Difference between revisions of "1994 AJHSME Problems/Problem 1"
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<math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{3}{8} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{7}{24}</math> | <math>\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{1}{4} \qquad \text{(C)}\ \dfrac{3}{8} \qquad \text{(D)}\ \dfrac{5}{12} \qquad \text{(E)}\ \dfrac{7}{24}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
C and D are the only answer choices where the numerator is close to half of the denominator. <math>\dfrac{3}{8} = .375</math> and <math>\dfrac{5}{12} \approx .41</math>. Thus the answer is <math>\boxed{\text{(D)}\ \frac{5}{12}}</math> | C and D are the only answer choices where the numerator is close to half of the denominator. <math>\dfrac{3}{8} = .375</math> and <math>\dfrac{5}{12} \approx .41</math>. Thus the answer is <math>\boxed{\text{(D)}\ \frac{5}{12}}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | Multiply every answer choice by <math>24</math>, the [[LCM]] of all of the answer choices. This gives | ||
| + | |||
| + | <math>\text{(A)}\ 8 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 7</math> | ||
| + | |||
| + | of which <math>\boxed{D}</math> is the largest. | ||
==See Also== | ==See Also== | ||
{{AJHSME box|year=1994|before=First <br /> Problem|num-a=2}} | {{AJHSME box|year=1994|before=First <br /> Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:47, 27 June 2023
Contents
Problem
Which of the following is the largest?
Solution 1
C and D are the only answer choices where the numerator is close to half of the denominator. 
and 
. Thus the answer is 
Solution 2
Multiply every answer choice by 
, the LCM of all of the answer choices. This gives
of which 
is the largest.
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
