1988 AMC 8 Problem 21

Attempt Problem 21 of the 1988 AMC 8 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 1988 AMC 8 solutions, or check the answer key.

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

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
Answer: C
Concepts:meanmedian (data)casework
Small hint:
The mean of the five numbers is 28+n5;\dfrac{28 + n}{5}; the median depends on where nn lands in the sorted order
Big hint:
Split into cases: the median is 6,6, is nn itself, or is 9,9, and solve the equation mean == median in each
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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