Difference between revisions of "1976 AHSME Problems/Problem 1"
(→Solution) |
(→Solution) |
||
Line 7: | Line 7: | ||
== Solution == | == 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=- | + | 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]]}} | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} |
Latest revision as of 17:34, 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 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 |