1955 AHSME Problems/Problem 5

Problem

$5y$ varies inversely as the square of $x$. When $y=16, x=1$. When $x=8, y$ equals:

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 128 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ 1024$

Solution

An inverse variation can be expressed in the form $xy = n$, where $n$ is any number (except perhaps zero). Since 5y varies inversely with the square of x, this particular one will be $5yx^2 = n$.

We can plug in 16 for y and 1 for x, which makes n 80. The equation is now $5yx^2 = 80$

When x = 8, we can solve for y using the equation $320y = 80$, which makes y $\textbf{(D)} \frac{1}{4}$

See Also

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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