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1994 AJHSME Problems/Problem 20: Difference between revisions

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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>
==Solution==
<cmath>\frac{W}{X} + \frac{Y}{Z} = \frac{WZ+XY}{XZ}</cmath>
Small fractions have small numerators and large denominators. To maximize the denominator, let <math>X=8</math> and <math>Z=9</math>.
<cmath>\frac{9W+8Y}{72}</cmath>
To minimize the numerator, let <math>W=1</math> and <math>Y=2</math>.
<cmath>\frac{9+16}{72} = \boxed{\text{(D)}\rightarrow \frac{25}{72}}</cmath>
==Solution 2==
To make the smallest fraction, you need the lowest numerator and the highest denominator. So, take the first <math>2</math> and last <math>2</math> digits of the set, which are <math>1,2</math> and <math>8,9</math>. Balance the equations to be "even". Since <math>1</math> is smaller than <math>2</math>, put it over <math>8</math>. You get <math>\frac{1}{8}+\frac{2}{9}=\frac{25}{72}</math>, or <math>\boxed{D}</math>.
-goldenn
==See Also==
{{AJHSME box|year=1994|num-b=19|num-a=21}}
{{MAA Notice}}

Latest revision as of 10:56, 27 June 2023

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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