1962 AHSME Problems/Problem 13

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

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

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
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. AMC logo.png

Invalid username
Login to AoPS