2001 AMC 10 Problem 1 — Solution | LIVE by Po-Shen Loh

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2001 AMC 10 Problem 1

Attempt Problem 1 of the 2001 AMC 10 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 10 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

The median of the list

$\begin{gathered} n,\ n+3,\ n+4,\ n+5,\ n+6, \\ \ n+8,\ n+10,\ n+12,\ n+15 \end{gathered}$

is $10.$ What is the mean?

$4$

$6$

$7$

$10$

$11$

Answer: E

Concepts:median (data)mean

Difficulty rating: 790

Solution:

The list has $9$ numbers in increasing order, so the median is the $5$ th term, $n+6.$ Setting $n+6=10$ gives $n=4.$

The sum of the terms is $9n$ $\small {}+(3+4+5+6+8+10+12+15)$ $=9n+63$ $=99,$ so the mean is $99/9=11.$

Thus, the correct answer is E.

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