Difference between revisions of "1991 AHSME Problems/Problem 1"
(Added solution and changed format of answer choices.) |
m (Fixed the box at the bottom) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 5: | Line 5: | ||
==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{ | + | 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>. |
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1991|num-b=1|num-a=2}} | ||
+ | |||
+ | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 05:46, 24 February 2018
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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.