2001 AMC 12 Problem 4

Below is the professionally curated solution for Problem 4 of the 2001 AMC 12, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 12 solutions, or check the answer key.

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Concepts:mean median (data)linear equation

Difficulty rating: 1150

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?

$5$

$20$

$25$

$30$

$36$

Solution:

Let $m$ be the mean. The least number is $m - 10,$ the greatest is $m + 15,$ and the middle number is the median $5.$ Their sum is $3m,$ so $(m - 10) + 5 + (m + 15) = 3m.$

This gives $m = 10,$ so the sum of the three numbers is $3m = 30.$

Thus, the correct answer is D.

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