1989 AJHSME Problems/Problem 17

Problem

The number $\text{N}$ is between $9$ and $17$. The average of $6$, $10$, and $\text{N}$ could be

$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution

We know that $9<N<17$ and we wish to bound $\frac{6+10+N}{3}=\frac{16+N}{3}$.

From what we know, we can deduce that $25<N+16<33$, and thus \[8.\overline{3}<\frac{N+16}{3}<11\]

The only answer choice that falls in this range is choice $\boxed{\text{B}}$

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions