Arithmetic

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GMAT Quantitative › Arithmetic

Questions 1 - 10

What is the mean of the following data set in terms of and ?

$\left {a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right }$

$\frac{3}{2} a +\frac{1}{2} x$

$\frac{3}{2} a -\frac{1}{2} x$

$\frac{3}{2} a$

Explanation

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

What is the median of the following numbers?

$\frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7}$

$\frac{9}{40}$

$\frac{1}{4}$

$\frac{223}{840}$

$\frac{1}{5}$

$\frac{2}{9}$

Explanation

The median of a data set with an even number of elements is the mean of its two middle elements, when ranked. The set is already ranked, so just find the mean of middle elements $\frac{1}{4}$ and $\frac{1}{5}$ :

$\frac{1}{2} \cdot \left (\frac{1}{4} + \frac{1}{5} \right ) = \frac{1}{2} \cdot \left (\frac{5}{20} + \frac{4}{20} \right )= \frac{1}{2} \cdot \frac{9}{20} = \frac{9}{40}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

The average of 10, 25, and 70 is 10 more than the average of 15, 30, and x. What is the missing number?

$30$

$25$

$15$

$20$

$35$

Explanation

The average of 10, 25, and 70 is 35: $\frac{10+25+70}{3}=35$

So the average of 15, 30, and the unknown number is 25 or, 10 less than the average of 10, 25, and 70 (= 35)

so $\frac{15+30+x}{3}=25$

$15+30+x=75$

$45+x=75$

$x=30$

What is the last digit in the base-eight representation of the number 735?

Explanation

Divide 735 by 8. The remainder is the last digit of the base-eight representation.

$735 \div 8 = 91 \textrm{ R } 7$

The correct choice is 7.

What is the mean of the following data set in terms of and ?

$\left {a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right }$

$\frac{3}{2} a +\frac{1}{2} x$

$\frac{3}{2} a -\frac{1}{2} x$

$\frac{3}{2} a$

Explanation

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

If $\dpi{100} \small x=\frac{1}{2}$ and $\dpi{100} \small y=\frac{1}{3}$ , which of the following is the smallest?