GMAT Math : Problem-Solving Questions

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Example Questions

Example Question #7 : Understanding Counting Methods

A= 1-3+5-7+9-11+...+9,997-9,999

B = 2-4+6-8+10-12...+9,998 -10,000

Which of the following statements is true?

Possible Answers:

B= A

B= A + 5,000

B= A + 2,500

B= A - 2,500

B= A - 5,000

Correct answer:

B= A

Explanation:

The finite series is obtained from by increasing each term by 1; since is an alternating series, this results in adding to :

(1-1) + (1-1) + (1-1) + ...(1-1) = 0

so B = A + 0 = A

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Example Question #8 : Understanding Counting Methods

Define set $A = \left \{ 1,2,3,4,5,6,7,8\right \}$ .

How many four-element subsets of include at least three even numbers?

Possible Answers:

Correct answer:

Explanation:

Only one subset of has four even numbers - that is the subset $\left \{ 2,4,6,8\right \}$ .

Forming a subset with three even numbers and one odd number can be restated as choosing which even number to leave out and which odd number to include. There are 4 choices each way, so the number of sets fitting this description is $4\times 4 = 16$ .

Therefore, the number of subsets with at least three even elements is 16 + 1 = 17 .

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Example Question #4 : Counting Methods

Twelve students are running for student council; each student will vote for four. Mick wants to vote for his sister Janine. How many ways can he cast his ballot so as to include Janine among his choices?

Possible Answers:

165

40,320

495

11,880

5,940

Correct answer:

165

Explanation:

Since one of Mick's choices is already decided, he will choose three people from a set of eleven without regard to order. This is a combination of three from a set of eleven; the number of such combinations is:

$C (11,3) = \frac{11!}{(11-3)!3!}= \frac{11!}{8!3!} = \frac{11 \times 10 \times 9}{3 \times 2\times 1} = \frac{990}{6} = 165$

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Example Question #5 : Counting Methods

Twelve students are running for student council; each student will vote for five. Claude does not want to vote for Gary or Mitch, neither of whom he likes. How many ways can Claude fill in the ballot so that he does not vote for Gary?

Possible Answers:

252

30,240

220

120

792

Correct answer:

252

Explanation:

Claude will choose five people from a set of ten - twelve minus the two he dislikes - without regard to order. This is a combination of five from a set of ten; the number of such combinations is:

$C (10,5) = \frac{10!}{(10-5)!\; 5!} = \frac{10!}{ 5!\; 5!} = \frac{3,628,800 }{ 120 \times 120} = 252$

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Example Question #1881 : Problem Solving Questions

The Department of Motor Vehicles wants to make all of the state's license numbers conform to two rules:

Rule 1: The license number must comprise three letters followed by four numerals.

Rule 2: Numerals can be repeated, but letters cannot.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

$P (26, 3)\times 10 ^{4}$

$C (26, 3)\times 10 ^{4}$

$P (26, 3)\times P (10,4)$

$26 ^{3} \times 10 ^{4}$

$C(26, 3)\times C (10,4)$

Correct answer:

$P (26, 3)\times 10 ^{4}$

Explanation:

Three different letters are selected from a group of 26, order being important - this is a permutation of three elements out of twenty-six. These permutations number P (26, 3) .

There are then four numerals, each chosen from a set of ten. Since repeats are allowed, the number of ways this can be done is $10 \times 10 \times 10 \times 10 = 10 ^{4}$

By the multiplication principle, there are

$P (26, 3)\times 10 ^{4}$

possible license plates.

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Example Question #12 : Counting Methods

The Department of Motor Vehicles wants to make all of the state's license numbers conform to three rules:

Rule 1: The license number must comprise three letters (A-Z), followed by three numerals (0-9), followed by a letter.

Rule 2: Repeats are allowed.

Rule 3: None of the letters can be an "O" or an "I".

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

$C(24,4) \times C (10,3)$

$24 ^{4} \times C (10,3)$

$24 ^{4} \times 8^{3}$

$24 ^{4} \times 10 ^{3}$

$C(24,4) \times 10 ^{3}$

Correct answer:

$24 ^{4} \times 10 ^{3}$

Explanation:

There are 24 ways to choose an allowed letter (since two of the 26 are excluded) and 10 ways to choose an allowed digit. Since repeats are allowed, and four letters and three digits will be chosen, there will be

$24 \times 24 \times 24 \times 10\times 10\times 10 \times 24 = 24 ^{4} \times 10 ^{3}$

possible license plate numbers.

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Example Question #1882 : Problem Solving Questions

The Department of Motor Vehicles wants to make all of the state's license numbers conform to four rules:

Rule 1: The license number must comprise two letters, followed by three digits, followed by two letters.

Rule 2: The letters "I" and "O" may not be used.

Rule 3: The digits "0" and "1" may not be used.

Rule 4: Repeats are allowed.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

$32 ^{7}$

$7^{32}$

$P(24,4) \times P \left ( 8,3 \right )$

$C(24,4) \times C \left ( 8,3 \right )$

$24 ^{4} \times 8 ^{3}$

Correct answer:

$24 ^{4} \times 8 ^{3}$

Explanation:

Each of the four letters can be selected twenty-four ways, since two letters are forbidden and since repeats are allowed. Similarly, each of the three digits can be selected eight ways for the same reason.

By the multiplication principle, the number of possible license plate numbers is

$24 \times 24 \times 8 \times 8 \times 8 \times 24 \times 24 = 24^{4} \times 8^{3}$

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Example Question #1883 : Problem Solving Questions

The Department of Motor Vehicles wants to make all of the state's license numbers conform to three rules:

Rule 1: Each license number must comprise six characters, each of which must be a letter or a digit.

Rule 2: No license number can comprise six letters or six digits.

Rule 3: Repeats are allowed.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Possible Answers:

P(36,6) - P(26,6) - P (10,6)

36! - 26! - 10!

$6 ^{36} - 6 ^{26} - 6^{10 }$

C (36,6) - C (26,6) - C (10,6)

$36 ^{6} - 26 ^{6} - 10^{6}$

Correct answer:

$36 ^{6} - 26 ^{6} - 10^{6}$

Explanation:

Each character of the licence plate number can be chosen from a set of 36 (26 letters plus 10 digits), and any character can be chosen multiple times. By the multiplication principle, this would allow $36 ^{6}$ different licence numbers. But licence numbers that comprise only letters and those that comprise only digits are disallowed, so by similar math, this reduces the possibilities by $26 ^{6}$ and $10 ^{6}$ , respectively.

Therefore, there are $36 ^{6} - 26 ^{6} - 10^{6}$ allowed licence plate numbers.

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Example Question #11 : Understanding Counting Methods

Jessica went shopping and bought herself shirts, skirts, and pairs of shoes. How many outfits can Jessica possibly create from what she just bought?

Possible Answers:

Correct answer:

Explanation:

She has 4 choices for the shirt, 2 choices for the skirt per shirt chosen, and 3 choices for the pair of shoes per combination of shirt and skirt:

4 x 2 x 3 = 24 possible outfits

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Example Question #11 : Counting Methods

Fred shuffles a deck of 52 cards. He then draws 1 card and keeps it. Then he draws a 2nd card. Then he puts it back into the deck and shuffles again. Fred then draws a 3rd card and keeps it

What's the probability that the 2nd and 3rd cards are the same?

Possible Answers:

1/104

1/51

1/52

None of the other answers

Correct answer:

1/51

Explanation:

Just before Fred draws the 2nd card, there are 51 cards in the deck, because he already took the 1st card out.

Then once he puts the 2nd card back in, and shuffles, there are still 51 cards in the deck; he did not put the 1st card back in.

So Fred has a 1 in 51 chance of drawing the same card again.

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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 #7 : Understanding Counting Methods

Example Question #8 : Understanding Counting Methods

Example Question #4 : Counting Methods

Example Question #5 : Counting Methods

Example Question #1881 : Problem Solving Questions

Example Question #12 : Counting Methods

Example Question #1882 : Problem Solving Questions

Example Question #1883 : Problem Solving Questions

Example Question #11 : Understanding Counting Methods

Example Question #11 : Counting Methods

All GMAT Math Resources

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