1988 AJHSME Problems/Problem 21

Problem

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

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ \text{more than }4$

Solution

The possible medians after $n$ is added are $6$, $n$, or $9$. Now we use casework.

Case 1: The median is $6$

In this case, $n<6$ and \[\frac{3+n+6+9+10}{5}=6 \Rightarrow n=2\] so this case contributes $1$.

Case 2: The median is $n$

We have $6<n<9$ and \[\frac{3+6+n+9+10}{5}=n \Rightarrow n=7\] so this case also contributes $1$.

Case 3: The median is $9$

We have $9<n$ and \[\frac{3+6+9+n+10}{5}=9 \Rightarrow 17\] so this case adds $1$.

In all there are $3\rightarrow \boxed{\text{C}}$ possible values of $n$.

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png