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

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

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 $\boxed{\text{B}}$

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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