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 | + | <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> | ||
==Solution== | ==Solution== | ||
− | {{ | + | |
+ | <cmath>R=c\cdot\frac{S}T</cmath> | ||
+ | |||
+ | for some constant <math>c</math>. | ||
+ | |||
+ | You know that | ||
+ | |||
+ | <cmath>\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,</cmath> | ||
+ | |||
+ | so | ||
+ | |||
+ | <cmath>c=\frac{4/3}{2/3}=2\,.</cmath> | ||
+ | |||
+ | When <math>R=\sqrt{48}</math> and <math>T=\sqrt{75}</math> we have | ||
+ | |||
+ | <cmath>\sqrt{48}=\frac{2S}{\sqrt{75}}\,,</cmath> | ||
+ | |||
+ | so | ||
+ | |||
+ | <cmath>S=\frac12\sqrt{48\cdot75}=30\,.</cmath> <math> \boxed{B} </math> | ||
+ | |||
+ | -- zixuan 12 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1962|before=Problem 12|num-a=14}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:19, 10 April 2023
Problem
varies directly as and inversely as . When and , . Find when and .
Solution
for some constant .
You know that
so
When and we have
so
-- zixuan 12
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |
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