# Difference between revisions of "1964 AHSME Problems/Problem 5"

## Problem

If $y$ varies directly as $x$, and if $y=8$ when $x=4$, the value of $y$ when $x=-8$ is:

$\textbf{(A)}\ -16 \qquad \textbf{(B)}\ -4 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 4k, k= \pm1, \pm2, \dots \qquad \\ \textbf{(E)}\ 16k, k=\pm1,\pm2,\dots$

## Solution

If $y$ varies directly as $x$, then $\frac{x}{y}$ will be a constant for every pair of $(x, y)$. Thus, we have:

$\frac{4}{8} = \frac{-8}{y}$

$y \cdot \frac{1}{2} = -8$

$y = -16$

$\boxed{\textbf{(A)}}$