GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #255 : Arithmetic

Two coins are loaded so that each comes up heads 60% of the time when tossed. If the coins are both tossed at the same time, what is the probability that the result will be one head and one tail?

Possible Answers:

0.6

0.24

0.48

0.52

0.5

Correct answer:

0.48

Explanation:

For each coin, the probability of a head is 0.6, and the probability of a tail is 0.4.

The probability of a head coming up on the first coin and a tail coming up on the second is $0.6 \cdot 0.4 = 0.24$ ; the same holds for the reverse case. Therefore, the probability of one head and one tail is

$\ 0.24 + 0.24 = 0.48$

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Example Question #256 : Arithmetic

A die is loaded so that it is equally likely to come up 1, 2, 3, 4, or 5, but twice as likely to come up 6 as it is likely to come up 5. If it is rolled twice, what is the probability that both rolls are even?

Possible Answers:

$\frac{16}{49}$

$\frac{9} {49}$

$\frac{1}{4}$

$\frac{4}{7}$

$\frac{7}{18}$

Correct answer:

$\frac{16}{49}$

Explanation:

Let be the probability that the die will come up 1. Then is also the common probability for each of the rolls of 2, 3, 4, or 5, and is the probability of the die coming up 6. To determine the value of , add these six probabilities and set the sum to 1, then solve:

$5 \cdot P + 2P = 1$

7P = 1

$P = \frac{1}{7}$

So 1, 2, 3, 4, and 5 each will come up with probability $\frac{1}{7}$ and 6 will cme up with probability $\frac{2}{7}$ .

An even number will come up with probability $\frac{1}{7} + \frac{1}{7} + \frac{2}{7} = \frac{4}{7}$ . Two even rolls will happen with probability $\frac{4}{7 } \cdot \frac{4}{7 } = \frac{16}{49}$ .

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Example Question #11 : Discrete Probability

Fifty marbles are put into a big box: ten red, ten yellow, ten green, and twenty blue. Which of the following actions would change the probability that a random draw would result in a blue marble?

Possible Answers:

Removing two yellow marbles, four green marbles, and four blue marbles.

Adding fifteen green marbles and ten blue marbles.

Removing half of the marbles of each color.

Adding three yellow marbles and two blue marbles.

Removing three marbles of each color.

Correct answer:

Removing three marbles of each color.

Explanation:

The ratio of blue marbles to non-blue marbles is 20:30 = 2:3 . To maintain the same probability of drawing a blue marble, this ratio must remain the same after the action.

We will look at all five actions.

If three yellow marbles and two blue marbles are added, the ratio is 22:33 = 2:3 .

If fifteen green marbles and ten blue marbles are added, the ratio is 30:45 = 2:3 .

If two yellow marbles, four green marbles, and four blue marbles are removed, the ratio is 16:24 = 2:3 .

If half of the marbles of each color are removed - that is, ten blue marbles and five of each of the other three colors - the ratio is 10:15 = 2:3 .

If three marbles of each color are removed, however, the ratio is $17: 21\neq 2:3$ . This is the correct choice.

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Example Question #12 : Calculating Discrete Probability

One hundred marbles - red, yellow, blue, or green - are placed in a box. There are an equal number of red and blue marbles, three times as many yellow marbles as green marbles, and five fewer green marbles than red marbles. What is the probability that a randomly drawn marble is NOT blue?

Possible Answers:

$\frac{11}{20}$

$\frac{17}{20}$

$\frac{3}{5}$

$\frac{3}{4}$

$\frac{4}{5}$

Correct answer:

$\frac{4}{5}$

Explanation:

If there are blue marbles, there are also red marbles, x-5 green marbles, and $3 \left (x-5 \right )$ yellow marbles. There are 100 marbles total, so we can solve for in the equation:

x + x + (x-5) + 3 (x-5) = 100

x + x + x-5 + 3 x-15= 100

6x -20 = 100

6x = 120

x = 20

20 of the 100 marbles are blue, so the probability that the marble is not blue is:

$\frac{100-20}{100} = \frac{80}{100} = \frac{4}{5}$

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Example Question #13 : Calculating Discrete Probability

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a multiple of 3?

Possible Answers:

$\frac{11}{32}$

$\frac{5}{16}$

$\frac{23}{64}$

$\frac{3}{8}$

$\frac{25}{64}$

Correct answer:

$\frac{11}{32}$

Explanation:

There are $8 \cdot 8= 64$ possible outcomes. The outcome can be between 2 and 16 inclusive; we count the number of rolls that result in any of the possible multiples of 3 - 3,6,9,12,15:

3: (1,2), (2,1)

6: (1,5) , (2,4), (3,3) , (4,2), (5,1)

9: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1)

12: (4,8), (5,7), (6,6) , (7,5), (8.4)

15: (7,8), (8.7)

22 out of 64 outcomes result in a multiple of 3, so the probability is $\frac{22}{64} = \frac{11}{32}$

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Example Question #14 : Discrete Probability

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a multiple of 5?

Possible Answers:

$\frac{7}{32}$

$\frac{15}{64}$

$\frac{13}{64}$

$\frac{3}{16}$

$\frac{11}{64}$

Correct answer:

$\frac{13}{64}$

Explanation:

There are $8 \cdot 8= 64$ possible outcomes. The sum can be between 2 and 16 inclusive; we count the number of rolls that result in any of the possible multiples of 5 - 5, 10, 15:

5: (1,4), (2,3), (3,2), (4,1)

10: (2,8), (3,7), (4,6), (5,5) ,(6,4), (7,3), (8,2)

15: (7,8), (8.7)

13 out of 64 outcomes result in a multiple of 3, so the probability is $\frac{13}{64}$ .

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Example Question #21 : Discrete Probability

Target_1

The above figure depicts a square target. It is divided into sixteen squares of equal area, one of which is in turn divided into two congruent triangles.

According to the rules of a game, if a dart is thrown at this target, the number of points scored for each color are as follows:

Gray: 1 point

Red: 2 points

Yellow: 4 points

Green: 7 points

Blue: 10 points.

Two darts are thrown at random at the target. Assuming that both darts hit the board, and assuming that there is no skill involved, what is the probability that more than 15 points will be scored?

Possible Answers:

$\frac{5}{512}$

$\frac{7}{ 1,024}$

$\frac{1}{ 256}$

$\frac{1}{ 128}$

$\frac{1}{16}$

Correct answer:

$\frac{7}{ 1,024}$

Explanation:

One of two things must happen for a score of 15 or better to be made: either both darts must land on blue (20 points), or one dart must land on green and one on blue, in either order (17 points). The next-highest possiblities are two green (14 points) or one yellow and one blue (14 points).

Since one half of a square out of sixteen is blue, the probability that a randomly thrown dart will hit blue is $\frac{\frac{1}{2}}{16} = \frac{1}{32}$ .

Since one and one-half squares out of sixteen are green, the probability that a randomly thrown dart will hit green is $\frac{\frac{3}{2}}{16} = \frac{3}{32}$ .

The probability that both darts will hit blue is $\frac{1}{32} \cdot \frac{1}{32} = \frac{1}{ 1,024}$ .

The probability that the first dart will hit blue and the second will hit green is $\frac{1}{32} \cdot \frac{3}{32} = \frac{3}{ 1,024}$ , which is also the probability that the reverse will happen.

Add these probabilities:

$\frac{1}{ 1,024} + \frac{3}{ 1,024}+ \frac{3}{ 1,024} = \frac{7}{ 1,024}$

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Example Question #22 : Discrete Probability

Two eight-sided dice from a role-playing game are thrown. Each die is fair and marked with the numbers 1 through 8. What is the probability that the sum of the dice will be a prime number?

Possible Answers:

$\frac{25}{64}$

$\frac{19}{64}$

$\frac{23}{64}$

$\frac{27}{64}$

$\frac{21}{64}$

Correct answer:

$\frac{23}{64}$

Explanation:

There are $8 \cdot 8= 64$ possible outcomes. The sum can be between 2 and 16 inclusive; we count the number of rolls that result in the sum being any of the possible prime numbers 2,3,5,7,11,13:

2: (1,1)

3: (1,2), (2,1)

5: (1,4), (2,3), (3,2), (4,1)

7: (1,6), (2,5), (3,4) , (4,3), (5,2), (6,1)

11: (3,8), (4,7), (5,6), (6,5), (7,4),(8,3)

13 : (5,8), (6,7), (7,6), (8,5)

23 out of 64 rolls result in prime sums, so the probability is $\frac{23}{64}$ .

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Example Question #23 : Discrete Probability

A box contains 12 balls, of which 5 are black, 4 are red, and 3 are white. If 2 balls are randomly selected from the box, one at a time without being replaced, what is the probability that the first ball selected will be red and the second ball selected will be white?

Possible Answers:

$\frac{1}{12}$

$\frac{1}{11}$

$\frac{1}{36}$

$\frac{7}{12}$

$\frac{1}{7}$

Correct answer:

$\frac{1}{11}$

Explanation:

The probability that the first ball selected will be red is $\frac{4}{12}=\frac{1}{3}$ .

The probability that the second ball selected will be white is $\frac{3}{11}$ .

We can solve by the multiplication principle since the two events happen together.

$\left ( \frac{1}{3} \right )\times\left ( \frac{3}{11} \right )=\frac{1}{11}$

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Example Question #1811 : Problem Solving Questions

Two fair eight-sided dice from a role-playing game are tossed; each die is marked with the numbers 1-8 on their faces. What is the probability that the difference of the two numbers will be 1?

Possible Answers:

$\frac{5}{18}$

$\frac{1}{8}$

$\frac{7}{64}$

$\frac{1}{4}$

$\frac{7}{32}$

Correct answer:

$\frac{7}{32}$

Explanation:

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \},\left \{ 2,3\right \}, \left \{ 3,4\right \} ,\left \{ 4,5\right \}, \left \{ 5,6\right \} , \left \{ 6,7\right \}, \left \{ 7,8\right \},$

$\left \{ 2,1\right \},\left \{ 3,2\right \}, \left \{ 4,3\right \} ,\left \{ 5,4\right \}, \left \{ 6,5\right \} , \left \{ 7,6\right \}, \left \{ 8,7\right \}$

This makes 14 favorable outcomes out of 64 outcomes total, so the probability is

$\frac{14}{64} = \frac{7}{32}$ .

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GMAT Math : Problem-Solving Questions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #255 : Arithmetic

Example Question #256 : Arithmetic

Example Question #11 : Discrete Probability

Example Question #12 : Calculating Discrete Probability

Example Question #13 : Calculating Discrete Probability

Example Question #14 : Discrete Probability

Example Question #21 : Discrete Probability

Example Question #22 : Discrete Probability

Example Question #23 : Discrete Probability

Example Question #1811 : Problem Solving Questions

All GMAT Math Resources

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