1991 AHSME Problems/Problem 1

Problem

If for any three distinct numbers $a$ , $b$ , and $c$ we define $f(a,b,c)=\frac{c+a}{c-b}$ , then $f(1,-2,-3)$ is

$\textbf {(A) } -2 \qquad \textbf {(B) } -\frac{2}{5} \qquad \textbf {(C) } -\frac{1}{4} \qquad \textbf {(D) } \frac{2}{5} \qquad \textbf {(E) } 2$

Solution

If we plug in $1$ as $a$ , $-2$ as $b$ , and $-3$ as $c$ in the expression $\frac{c+a}{c-b}$ , then we get $\frac{-3+1}{-3-(-2)}=\frac{-2}{-1}=2$ , which is choice $\boxed{\textbf{A}}$ . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1991 AHSME Problems/Problem 1

Problem

Solution