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

(Solution)
m (Solution 2)
 
Line 24: Line 24:
  
 
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: 1,2 and 8,9. Balance the equations to be "even". Since 1 is smaller than 2, put it over 8. You get 1/8 + 2/9 = 25/72, or D
 
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: 1,2 and 8,9. Balance the equations to be "even". Since 1 is smaller than 2, put it over 8. You get 1/8 + 2/9 = 25/72, or D
 +
 +
-goldenn
  
 
==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}}

Latest revision as of 00:17, 11 September 2019

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: 1,2 and 8,9. Balance the equations to be "even". Since 1 is smaller than 2, put it over 8. You get 1/8 + 2/9 = 25/72, or D

-goldenn

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS