Difference between revisions of "1976 AHSME Problems/Problem 1"
(→Problem 1) |
(→Solution) |
||
| (4 intermediate revisions by one other user not shown) | |||
| Line 4: | Line 4: | ||
<math>\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad \textbf{(E) }3</math> | <math>\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad \textbf{(E) }3</math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | The reciprocal of <math>(1-x)</math> is <math>\frac{1}{1-x}</math>, so our equation is <cmath>1-\frac{1}{1-x}=\frac{1}{1-x},</cmath> which is equivalent to <math>\frac{1}{1-x}=\frac{1}{2}</math>. So, <math>1-x=2</math> and <math>x=-1\Rightarrow \textbf{(B)}</math>.~MathJams | ||
| + | |||
| + | |||
| + | |||
| + | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Latest revision as of 16:34, 29 November 2020
Problem 1
If one minus the reciprocal of 
equals the reciprocal of , then 
equals
Solution
The reciprocal of is 
, so our equation is ![\[1-\frac{1}{1-x}=\frac{1}{1-x},\]](http://latex.artofproblemsolving.com/f/f/3/ff381ea1ef9861150899951ec43cba92a5a1eb86.png)
which is equivalent to 
. So, 
and 
.~MathJams
| 1976 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by 1975 AHSME |
Followed by 1977 AHSME | |
| 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 | ||
