GMAT Math : Arithmetic

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Example Questions

Example Question #1791 : Gmat Quantitative Reasoning

and are distinct positive integers. a+b is an odd quantity. Which of the following must be even?

(You may assume all of these are positive quantities.)

(a) (a-1)(b+1)

(b) (ab-1)(ab+1)

Possible Answers:

None of these

(b) and (c) only

(a) and (c) only

(a) and (b) only

(a), (b), and (c)

Correct answer:

(a) and (c) only

Explanation:

a+b is an odd quantity if and only if one of and is even and the other is odd. We can assume without loss of generality that is the even quantity and is the odd quantity, since a similar argument holds if is even.

(a) is odd, so b+1 is even. (a-1)(b+1) , which has an even factor, is even.

(b) is even and is odd, so is even, and ab -1 and ab+ 1 are both odd. (ab-1)(ab+1) , the product of odd factors, is odd.

The correct response is (a) and (c) only.

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Example Question #71 : Understanding The Properties Of Integers

Which of the following is not an integer?

Possible Answers:

$\pi$

6789

Correct answer:

$\pi$

Explanation:

Which of the following is not an integer?

The definition of an integer is, "All positive or negative whole numbers, including zero."

Therefore, our answer must pi, because pi is not a whole number. All other options fit the defintion of integers.

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Example Question #1792 : Gmat Quantitative Reasoning

Which of the following is an integer?

Possible Answers:

$45\frac{4}{5}$

-2.5

-55.55

Correct answer:

Explanation:

An integer is an positive or negative whole number, including zero. Eliminate all options which are not whole numbers and you are left with 0!

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Example Question #73 : Understanding The Properties Of Integers

is a positive integer and has an even number of prime factors. What is ?

Possible Answers:

169

Correct answer:

169

Explanation:

We know that is a positive integer with an even factors of prime numbers, for example could be $3^{2}$ or could be $2^{4}$ , in other words, must be a perfect square. The only possible answer is 169 , the perfect square of .

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Example Question #1 : Calculating Discrete Probability

Race cars for a particular race are numbered sequentially from 12 to 115. What is the probability that a car selected at random will have a tens digit of 1?

Possible Answers:

$\frac{1}{13}$

$\frac{13}{104}$

$\frac{8}{103}$

$\frac{7}{52}$

$\frac{14}{103}$

Correct answer:

$\frac{7}{52}$

Explanation:

There are 104 integers from 12 to 115 inclusive. There are 8 integers from 12 to 19 and 6 integers from 110 to 115 for a total of 14 integers with a tens digit of 1. The probability of selecting a car with a tens digit of 1 is $\frac{14}{104}$ = $\frac{7}{52}$ .

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Example Question #2 : Discrete Probability

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

Possible Answers:

$\dpi{100} \small \frac{7}{15}$

$\dpi{100} \small \frac{4}{53}$

$\dpi{100} \small \frac{47}{80}$

$\dpi{100} \small \frac{3}{13}$

$\dpi{100} \small \frac{1}{2}$

Correct answer:

$\dpi{100} \small \frac{47}{80}$

Explanation:

Prob(new OR yellow) = P(new) + P(yellow) - P(new AND yellow)

Prob(new) = $\dpi{100} \small \frac{45}{100}$

Prob(yellow) = $\dpi{100} \small \frac{25}{100}$

Prob(new AND yellow) = $\dpi{100} \small \frac{45}{100}\times \frac{25}{100}$

so P(new OR yellow) = $\dpi{100} \small \frac{45}{100}+\frac{25}{100}-\frac{45}{100}\times \frac{25}{100}$

$\dpi{100} \small =\frac{70}{100}-\frac{9}{80}$

$\dpi{100} \small =\frac{7}{10}-\frac{9}{80}$

$\dpi{100} \small =\frac{56}{80}-\frac{9}{80}=\frac{47}{80}$

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Example Question #1 : Discrete Probability

Flight A is on time for 93% of flights. Flight B is on time for 89% of flights. Flight A and B are both on time 87% of the time. What is the probabiity that at least one flight is on time?

Possible Answers:

0.95

0.87

0.89

1.82

0.93

Correct answer:

0.95

Explanation:

P(Flight A is on time) = 0.93

P(Flight B is on time) = 0.89

P(Flights A and B are on time) = 0.87

Then P(A OR B) = P(A) + P(B) - P(A AND B) = 0.93 + 0.89 – 0.87 = 0.95

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Example Question #4 : Discrete Probability

At a school fair, there are 25 water balloons. 10 are yellow, 8 are red, and 7 are green. You try to pop the balloons. Given that you first pop a yellow balloon, what is the probability that the next balloon you hit is also yellow?

Possible Answers:

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{9}{25}$

$\frac{3}{8}$

$\frac{5}{12}$

Correct answer:

$\frac{3}{8}$

Explanation:

At the start, there are 25 balloons and 10 of them are yellow. You hit a yellow balloon. Now there are 9 yellow balloons left out of 24 total balloons, so the probability of hitting a yellow next is

$\frac{9}{24}=\frac{3}{8}$ .

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Example Question #5 : Discrete Probability

A bag has 7 blue balls and 3 red balls. 2 balls are to be drawn successively and without replacement. What is the probability that the first ball is red and the second ball is blue?

Possible Answers:

$\frac{21}{100}$

$\frac{7}{9}$

$\frac{7}{30}$

$\frac{3}{10}$

$\frac{1}{15}$

Correct answer:

$\frac{7}{30}$

Explanation:

We first have 7 blue and 3 red out of 10, so P(1st ball is red) = $\frac{3}{10}$ . Now, we have pulled a red ball out of the bag, leaving us with 7 blue and 2 red out of 9 total balls. Then P(2nd ball is blue) = $\frac{7}{9}$ . Put this together, P(1st red AND 2nd blue) = $\frac{3}{10}\cdot \frac{7}{9}=\frac{7}{30}$ .

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Example Question #6 : Discrete Probability

4 cards are to be dealt successively and without replacement from an ordinary deck of 52 cards. What is the probability of receiving, in order, a spade, a heart, a diamond, and a club?

Possible Answers:

$\frac{13}{52 }\cdot \frac{12}{52}\cdot \frac{11}{52}\cdot \frac{10}{52}$

$\frac{4}{52 }\cdot \frac{4}{51}\cdot \frac{4}{50}\cdot \frac{4}{49}$

$\frac{13}{52 }\cdot \frac{13}{52}\cdot \frac{13}{52}\cdot \frac{13}{52}$

$\frac{13}{52 }\cdot \frac{12}{51}\cdot \frac{11}{50}\cdot \frac{10}{49}$

$\frac{13}{52 }\cdot \frac{13}{51}\cdot \frac{13}{50}\cdot \frac{13}{49}$

Correct answer:

$\frac{13}{52 }\cdot \frac{13}{51}\cdot \frac{13}{50}\cdot \frac{13}{49}$

Explanation:

There are 4 suits, and each suit has 13 cards, so P(spade) = $\frac{13}{52}$ .

Now there are 51 cards left: 12 spades, 13 hearts, 13 diamonds, and 13 clubs, so now P(heart) = $\frac{13}{51}$ .

Once again, there are now 50 cards: 12 spades, 12 hearts, 13 diamonds, and 13 clubs, so now P(diamond) = $\frac{13}{50}$ .

Before our last draw we have 49 cards left: 12 spades, 12 hearts, 12 diamonds, and 13 clubs, so P(club) = $\frac{13}{49}$ .

Putting these four probabilities together gives us our answer:

P(spade AND heart AND diamond AND club) = $\frac{13}{52 }\cdot \frac{13}{51}\cdot \frac{13}{50}\cdot \frac{13}{49}$

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GMAT Math : Arithmetic

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1791 : Gmat Quantitative Reasoning

Example Question #71 : Understanding The Properties Of Integers

Example Question #1792 : Gmat Quantitative Reasoning

Example Question #73 : Understanding The Properties Of Integers

Example Question #1 : Calculating Discrete Probability

Example Question #2 : Discrete Probability

Example Question #1 : Discrete Probability

Example Question #4 : Discrete Probability

Example Question #5 : Discrete Probability

Example Question #6 : Discrete Probability

All GMAT Math Resources

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