# 1994 AJHSME Problems/Problem 20

## Problem

Let $W,X,Y$ and $Z$ be four different digits selected from the set

$\{ 1,2,3,4,5,6,7,8,9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\text{(A)}\ \dfrac{2}{17} \qquad \text{(B)}\ \dfrac{3}{17} \qquad \text{(C)}\ \dfrac{17}{72} \qquad \text{(D)}\ \dfrac{25}{72} \qquad \text{(E)}\ \dfrac{13}{36}$

## Solution

$$\frac{W}{X} + \frac{Y}{Z} = \frac{WZ+XY}{XZ}$$

Small fractions have small numerators and large denominators. To maximize the denominator, let $X=8$ and $Z=9$.

$$\frac{9W+8Y}{72}$$

To minimize the numerator, let $W=1$ and $Y=2$.

$$\frac{9+16}{72} = \boxed{\text{(D)}\ \frac{25}{72}}$$

## See Also

 1994 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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

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