A list consists of six numbers: 3, 7, 10, 14, 18, and x. If the arithmetic mean of these s… — GRE General Test

A list consists of six numbers: $3$ , $7$ , $10$ , $14$ , $18$ , and $x$ . If the arithmetic mean of these six numbers is equal to their median, which of the following could be the value of $x$ ? Indicate all such values.

$-1$ Answer
B
$5$
$11$ Answer
D
$14$
$20$ Answer

Answer

The valid values for

x

are

-1

11

, and

20

The values

-1

11

, and

20

each yield a dataset where the arithmetic mean equals the median:

-1

gives a mean and median of

8.5

11

gives a mean and median of

10.5

, and

20

gives a mean and median of

12

Step-by-Step Solution

Express the arithmetic mean in terms of

x

Mean =

\frac{3 + 7 + 10 + 14 + 18 + x}{6} = \frac{52 + x}{6}

The mean of a dataset of

n

numbers is the sum of all elements divided by

n

Analyze the median based on the position of

x

relative to the sorted known values

3, 7, 10, 14, 18

Case 1:

x \le 7 \implies

median =

\frac{7+10}{2} = 8.5

.
Case 2:

7 < x < 14 \implies

median =

\frac{x+10}{2}

(for

7 < x \le 10

) or

\frac{10+x}{2}

(for

10 < x < 14

).
Case 3:

x \ge 14 \implies

median =

\frac{10+14}{2} = 12

For an even number of elements (

n=6

), the median is the average of the 3rd and 4th terms in ascending order.

Set the mean equal to the median for each case and solve for

x

Case 1:

\frac{52+x}{6} = 8.5 \implies 52+x = 51 \implies x = -1

(valid since

-1 \le 7

).
Case 2:

\frac{52+x}{6} = \frac{10+x}{2} \implies 52+x = 30+3x \implies 2x = 22 \implies x = 11

(valid since

7 < 11 < 14

).
Case 3:

\frac{52+x}{6} = 12 \implies 52+x = 72 \implies x = 20

(valid since

20 \ge 14

This identifies all values of

x

satisfying the problem constraint.

Key Concept

Evaluating mean and median of a dataset containing an unknown variable across different intervals of the variable's possible values.

Question

Answer

Step-by-Step Solution

Key Concept

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