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

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2001 AMC 10 Problem 1

Below is the professionally curated solution for Problem 1 of the 2001 AMC 10, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:median (data)mean

Difficulty rating: 790

1.

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?

$4$

$6$

$7$

$10$

$11$

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 $9n+(3+4+5+6+8+10+12+15)=9n+63=99,$ so the mean is $99/9=11.$

Thus, the correct answer is E.

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