Difference between revisions of "1962 AHSME Problems"

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[[1962 AHSME Problems/Problem 4|Solution]]
 
[[1962 AHSME Problems/Problem 4|Solution]]
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==See Also==
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 22:23, 10 November 2013

Problem 1

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

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

Solution

Problem 2

The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}$ (Error compiling LaTeX. ! Missing } inserted.) is equal to:

$\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$

Solution

Problem 3

The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:

$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$

Solution

Problem 4

If $8^x = 32$, then x equals:

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ \frac{5}{3}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{3}{5}\qquad\textbf{(E)}\ \frac{1}{4}$

Solution







See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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