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2001 AMC 10 Problem 16

2001 AMC 10 Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 2 of 5 · about 2 min · position P16-20

The mean of three numbers is $10$ more than the least of the numbers and $15$ less than the greatest. The median of the three numbers is $5$ . What is their sum?

A)
$5$
B)
$20$
C)
$25$
D)
$30$
E)
$36$

Answer

Key insight

Write the least and greatest as m - 10 and m + 15 in terms of the mean m; with median 5 the sum 2m + 10 must equal 3m.

Solution

Let the mean be $m$ . Then the least number is $m - 10$ , the greatest is $m + 15$ , and the middle number is the median, $5$ .

The three numbers add to $3m$ (that is what "mean $m$ " means), so
$(m - 10) + 5 + (m + 15) = 3m \quad\Longrightarrow\quad 2m + 10 = 3m \quad\Longrightarrow\quad m = 10.$
The sum is $3m = 30$ . As a check, the numbers are $0, 5, 25$ , whose mean is $10$ , with $10 = 0 + 10 = 25 - 15$ .

The answer is $\boxed{\textbf{(D)}\ 30}$ .

Why this works

Expressing every quantity in terms of a single unknown (the mean) keeps the problem to one equation. The definition "sum equals three times the mean" is used twice: once to write the sum, once to read off the final answer, so no individual number has to be found.

The trap

Answering the mean (10) instead of the sum, or forgetting that for three numbers the median is the middle value.

Common mistakes

Answering the mean (10) instead of the sum, or forgetting that for three numbers the median is the middle value.
Setting the least number equal to $m + 10$ (reversing "10 more than the least"), which gives $m = 20$ and a sum of $60$ .

Techniques

Set up the equation/formula and compute; no special trick needed