Difference between revisions of "1988 AJHSME Problems/Problem 21"

Revision as of 14:34, 20 April 2009

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 have cases.

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 $n$ .

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Difference between revisions of "1988 AJHSME Problems/Problem 21"

Revision as of 14:34, 20 April 2009

Problem

Solution

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