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

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==Problem==
 
==Problem==
<math>R</math> varies directly as <math>S</math> and inverse as <math>T</math>. When <math>R = \frac{4}{3}</math> and <math>T = \frac {9}{14}</math>, <math>S = \frac37</math>. Find <math>S</math> when <math>R = \sqrt {48}</math> and <math>T = \sqrt {75}</math>.
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<math>R</math> varies directly as <math>S</math> and inversely as <math>T</math>. When <math>R = \frac{4}{3}</math> and <math>T = \frac {9}{14}</math>, <math>S = \frac37</math>. Find <math>S</math> when <math>R = \sqrt {48}</math> and <math>T = \sqrt {75}</math>.
  
 
<math> \textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60 </math>
 
<math> \textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60 </math>
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You know that
 
You know that
  
<cmath>\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=\frac23\,,</cmath>
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<cmath>\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,</cmath>
  
 
so
 
so

Latest revision as of 21:19, 10 April 2023

Problem

$R$ varies directly as $S$ and inversely 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

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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