# Difference between revisions of "1991 AHSME Problems/Problem 1"

(Added solution and changed format of answer choices.) |
|||

Line 1: | Line 1: | ||

+ | ==Problem== | ||

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

− | (A) | + | <math> \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 </math> |

+ | |||

+ | ==Solution== | ||

+ | If we plug in <math>1</math> as <math>a</math>, <math>-2</math> as <math>b</math>, and <math>-3</math> as <math>c</math> in the expression <math>\frac{c+a}{c-b}</math>, then we get <math>\frac{-3+1}{-3-(-2)}=\frac{-2}{-1}=2</math>, which is choice <math>\boxed{\textbf{A}}</math>. | ||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 13:51, 22 July 2014

## Problem

If for any three distinct numbers , , and we define , then is

## Solution

If we plug in as , as , and as in the expression , then we get , which is choice . The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.