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

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(Solution)
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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> \boxed{\text{(A)}\ \ </math>A+B$
  
 
==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 14:01, 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 $Y=2$.

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

Problem

Mr. Langry had two variables, $A$ and $B$. What is $A+B$?

$\text{(A)}\ $A+B$\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

The answer is simple, it is $\boxed{\text{(A)}\ $ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)A+B$

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

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