GRE Math Practice Test 2

Test 2 GRE

MATH

For Complete GRE MATH Preparation Material Visit http://studymaterialcollection.blogspot.com/2015/12/complete-gre-math-preparation-material.html

GRE Math Tests

34

Questions: 24 Time: 45 minutes

[Multiple-choice Question – Select One or More Answer Choices] 1. Mr. Smith's average annual income in each of the years 2006 and 2007 is positive x dollars. His

average annual income in each of the years 2008, 2009, and 2010 is positive y dollars. He did earn at least some money each year. Which of the following could be his average annual income for the four continuous years 2006 through 2009?

(A) x/2 (B) x/2 + y/2 (C) x/2 + y (D) x/2 + 3y/4 (E) x/2 + 3y/5 (F) 3x/5 + y/2

2. Column A For any positive integer n, (n)

represents the number of factors of n, inclusive of 1 and itself. a and b are unequal prime numbers.

Column B

(a) + (b) (a b)

[Multiple-choice Question – Select One or More Answer Choices] 3. a and b are multiples of 20, b and c are multiples of 30, and a and c are multiples of 40. Which of the

following must be individually divisible by a, b, and c?

(A) 10 (B) 20 (C) 30 (D) 40 (E) 120 (F) 240

Test 2—Questions

35

[Multiple-choice Question – Select One or More Answer Choices] 4. Which of the following could be the fraction of numbers that are divisible by both 7 and 10 in a set of

71 consecutive integers?

(A) 1/71 (B) 2/71 (C) 3/71 (D) 4/71 (E) 5/71

[Multiple-choice Question – Select One or More Answer Choices] 5. In the triangle, what is the value of x?

(A) 25 (B) 25 + y (C) 25 – y (D) 55 (E) 60

60° + x°

y° + x°

70° – y°

GRE Math Tests

36

[Multiple-choice Question – Select One or More Answer Choices] 6. In the figure, O is the center of the circle. Which of the following must be true about the perimeter of the triangle shown?

(A) Less than 10 (B) Greater than 20 (C) Greater than 30 (D) Less than 40 (E) Less than 40 and greater than 20

10

O

A B

Test 2—Questions

37

[Multiple-choice Question – Select One Answer Choice Only] 7. In the figure, ABCD is a rectangle inscribed in the circle shown. What is the length of the smaller arc

DC ?

(A) /4 (B) /3 (C) 2 /3 (D) 3 /4 (E) 4 /3

A B

C D 2

4


Do You Want to Search the Specific:Fulbright Scholarships, JOB, Online Work or Online Job, Free Video Tutorial, Free Video Lectures for DifferentSubjects and Free Online Courses Visit the SEARCH ENGINE’s Collection page http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html

GRE Math Tests

38

[Multiple-choice Question – Select One Answer Choice Only] 8. In the figure, y =

(A) –12 (B) –3 (C) 1 (D) 35 (E) 12

[Multiple-choice Question – Select One or More Answer Choices] 9. In triangle ABC, CA = 4, and CB = 6. Which of the following could be the area of triangle ABC ?

(A) 7 (B) 8 (C) 9 (D) 12 (E) 15

A

B

C D

E

y 9 – y

x°

x°

5 + y

4 + 2y

Test 2—Questions

39

[Multiple-choice Question – Select One Answer Choice Only] 10. Suppose five circles, each 4 inches in diameter, are cut from a rectangular strip of paper 12 inches

long. If the least amount of paper is to be wasted, what is the width of the paper strip?

(A) 5 (B)

4 + 2 3 (C) 8 (D)

4 1+ 3( ) (E) not enough information

11. Column A (4 – 5x)2 = 1 Column B (4 – 5x) + (4 – 5x)2 1

12. Column A 0 < x < y Column B x + 1/x y + 1/y

GRE Math Tests

40

[Multiple-choice Question – Select One or More Answer Choices] 13. If x = a + 2 and b = x + 1, then which of the following must be false?

(A) a > b (B) a < b (C) a = b (D) a = b2 (E) b = a2

[Multiple-choice Question – Select One or More Answer Choices] 14. The difference between two angles of a triangle is 24 . If all three angles are numerically double-digit

integers, then which of the following could be the measure of the third angle?

(A) 10 (B) 12 (C) 59 (D) 74 (E) 98

15. Column A The monthly incomes of A and B

are in the ratio 3 : 2, and the monthly expenditures of the two are in the ratio 4 : 3. The income of both is greater than the expenditures, and the Savings from income is defined as the Income – Expenditure.

Column B

Savings of A Savings of B

Test 2—Questions

41

[Multiple-choice Question – Select One or More Answer Choices] 16. If x – 3 = 10/x and x> 0, then x could equal which of the following?

(A) –2 (B) –1 (C) 3 (D) 5 (E) 10

[Multiple-choice Question – Select One Answer Choice Only] 17. If r and s are two positive numbers and r is 25% greater than s, what is the value of the ratio r/s ?

(A) 0.75 (B) 0.8 (C) 1 (D) 1.2 (E) 1.25

[Multiple-choice Question – Select One Answer Choice Only]

18. If 1/x + 1/y = 1/3, then xyx + y

=

(A) 1/5 (B) 1/3 (C) 1 (D) 3 (E) 5

GRE Math Tests

42

19. Column A A stores sells Brand A items and Brand B items at 75% and 80%, respectively, of list prices. During festival season, discounts of 20% and 25%, on Brand A and Brand B, respectively, are offered. Spinach tins of Brand A and Brand B have the same list price.

Column B

Cost of spinach tins of Brand A during festival season

Cost of spinach tins of Brand B during festival season

[Numeric Entry Question] 20. 2ab5 is a four-digit number divisible by 25. If the number ab formed from the two digits a and b is a

multiple of 13, then the three-digit number 2ab =


Do You Want to Search the Specific:Fulbright Scholarships, JOB, Online Work or Online Job, Free Video Tutorial, Free Video Lectures forDifferent Subjects and Free Online Courses Visit the SEARCH ENGINE’s Collection page http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html

Test 2—Questions

43

[Multiple-choice Question – Select One or More Answer Choices] 21. Two data sets S and R are defined as follows:

Data set S: 28, 30, 25, 28, 27 Data set R: 22, 19, 15, 17, 21, 25

Which of the following are the medians of the sets S and R or the difference between the medians?

(A) 8 (B) 10 (C) 20 (D) 25 (E) 28

22. Column A x > y > 0 Column B

x – y

x3

+y3

[Multiple-choice Question – Select One Answer Choice Only] 23. Seven years ago, Scott was 3 times as old as Kathy was at that time. If Scott is now 5 years older than

Kathy, how old is Scott?

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

GRE Math Tests

44

24. The University of Maryland, University of Vermont, and Emory University have each 4 soccer players. If a team of 9 is to be formed with an equal number of players from each university, how many possible teams are there?

(A) 3 (B) 4 (C) 12 (D) 16 (E) 64

Test 2—Answers and Solutions

45

Answers and Solutions Test 2:

Question Answer

1. E, F 2. C 3. E, F 4. A, B 5. A 6. B, D, E 7. C 8. C 9. D, E 10. B 11. D 12. D 13. A, C, D 14. B, D 15. A 16. D 17. D 18. D 19. C 20. 252 21. A, C, E 22. D 23. E 24. E

1. Since Mr. Smith's average annual income in each of the two years 2006 and 2007 is x dollars, his total income in the two years is 2 x = 2x. Since Mr. Smith's average annual income in each of the three years 2008 and 2010 is y dollars, his total income in the three years is 3 y = 3y. Now, average income from 2006 through 2009 is [(2 x amount earned between 2006, 2007) + (Amount earned between 2008, 2009, which could be between

0 and 3y)]/4 years =

Between Min = (2x + 0)/4 = x/2 and Max = (2x + 3y)/4 = x/2 + 3y/4 The possible incomes in the range NOT including both Min and Max values above are Choices. Select such choices. Since, the range is x/2 through x/2 + 3y/4 NOT including both, Reject choices (A) and (D), which are the extremes Min and Max. Since y/2 < 3y/4, x/2 + y/2 < x/2 + 3y/4; Choice (B) < Max. x/2 + y/2 is clearly greater than just x/2. Choice (B) > Min. Accept. Since y > 3y/4, Choice(C) = x/2 + y > x/2 + 3y/4; Choice (C) > Max. Reject. Choice (E): x/2 + 3y/5is similarly > Min and < Max. Accept.

GRE Math Tests

46

Choice (F): 3x/5 + y/2; 3x/5 could be in the range and y/2 could be in the range and their sum is not necessarily out of range. Hence, select Choice (F) as well. Hence, the answer is (E) and (F). 2. The only factors of a prime number are 1 and itself. Hence, (any prime number) = 2. So, (a) = 2 and

(b) = 2 and therefore Column A equals (a) + (b) = 2 + 2 = 4. Now, the factors of ab are 1, a, b, and ab itself. Since a and b are different, the total number of factors of ab is 4. In other words, (a b) = 4. Hence, Column B also equals 4. Since the columns are equal, the answer is (C). 3. We are given that “a and b are multiples of 20, b and c are multiples of 30, and a and c are multiples of 40.” We can summarize this information as follows:

a is multiple of 20; a is multiple of 40. Hence, a = 40l [40 is the LCM of 20 and 40] b is multiple of 20; b is multiple of 30. Hence, b = 60m [60 is the LCM of 20 and 30] c is multiple of 30; c is multiple of 40. Hence, a = 120n [120 is the LCM of 30 and 40]

where l, m, n are integers. Now, a number divisible by a, b, and c individually must be a multiple of the least common multiple of 40, 60, and 120, which is 120. Hence, the answers are (E) 120 and (F) 240. 4. The numbers that are divisible by both 7 and 10 must be divisible by their least common multiple 7 10 = 70 also. The numbers divisible by 70 occur once every 70 consecutive integers. Now, if for example, the set is 70 through 140, then two numbers, 70 and 140, in the set are divisible by 70. Here, the fraction is 2/71. Select (B). But if for example, the set is 71 through 141, then only one number, 140, is divisible by 70. Here the fraction is 1/71. Select (A). The possible choices are (A) and (B). 5. The angle sum of a triangle is 180°. Hence,

(y + x) + (60 + x) + (70 – y) = 180 Simplifying the equation yields 2x + 130 = 180. Solving for x yields x = 25. The answer is (A)—Only one choice must be chosen. Note: Sometimes when a problem says, "Select One or More Answer Choices," there will be only one answer. When you find just one answer, double-check your work.


47

6. In AOB, OA and OB are radii of the circle. Hence, both equal 10 (since OA = 10 in the figure).

Now, the perimeter of a triangle equals the sum of the lengths of the sides of the triangle. Hence,

Perimeter of AOB =

OA + OB + AB =

10 + 10 + AB =

20 + AB In a triangle, the length of any side is less than the sum of the lengths of the other two sides and is greater than their difference. This makes AB less than AO + OB (= 10 + 10 = 20) and AB greater than 10 – 10 = 0. So, the range of AB is 0 through 20, not including either number. Hence, the range of 20 + AB is evaluated as the inequality

20 < 20 + AB < 20 + 20

20 < Perimeter < 40 Logically, the choices that apply are (B), (D), and (E). 7. In the figure, BD is a diagonal of the rectangle inscribed in the circle. Hence, BD is a diameter of the circle. So, the midpoint of the diagonal must be the center of the circle, and the radius must equal half the length of the diameter: BD/2 = 4/2 = 2. Now, joining the center of the circle, say, O to the point C yields the following figure:

Now, in DOC since OD and OC are radii of the circle, both equal 2. So, OD = OC = 2. Since DC also equals 2 (in figure given), OD = OC = DC = 2 (all three sides are equal). Hence, the triangle is equilateral and each angle must equal 60°, including DOC. Now, the circumference of the given circle equals 2 radius = 2 (2) = 4 . The fraction of the complete angle that the arc DC makes in the circle is 60°/360° = 1/6. The arc length would also be the same fraction of the circumference of the circle. Hence, the arc length equals 1/6 4 = 2 /3. The answer is (C).

A B

C D 2

O

2 2


Do You Want to Search the Specific:Fulbright Scholarships, JOB, Online Work or Online Job, Free Video Tutorial, Free Video Lectures forDifferent Subjects and Free Online Courses Visit the SEARCH ENGINE’s Collection page http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html

GRE Math Tests

48

8. In the figure, y =

(A) –12 (B) –3 (C) 1 (D) 35 (E) 12

In the figure, in the triangles ABC and EDC, we have

ACB = DCE Common Angle BAC = DEC = x

Therefore, triangles ABC and EDC are similar. Equating corresponding side ratios yields

AC/BC = EC/DC (y + 9 – y)/(5+ y) = (5+ y + 4 + 2y)/(9 – y) 9/(5 + y) = (9 + 3y)/(9 – y) 9(9 –y) = (9 + 3y)(5 + y) cross-multiplication 81 – 9y = 45 + 9y + 15y + 3y2 81 = 45 + 9y + 15y + 9y + 3y2

27 = 15 + 3y + 5y + 3y + y2 dividing by 3 12 = 11y + y2 0 = –12 + 11y + y2

y2 + 11y – 12 = 0 y2 + 12y – y – 12 = 0 y(y + 12) – (y + 12) = 0 (y – 1)(y + 12) = 0 y = 1 or –12 AD = y = 1. Eliminate –12 because the side AD = y cannot have a negative length.

The answer is (C) only.

A

B

C D

E

y 9 – y

x°

x°

5 + y

4 + 2y


49

9. We are given that in triangle ABC, CA = 4 and CB = 6. The area of the triangle is greatest when the sides form a right angle, and least when the sides align in a line. When the sides form a right-angle, the area is 1/2 base height = 1/2 4 6 = 12; and when they are aligned in a line, the area is 0 because the height is zero. Hence, the permissible range is 0 through 12. Select choices in this range. The answer is (D) and (E).

10. Since this is a hard problem, we can eliminate (E), “not enough information.” And because it is too easily derived, we can eliminate (C), (8 = 4 + 4). Further, we can eliminate (A), 5, because answer-choices (B) and (D) are a more complicated. At this stage we cannot apply any more elimination rules; so if we could not solve the problem, we would guess either (B) or (D).

Let’s solve the problem directly. The drawing below shows the position of the circles so that the paper width is a minimum.

Now, take three of the circles in isolation, and connect the centers of these circles to form a triangle:

Since the triangle connects the centers of circles of diameter 4, the triangle is equilateral with sides of length 4.

Drawing an altitude gives

4

4

4

GRE Math Tests

50

Applying the Pythagorean theorem to either right triangle gives h2 + 22 = 42 Squaring yields h2 + 4 = 16 Subtracting 4 from both sides of this equation yields h2 = 12 Taking the square root of both sides yields

h = 12 = 4 3 Removing the perfect square 4 from the radical yields

h = 2 3 Summarizing gives

Adding to the height,

h = 2 3 , the distance above the triangle and the distance below the triangle to the edges of the paper strip gives

width = 2+ 2( ) + 2 3 = 4 + 2 3 The answer is (B). 11. (4 – 5x)2 equals 1 when either 4 – 5x = –1 or +1. If 4 – 5x equals –1, then

(4 – 5x) + (4 – 5x)2 =

–1 + 1 =

0 And if 4 – 5x equals 1, then

(4 – 5x) + (4 – 5x)2 =

1 + 1 =

2 In the first case, Column A is less than Column B; and in the second case, Column B is less than Column A. Hence, we have a double case, and the answer is (D).

44h

2 2

h

2

2


51

12. If x = 1 and y = 2, then the columns become

1 + 1/1 2 + 1/2

2 2 1/2 In this case, Column B is greater. If x = 1/2 and y = 1, the columns become

12

+112

1+11

2 12

2

In this case, Column A is greater. Hence, we have a double case, and the answer is (D). 13. We have the equations x = a + 2 and b = x + 1. Substituting the first equation into the second equation yields b = (a + 2) + 1 = a + 3. This equation shows that b is 3 units greater than a. Hence, b must be greater than a. So, (A) and (C) are false and therefore correct answers (remember, we are looking for statements that must be false). Next, assume a = b2 [Choice (D)]. Then a must be positive. If a is between 0 and 1, b is also between 0 and 1, which violates b = a + 3. Here, our assumption fails. If a were greater than 1, then b equals a, which must be less than a—but we know that b is greater than a. Again, our assumption fails. Accept choice (D). Next, assume b = a2 [Choice (E)]. This could be true. For example, consider b = a + 3 = a2. Subtracting a and 3 from both sides yields a2 – a – 3 = 0. The quadratic equation is in the form ax2 + bx + c = 0. Since the discriminant of the quadratic equation which is b2 – 4ac equals (–1)2 – 4(1)(–3) = 1 + 12 = 13, is positive, we will have valid real solutions (Note, the GRE is only interested in Real Numbers), which means that b = a2 is possible with given information. Hence, reject this choice (remember, we are looking for answers that must be false). The answer consists of (A), (C), and (D).

GRE Math Tests

52

14. Let a, b, and c be the angles. Their sum is 180: a + b + c = 180. Let a and b be the angles differing by 24 , with a being the greater angle. So, we have a – b = 24. Subtracting the equations yields

a + b + c – (a – b) = 180 – 24 2b + c = 156 2b = 156 – c b = (156 – c)/2 = 78 – c/2

Since b is at least a 2-digit number, c must be in the range between 10 and 98, including both. So,

78 – 10/2 > b > 78 – 98/2; 73 > b > 29; 72 > b > 30 (EVEN integers in the range). Since a = b + 24, the range of a is

72 + 24 > a > 30 + 24; 96 > a > 54 (EVEN integers in the range); Hence, the range of a + b is

96 + 72 > a + b > 54 + 30; 168 > a + b > 84 (EVEN integers in the range); Now, c = 180 – (a + b) yields the range for c as

180 – 168 < c <180 – 84; 12 < c < 96 (EVEN integers in the range); Thus, the third-angle could be any EVEN integer from 12 through 96. Select (B) and (D). 15. Let the income of A and B be 3s and 2s, respectively, and let their expenditures be 4t and 3t. Then since savings is defined as income minus expenditure, Column A = the saving of A = 3s – 4t, and Column B = saving of B = 2s – 3t. Column A is greater than Column B when 3s – 4t > 2s – 3t, or s > t. Since money is positive, s and t are positive. Since income is greater than expenditure (given), the Income of B = 2s > Expenditure of B = 3t. Hence, s > 3t/2. Clearly, s > t. Since we know that Column A > Column B when s > t, the answer is (A). 16. We have the equation x – 3 = 10/x. Multiplying the equation by x yields x2– 3x = 10. Subtracting 10 from both sides yields x2 – 3x – 10 = 0. Factoring the equation yields (x – 5)(x + 2) = 0. The possible solutions are 5 and –2. The only solution that also satisfies the given inequality x > 0 is x = 5. The answer is (D). 17. We are given that r is 25% greater than s. Hence, r = s + 25%s = s + 0.25s = (1 + 0.25)s = 1.25s. So, r/s = 1.25s/s = 1.25. The answer is (E).


Do You Want to Search the Specific:Fulbright Scholarships, JOB, Online Work or Online Job, Free Video Tutorial, Free VideoLectures for Different Subjects and Free Online Courses Visit the SEARCH ENGINE’s Collection page http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html


53

18. Multiplying the given equation 1/x + 1/y = 1/3 by xy yields

y + x = xy/3, or x + y = xy/3

Multiplying both sides of the equation x + y = xy/3 by 3/(x + y) yields

xyx + y

= 3

The answer is (D). 19. Since the spinach tins of either brand have the same list price, let each be x dollars. Now, 75% of x is (75/100)x = 3x/4, and 80% of x is (80/100)x = 4x/5. So, spinach tins of brands A and B are sold at 3x/4 and 4x/5, respectively. After festival discounts of 20% and 25% on the respective brand items, the cost of spinach tins drop to

1 20100

3x4

=45

3x4

=3x5

and

1 25100

4x5

=

34

4x5

=3x5

Hence, the columns are equal, and the answer is (C). 20. We have that the number 2ab5 is divisible by 25. Any number divisible by 25 ends with the last two digits as 00, 25, 50, or 75. So, b5 should equal 25 or 75. Hence, b = 2 or 7. We also have that ab is divisible by 13. The multiples of 13 are 13, 26, 39, 52, 65, 78, and 91. Among these, the only number ending with 2 or 7 is 52. Hence, ab = 52. Then, 2ab = 252. Enter 252 in the grid. 21. The definition of median is “When a set of numbers is arranged in order of size, the median is the middle number. If a set contains an even number of elements, then the median is the average of the two middle elements.” Data set S (arranged in increasing order of size) is 25, 27, 28, 28, 30. The median of the set is the third number 28. Data set R (arranged in increasing order of size) is 15, 17, 19, 21, 22, 25. The median is the average of the two middle numbers (the 3rd and 4th numbers): (19 + 21)/2 = 40/2 = 20. The difference of 28 and 20 is 8. The correct answers are (A), the difference between the medians, (C), the median of the set R, and (E), the median of the set S.

22. If x = 2 and y = 1, then x – y = 2 – 1 = 1 =

33

=23

+13

=x3

+y3

. In this case, the columns are equal.

If x = 3 and y = 1, then x – y = 3 – 1 = 2

33

+13

=43

=x3

+y3

. In this case, the columns are not equal and

therefore the answer is (D).

GRE Math Tests

54

23. Let S be Scott’s age and K be Kathy’s age. Then translating the sentence “If Scott is now 5 years older than Kathy, how old is Scott” into an equation yields

S = K + 5 Now, Scott’s age 7 years ago can be represented as S = –7, and Kathy’s age can be represented as K = –7. Then translating the sentence “Seven years ago, Scott was 3 times as old as Kathy was at that time” into an equation yields S – 7 = 3(K – 7). Combining this equation with S = K + 5 yields the system:

S – 7 = 3(K – 7) S = K + 5

Solving this system gives S = 14 . The answer is (E). 24. The selection from the 3 universities can be done in 4 4 4 = 43 = 64 ways. The answer is (E).

For Complete GRE MATH Preparation Material

Visit

http://studymaterialcollection.blogspot.com/2015/12/complete-gre-math-preparation-material.html

Do You Want to Search the Specific:

Fulbright Scholarships, JOB, Online Work or Online Job, Free Video Tutorial, Free Video

Lectures for Different Subjects and Free Online Courses, GRE, GMAT, IELTS, TOEFL That

Assist You in Your Professional Life or Field.

Then Below is The Link For you Where You Get the Several Free and Smart Search Engine’s

Links That Are Specifically Design According to Your Requirement.

Visit and Use These Specific Search Engine’s That Design for Your Requirement

Click below To Visit the SEARCH ENGINE’s Collection page

http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html

==== ++++ ==== ++++ ==== ++++ ==== ++++ ==== ++++ ==== ++++ ====

If Want Complete Data for GRE MATH Preparation Then You May Like To Visit

Our Site

Complete GRE MATH Preparation Material



http://searchenginecollectionpage.blogspot.com/2015/10/SearchEngineCollection.html


Documents

GRE Math Practice Test 2