GRE_Math_Bible_eBook

Miscellaneous Problems 491

Questions 11 and 12 refer to the discussion below:

A manufacturer sells goods at $4 per unit to stockists after a 10% profit. The stockists then sell the goods to distributors at 25% profit. The distributor adds a 20% profit on it and sells it to a retailer.

11.At what price (per unit) did the retailer purchase the goods?

(A)0.6

(B)6

(C)6.3

(D)6.6

(E)7

12.If the retailer sells the goods to the end customer at 10% profit, then what does each unit of the goods cost to a customer?

(A)4

(B)5

(C)6

(D)6.6

(E)7

13.The letters of the word JOHNY can be jumbled in 120 ways. In how many of them does the letter ‘H’ appear in the middle?

(A)1

(B)20

(C)24

(D)26

(E)30

14.Craig invited four friends to watch a TV show. He arranged 5 seats in a row. The number of ways he and his four friends can sit in the row is n. In how many of these ways can Craig sit in the middle?

(A)n

(B)n/2

(C)n/3

(D)n/4

(E)n/5

15.Eric and Ortega and their teammates watch a movie. They all sit in a row, and they can sit in n different ways. In how many of the ways can Eric sit to the right of Ortega?

(A)n

(B)n/2

(C)n/3

(D)n/4

(E)n/5

16.If distinct numbers x, y, z, and p are chosen from the numbers –2, 2, 1/2, –1/3, what is the largest

x2y possible value of the expression z − p ?

(A)26/4

(B)34/5

(C)38/5

(D)48/5

(E)52/5

492GRE Math Bible

17.A bank pays interest to its customers on the last day of the year. The interest paid to a customer is calculated as 10% of the average monthly balance maintained by the customer. John is a customer at the bank. On the last day, when the interest was accumulated into his account, his bank balance doubled to $5680. What is the average monthly balance maintained by John in his account during the year?

(A)2840

(B)5680

(C)6840

(D)7540

(E)28400

Use the table shown to answer the next two questions.

18.Which one of the following diets or the combination of diets supplies more protein for every dollar?

(A)Food A

(B)Food B

(C)Food C

(D)60% food B and 40% food C

(E)60% food A and 40% food C

19.Which diet supplies more protein than fat in each gram?

(A)Food A

(B)Food B

(C)Foods A and B in the ratio 60 : 40

(D)Foods B and C in the ratio 60 : 40

(E)Foods A and C in the ratio 60 : 40

20.Forty tiles of dimensions 1 foot × 2 foot each are required to completely cover a floor. How many tiles of dimensions 2 foot × 4 foot each would be required to completely cover the same floor?

(A)10

(B)20

(C)80

(D)160

(E)320

21.If s and t are positive integers and s/t = 39.12, then which one of the following could t equal?

(A)8

(B)13

(C)15

(D)60

(E)75

494GRE Math Bible

Very Hard

28.In 2003, there are 28 days in February and there are 365 days in the year. In 2004, there are 29 days in February and there are 366 days in the year. If the date March 11, 2003 is a Tuesday, then which one of the following would the date March 11, 2004 be?

(A)Monday

(B)Tuesday

(C)Wednesday

(D)Thursday

(E)Sunday

29.There are 5 packers A, B, C, D, and E in a company. The five packers A, B, C, D and E charge $66, $52, $46, $32, and $28, respectively, to pack each item. The time taken by the packers to pack one item is 20 minutes, 24 minutes, 30 minutes, 40 minutes and 48 minutes, respectively. All the items are sold at the end of the day. Each item earns a profit of 100 dollars, and the packers are paid from this profit. If each packer works 8 hours a day, which packer contributes the most to the net profit of the company?

(A)Packer A

(B)Packer B

(C)Packer C

(D)Packer D

(E)Packer E

Miscellaneous Problems 495

Answers and Solutions to Problem Set CC

Easy

1.We have that the value of a share of Company A increased by 13 dollars and that of Company B decreased by 8 dollars. Hence, the net increase in the combined value of the two shares is 13 – 8 = 5. The answer is (D).

2.Neel purchased the shavers at $4 a piece and sold them at $6 a piece. Hence, he gained 6 – 4 = 2 dollars on each piece. On n such pieces, he gained (n items)(2 dollars on each item) = 2n dollars. Hence, Column A equals 2n.

Nick purchased the same items at $6 each and sold them at $8 each. Hence, he gained 8 – 6 = 2 dollars on each piece. On the n pieces, he gained (n items)(2 dollars on each item) = 2n dollars. Hence, Column B also equals 2n.

Since Column A equals Column B, the answer is (C).

Medium

3. Neel purchased the shavers at $4 a piece. Hence, n pieces cost him 4n dollars.

Nick purchased the shavers from Neel at $6 a piece. Hence, n pieces cost him 6n dollars.

Column A: The percentage of profit made by Neel equals profitcost 100 = 42nn 100 = 12 100 = 50%. Column B: The percentage of profit made by Nick equals profitcost 100 = 26nn 100 = 13 100 ≈ 33.3%.

Hence, Column A is greater than Column B, and the answer is (A).

4. Each answer-choice has two factors. The first factor of each answer-choice varies from 6.00 to 6.04, and the second factor varies from 0.16 to 0.20. The percentage change in the first factor is very small (0.67%) compared to the large (almost 25%) change in the second factor. Hence, we can approximate the first factor with 6.00, and the answer-choice that has the greatest second factor [choice (A)] is the biggest. Hence, the answer is (A).

Method II:

All the answer-choices are positive. Hence, we can use the ratios of the answer-choices to find which choice is the greatest:

Positive Numerator Lesser Positive Denominator >1

Hence, Choice (A) > Choice (D). Reject Choice (D).

496 GRE Math Bible

Positive Numerator Lesser Positive Denominator >1

Hence, Choice (A) > Choice (E). Reject Choice (E).

The answer is (A).

5.Since the value of the fund doubled each year for the last 10 years, its value would halve each year going back for the period. Hence, the answer is (B).

6.We know that the square of a sum of n positive numbers is always greater than the sum of the squares of the n numbers. For example, 25 = (2 + 3)2 > 22 + 32 = 13. So, (15 + 17 + 19)2 is greater than 152 + 172 + 192. Hence, Column B is greater than Column A, and the answer is (B).

7.We are given that the average temperature from January through August is 36°C.

The average of a set of numbers always lies between the smallest number and the greatest number in the set. The minimum and the maximum temperatures between September and December are 26°C and 36°C, respectively. Hence, the average temperature of the period lies between 26°C and 36°C. So, the average temperature for the period September through December is less than 36°C.

So, the overall temperature for the year is less than 36°C. Hence, Column A is less than Column B, and the answer is (B).

8. The period February 28, 1999 through February 28, 2000 (not including the former date) does not include the complete month of February 2000 (which actually had 29 days). Hence, the length of the period is exactly 365 days (equal to the length of a normal year). Now, dividing 365 by 7 (the number of days in a week) yields a quotient of 52 and a remainder of 1. Hence, the exact length of the period is 52 weeks and one day. Hence, the day February 28 of the year 2000 would advance by one day over the date February 28 of the year 1999. Hence, since February 28, 1999 is a Sunday, the date February 28, 2000 is a Monday. The answer is (A).

Hard

9. Let the amounts received by Park, Jack, and Galvin be P, J, and G, respectively.

Since the prize money of $120 was distributed to Park, Jack, and Galvin, the amount that Jack and Galvin together received equals 120 – (the amount received by Park) = 120 – P.

Since we are given that Park received 3/10 of what Jack and Galvin together received, we have the equation

P = (3/10)(120 – P)

P = 3/10 120 – 3/10 P P + 3/10 P = 3/10 120 13/10 P = 3/10 120

P = 3/13 120 = Column A

Similarly, since we are given that Jack received 3/11 of what Park and Galvin together received (120 – J), we have the equation

J = (3/11)(120 – J)

J = 3/11 120 – 3/11 J J + 3/11 J = 3/11 120 14/11 J = 3/11 120 J = 3/14 120 = Column B

Since 3/13 120 is greater than 3/14 120, Column A > Column B and the answer is (A).

Miscellaneous Problems 497

10. Marc contributed 2/5 of what his three friends contributed together, while Jack contributed only 1/3 of what his three friends contributed together. Clearly, Marc must have contributed more than Jack. Let’s see this in detail:

Let the total contribution of the four friends be T, the contribution by Jack be J, and the contribution by Marc be M.

Now, the contribution by the three friends other than Jack is T – J. Jack contributed 1/3 of this. Hence, we have J = (1/3)(T – J), or J = T/4 (by solving for J).

Also, the contribution by the three friends other than Marc is T – M. As given, Marc contributed 2/5 of this. Hence, we have M = (2/5)(T – M), or M = 2T/7 (by solving for M).

Now, 2T/7 is greater than T/4 and therefore Marc contributed more. The answer is (B).

11. The supply chain can be visually mapped as shown:

The manufacturer sells the goods to stockists at $4 per unit.

The stockist now sells the goods to distributors at a profit of 25%. By the formula, Selling price = (Cost price)(1 + profit percent/100), the selling price of the stockist (which also equals the cost price to the distributors) is 4(1 + 25/100) = 4(1 + 1/4) = 4(5/4) = $5.

Then the distributor sells the goods to retailers after a profit of 20%. Again since the Selling price = (Cost price)(1 + profit percent/100), the selling price of the distributor is 5(1 + 20/100) = 5(1 + 1/5) = 5(6/5) = $6. This is also the cost price of the retailer. Hence, the answer is (B).

12.The cost price to the end customer is the selling price of the retailer. The goods cost $6 per unit to the retailer (from the solution of the previous question). He sells it to the end customer after 10% profit. Hence, by the formula, selling price = (cost price)(1 + profit percent/100) = $6(1 + 10/100) = $6.6. The answer is

(D).

13.To form a word by jumbling the five-lettered word JOHNY, the letter ‘H’ can be placed in any one of the five relative positions with equal probability. Hence, one fifth of the jumbled words will have the letter

‘H’ in the middle. We know that a total of 120 words can be formed by jumbling the word JOHNY. Hence, one-fifth of them (1/5 × 120 = 24 words) have the letter ‘H’ in the middle. The answer is (C).

Method II

Let the letters of the words be represented by 5 compartments:

Placing H in the middle compartment gives

H

Now, there are 4 letters remaining for the first position (J, O, N, Y):

Since one of these letters will be used for the first position, there are 3 letters available for the second position:

4 3 H

Similarly, there are 2 letters available for the fourth position and 1 letter for the fifth position:

4 3 H 2 1

Miscellaneous Problems 499

19. The protein-fat ratio for the foods here equals the ratio of “% of protein” to “% of fat.” The ratios for foods A, B and C are, respectively, 10/30 = 1/3, 20/20 = 1/1, 30/35 = 6/7. In all of these, protein is less than fat.

Choice (C): Now, considering the combination in choice (C), food A to B ratio is 60 : 40. Hence, choose 6 gm of food A and 4 gm of food B, so that foods are in this ratio.

Food A has 10% protein and 30% fat. 10% of 6 gm is 0.6 and 30% of 6 is 1.8. Food B has 20% protein and 10% fat. 20% of 4 gm is 0.8 and 10% of 4 is 0.4. The net sum: Protein = 0.6 (from A) + 0.8 (from B) = 1.4.

The net sum: Fat = 1.8 (from A) + 0.4 (from B) = 2.2. The food has less protein than Fat. Reject the choice.

Choice (D): Now, considering the combination in choice (D), food B to C ratio is 60 : 40. Hence, choose 6 gm of food B and 4 gm of food C, so that foods are in this ratio.

Food B has 20% protein and 10% fat. 20% of 6 gm is 1.2 and 10% of 6 is 0.6.

Food C has 30% protein and 35% fat. 30% of 4 gm is 1.2 and 35% of 4 is 1.4.

The net sum: Protein = 1.2 (from B) + 1.2 (from C) = 2.4.

The net sum: Fat = 0.6 (from B) + 1.4 (from C) = 2.0.

The food has more protein than Fat. This is a correct choice.

Choice (E): Both the foods A and C have higher fat percentage than protein percentage. So, any combination of the food will not result in a higher protein than fat. Reject the choice.

The answer is (D).

20.The area of a 2 × 4 tile is 4 times as large as the area of a 1 × 2 tile. Hence, we need only 1/4 as many large tiles to cover the same area as the 40 small tiles. Hence, 40/4 = 10. The answer is (A).

21.We have that s/t = 39.12. Solving for s yields s =

39.12t =

(39 + 0.12)t =

39t + 0.12t =

39 × (a positive integer) + 0.12t = (a positive integer) + 0.12t

s is a positive integer only when 0.12t is also a positive integer. Now, 0.12t equals 12/100 × t = 3/25 × t and would result in an integer only when the denominator of the fraction (i.e., 25) is canceled out by t. This happens when t is a multiple of 25. The answer is (E), the only answer-choice that is a multiple of 25.

22. Let m columns and n rows be formed from the tiles of size 16 × 24. Let the columns be formed by the 16 inch sides and the rows be formed by the 24 inch sides. Then the total length of all the rows is 16m, and the total length of all the columns is 24n.

Since the result is a square, 16m = 24n or m/n = 24/16 = 3/2. The minimum possible values of m and n are 3 and 2, respectively. Hence, the total number of tiles required is mn = 3 × 2 = 6. The answer is (A).

Method II

Since the tiles form a square, a side of the square formed must be a multiple of both 16 and 24. The least such number is 48. Since 48/16 = 3, there are three columns of length 16. Since 48/2 = 24, there are two rows of length 24. Hence, the total number of tiles required is 3 × 2 = 6. The answer is (A).

23. The consignment of c flowers cost Nancy d dollars. Hence, effectively, each flower cost her d/c dollars. Now, she made bouquets of d flowers each and sold them at the price of c dollars. Hence, effectively, the selling price of each flower is c/d dollars. Since, overall she made a profit, the selling price per flower must be greater than the cost per flower. Hence, we have c/d > d/c. Multiplying both sides of the inequality by cd yields c2 > d2. Square rooting both sides yields c > d. Hence, Column A is greater than Column B, and the answer is (A).

500 GRE Math Bible

24. The consignment of c flowers cost Nancy d dollars. Hence, effectively, each flower cost her d/c dollars. Hence, Column A equals d/c. Now, she made bouquets of d flowers each, and sold them at the price of c dollars. Hence, effectively, the selling price of each flower is c/d dollars. Since, overall she made a profit, the selling price per flower must be greater than the cost per flower. Hence, we have c/d > d/c. Multiplying both sides of the inequality by cd yields c2 > d2. Square rooting both sides yields c > d. Dividing both sides of the inequality c > d by c yields 1 > d/c. Hence, Column A is less than Column B, and the answer is (B).

25. Since 72 bananas cost 720 cents, each banana costs 720/72 = 10 cents.

When price is reduced by 1, the new price is 10 – 1 = 9 cents. Hence, 720 cents can buy 80 (= 720/9) bananas. Equating this to 72 + x yields 72 + x = 80, or x = 80 – 72 = 8.

When price is reduced by 2, the new price is 10 – 2 = 8 cents. Hence, 720 cents can buy 90 (= 720/8) bananas. Equating this to 72 + y yields 72 + y = 90, or y = 90 – 72 = 18.

Now, Column A equals 2x = 2(8) = 16, and this is less than 18 (= Column B). The answer is (B).

26. Let the cost of each apple be a, and the cost of each orange be b. Since 100 apples and 150 oranges cost the same as 50 apples and 225 oranges, we have the equation

100a + 150b = 50a + 225b 50a = 75b

a = (75/50)b = 3b/2 2a/3 = b

Hence, the apples cost 3/2 times as much as the oranges. Now, 100 apples cost 100a, and 150 oranges cost 150b = 150(2a/3) [Using the known equation 2a/3 = b] = 100a. Hence, both columns equal 100a, and the answer is (C).

27.Column A = The sum of the product of each element in A with each element in B

=(–2) × –4 + (–2) × –2 + (–2) × 0 + (–2) × 2 + (–2) × 4

+(–1) × –4 + (–1) × –2 + (–1) × 0 + (–1) × 2 + (–1) × 4

+0 × –4 + 0 × –2 + 0 × 0 + 0 × 2 + 0 × 4

+1 × –4 + 1 × –2 + 1 × 0 + 1 × 2 + 1 × 4

+2 × –4 + 2 × –2 + 2 × 0 + 2 × 2 + 2 × 4

=(–2){–4 – 2 + 0 + 2 + 4}

+(–1){–4 – 2 + 0 + 2 + 4}

+0{–4 – 2 + 0 + 2 + 4}

+1{–4 – 2 + 0 + 2 + 4}

+2{–4 – 2 + 0 + 2 + 4}

=–2 × 0 + (–1) × 0 + 0 × 0 + 2 × 0 + 4 × 0

=0 = Column B

The answer is (C).

Very Hard

28. The period March 11, 2003 through March 11, 2004, not including the former date, includes the complete month of February 2004. So, the length of the period is 366 days (equal to the length of a leap year). The number 366 has a quotient of 52 and a remainder of 2 when divided by 7. Hence, the length of the period is 52 weeks and 2 days. So, the date March 11, 2004 is 2 days advanced over the date March 11, 2003, which is given to be a Tuesday. The second day after a Tuesday is a Thursday. Hence, March 11, 2004 is a Thursday. The answer is (D).