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2001 AMC 10 Problem 1

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2001 AMC 10 Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 1 of 5 · about 1 min · position P1-10

The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?

A)
$4$
B)
$6$
C)
$7$
D)
$10$
E)
$11$

Answer

Key insight

The list is already sorted, so the median is the fifth entry n+6; the mean is n plus the average of the offsets, 63/9 = 7.

Solution

The nine entries are listed in increasing order, so the median is the fifth one, $n + 6$ . Setting $n + 6 = 10$ gives $n = 4$ .

For the mean, add the offsets: $0 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 = 63$ . The total of the list is $9n + 63$ , so the mean is
$\frac{9n + 63}{9} = n + 7 = 4 + 7 = 11.$

The answer is $\boxed{\textbf{(E)}\ 11}$ .

Why this works

Every entry is $n$ plus a constant, so the mean is $n$ plus the mean of the constants and the median is $n$ plus the median of the constants. Shifting every data value by the same amount shifts the mean and median by exactly that amount; recognizing this saves rewriting the whole list.

Alternative approach

Skip $n$ entirely: the median offset is $6$ and the mean offset is $7$ , so the mean is exactly $1$ more than the median, i.e. $10 + 1 = 11$ .

The trap

Stopping at n = 4 or answering the median 10 instead of the mean, or miscounting which entry is the middle one.

Common mistakes

Stopping at n = 4 or answering the median 10 instead of the mean, or miscounting which entry is the middle one.
Adding the offsets incorrectly (there are nine terms and the sum is $63$ , not $60$ or $66$ ).

Techniques

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