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 == | ||
| + | {{AHSME box|year=1991|num-b=1|num-a=3}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:14, 28 September 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 
.
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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. 
