GRE_Math_Bible_eBook
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Equations 271
24. We are given that a > 0 and (a + 2)(a – 3)(a + 4) = 0. Hence, the possible solutions are
a + 2 = 0; a = –2, a is not greater than 0, so reject. a – 3 = 0; a = 3, a is greater than 0, so accept.
a + 4 = 0; a = –4, a is not greater than 0, so reject. The answer is (C).
Hard
25. We are given the system of equations:
x + y = 7 x2 + y2 = 25
Solving the top equation for y yields y = 7 – x. Substituting this into the bottom equation yields
x2 + (7 – x)2 = 25
x2 + 49 – 14x + x2 = 25 2x2 – 14x + 24 = 0
x2 – 7x + 12 = 0 (x – 3)(x – 4) = 0
x – 3 = 0 or x – 4 = 0 x = 3 or x = 4
If x = 3, then y = 7 – 3 = 4. If x = 4, then y = 7 – 4 = 3. In either case, x3 + y3 = 33 + 43 = 27 + 64 = 91. The answer is (E).
26. The given system of equations is
x + l = 6 x – m = 5 x + p = 4 x – q = 3
Subtracting the second equation from the first one yields
(x + l) – (x – m) = 6 – 5 |
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Subtracting the fourth equation from the third one yields
(x + p) – (x – q) = 4 – 3 |
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Adding equations (1) and (2) yields
(l + m) + (p + q) = 1 + 1 = 2. l + m + p + q = 2
The answer is (A).
27. We are given the two equations 1/m + 1/n = 4/3 and mn = 3. From the second equation, we have n = 3/m. Substituting this in the equation 1/m + 1/n = 4/3 yields 1/m + m/3 = 4/3. Multiplying the equation by 3m yields m2 – 4m + 3 = 0. The two possible solutions of this equation are 1 and 3.
When m = 1, n = 3/m = 3/1 = 3 and the expression 0.1 + 0.11/m + 0.11/n equals 0.1 + 0.11/1 + 0.11/3; and when m = 3, n = 3/m = 3/3 = 1 and the expression 0.1 + 0.11/m + 0.11/n equals 0.1 + 0.11/3 + 0.11/1.
In either case, the expressions equal 0.1 + 0.11/3 + 0.11/1 = 0.2 + 0.11/3. Hence, the answer is (A).
272GRE Math Bible
28. We are given the two equations:
(x – 2y)(x + 2y) = 5 (2x – y)(2x + y) = 35
Applying the Difference of Squares formula, (a + b)(a – b) = a2 – b2, to the left-hand sides of each equation yields
x2 – (2y)2 = 5 (2x)2 – y2 = 35
Simplifying these two equations yields
x2 – 4y2 = 5 4x2 – y2 = 35
Subtracting the bottom equation from the top one yields
(x2 – 4y2) – (4x2 – y2) = 5 – 35 –3x2 – 3y2 = –30
–3(x2 + y2) = –30
x2 + y2 = –30/–3 = 10
Now, Column A – Column B = (2x2 – y2) – (x2 – 2y2) = x2 + y2 = 10. Hence, Column A is 10 units greater than Column B. The answer is (A).
29. Let each part of the given equation |
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Simplifying, we get x = t(a + b – 2c) = at + bt – 2ct, y = t(b + c – 2a) = bt + ct – 2at, and z = t(c + a – 2b) = ct + at – 2bt.
Hence, x + y + z = (at + bt – 2ct) + (bt + ct – 2at) + (ct + at – 2bt) = 0. The answer is (A).
30. Let each expression in the equation equal k. Then we have |
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= (b – c)k, y = (c – a)k, z = (a – b)k. Now, ax + by + cz equals |
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a(b – c)k + b(c – a)k + c(a – b)k = |
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31. Let each expression in the equation |
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Simplifying yields
x = k(b – c) = kb – kc y = k(c – a) = kc – ka z = k(a – b) = ka – kb
Hence, x + y + z = (kb – kc) + (kc – ka) + (ka – kb) = 0. The answer is (A).
Equations 273
32. Solving the given equation for y yields y = 750 – x. Now, let’s substitute this into each answer-choice. The one that returns a numeric value for x is the answer.
Choice (A): x + 2y = d; x + 2(750 – x) = d; x + 1500 – 2x = d; x = 1500 – d; Since d is unknown, the value of x cannot be calculated. Reject.
Choice (B): 2x + 4y = 2d; 2x + 4(750 – x) = 2d; 2x + 3000 – 4x = 2d; –2x + 3000 = 2d; x = 1500 – d; Since d is unknown, the value of x cannot be calculated. Reject.
Choice (C): 2x + 2y = 1500; 2x + 2(750 – x) = 1500; 2x + 1500 – 2x = 1500; 0 = 0, a known fact. No derivation is possible from this. Hence, the value of x cannot be calculated. Reject.
Choice (D): 3x = 2250 – 3y; 3x = 2250 – 3(750 – x) = 3x; 3x = 3x; A known result. Hence, no derivation is possible. Reject.
Choice (E): 2x + y = 15; 2x + 750 – x = 15; x + 750 = 15; x = 15 – 750 = –735. We have numeric value for x here. Hence, accept it.
The answer is (E).
Very Hard
33. We are given three equations ax = b , by = c, and cz = a. From the first equation, we have b = ax. Substituting this in the second equation gives (ax)y = c. We can replace a in this equation with cz (according to the third equation cz = a):
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The answer is (B).
Average Speed 
276GRE Math Bible
Medium
4.A group of 30 employees of Cadre A has a mean age of 27. A different group of 70 employees of Cadre B has a mean age of 23. What is the mean age of the employees of the two groups together?
(A)23
(B)24.2
(C)25
(D)26.8
(E)27
5.The difference between two angles of a triangle is 24°. The average of the same two angles is 54°. Which one of the following is the value of the greatest angle of the triangle?
(A)45°
(B)60°
(C)66°
(D)72°
(E)78°
6.The average length of all the sides of a rectangle equals twice the width of the rectangle. If the area of the rectangle is 18, what is its perimeter?
(A)6
6
(B)8
6
(C)24
(D)32
(E)48
7.In quadrilateral ABCD, A measures 20 degrees more than the average of the other three angles of the quadrilateral. Then A =
(A)70°
(B)85°
(C)95°
(D)105°
(E)110°
8.The five numbers 1056, 1095, 1098, 1100, and 1126 are represented on a number line by the points A, B, C, D, and E, respectively, as shown in the figure. Which one of the following points represents the average of the five numbers?
(A)Point A
(B)Point B
(C)Point C
(D)Point D
(E)Point E
A(1056) B(1095) C(1098) D(1100) E(1126)
9.The arithmetic mean (average) of m and n is 50, and the arithmetic mean of p and q is 70. What is the arithmetic mean of m, n, p, and q?
(A)55
(B)65
(C)60
(D)120
(E)130
278GRE Math Bible
Very Hard
17.40% of the employees in a factory are workers. All the remaining employees are executives. The annual income of each worker is $390. The annual income of each executive is $420. What is the average annual income of all the employees in the factory together?
(A)390
(B)405
(C)408
(D)415
(E)420
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Averages 279
Answers and Solutions to Problem Set P
Easy
1. 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.”
From the number line M = 16, P = 18, Q = 20, R = 21, and S = 25. The numbers arranged in order are 16, 18, 20, 21, and 25. The median is 20. Since Q = 20, the answer is (C).
2. The average of 13 and 23 is (13 + 23)/2 = 36/2 = 18, so the last digit is 8.
The average of 113 and 123 is (113 + 123)/2 = 236/2 = 118, so the last digit is also 8.
Hence, Column A equals Column B, and the answer is (C).
3. Column A: The mean rainfall for the 8 months is the sum of the eight rainfall measurements divided by 8:
2 + 4 + 4 + 5+ 7 + 9 +10 +11 = 6.5 8
Column B: 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. The average of 5 and 7 is 6, which is the median of the set. Hence, Column A is greater than Column B, and the answer is (A).
Medium
4. Cadre A has 30 employees whose mean age is 27. Hence, the sum of their ages is 30 27 = 810. Cadre B has 70 employees whose mean age is 23. Hence, the sum of their ages is 23 70 = 1610. Now, the total sum of the ages of the 100 (= 30 + 70) employees is 810 + 1610 = 2420. Hence, the average age is
The sum of the ages divided by the number of employees =
2420/100 =
24.2
The answer is (B).
5. Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24°, so a – b = 24. Since the average of the two angles is 54°, we have (a + b)/2 = 54. Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields
a + (a 24) = 54 2
2a 24 = 54
2
2a – 24 = 54 2
2a – 24 = 108
2a = 108 + 24
2a = 132 a = 66
Also, b = a – 24 = 66 – 24 = 42.
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is 180°, a + b + c = 180. Plugging the previous results into the equation yields 66 + 42 + c = 180. Solving for c yields c = 72. Hence, the greatest of the three angles a, b and c is c, which equals 72°. The answer is (D).
280 GRE Math Bible
6. The perimeter of a rectangle is twice the sum of its length and width. Hence, if l and w are length and width, respectively, of the given rectangle, then the perimeter of the rectangle is 2(l + w). Also, the average side length of the rectangle is 1/4 times the sum. So, the average side length is 2(l + w)/4 = l/2 + w/2.
Now, we are given that the average equals twice the width. Hence, we have l/2 + w/2 = 2w. Multiplying the equation by 2 yields l + w = 4w and solving for l yields l = 3w.
Now, the area of the rectangle equals length width = l w = 18 (given). Plugging 3w for l in the equation yields 3w w = 18. Dividing the equation by 3 yields w2 = 6, and square rooting both sides yields w = 
6 . Finally, the perimeter equals 2(l + w) = 2(3w + w) = 8w = 8
6 . The answer is (B).
7. Setting the angle sum of the quadrilateral to 360° yields A + B + C + D = 360. Subtracting A from both sides yields B + C + D = 360 – A. Forming the average of the three angles B, C, andD yields (B + C + D)/3 and this equals (360 – A)/3, since we know that B + C + D = 360 – A. Now, we are given that A measures 20 degrees more than the average of the other three angles. Hence, A = (360 – A)/3 + 20. Solving the equation for A yields A = 105. The answer is (D).
8. The average of the five numbers 56, 95, 98, 100, and 126 is
56 + 95+ 98 +100 +126 = 5
475
5 =
95
Hence, the average of the five numbers 1056 (= 1000 + 56), 1095 (= 1000 + 95), 1098 (= 1000 + 98), 1100 (= 1000 + 100), and 1126 (= 1000 + 126) must be 1000 + 95 = 1095. The point that represents the number on the number line is point B. Hence, the answer is (B).
9. The arithmetic mean of m and n is 50. Hence, (m + n)/2 = 50. Multiplying the equation by 2 yields m + n = 100.
The arithmetic mean of p and q is 70. Hence, (p + q)/2 = 70. Multiplying the equation by 2 yields p + q = 140.
Now, the arithmetic mean of m, n, p, and q is
m + n + p + q = 4
(m + n) + (p + q) =
4
100 +140 =
4
240
4 =
60
The answer is (C).
10. The average of the elements in the original set S is (0 + 2 + 4 + 5 + 9)/5 = 20/5 = 4. If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}. The average of the elements is (0 + 2 + 5 + 9)/4 = 16/4 = 4. The answer is (C).