GRE/ GMAT : Sample Questions with Solution

1. There are 750 male and female participants in a meeting. Half the female participants and one-quarter of the male participants are Democrats. One-third of all the participants are Democrats. How many of the Democrats are female?

(A) 75

(B) 100

(C) 125

(D) 175

(E) 225

Let m be the number of male participants and f be the number of female participants in the meeting. The total number of participants is given as 750. Hence, we have

m + f = 750

Now, we have that half the female participants and one-quarter of the male participants are Democrats. Let d equal the number of the Democrats. Then we have the equation

f/2 + m/4 = d

Now, we have that one-third of the total participants are Democrats. Hence, we have the equation d = 750/3 = 250

Solving the three equations yields the solution f = 250, m = 500, and d = 250. The number of female democratic participants equals half the female participants equals 250/2 = 125. The answer is (C).

2. A prize of $200 is given to anyone who solves a hacker puzzle independently. The probability that Tom will win the prize is 0.6, and the probability that John will win the prize is 0.7. What is the probability that both will win the prize?

(A) 0.35

(B) 0.36

(C) 0.42

(D) 0.58

(E) 0.88

Let P(A) = The probability of Tom solving the problem = 0.6, and let P(B) = The probability of John solving the problem = 0.7. Now, since events A and B are independent (Tom’s performance is independent of John’s performance and vice versa), we have

P(A and B) = P(A) ×  P(B) = 0.6 ×  0.7 = 0.42

The answer is (C).

3. In how many ways can 3 boys and 2 girls be selected from a group of 6 boys and 5 girls?

(A) 10

(B) 20

(C) 50

(D) 100

(E) 200

We have two independent actions to do:

1) Select 3 boys from 6 boys.

2) Select 2 girls from 5 girls.

Selection is a combination problem since selection does not include ordering. Hence, by Model 2, the number of ways is

(6C3 ways for boys) . (5C2 ways for girls) = (6!/3!.3!) . (5!/2!.3!)

= 20 .10 = 200

The answer is (E).

4. In how many ways can a committee of 5 members be formed from 4 women and 6 men such that at least 1 woman is a member of the committee?

(A) 112

(B) 156

(C) 208

(D) 246

(E) 252

Forming members of committee is a selection action and therefore this is a combination problem. Whether you select A first and B next or vice versa, it will only be said that A and B are members of the committee.

The number of ways of forming the committee of 5 from 4 + 6 = 10 people is 10C5. The number of ways of forming a committee with no women (5 members to choose from 6 men) is 6C5. Hence, the number of ways of forming the combinations is

10C5 - 6C5 = 10!/5!.5!  - 6!/5! = 252 - 6 = 246

The answer is (D).

5. Which one of the following products has the greatest value?

(A) 6.00 × 0.20

(B) 6.01 × 0.19

(C) 6.02 × 0.18

(D) 6.03 × 0.17

(E) 6.04 × 0.16

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).