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> | ||
Latest revision as of 21:19, 10 April 2023
Problem
varies directly as 
and inversely as 
. When 
and 
, 
. Find when 
and 
.
Solution
![\[R=c\cdot\frac{S}T\]](http://latex.artofproblemsolving.com/c/a/3/ca38721927c8334991b8e7bd3a838607b30541cb.png)
for some constant 
.
You know that
![\[\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,\]](http://latex.artofproblemsolving.com/9/2/e/92eb3b5f92fdc3051783633a99b872b71d9d5b81.png)
so
![\[c=\frac{4/3}{2/3}=2\,.\]](http://latex.artofproblemsolving.com/c/f/b/cfb30b546e4a958b2c966d8a21272848adf8a5e2.png)
When 
and 
we have
![\[\sqrt{48}=\frac{2S}{\sqrt{75}}\,,\]](http://latex.artofproblemsolving.com/2/1/1/21165b56eecfc4b318d5b85836f213b77ba03b1a.png)
so
![\[S=\frac12\sqrt{48\cdot75}=30\,.\]](http://latex.artofproblemsolving.com/6/2/a/62abf441dace91999775cc289ae81c722be53d87.png)
-- zixuan 12
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| 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. 
