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 $\boxed{\textbf{E}}$ .

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	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 • 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.

Page

Toolbox

Search

1991 AHSME Problems/Problem 1

Problem

Solution

See also