Arithmetic Mean

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GMAT Quantitative › Arithmetic Mean

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:

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

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

$15+30+x=75$

$45+x=75$

$x=30$

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 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:

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

The average of the following 6 digits is 75. What is a possible value of ?

80, 78, 78, 70, 71,

Explanation

Therefore, the sum of all 6 digits must equal 450.

Subtract 377 from both sides.

Find such that the arithmetic mean of is equal to the arithmetic mean of

Explanation

The formula for the arithmetic mean is:

Mean= $\frac{x_{1}+x_{2}+...+x_{n}}{n}$

We can then write:

$\frac{n+13+27+35}{4}=\frac{20+16+41+23}{4}$

What is the mean of this data set?

$\left { 1, 0.1, 0.01, 0.001, 0.0001, 0.00001 \right }$

Explanation

Add the numbers and divide by 6:

$\left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$= 1.11111\div 6 = 0.185185$

Find such that the arithmetic mean of is equal to the arithmetic mean of

Explanation

The formula for the arithmetic mean is:

Mean= $\frac{x_{1}+x_{2}+...+x_{n}}{n}$

We can then write:

$\frac{n+13+27+35}{4}=\frac{20+16+41+23}{4}$

If and , then give the mean of , , , , and .

Insufficient information is given to answer this question.

Explanation

The mean of , , , , and is $\frac{1}{5} \left (a + b + c + d + e \right )$

If you add both sides of each equation:

$\underline{5a + 4b + 3c + 2d + e = 5,300}$

$6\left (a + b + c + d + e \right )= 8,700$

Equivalently,

$6\left (a + b + c + d + e \right ) \div 6= 8,700 \div 6$

$\frac{1}{5} \left (a + b + c + d + e \right )= \frac{1}{5} \cdot 1,450 = 290$ ,

making 290 the mean.

When assigning a score for the term, a professor takes the mean of all of a student's test scores except for his or her lowest score.

On the first five exams, Donna has achieved the following scores: 76, 84, 80, 65, 91. There is one more exam in the course. Assuming that 100 is the maximum possible score, what is the range of possible final averages she can achieve (nearest tenth, if applicable)?

Minimum 79.2; maximum 86.2

Minimum 66; maximum 71.8

Minimum 66; maximum 99.2

Minimum 79.2; maximum 99.2

Minimum 80, maximum 84