GMAT Math : Arithmetic

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Example Questions

Example Question #52 : Arithmetic Mean

Give the arithmetic mean of and .

Statement 1: A trapezoid with bases of length and and height 20 has area 1,000.

Statement 2: If and are, respectively, the first and second terms of a geometric sequence with positive terms, the third term is 810.

Possible Answers:

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

Correct answer:

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

Explanation:

The arithmetic mean of two numbers is half their sum.

Assume Statement 1 alone. The area of a trapezoid can be calculated from its height and the lengths of its bases and as follows:

$A= \frac{B+b}{2} \cdot h$

Setting A = 1,000 , h = 20 , B = X , and b = Y , this becomes

$1,000= \frac{X+Y}{2} \cdot 20$

$1,000 \div 20 = \frac{X+Y}{2} \cdot 20 \div 20$

$\frac{X+Y}{2} = 50$

Therefore, the arithmetic mean of and is, by definition, 50.

Assume Statement 2 alone. A geometric sequence is one in which the ratio of each term to the previous term is constant - a common ratio. Consider these two sequences:

10, 90, 810,...

8.1, 81, 810,...

Both sequences have 810 as their third term. Both are geometric; the common ratio of the first sequence is 9, and that of the second is 10. The arithmetic means of the first two terms, however, are different. In the first sequence it is

$\frac{10+90}{2} = 50$

In the second sequence, it is

$\frac{8.1+81}{2} = 44.55$

Therefore, Statement 2 alone gives insufficient information.

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Example Question #521 : 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".

A carnival wants to set up a game by which a player can pay to draw a ball and win an amount of money equal to the number marked on the ball drawn. What should the carnival charge per play in order to make the expected value $1 per game in the carnival's favor?

Possible Answers:

$\$2.00$

$\$5.00$

$\$3.00$

$\$4.00$

$\$ 1.00$

Correct answer:

$\$5.00$

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 frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$

The expected value of the payment that the church will have to make per game can be calculated by multiplying the frequency of each outcome by the respective payment:

$1: \frac{10}{55} \cdot 1 = \frac{10}{55}$

$2: \frac{9}{55} \cdot 2 = \frac{18}{55}$

$3: \frac{8}{55} \cdot 3 = \frac{24}{55}$

$4: \frac{7}{55} \cdot 4 = \frac{28}{55}$

$5: \frac{6}{55} \cdot 5 = \frac{30}{55}$

$6: \frac{5}{55} \cdot 6 = \frac{30}{55}$

$7: \frac{4}{55} \cdot 7= \frac{28}{55}$

$8: \frac{3}{55} \cdot 8 = \frac{24}{55}$

$9: \frac{2}{55} \cdot 9 = \frac{18}{55}$

$1: \frac{1}{55} \cdot 10 = \frac{10}{55}$

...and then adding these products.

$\frac{10}{55}+ \frac{18}{55}+ \frac{24}{55}+ \frac{28}{55}+ \frac{30}{55}+ \frac{30}{55}+ \frac{28}{55}+ \frac{24}{55}+ \frac{18}{55}+ \frac{10}{55}$

$= \frac{220}{55}$

= 4

The carnival should expect to pay a prize of $4 per draw, so to expect an average profit of $1 per game, it should charge $5 per play.

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Example Question #521 : Arithmetic

10,25,4,15,99,10,9,13,11

What is the median of the data set?

Possible Answers:

10.5

21.8

Correct answer:

Explanation:

Write the numbers in order from smallest to largest. The median is the middle number, which is 11.

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Example Question #2 : Calculating Median

What is the median of the following numbers?

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

Possible Answers:

$\frac{1}{5}$

$\frac{9}{40}$

$\frac{2}{9}$

$\frac{223}{840}$

$\frac{1}{4}$

Correct answer:

$\frac{9}{40}$

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 $\frac{1}{5}$ :

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

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Example Question #2073 : Gmat Quantitative Reasoning

Give the median of the following data set in terms of :

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

Possible Answers:

a - 1

a+2

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

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

Correct answer:

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

Explanation:

The data set can be arranged from least to greatest as follows:

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

The median of a data set with eight elements is the mean of its fourth-highest and fourth-lowest elements, which are a-1 and a + 2 . Add and divide by two:

$\frac{(a-1) + (a+2)}{2} = \frac{2a+1}{2} = a + \frac{1}{2}$

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Example Question #1 : Calculating Median

Janice's course score in her statistics class depends on her six hourly tests. The mean of her best five scores and the median of her best five scores are both calculated, and she is assigned the better of the two.

Janice's first five tests were 90, 92, 80, 75, and 86. What score, at minimum, does she need to be assured a course score of 85 or better for the term?

Possible Answers:

Janice is already assured of a course average of at least 85 for the term.

Correct answer:

Janice is already assured of a course average of at least 85 for the term.

Explanation:

The median of her current five scores is the third-highest, or 86. Even if she does not take the sixth examination, this median will stand, and even if her mean is less, she has already achieved a score of 86 or better.

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Example Question #522 : Arithmetic

A data set with nine elements has median 50. A new data set is formed with these nine elements, plus two new elements, 40 and 73.

What is the median of this new data set?

Possible Answers:

It is impossible to tell from the information given

$53 \frac{2}{3}$

Correct answer:

Explanation:

The median of the original data set, which has nine elements, is the fifth-highest element; here it is 50. The median of a data set with eleven elements is its sixth-highest element; since one of the elements added is greater than 50 and one is less than 50, 50 becomes the sixth-highest element in the new set, and it remains the median.

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Example Question #1 : Median

Consider the data set

$\left \{ 24,61, 27, 57, 34, 37, 63, 50, N\right \}$

where is a positive integer.

What are the possible values of the median of the set?

Possible Answers:

37 or 50

37, 43.5, or 50

Any integer from 37 to 50 inclusive

38 or 49

Any integer from 38 to 49 inclusive

Correct answer:

Any integer from 37 to 50 inclusive

Explanation:

Arrange the eight known values from least to greatest.

$\left \{ 24, 27, 34, 37, 50, 57,61, 63\right \}$

If the unknown positive integer is added to the set, then the median of the resulting nine-element set is the fifth-highest.

Case 1: $N \leq 37$

The fifth-highest element is then 37.

Case 2: 37 < N < 50

The fifth-highest element is then .

Case 3: $N \geq 50$

The fifth-highest element is then 50.

Therefore, it is possible for the median to be any integer from 37 to 50 inclusive.

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Example Question #1 : Calculating Median

Consider the data set

$\left \{ 50, 55, 59, 66, 72, 74, 77, M, N\right \}$

For the median to be 70, what must be true about and ?

Possible Answers:

One must be equal to 70; the other must be greater than 70.

One must be equal to 70; the other must be greater than or equal to 70.

Both must be equal to 70.

One must be equal to 70; the other must be less than 70.

One must be equal to 70; the other must be less than or equal to 70.

Correct answer:

One must be equal to 70; the other must be greater than or equal to 70.

Explanation:

There are nine elements, so the median is the fifth-highest element. For this fifth-highest element to be 70, first of all, 70 must be in the set; since none of the known elements are equal to 70, then one of the two unknowns must be 70.

Assume without loss of generality that M = 70 . Then four of the elements are already known to be less than 70. Since four elements must be greater than or equal to 70, must be one of them.

Therefore, the correct choice is that one must be equal to 70 and the other must be greater than or equal to 70.

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Example Question #523 : Arithmetic

What is the median of the following set:

{5,9,21,36,45,7,4}

Possible Answers:

Correct answer:

Explanation:

Put the numbers in order from least to greatest:

{4,5,7,9,21,36,45}

The middle number is the median.

median = 9

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GMAT Math : Arithmetic

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #52 : Arithmetic Mean

Example Question #521 : Arithmetic

Example Question #521 : Arithmetic

Example Question #2 : Calculating Median

Example Question #2073 : Gmat Quantitative Reasoning

Example Question #1 : Calculating Median

Example Question #522 : Arithmetic

Example Question #1 : Median

Example Question #1 : Calculating Median

Example Question #523 : Arithmetic

All GMAT Math Resources

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