1951 AHSME Problems/Problem 28

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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