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

(Created page with "==Problem== Let <math>W,X,Y</math> and <math>Z</math> be four different digits selected from the set <math>\{ 1,2,3,4,5,6,7,8,9\}.</math> If the sum <math>\dfrac{W}{X} + \dfra...")
 
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<math>\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}</math>
 
<math>\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}</math>
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==Solution==
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<cmath>\frac{W}{X} + \frac{Y}{Z} = \frac{WZ+XY}{XZ}</cmath>
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Small fractions have small numerators and large denominators. To maximize the denominator, let <math>X=8</math> and <math>Z=9</math>.
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<cmath>\frac{9W+8Y}{72}</cmath>
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To minimize the numerator, let <math>W=1</math> and <math>Y=2</math>.
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<cmath>\frac{9+16}{72} = \boxed{\text{(D)}\ \frac{25}{72}}</cmath>
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==See Also==
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{{AJHSME box|year=1994|num-b=19|num-a=21}}

Revision as of 02:04, 23 December 2012

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 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