# 1996 AHSME Problems/Problem 5

## 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.

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