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Difference between revisions of "1996 AHSME Problems/Problem 5"

(Created page with "==See also== {{AHSME box|year=1996|num-b=4|num-a=6}}")
 
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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}}

Revision as of 19:36, 18 August 2011

Problem

Given that $0 < a < b < c < d$, which of the following is the largest?

$\text{(A)}\ \frac{a+b}{c+d} \qquad\text{(B)}\ \frac{a+d}{b+c} \qquad\text{(C)}\ \frac{b+c}{a+d} \qquad\text{(D)}\ \frac{b+d}{a+c} \qquad\text{(E)}\ \frac{c+d}{a+b}$

Solution 1

Assuming that one of the above fractions is indeed always the largest, try plugging in $a=1, b=2, c=3, d=4$, since those are valid values for the variables given the constraints of the problem. The options become:

$\text{(A)}\ \frac{1+2}{3+4} \qquad\text{(B)}\ \frac{1+4}{2+3} \qquad\text{(C)}\ \frac{2+3}{1+4} \qquad\text{(D)}\ \frac{2+4}{1+3} \qquad\text{(E)}\ \frac{3+4}{1+2}$

Simplified, the options are $\frac{3}{7}, 1, 1, \frac{3}{2}, \frac{7}{3}$, respectively. Since $\frac{7}{3}$ is the only option that is greater than $2$, the answer is $\boxed{E}$.

Solution 2

To make a fraction large, you want the largest possible numerator, and the smallest possible denominator. Since $c$ and $d$ are the two largest numbers, they should go in the numerator as a sum. Since $a$ and $b$ are the smallest numbers, they should go in the denominator as a sum. Thus, the answer is $\boxed{E}$.

You can compare option $E$ with every other fraction: all numerators are smaller than $E$'s numerator, and all denominators are larger than $E$'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
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