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

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<cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath>
 
<cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath>
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Problem
 
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Mr. Langry had two variables, <math>A</math> and <math>B</math>. What is <math>A+B</math>?
 
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<math>\text{(A)}\ \ </math>A+B<math> \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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The answer is simple, it is <math></math> \boxed{\text{(A)} \ \ <math>A+B</math> <math></math>
 
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1994|num-b=19|num-a=21}}
 
{{AJHSME box|year=1994|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:04, 30 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 $Y=2$.

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

See Also

1994 AJHSME (Problems • Answer Key • Resources)
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

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