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

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==Solution==
 
==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>.
 
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>.
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== See also ==
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{{AHSME box|year=1991|num-b=1|num-a=3}} 
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[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:14, 28 September 2014

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 $\boxed{\textbf{A}}$.


See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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