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

Revision as of 21:22, 9 January 2021

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

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.

@@ Line 5: / Line 5: @@
 ==Solution==
-<math> \boxed{B} </math>
+<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=\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}}

Page

Toolbox

Search

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

Revision as of 21:22, 9 January 2021

Problem

Solution

See Also