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.

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

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

Let

m

be the mean; the least number is

m - 10

and the greatest is

m + 15

The median is the middle number

5,

(m - 10) + 5 + (m + 15) = 3m

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,

(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.

Problem 4 in other years