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

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== See also ==
 
== See also ==
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[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
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{{MAA Notice}}

Latest revision as of 05:46, 24 February 2018

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

See also

1991 AHSME (ProblemsAnswer KeyResources)
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

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