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

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==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
  
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(A) <math>-2</math>  (B) <math>-\frac{2}{5}</math> (C) <math>-\frac{1}{4}</math> (D) <math>\frac{2}{5}</math> (E) <math>2</math>
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<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>
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==Solution==
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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{E}}</math>.
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== See also ==
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{{AHSME box|year=1991|num-b=1|num-a=2}} 
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[[Category: Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 04:46, 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 (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

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