GRE_Math_Bible_eBook

Sequences & Series 401

7.In the sequence an, the nth term is defined as (an – 1 – 1)2. If a1 = 4, then what is the value of a2 ?

(A)2

(B)3

(C)4

(D)5

(E)9

Hard

8.A worker is hired for 7 days. Each day, he is paid 10 dollars more than what he is paid for the preceding day of work. The total amount he was paid in the first 4 days of work equaled the total amount he was paid in the last 3 days. What was his starting pay?

(A)90

(B)138

(C)153

(D)160

(E)163

9.A sequence of numbers is represented as a1, a2, a3, …, an. Each number in the sequence (except the first and the last) is the mean of the two adjacent numbers in the sequence. If a1 = 1 and a5 = 3, what is the value of a3 ?

(A)1/2

(B)1

(C)3/2

(D)2

(E)5/2

10.A series has three numbers a, ar, and ar2. In the series, the first term is twice the second term. What is the ratio of the sum of the first two terms to the sum of the last two terms in the series?

(A)1 : 1

(B)1 : 2

(C)1 : 4

(D)2 : 1

(E)4 : 1

11.The sequence of numbers a, ar, ar2, and ar3 are in geometric progression. The sum of the first four terms in the series is 5 times the sum of first two terms and r –1 and a 0. How many times larger is the fourth term than the second term?

(A)1

(B)2

(C)4

(D)5

(E)6

12.In the sequence an, the nth term is defined as (an – 1 – 1)2. If a3 = 64, then what is the value of a2?

(A)2

(B)3

(C)4

(D)5

(E)9

404 GRE Math Bible

10. Since “the first term in the series is twice the second term,” we have a = 2(ar). Canceling a from both sides of the equation yields 1 = 2r. Hence, r = 1/2.

Hence, the three numbers a, ar, and ar2 become a, a(1/2), and a(1/2)2, or a, a/2, and a/4.

The sum of first two terms is a + a/2 and the sum of the last two terms is a/2 + a/4. Forming their ratio yields

11. In the given progression, the sum of first two terms isa + ar, and the sum of first four terms is a + ar + ar2 + ar3. Since “the sum of the first four terms in the series is 5 times the sum of the first two terms,” we have

a + ar + ar2 + ar3 = 5(a + ar)

Now, the fourth term is ar3/ar = r2 = 4 times the second term. Hence, the answer is (C).

Since, we know that a2 is the result of the square of number [a2 = (a1 – 1)2], it cannot be negative. Hence, pick the positive value 9 for a2. The answer is (E).

406 GRE Math Bible

Event One: 3 times

}Total number of events: m. k = 3. 2 = 6

Event Two: 2 times for each occurrence of Event One

Example 3: A drum contains 3 to 5 jars each of which contains 30 to 40 marbles. If 10 percent of the marbles are flawed, what is the greatest possible number of flawed marbles in the drum?

There is at most 5 jars each of which contains at most 40 marbles; so by the fundamental counting principle, there is at most 5 40 = 200 marbles in the drum. Since 10 percent of the marbles are flawed, there is at most 20 = 10% 200 flawed marbles. The answer is (D).

MISCELLANEOUS COUNTING PROBLEMS

Example 4: In a legislative body of 200 people, the number of Democrats is 50 less than 4 times the number of Republicans. If one fifth of the legislators are neither Republican nor Democrat, how many of the legislators are Republicans?

Let D be the number of Democrats and let R be the number of Republicans. "One fifth of the legislators are neither Republican nor Democrat," so there are 15 200 = 40 legislators who are neither Republican nor

Democrat. Hence, there are 200 – 40 = 160 Democrats and Republicans, or D + R = 160. Translating the clause "the number of Democrats is 50 less than 4 times the number of Republicans" into an equation yields D = 4R – 50. Plugging this into the equation D + R = 160 yields

4R – 50 + R = 160

5R – 50 = 160

5R = 210

R = 42

The answer is (A).

Example 5: Speed bumps are being placed at 20 foot intervals along a road 1015 feet long. If the first speed bump is placed at one end of the road, how many speed bumps are needed?

(A) 49 (B) 50 (C) 51 (D) 52 (E) 53

101520 = 50.75, or 50 full

sections in the road. If we ignore the first speed bump and associate the speed bump at the end of each section with that section, then there are 50 speed bumps (one for each of the fifty full sections). Counting the first speed bump gives a total of 51 speed bumps. The answer is (C).

SETS

A set is a collection of objects, and the objects are called elements of the set. You may be asked to form the union of two sets, which contains all the objects from either set. You may also be asked to form the intersection of two sets, which contains only the objects that are in both sets. For example, if Set A = {1, 2, 5} and Set B = {5, 10, 21}, then the union of sets A and B would be {1, 2, 5, 10, 21} and the intersection would be {5}.

Counting 409

Very Hard

13.A survey of n people in the town of Eros found that 50% of them prefer Brand A. Another survey of 100 people in the town of Angie found that 60% prefer Brand A. In total, 55% of all the people surveyed together prefer Brand A. What is the total number of people surveyed?

(A)50

(B)100

(C)150

(D)200

(E)250

410 GRE Math Bible

Answers and Solutions to Problem Set Y

Easy

1. Let j be the total number of marbles in the jar. Then 60%j must be red (given), and the remaining 40%j must be green (given). Now,

Column A equals 40% of the red marbles = 40%(60%j) = .40(.60j) = .24j.

Column B equals 60% of the green marbles = 60%(40%j) = .60(.40j) = .24j.

Since both columns equal .24j, the answer is (C).

Medium

2. Let the number of pigeons be p and the number of rabbits be r. Since the head count together is 12,

Since each pigeon has 2 legs and each rabbit has 4 legs, the total leg count is

Dividing equation (2) by 2 yields p + 2r = 16. Subtracting this equation from equation (1) yields

(p + r) – (p + 2r) = 12 – 16 p + r – p – 2r = –4

r = 4

Substituting this into equation (1) yields p + 4 = 12, which reduces to p = 8.

Hence, the number of pigeons is p = 8, and the number of rabbits is r = 4. The answer is (D).

3. Any number divisible by both 7 and 10 is a common multiple of 7 and 10. The least common multiple of 7 and 10 is 70. Hence, Column A reduces to the fraction of numbers from 0 through 1000 that are divisible by 70.

Similarly, any number divisible by both 5 and 14 is a common multiple of 5 and 14. The LCM of 5 and 14 is 70. Hence, Column B reduces to the fraction of numbers from 0 through 1000 that are divisible by 70.

Since the statement in each column is now the same, the columns are equal and the answer is (C).

Hard

4. The sum of 13/n, 18/n, and 29/n is 13+18 + 29 = 60 . Now, if 60/ n is to be an integer, n must be a factor n n

of 60. Since the factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60, there are 12 possible values for n. The answer is (E).

5.The sum of 3n, 9n, and 11n is 23n. Since this is to be greater than 200, we get the inequality 23n > 200. From this, we get n > 200/23 8.7. Since n is an integer, n > 8. Now, we are given that 5 n 20. Hence, the values for n are 9 through 20, a total of 12 numbers. The answer is (C).

6.We are given that that 59% of the employees E are workers. Since the factory consists of only workers, executives, and clerks, the remaining 100 – 59 = 41% of the employees must include only executives and clerks. Since we are given that the number of executives is 460 and the number of clerks is 360, which sum to 460 + 360 = 820, we have the equation (41/100)E = 820, or E = 100/41 820 = 2000. The answer is (B).