Difference between revisions of "1994 AJHSME Problems/Problem 20"
Mrdavid445 (talk | contribs) (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)}\ \frac{25}{72}}</cmath> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AJHSME box|year=1994|num-b=19|num-a=21}} |
Revision as of 02:04, 23 December 2012
Problem
Let and be four different digits selected from the set
If the sum is to be as small as possible, then must equal
Solution
Small fractions have small numerators and large denominators. To maximize the denominator, let and .
To minimize the numerator, let and .
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 |