GRE_Math_Bible_eBook

Word Problems 391

13.Mike ran 10 miles at 10 miles per hour (Time = Distance/Rate = 10 miles/10 miles per hour = 1 hour). He ran at 5 miles per hour for the remaining 20 miles (Time = Distance/Rate = 20 miles/5 miles per hour = 4 hrs). The net length of the Marathon track is 30 miles, and the total time taken to cover the track is 5 hours. Hence, the answer is (E).

14.Suppose Fritz took t hours to complete the 30-mile Marathon. Then as given, Fritz ran at 10 miles per hour for t/3 hours and 5 miles per hour for the remaining 2t/3 hours. Now, by formula, Distance =

Rate Time, the net distance covered would be (10 miles per hour) t/3 + (5 miles per hour) 2 t/3 = (10/3 + 10/3)t = 30 miles. Solving the equation for t yields t = 90/20 = 4.5 hours. The answer is (D).

15.We have that water enters the ship at 2 tons per minute and the pumps remove the water at 1.75 tons per minute. Hence, the effective rate at which water is entering the ship is 2 – 1.75 = 0.25 tons per minute. Since it takes an additional 120 tons of water to sink the ship, the time left is (120 tons)/(0.25 tons per minute) = 120/0.25 = 480 minutes. The answer is (A).

16.Let the original price of each orange be x dollars. Remember that Quantity = Amount ÷ Rate. Hence, we can purchase 12/x oranges for 12 dollars. After a 40% drop in price, the new price is x(1 – 40/100) = 0.6x dollars per orange. Hence, we should be able to purchase 12/(0.6x) = 20/x oranges for the same 12 dollars. The excess number of oranges we get (for $12) from the lower price is 20/x – 12/x = (1/x)(20 – 12) = (1/x)(8) = 8/x = 4 (given). Solving the equation 8/x = 4 for x yields x = 2. Hence, the number of oranges that can be purchased for 24 dollars at original price x is 24/2 = 12. The answer is (B).

17. Since 6 mangoes are returnable for 9 oranges, if each mango costs m and each orange costs n, then 6m = 9n, or 2m = 3n. Solving for n yields, n = 2m/3. Now, since 50 mangoes and 30 oranges together cost 42 dollars,

50m + 30n = 42

50m + 30(2m/3) = 42 m(50 + 30 2/3) = 42 m(50 + 20) = 42 70m = 42

m = 42/70 = 6/10 = 0.6

The answer is (E).

18. One pound of rice costs 0.33 dollars. A dozen eggs cost as much as one pound of rice, and a dozen has 12 items. Hence, 12 eggs cost 0.33 dollars.

Now, since half a liter of kerosene costs as much as 8 eggs, one liter must cost 2 times the cost of 8 eggs, which equals the cost of 16 eggs.

Now, suppose 16 eggs cost x dollars. We know that 12 eggs cost 0.33 dollars. So, forming the proportion yields

The answer is (D).

392 GRE Math Bible

19. Suppose x and y are the amounts received in dollars by the elder and the younger son, respectively. Then the amount the father had is x + y.

The elder son received 3/5 of the amount. Expressing this as an equation yields

x = (3/5)(x + y) x = (3/5)x + (3/5)y (2/5)x = (3/5)y

x = (3/2)y

Hence, x + y, the amount father had, equals 3y/2 + y = 5y/2 = (5/2)(30,000) [Given that the younger son received 30,000 dollars] = 75,000, and the answer is (D).

20.We have that her average speed is 50 mph. The formula for the Average Speed is Distance Traveled ÷ Time Taken. Hence, we have the equation 50 mph = Distance Traveled ÷ 1 hour. Solving this equation yields Distance Traveled = 50 miles. Hence, the answer is (C).

21.Let t be the entire time of the trip.

We have that the car traveled at 80 mph for t/2 hours and at 40 mph for the remaining t/2 hours. Remember that Distance = Speed Time. Hence, the net distance traveled during the two periods equals 80 t/2 + 40 t/2. Now, remember that

Average Speed =

Net Distance = Time Taken

80 2t + 40 2t = t

80 12 + 40 12 =

40 + 20 =

60

The answer is (D).

22. Since Mr. Smith's average annual income in each of the two years 1966 and 1967 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 next three years 1968 through 1970 is y dollars, his total income in the three years is 3 y = 3y.

Hence, the net income in the five continuous years is 2x + 3y.

Hence, the average income in the five years is

(the net income) ÷ 5 = (2x + 3y)/5 = 2x/5 + 3y/5

The answer is (A).

23. Hose A takes 5 minutes to fill one tank. To fill 6 tanks, it takes 6 5 = 30 minutes. Hose B takes 6 minutes to fill one tank. Hence, in the 30 minutes, it would fill 30/6 = 5 tanks. The answer is (C).

Word Problems 393

24. 12 trophies cost 60 dollars, so each trophy costs 60/12 = 5 dollars.

If the price decreases by 1, the new price is 5 – 1 = 4 dollars. Hence, 60 dollars can now buy 15 (= 60/4) trophies. Equating this to 12 + x yields 12 + x = 15, or x = 3.

If the price increases by 1, the new price is 5 + 1 = 6 dollars. Hence, 60 dollars can now buy 10 (= 60/6) trophies. Equating this to 12 – y yields 12 – y = 10, or y = 2.

Now, Column A equals x = 3, and Column B equals y = 2. Hence, Column A > Column B, and the answer is (A).

Very Hard

25. Let m and n be the costs of the equities of type A and type B, respectively. Since the costs are integers (given), m and n must be positive integers.

We have that 4 equities of type A and 5 equities of type B together cost 27 dollars. Hence, we have the equation 4m + 5n = 27. Since m is a positive integer, 4m is a positive integer; and since n is a positive integer, 5n is a positive integer. Let p = 4m and q = 5n. So, p is a multiple of 4 and q is a multiple of 5 and p + q = 27. Subtracting q from both sides yields p = 27 – q [(a positive multiple of 4) equals 27 – (a positive multiple of 5)]. Let’s seek such a solution for p and q:

If q = 5, p = 27 – 5 = 22, not a multiple of 4. Reject. If q = 10, p = 27 – 10 = 17, not a multiple of 4. Reject.

If q = 15, p = 27 – 15 = 12, a multiple of 4. Acceptable. So, n = p/4 = 3 and m = q/5 = 3.

The following checks are not actually required since we already have an acceptable solution.

If q = 20, p = 27 – 20 = 7, not a multiple of 4. Reject.

If q = 25, p = 27 – 25 = 2, not a multiple of 4. Reject.

If q 30, p 27 – 30 = –3, not positive. Reject.

Hence, the cost of 2 equities of type A and 3 equities of type B is 2m + 3n = 2 3 + 3 3 = 15. The answer is (A).

26. Let m coins of 0.5 dollars each and n coins of 0.7 dollars each add up to 4.6 dollars. Then, we have the equation 0.5m + 0.7n = 4.6. Multiplying both sides by 10 to eliminate the decimals yields 5m + 7n = 46. Since m is a positive integer, 5m is positive integer; and since n is a positive integer, 7n is a positive integer. Let p = 5m and q = 7n. So, p is a multiple of 5 and q is a multiple of 7 and p + q = 46. Subtracting q from both sides yields p = 46 – q [(a positive multiple of 5) equals 46 – (a positive multiple of 7)]. Let’s seek such solution for p and q:

If q = 7, p = 46 – 7 = 39, not a multiple of 5. Reject.

If q = 14, p = 46 – 14 = 32, not a multiple of 5. Reject.

If q = 21, p = 46 – 21 = 25, a multiple of 5. Acceptable. So, n = q/7 = 3 and m = p/5 = 5. The following checks are not actually required since we already have an acceptable solution.

If q = 28, p = 46 – 28 = 18, not a multiple of 5. Reject.

If q = 35, p = 46 – 35 = 11, not a multiple of 5. Reject.

If q = 42, p = 46 – 42 = 4, not a multiple of 5. Reject.

If q 49, p = 46 – 49 = –3, not positive. Reject.

The answer is (E).

394 GRE Math Bible

27. Let the length of the train be l and the length of the bridge be b. We are given that the train crosses the bridge in 15 seconds, and it crosses the same bridge in 10 seconds when its length is halved. In each case, the speed is the same. Hence, let's derive expressions for speed for each case and equate. The distance traveled by the train in crossing the bridge is

(length of bridge) + (length of train)

In the first case, distance traveled is b + l, and, in the second case, with train length halved, distance traveled is b + l/2. In the first case, time taken is 15 seconds, and in the second case, time taken is 10 seconds. By the formula Speed = Distance/Time, the speed in the first case is (b + l)/15 and in the second case is (b + l/2)/10. Since the speed in both cases is the same, we have

2(b + l) = 3(b + l/2) 2b + 2l = 3b + 3l/2 l/2 = b

l = 2b

Hence, the length of the train is twice the length of the bridge. So, Column B is greater, and the answer is

(B).

28. If d is the distance to the office from home and if v is his usual speed, then the time taken is t = d/v (by the formula Time = Distance ÷ Speed). Hence, we only need to find the ratio d/v.

Now, we are given that on a day after starting 30 minutes late and driving 25% slower [i.e., at speed (1 – 25/100)v = 0.75], he reached his office 50 minutes late.

Here, there are two delays: One delay is the “30 minute delay in starting from home” and the other is from driving slower than his regular speed. The total delay is 50 minutes. Hence, the delay caused by driving slower is 50 – 30 = 20.

Now, usually, Richards would have taken the time d/v to reach the office. But, on this day, he drove at a speed 25% slower [Speed = v (1 – 25/100) = 0.75v]. Hence, the time he would have taken for the trip equals d/0.75v, and the component of the delay from driving slower is

d = 20 minutes (given) 3v

Multiplying both sides by 3 yields d/v = 60. The answer is (C).

29. The cost of wheat at Katrina’s store is $3 per pound. After the 10% discount (festival season discount) the cost of the wheat would be, by the known formula the original price (1 – discount percent/100) = 3(1 – 10/100) = 3(90/100) = 2.7 dollars per pound.

Since her faulty balance was reading 0.9 pounds for a pound, she was unknowingly selling 1 pound in the name of 0.9 pounds This is equivalent to selling 1/0.9 = 10/9 pounds in the name of one pound. Hence, Katrina was trying to sell the wheat at 2.7 dollars per pound, but effectively she was selling the wheat at 2.7 dollars per (10/9 pounds) = 2.7 9/10 = 2.43 dollars a pound. Since she earned neither a profit nor a loss, she must have purchased the wheat at this same cost from the wholesaler. Hence, the answer is (A).

Word Problems 395

30. We have that the number of employees in the technical division is 15 and the number of employees in the recruitment division is 10. Each technical person is given 15 shares, and each recruitment person is given 10 shares. Hence, the net shares given equals 15 15 + 10 10 = 225 + 100 = 325. Each share is worth 10 dollars, so the net worth of the shares is 325 10 = 3,250. The answer is (E).

Note: In the problem, employees belonging to the technical community get 15 shares, employees in the recruitment community get 10 shares, and the employees in both communities get 25 (= 15 + 10) shares. Effectively, independent treatment is given to the two communities. Hence, we did not give any special consideration to employees who are in both communities.

31. Let d be the distance between the towns A and B. 65% of this distance (= 65% of d = 65/100 d = 0.65d) was traveled at 65 mph and the remaining 100 – 65 = 35% of the distance was traveled at v mph. Now, Remember that Time = Distance ÷ Rate. Hence, the time taken by the car for the first 65% distance is 0.65d/65 = d/100, and the time taken by the car for the last 35% distance is 0.35d/v. Hence, the total time taken is d/100 + 0.35d/v = d(1/100 + 0.35/v).

Now, remember that Average Speed = Total Distance Traveled .

1= 50 1 + 50 0.35 100 v

1= 12 + 50 0.35v

1 = 50 0.35

2 v

v = 2 50 0.35 = 100 0.35 = 35 The answer is (E).

396 GRE Math Bible

32. The string is cut into three along its length. Let l be the length of the smallest piece. Then the length of the longest piece is 3l, and the total length of the three pieces is 35 cm. The length of the longest and shortest pieces together is l + 3l = 4l. Hence, the length of the third piece (medium-size piece) must be 35 – 4l. Arranging the lengths of the three pieces in increasing order of length yields the following inequality:

l < 35 – 4l < 3l 5l < 35 < 7l

5l < 35 and 35 < 7l l < 7 and 5 < l

5 < l < 7

20 < 4l < 28 –20 > –4l > –28

35 – 20 > 35 – 4l > 35 – 28 15 > 35 – 4l > 7

15 > The length of the medium-size piece > 7 15 > Column A > 7

Column B > Column A > 7

Hence, the answer is (B).

Method II:

Had the length of the medium-size piece been greater than or equal to 15, the length of the longest-size piece would be greater than 15 and the length of the smallest piece, which equals 1/3 the length of the longest piece, would be greater than 15/3 = 5. Hence, the net sum of the three lengths exceeds 15 + 15 + 5 (= 35). Since this is impossible, our assumption that the length of the medium-sized piece is greater than or equal to 15 is false. Hence, it is less than 15 and therefore Column A is less than Column B. The answer is

(B).

Sequences & Series 399

Advanced concepts: (Sequence Formulas)

Note, none of the formulas in this section are necessary to answer questions about sequences on the GRE.

Since each term of a geometric progression “is generated by multiplying the preceding term by a fixed number,” we get the following:

The sum of the first n terms of an geometric sequence is

a(1 r n )

1 r

SERIES

A series is simply the sum of the terms of a sequence. The following is a series of even numbers formed from the sequence 2, 4, 6, 8, . . . :

2 + 4 + 6 + 8 + . . .

A term of a series is identified by its position in the series. In the above series, 2 is the first term, 4 is the second term, etc. The ellipsis symbol (. . .) indicates that the series continues forever.

We are given a formula for the sum of the squares of the first n positive integers. Pluggingn = 9 into this formula yields

3 = 2(3) + (2 3 1) + (2 3 2) + (2 3 3) + (2 3 4) + (2 3 5) = 6 + 5+ 4 + 3+ 2 + 1 = 21

2 = 2(2) + (2 2 1) + (2 2 2) + (2 2 3) = 4 + 3 + 2 + 1= 10

Hence, 3 2 = 21 10 = 210, and the answer is (C).

400GRE Math Bible

Problem Set X:

Medium

1.In a sequence, the nth term an is defined by the rule (an - 1 – 3)2, a1 = 1. What is the value of a4 ?

(A)1

(B)4

(C)9

(D)16

(E)25

2.If the nth term in a sequence of numbers a0, a1, a2, …, an is defined to equal 2n + 1, then what is the numerical difference between the 5th and 6th terms in the sequence?

(A)1

(B)2

(C)4

(D)5

(E)6

3.A sequence of numbers a1, a2, a3, …, an is generated by the rule an+1 = 2an. If a7 – a6 = 96, then what is the value of a7 ?

(A)48

(B)96

(C)98

(D)192

(E)198

4.A sequence of positive integers a1, a2, a3, … , an is given by the rule an+1 = 2an + 1. The only even number in the sequence is 38. What is the value of a2 ?

(A)11

(B)25

(C)38

(D)45

(E)77

5.The sum of the first n terms of an arithmetic series whose nth term is n can be calculated by the formula n(n + 1)/2. Which one of the following equals the sum of the first eight terms in a series whose nth term is 2n ?

(A)24

(B)48

(C)56

(D)72

(E)96

6.The sum of the first n terms of a series is 31, and the sum of the first n – 1 terms of the series is 20. What is the value of nth term in the series?

(A)9

(B)11

(C)20

(D)31

(E)51