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Difference between revisions of "1951 AHSME Problems/Problem 28"

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==Problem==
 
==Problem==
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The pressure <math>(P)</math> of wind on a sail varies jointly as the area <math>(A)</math> of the sail and the square of the velocity <math>(V)</math> of the wind. The pressure on a square foot is <math>1</math> pound when the velocity is <math>16</math> miles per hour. The velocity of the wind when the pressure on a square yard is <math>36</math> pounds is:
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<math> \textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph} </math>
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==Solution==
 
==Solution==
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Unsolved
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Because <math>P</math> varies jointly as <math>A</math> and <math>V^2</math>, that means that there is a number <math>k</math> such that <math>P=kAV^2</math>. You are given that <math>P=1</math> when <math>A=1</math> and <math>V=16</math>. That means that <math>1=k(1)(16^2) \rightarrow k=\frac{1}{256}</math>. Then, substituting into the original equation with <math>P=36</math> and <math>A=9</math> (because a square yard is <math>9</math> times a square foot), you get <math>4=\frac{1}{256}(V^2)</math>. Solving for <math>V</math>, we get <math>V^2=1024</math>, so <math>V=32</math>. Hence, the answer is <math>\boxed{C}</math>.
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=27|num-a=29}}  
 
{{AHSME 50p box|year=1951|num-b=27|num-a=29}}  

Latest revision as of 18:45, 20 November 2014

Problem

The pressure $(P)$ of wind on a sail varies jointly as the area $(A)$ of the sail and the square of the velocity $(V)$ of the wind. The pressure on a square foot is $1$ pound when the velocity is $16$ miles per hour. The velocity of the wind when the pressure on a square yard is $36$ pounds is:

$\textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph}$

Solution

Because $P$ varies jointly as $A$ and $V^2$, that means that there is a number $k$ such that $P=kAV^2$. You are given that $P=1$ when $A=1$ and $V=16$. That means that $1=k(1)(16^2) \rightarrow k=\frac{1}{256}$. Then, substituting into the original equation with $P=36$ and $A=9$ (because a square yard is $9$ times a square foot), you get $4=\frac{1}{256}(V^2)$. Solving for $V$, we get $V^2=1024$, so $V=32$. Hence, the answer is $\boxed{C}$.

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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