Difference between revisions of "1996 AHSME Problems/Problem 5"
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| + | ==Problem== | ||
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| + | Given that <math> 0 < a < b < c < d </math>, which of the following is the largest? | ||
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| + | <math> \text{(A)}\ \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\ \frac{b+c}{a+d} \qquad\text{(D)}\ \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b} </math> | ||
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| + | ==Solution 1== | ||
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| + | Assuming that one of the above fractions is indeed always the largest, try plugging in <math>a=1, b=2, c=3, d=4</math>, since those are valid values for the variables given the constraints of the problem. The options become: | ||
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| + | <math> \text{(A)}\ \frac{1+2}{3+4} \qquad\text{(B)}\ \frac{1+4}{2+3} \qquad\text{(C)}\ \frac{2+3}{1+4} \qquad\text{(D)}\ \frac{2+4}{1+3} \qquad\text{(E)}\ \frac{3+4}{1+2} </math> | ||
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| + | Simplified, the options are <math>\frac{3}{7}, 1, 1, \frac{3}{2}, \frac{7}{3}</math>, respectively. Since <math>\frac{7}{3}</math> is the only option that is greater than <math>2</math>, the answer is <math>\boxed{E}</math>. | ||
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| + | ==Solution 2== | ||
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| + | To make a fraction large, you want the largest possible numerator, and the smallest possible denominator. Since <math>c</math> and <math>d</math> are the two largest numbers, they should go in the numerator as a sum. Since <math>a</math> and <math>b</math> are the smallest numbers, they should go in the denominator as a sum. Thus, the answer is <math>\boxed{E}</math>. | ||
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| + | You can compare option <math>E</math> with every other fraction: all numerators are smaller than <math>E</math>'s numerator, and all denominators are larger than <math>E</math>'s denominator. | ||
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==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=4|num-a=6}} | {{AHSME box|year=1996|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:07, 5 July 2013
Contents
Problem
Given that 
, which of the following is the largest?
Solution 1
Assuming that one of the above fractions is indeed always the largest, try plugging in 
, since those are valid values for the variables given the constraints of the problem. The options become:
Simplified, the options are 
, respectively. Since 
is the only option that is greater than 
, the answer is 
.
Solution 2
To make a fraction large, you want the largest possible numerator, and the smallest possible denominator. Since 
and 
are the two largest numbers, they should go in the numerator as a sum. Since 
and 
are the smallest numbers, they should go in the denominator as a sum. Thus, the answer is .
You can compare option 
with every other fraction: all numerators are smaller than 's numerator, and all denominators are larger than 's denominator.
See also
| 1996 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
