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1989 AJHSME Problems/Problem 17

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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 (ProblemsResources)
Preceded by
Problem 16
Followed by
Problem 18
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