Difference between revisions of "1994 AJHSME Problems/Problem 20"

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Revision as of 12:17, 18 October 2016

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21soccer
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

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