Difference between revisions of "1991 AHSME Problems/Problem 3"
(Created page with "<math>(4^{-1}-3^{-1})^{-1}=</math> (A) <math>-12</math> (B) <math>-1</math> (C) <math>\frac{1}{12}</math> (D) <math>1</math> (E) <math>12</math>") |
(Added a solution with explanation) |
||
| (3 intermediate revisions by 2 users not shown) | |||
| Line 1: | Line 1: | ||
| + | == Problem == | ||
| + | |||
<math>(4^{-1}-3^{-1})^{-1}=</math> | <math>(4^{-1}-3^{-1})^{-1}=</math> | ||
(A) <math>-12</math> (B) <math>-1</math> (C) <math>\frac{1}{12}</math> (D) <math>1</math> (E) <math>12</math> | (A) <math>-12</math> (B) <math>-1</math> (C) <math>\frac{1}{12}</math> (D) <math>1</math> (E) <math>12</math> | ||
| + | |||
| + | == Solution == | ||
| + | <math>\fbox{A}</math> This is <math>\frac{1}{\frac{1}{4} - \frac{1}{3}} = \frac{1}{\frac{-1}{12}} = -12.</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1991|num-b=2|num-a=4}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:14, 23 February 2018
Problem
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
This is 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
