | 68

Question 2 is another classic sort and scan. Sort by Fragrance to align all of the Orange products together, and then scan them along the % Change column. All signs are positive, so none of the Orange products experienced a decline in Unit Sales. The answer is “true.”

The table above gives sale information for the 20 bathroom cleaners in the United States in 2010. For each product, the table describes the brand of the product, the product type, fragrance, unit sales, percent change in unit sales since 2009, total dollar sales, percent change in dollar sales since 2009, average price of each unit sold, and the dollar change in price since 2009.

Calculator

Brand Type Fragrance Units % Change Average Price $ Change% ChangeDollars

Product Unit Sales Dollar Sales Price

Clean it!

FragranceSort by

Spray

Aerosol

Powder

Aerosol

Spray

Aerosol

Spray

Spray

Aerosol

Aerosol

Spray

Aerosol

Spray

Spray

Aerosol

Spray

Aerosol

Aerosol

Spray

Powder

Clean it!

Deluxe

Deluxe

Deluxe

Dirt Blaster

Dirt Blaster

Incredible

Incredible

Magic Clean

Magic Clean

Mrs. Grime

Mrs. Grime

Super Clean

Super Clean

Tornado

Tornado

Ultra Shine

Ultra Shine

Ultra Shine

Orange

Orange

Lemon

Orange

Orange

Orange

Orange

Lemon

Unscented

Lemon

Unscented

Lemon

Unscented

Fresh

Fresh

Fresh

Unscented

Fresh

Fresh

Lemon

4,768,920

6,695,560

525,040

4,934,600

4,432,700

2,440,870

2,179,130

147,470

3,654,370

1,569,200

6,673,490

5,208,300

4,242,520

8,674,230

1,248,640

1,694,650

1,537,820

5,010,090

2,235,370

8,661,260

7.1%

3.9%

-16.7%

2.9%

0.9%

6.9%

6.0%

-18.3%

-11.1%

-4.0%

-1.3%

-4.1%

-2.5%

10.8%

8.2%

4.4%

-7.9%

3.1%

2.1%

3.3%

$ 2.97

$ 3.24

$ 3.10

$ 2.61

$ 3.66

$ 2.93

$ 2.80

$ 2.90

$ 3.16

$ 2.94

$ 4.12

$ 4.30

$ 3.09

$ 3.86

$ 4.33

$ 3.88

$ 3.43

$ 3.08

$ 3.14

$ 3.99

$ -0.10

$ 0.31

$ -0.29

$ 0.40

$ 0.49

$ -0.45

$ 0.20

$ -0.25

$ 0.31

$ 0.36

$ 0.37

$ 0.36

$ 0.19

$ 0.39

$ -0.05

$ 0.36

$ 0.22

$ 0.25

$ -0.01

$ 0.31

3.6%

14.8%

-23.8%

21.5%

16.4%

-7.3%

14.1%

-24.7%

-1.4%

9.3%

8.4%

4.6%

3.8%

23.2%

6.9%

15.0%

-1.5%

12.2%

1.7%

12.0%

$ 14,163,690.00

$ 21,693,610.00

$ 1,627,620.00

$ 12,879,300.00

$ 16,223,680.00

$ 7,151,740.00

$ 6,101,560.00

$ 427,660.00

$ 11,547,800.00

$ 4,613,440.00

$ 27,494,770.00

$ 22,395,690.00

$ 13,109,380.00

$ 33,482,520.00

$ 5,406,610.00

$ 6,575,240.00

$ 5,274,720.00

$ 15,431,070.00

$ 7,019,060.00

$ 34,558,420.00

AWA/IR | 77

LESSON

Precision in Language

The Graphics Interpretation question format lends itself well to the classic GMAT testmaker trick of misdirection. By making you think that you are solely focusing on numbers and graphics, the author can sneakily slip a Critical Reasoning– or Reading Comprehension–style precision in language trick past you. Recognize, then, that many questions will hinge on the wording of the question or of the legend of the graph. Consider the question:

The graph above represents the results of a survey taken by focus groups, each shown a different version of an upcoming commercial for a sports car. Each version had a different amount of content showing the car in fast-driving scenes.

After the group viewed the commercial, each member was asked two questions: “Did you feel that the car was sporty?” and “How likely are you to purchase this car?”

The position of the circles above indicate how much fast-driving content (F) was in the version of the commercial show and the % of viewers who found the car to be “sporty” (S). The relative size of the circles indicates how likely the members of the group were to purchase the car (L). The bigger the circle, the more likely they were to purchase.

SECTION 2: INTEGRATED REASONING

Graphics Interpretation2

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

joker125

Typewritten Text

AWA/IR | 79

LESSON

Sample Questions

The graph above is a scatter plot with 35 points, each representing the population of a city and the number of auto thefts in that city, per person, per year. All 35 measurements were made in the year 2010, counting the number of auto thefts during the year and the number of residents in each city as of January 1, 2010. The dashed line runs through points (0, 0) and (35, 1200).

Use the drop-down menus to fill in the blanks in each of the following statements based on the information given by the graph.

1. The number of cities that had at least 20 auto thefts per 1,000 is closest to ____% of the total cities measured.

(A) 16 (B) 33 (C) 50

2. Every city with a population of no more than 600,000 had no more than _________ auto thefts per 1,000 people.(A) 10 (B) 20 (C) 30

3. There is a(n) _______________ relationship between a city’s population and its number of auto thefts per 1,000 people.

(A) positive (B) negative (C) equivalent


Graphics Interpretation2

AWA/IR | 91

LESSON

Two-Part Analysis #2

Joseph: Health insurance premiums are growing at an alarming rate. This is, in part, because many hospitals and clinics bill for unnecessary diagnostics and tests that inflate the subsequent amount that insurers pay out to them. These expenses are then passed on to consumers in the form of increased insurance premiums. Therefore, reducing the number of unnecessary tests performed by health care providers will be effective in controlling growing health insurance premiums.

Ronald: Often times, the unnecessary diagnostics that you speak of are the result of decisions made by doctors on behalf of their patients. Doctors will often choose the diagnostics that will allow them to bill insurers for more money, but may not necessarily benefit the patient in a meaningful way or influence the course of treatment chosen. As a result, in order to succeed in reducing the number of unnecessary tests, we should allow the patients to decide which course of diagnostics they would like to undergo.

In the table below, identify the assumptions upon which each person’s argment depends. Make only one selection in each column, one for Joseph and one for Ronald.

Doctors are generally able to determine, with great reliability, which diagnostic procedures and tests would yield the most effective results.

Tests and diagnostic procedures do not make up an insignificant portion of the bills that are sent to insurers.

Insurance companies in other industries, such as auto and home, have been able to reduce costs by reducing the number of unneces-sary repairs and replacements on claims for automobiles and homes.

Patients are not just as likely as doctors to choose the most expensive diagnostics and tests.

Health insurance premiums have increased twice as fast in the past five years than they have over an average of the past 25 years.


Two-Part Analysis2

AWA/IR | 97

LESSON

Note that a calculator might be tempting in this case, but that each calculation requires you to key at least nine digits—a time-consuming process that raises your potential for typo-based error. An eye for both logical setup and relative math can guide you through this process efficiently and confidently. First, note the correct relationship—the lowest percentage savings, or the lowest savings-to-budget ratio. Your goal, then, is to test the ratios of the number in the left column to those in the right column, looking for the smallest ratio. Your “baseline” for Andersonville is approximately

848, or

16 . In

relation to 16 , you know that the numerator is a little more than 8 and the denominator

is a little less than 48, so the overall ratio is going to be slightly greater than 16 . You can

denote this quickly on your noteboard with a + sign or a > sign to help you recognize the direction of your estimates.

(A) > 8

48, so >16

(B) < 20

140, so < 17 (the current “leader” in smallest ratio)

(C) < 4

20, so < 15

(D) > 3

16, and since you’re comparing against 17 (or

321) you know that this

is greater

(E) > 1352, so >

14

The correct answer choice must be B, and if you’ve employed the above estimates you won’t have had to perform any true calculations to get there. Bronxtown had the lowest percentage savings.


Relative Math2

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