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 $Y=2$ .

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

Solution 2

To make the smallest fraction, you need the lowest numerator and the highest denominator. So, take the first $2$ and last $2$ digits of the set, which are $1,2$ and $8,9$ . Balance the equations to be "even". Since $1$ is smaller than $2$ , put it over $8$ . You get $\frac{1}{8}+\frac{2}{9}=\frac{25}{72}$ , or $\boxed{D}$ .

-goldenn

1994 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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1994 AJHSME Problems/Problem 20

Contents

Problem

Solution

Solution 2

See Also