2001 AMC 12 Problem 4

Attempt Problem 4 of the 2001 AMC 12 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 12 solutions, or check the answer key.

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4.

The mean of three numbers is 1010 more than the least of the numbers and 1515 less than the greatest. The median of the three numbers is 5.5. What is their sum?

55

2020

2525

3030

3636

Answer: D
Concepts:meanmedian (data)linear equation
Difficulty rating: 1150
Solution:

Let mm be the mean. The least number is m10,m - 10, the greatest is m+15,m + 15, and the middle number is the median 5.5. Their sum is 3m,3m, so (m10)+5+(m+15)=3m. (m - 10) + 5 + (m + 15) = 3m.

This gives m=10,m = 10, so the sum of the three numbers is 3m=30.3m = 30.

Thus, the correct answer is D.

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