1988 AMC 8 Problem 21

Below is the professionally curated solution for Problem 21 of the 1988 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 8 solutions, or check the answer key.

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Concepts:meanmedian (data)casework

Difficulty rating: 1260

21.

A fifth number, n,n, is added to the set of numbers {3,6,9,10}\{3, 6, 9, 10\} to make the mean of the set of five numbers equal to its median. The number of possible values for nn is

11

22

33

44

more than 44

Solution:

The five numbers have sum 28+n28 + n and mean 28+n5.\dfrac{28 + n}{5}. The median is the middle value when sorted, which is 6,6, n,n, or 99 depending on the size of n.n.

If the median is 66: 28+n5=6\dfrac{28 + n}{5} = 6 gives n=2,n = 2, which is indeed less than 6.6. If the median is nn: 28+n5=n\dfrac{28 + n}{5} = n gives n=7,n = 7, which is between 66 and 9.9. If the median is 99: 28+n5=9\dfrac{28 + n}{5} = 9 gives n=17,n = 17, which is indeed greater than 9.9.

Each case yields a valid value, so nn can be 2,2, 7,7, or 1717 — three possible values.

Thus, the correct answer is C .

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