Difference between revisions of "1962 AHSME Problems/Problem 13"

Revision as of 21:17, 10 April 2023

Problem

$R$ varies directly as $S$ and inverse as $T$ . When $R = \frac{4}{3}$ and $T = \frac {9}{14}$ , $S = \frac37$ . Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$ .

$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$

Solution

$R=c\cdot\frac{S}T$

for some constant $c$ .

You know that

$\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,$

so

$c=\frac{4/3}{2/3}=2\,.$

When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have

$\sqrt{48}=\frac{2S}{\sqrt{75}}\,,$

so

$S=\frac12\sqrt{48\cdot75}=30\,.$ $\boxed{B}$

-- zixuan 12

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 You know that
-<cmath>\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=\frac23\,,</cmath>
+<cmath>\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,</cmath>
 so

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Difference between revisions of "1962 AHSME Problems/Problem 13"

Revision as of 21:17, 10 April 2023

Problem

Solution

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