GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #303 : Arithmetic

Target

The upper half of the above target has radius twice that of the lower half.

A blindfolded man throws a dart at random at the above target. Disregarding any skill factor, and assuming that the dart hits the target, what is the probability that it will land inside a red region?

Possible Answers:

$\frac{5}{12}$

$\frac{3}{8}$

None of the other responses gives the correct answer.

$\frac{7}{18}$

$\frac{11}{30}$

Correct answer:

$\frac{11}{30}$

Explanation:

For the sake of simplicity, we will assume the smaller half of the target has radius 1, and, consequently, the larger half has radius 2.

The upper semicircle has total area

$A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi$

The lower semicircle has total area

$A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi$

The total area of the target is

$A = A_{1}+ A_{2} = 2 \pi + \frac{1}{2} \pi = \frac{5}{2} \pi$

Each of the six sectors of the upper semicircle has area one sixth of its area, or

$\frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi$

Each of the two sectors of the lower semicircle has area one half of its area, or

$\frac{ 1}{2}\cdot \frac{ 1}{2} \pi = \frac{1}{4} \pi$

The red regions comprise two sectors of the upper semicircle and one of the lower, so their total area is

$\frac{1}{3} \pi + \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{4}{12} \pi + \frac{4}{12} \pi + \frac{3}{12} \pi = \frac{11}{12} \pi$

The probability of the dart hitting one of these red regions is therefore

$\frac{ \frac{11}{12} \pi}{\frac{5}{2} \pi} = \frac{ \frac{11}{12} \pi}{\frac{30}{12} \pi} = \frac{11}{30}$

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Example Question #304 : Arithmetic

The upper semicircle of the above spinner has radius twice that of the lower semicircle.

If the above spinner is spun, what is the probability that the arrow will stop inside a yellow region?

Possible Answers:

$\frac{1}{5}$

$\frac{5}{18}$

$\frac{1}{3}$

$\frac{7}{30}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{3}$

Explanation:

The radii of the semicircles of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count.

The yellow regions comprise one sector of the top semicircle and one sector of the bottom one. The former is one-sixth of a semicircle and measures $\frac{1}{6} \times 180^{\circ } = 30^{\circ }$ ; the latter is a fourth of a circle and measures $\frac{1}{4} \times 360^{\circ } = 90^{\circ }$ . Therefore, the total of the measures of the central angles is

$30^{\circ }+ 90 ^{\circ }= 150^{\circ }$ ,

and the probability that the arrow will stop in one of these regions is

$\frac{120}{360} = \frac{120 \div 120 }{360\div 120 } = \frac{1}{3}$ .

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Example Question #311 : Arithmetic

Some balls are placed in a large box, which include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1".

What is the probability that a ball randomly drawn from this box will be marked with an even number?

Possible Answers:

$\frac{5}{9}$

$\frac{5}{11}$

$\frac{6}{11}$

$\frac{1}{2}$

$\frac{4}{9}$

Correct answer:

$\frac{5}{11}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 10 .

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The even numbers are 2, 4, 6, 8, and 10. According to the pattern, there are nine balls marked "2", seven marked "4", five marked "6", three marked "8", and one marked "10". Therefore, there are

9+5+7+3+1 = 25

balls marked with even numbers.

The probability of drawing a ball with an even number is therefore

$\frac{25}{55} = \frac{5}{11}$

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Example Question #312 : Arithmetic

Some balls are placed in a large box, which include one ball marked , two balls marked , and continuing up to balls marked .

What is the probability that a ball randomly drawn from this box will be marked with a prime number?

Possible Answers:

$\frac{2}{5}$

$\frac{18}{55}$

$\frac{1}{2}$

$\frac{17}{55}$

$\frac{17}{45}$

Correct answer:

$\frac{17}{55}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 10 .

Since,

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The prime numbers in the range of $\textup{1 to 10}$ are $\textup{ 2, 3, 5, and 7}$ . The number of balls with each number corresponds to the number itself, so the number of balls with prime numbers will be

2+3+5+7 = 17 .

The probability of drawing one of them is $\frac{17}{55}$ .

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Example Question #313 : Arithmetic

Spinner 2

The upper semicircle of the above spinner has radius twice that of the lower semicircle.

Suppose the arrow is spun. Let be the probability that the arrow stops when pointing to a green region; let and be similarly defined for a red or yellow region. Which of the following is a true statement?

Possible Answers:

Y < R = G

Y < R < G

R = G < Y

Y < G< R

G < Y < R

Correct answer:

G < Y < R

Explanation:

The radii of the halves of the spinner - and the areas of the sectors - are actually irrelevant to the problem; it is the measures of their central angles that count. In order to compare the probabilities of the arrow stopping in regions of the three colors, it suffices to compare the totals of the measures of the central angles of the different colors - we will call them G', R', Y' .

Each sector of the larger semicircle is one-sixth of a $180^{\circ }$ sector and has central angle

$\frac{1}{6} \times 180^{\circ } = 30^{\circ }$ .

Each sector of the smaller semicircle is a quarter of a circle, and has central angle

$\frac{1}{4} \times 360^{\circ } = 90^{\circ }$ .

The green regions comprise three sectors of the top semicircle, so the total of their central angles is

$G' = 3 \times 30^{\circ } = 90^{\circ }$ .

The yellow regions comprise one sector of the top semicircle and one sector of the bottom one, so the total of their central angles is

$Y' =30^{\circ } +90^{\circ } = 120^{\circ }$ .

The red regions comprise two sectors of the top semicircle and one sector of the bottom one, so the total of their central angles is

$R' = 2 \times 30^{\circ } + 90 ^{\circ } = 60^{\circ } +90^{\circ } = 150^{\circ }$ .

G'<Y'<R' ; it follows that G<Y<R .

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Example Question #1861 : Problem Solving Questions

Some balls are placed in a large box, which include ball marked "A", balls marked "B", balls marked "C", and so on up to balls marked "Z".

A ball is drawn at random. Give the probability that the ball will be marked "O".

Possible Answers:

$\frac{5}{117}$

$\frac{3}{130}$

$\frac{3}{65}$

$\frac{3}{70}$

$\frac{5}{234}$

Correct answer:

$\frac{5}{117}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 26 .

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 26 = \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351$ .

Since "O" is the fifteenth letter of the alphabet, there are balls out of 351 marked with "O". Therefore, the probability that a randomly-drawn ball will be one of these is

$\frac{15}{351} = \frac{15\div 3 }{351 \div 3 } = \frac{5}{117}$

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Example Question #71 : Calculating Discrete Probability

Target

The upper semicircle of the above target has radius twice that of the lower semicircle.

A blindfolded woman throws a dart at random at the above target. Let be the probability that the dart lands inside a green region; let and be similarly defined for a red or yellow region. Which of the following is a true statement?

(Assume that skill is not a factor, and that the dart hits the target.)

Possible Answers:

Y < R < G

Y < R =G

G = Y< R

Y< G < R

G < Y< R

Correct answer:

Y < R < G

Explanation:

Since the areas of the sectors are what are important here, we need only rank the total areas of the different colors, which we will call G', R', Y' .

For the sake of simplicity, we will assume the smaller semicircle of the target has radius ; consequently, the larger semicircle has radius .

The larger semicircle has total area

$A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi$

The smaller semicircle has total area

$A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi$

Each of the six sectors of the larger semicircle has area $\frac{1}{6}$ of its area, or

$\frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi$

Each of the two sectors of the smaller semicircle has area $\frac{1}{2}$ of its area, or

$\frac{ 1}{2}\cdot \frac{ 1}{2} \pi = \frac{1}{4} \pi$

The total of the areas of the green sectors - three sectors of the larger semicircle - is

$G' = 3 \cdot \frac{1}{3} \pi = \pi$

The total of the areas of the red sectors - two sectors of the larger semicircle and one of the smaller - is

$R' = 2 \cdot \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{2}{3} \pi + \frac{1}{4} \pi = \frac{8}{12} \pi + \frac{3}{12} \pi = \frac{11}{12} \pi$

The total of the areas of the yellow sectors - one sector of the larger semicircle and one of the smaller - is

$Y' = \frac{1}{3} \pi + \frac{1}{4} \pi = \frac{4}{12} \pi + \frac{3}{12} \pi = \frac{7}{12} \pi$

Y' < R' < G' ; it follows that Y < R < G .

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Example Question #71 : Discrete Probability

Some balls are placed in a large box, which include one ball marked "A", two balls marked "B", three balls marked "C", and so forth up to twenty-six balls marked "Z".

A ball is drawn at random. Give the odds against the ball being marked "I".

Possible Answers:

$\textup{78 to 1}$

$\textup{39 to 1}$

$\textup{76 to 1}$

$\textup{38 to 1}$

$\textup{77 to 1}$

Correct answer:

$\textup{38 to 1}$

Explanation:

The total number of balls in the box will be

1 + 2 + 3 + ... + 26 .

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 26 = \frac{26 (26 + 1)}{2} = \frac{26 \cdot 27}{2} = 351$ .

Since "I" is the ninth letter of the alphabet, there are nine balls marked "I", and 351-9 = 342 other balls. Therefore, the odds against drawing a ball marked "I" are 342 to 9, or

$\frac{342}{9} = \frac{342 \div 9}{9 \div 9} = \frac{38}{1}$

The correct choice is 38 to 1.

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Example Question #73 : Calculating Discrete Probability

Target

The upper semicircle of the above target has radius twice that of the lower semicircle.

A blindfolded woman throws a dart at random at the above target. Disregarding any skill factor, and assuming that the dart hits the target, what are the odds against the dart landing inside the purple region?

Possible Answers:

$\textup{11 to 1}$

$\textup{13 to 2}$

$\textup{7 to 1}$

$\textup{6 to 1}$

$\textup{11 to 2}$

Correct answer:

$\textup{13 to 2}$

Explanation:

For the sake of simplicity, we will assume the smaller semicircle of the target has radius 1, and, consequently, the larger semicircle has radius 2.

The larger semicircle has total area

$A_{1} = \frac{1}{2} \cdot \pi r_{1}^{2} = \frac{1}{2} \cdot \pi \cdot 2 ^{2} = \frac{1}{2} \cdot 4 \pi = 2 \pi$

The smaller semicircle has total area

$A_{2} = \frac{1}{2} \cdot \pi r_{2} ^{2} = \frac{1}{2} \cdot \pi \cdot 1 ^{2} = \frac{1}{2} \pi$

The total area of the target is

$A = A_{1}+ A_{2} = 2 \pi + \frac{1}{2} \pi = \frac{5}{2} \pi$

The purple sector has area one-sixth that of the larger semicircle, or

$\frac{ 1}{6}\cdot 2 \pi = \frac{1}{3} \pi$

The area that is not purple is

$\frac{5}{2} \pi - \frac{1}{3} \pi = \frac{15}{6} \pi - \frac{2}{6} \pi = \frac{13}{6} \pi$

The odds against the dart landing in the purple region are

$\frac{\frac{13}{6} \pi}{\frac{1}{3} \pi} = \frac{\frac{13}{6} \pi \cdot \frac{6}{\pi}}{\frac{1}{3} \pi \cdot \frac{6}{\pi}} = \frac{13}{2}$ - that is, 13 to 2.

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Example Question #71 : Discrete Probability

Eduardo has a standard deck of cards plus two jokers (54 cards in all). What are the odds that he draws a joker, followed by a red card that isn't a king? Assume replacement and that there is one red joker and one black joker.

Possible Answers:

0.17

0.017

0.083

0.184

Correct answer:

0.017

Explanation:

This is a probability question. Probability can be found via the following:

$P=\frac{\#\ of\ desired\ outcomes}{total\ number\ of\ outcomes}$

To find the probability of multiple events, we find the probability of each event and multiply them together.

The probability of the first event is as follows:

$P_1=\frac{2}{54}=\frac{1}{27}$

Note that there are 54 cards and only two jokers.

The second event will still have the same denominator as the first event. The numerator however, will need to change.

"...a red card that isn't a king..." So, our deck would normally be half red. However, two of those cards will be kings, so the number of desired outcomes should be 25

$P_2=\frac{25}{54}$

Next, we need to multiple our two probabilities together.

$P_{tot}=\frac{25}{54}*\frac{1}{27}=\frac{25}{1458}\approx0.017$

So our answer is:

0.017

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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 #303 : Arithmetic

Example Question #304 : Arithmetic

Example Question #311 : Arithmetic

Example Question #312 : Arithmetic

Example Question #313 : Arithmetic

Example Question #1861 : Problem Solving Questions

Example Question #71 : Calculating Discrete Probability

Example Question #71 : Discrete Probability

Example Question #73 : Calculating Discrete Probability

Example Question #71 : Discrete Probability

All GMAT Math Resources

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