GRE / GMAT : Sample Questions with Solution

1. Define x* by the equation x* = ∏/x . Then ((-∏)*)* =

    a) -1/∏   b) -1/2            c) -∏          d) 1/∏          e)∏

    Sol:  It is given that x* = ∏/x . Then ((-∏)*)* =( -∏/∏)* = ∏/-1 = -∏

          So answer is C.

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2. 2ab5 is a four-digit number divisible by 25. If the number formed from the two digits ab is a multiple of 13, then ab =

(A) 10

(B) 25

(C) 52

(D) 65

(E) 75

Sol: . We have that the number 2ab5 is divisible by 25. Any number divisible by 25 ends with the last two digits 00, 25, 50, or 75. So, b5 should equal 25 or 75. Hence, b = 2 or 7. Since a is now free to take any digit from 0 through 9, ab can have multiple values.

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. The answer is (C).

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3. What is the remainder when 7² . 8² is divided by 6?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Sol: 7². 8²= (7 . 8)²= 56².

The number immediately before 56 that is divisible by 6 is 54. Now, writing 56² as (54 + 2)², we have

56² = (54 + 2)²

= 54² + 2(2)(54) + 2² by the formula (a + b)²= a² + 2ab + b²

= 54[54 + 2(2)] + 2²

= 6 × 9[54 + 2(2)] + 4 here, the remainder is 4

Since the remainder is 4, the answer is (D).

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4. If 42.42 = k(14 + 7/50), then what is the value of k?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

  The given equation is

42.42 = k(14 + 7/50)

42.42 = k(14 + 14/100)

42.42 = k(14 + 0.14)

42.42 = k(14.14)

42.42/14.14 = k

3 = k

The answer is (C).

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5. If x + y = 7 and x² + y² = 25, then which one of the following equals the value of x³ + y³ ?

(A) 7

(B) 25

(C) 35

(D) 65

(E) 91

 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).

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GRE / GMAT : Sample Questions with Solution

1. Hose A can fill a tank in 5 minutes, and Hose B can fill the same tank in 6 minutes. How many tanks would Hose B fill in the time Hose A fills 6 tanks?

(A) 3

(B) 4

(C) 5

(D) 5.5

(E) 6

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).

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2. A car traveled at 80 mph for the first half (by time) of a trip and at 40 mph for the second half of the trip. What is the average speed of the car during the entire trip?

(A) 20

(B) 40

(C) 50

(D) 60

(E) 80

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 × t/2 + 40 × t/2/ t

=80 × ½ + 40 × 1/2

=40 + 20 = 60

The answer is (D).

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3.A father distributed his total wealth to his two sons. The elder son received 3/5 of the amount. The younger son received $30,000. How much wealth did the father have?

(A) 15,000

(B) 45,000

(C) 60,000

(D) 75,000

(E) 89,000

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).

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4. John has $42. He purchased fifty mangoes and thirty oranges with the whole amount. He then chose to return six mangoes for nine oranges as both quantities are equally priced. What is the price of each Mango?

(A) 0.4

(B) 0.45

(C) 0.5

(D) 0.55

(E) 0.6

 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).

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5. A ship is sinking and 120 more tons of water would suffice to sink it. Water seeps in at a constant rate of 2 tons a minute while pumps remove it at a rate of 1.75 tons a minute. How much time in minutes has the ship to reach the shore before is sinks?

(A) 480

(B) 560

(C) 620

(D) 680

(E) 720

 

  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).

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