Difference between revisions of "1962 AHSME Problems/Problem 1"

Latest revision as of 21:11, 3 October 2014

The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to:

$\textbf{(A)}\ \frac{4y-1}{8}\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{15}{2}\qquad\textbf{(D)}\ \frac{15}{8}\qquad\textbf{(E)}\ \frac{1}{8}$

We simplify the expression to yield:

$\dfrac{1^{4y-1}}{5^{-1}+3^{-1}}=\dfrac{1}{5^{-1}+3^{-1}}=\dfrac{1}{\dfrac{1}{5}+\dfrac{1}{3}}=\dfrac{1}{\dfrac{8}{15}}=\dfrac{15}{8}$ .

Thus our answer is $\boxed{\textbf{(D)}\ \frac{15}{8}}$ .

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

@@ Line 6: / Line 6: @@
 ==Solution==
-''Unsolved''
+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>.
+Thus our answer is <math>\boxed{\textbf{(D)}\ \frac{15}{8}}</math>.
 ==See Also==
+{{AHSME 40p box|year=1962|before=First Question|num-a=2}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}