Difference between revisions of "1962 AHSME Problems/Problem 1"
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| + | ==Problem== | ||
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The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to: | The expression <math>\frac{1^{4y-1}}{5^{-1}+3^{-1}}</math> is equal to: | ||
<math> \textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8} </math> | <math> \textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8} </math> | ||
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| + | ==Solution== | ||
| + | We simplify the expression to yield: | ||
| + | |||
| + | <math> \dfrac{1^{4y-1}}{5^{-1}+3^{-1}}=\dfrac{1}{5^{-1}+3^{-1}}=\dfrac{1}{\dfrac{1}{5}+\dfrac{1}{3}}=\dfrac{1}{\dfrac{8}{15}}=\dfrac{15}{8} </math>. | ||
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| + | Thus our answer is <math>\boxed{\textbf{(D)}\ \frac{15}{8}}</math>. | ||
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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|before=First Question|num-a=2}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:11, 3 October 2014
Problem
The expression 
is equal to:
Solution
We simplify the expression to yield:
.
Thus our answer is 
.
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
