# 1999 AHSME Problems/Problem 3

## Problem

The number halfway between $1/8$ and $1/10$ is $\mathrm{(A) \ } \frac 1{80} \qquad \mathrm{(B) \ } \frac 1{40} \qquad \mathrm{(C) \ } \frac 1{18} \qquad \mathrm{(D) \ } \frac 1{9} \qquad \mathrm{(E) \ } \frac 9{80}$

## Solutions

### Solution 1

To find the number halfway between $\frac{1}{8}$ and $\frac{1}{10}$, simply take the arithmetic mean, which is $\frac{\frac{1}{8}+\frac{1}{10}}{2}=\frac{\frac{9}{40}}{2}=\frac{9}{80}.$

Thus the answer is choice $\boxed{E}.$

### Solution 2

Note that $\frac{1}{10} = 0.1$ and $\frac{1}{8} = 0.125$. Thus, the answer must be greater than $\frac{1}{10}$.

Answers $A$, $B$, and $C$ are all less than $\frac{1}{10}$, so they can be eliminated.

Answer $D$ is equivalent to $0.\overline{1}$, which is $0.0\overline{1}$ away from $0.1$, and is $0.013\overline{8}$ away from $0.125$. These distances are not equal, eliminating $D$.

Thus, $\boxed{E}$ must be the answer. Computing $\frac{9}{80} = 0.1125$ as a check, we see that it is $0.1125 - 0.1 = 0.0125$ away from $\frac{1}{10}$, and similarly it is $0.125 - 0.1125 = 0.0125$ away from $\frac{1}{8}$.

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