# Difference between revisions of "1967 AHSME Problems/Problem 19"

## Problem

The area of a rectangle remains unchanged when it is made $2 \frac{1}{2}$ inches longer and $\frac{2}{3}$ inch narrower, or when it is made $2 \frac{1}{2}$ inches shorter and $\frac{4}{3}$ inch wider. Its area, in square inches, is: $\textbf{(A)}\ 30\qquad \textbf{(B)}\ \frac{80}{3}\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ \frac{45}{2}\qquad \textbf{(E)}\ 20$

## Solution

We are given $xy = (x+\frac{5}{2})(y-\frac{2}{3}) = (x - \frac{5}{2})(y + \frac{4}{3})$

FOILing each side gives: $xy = xy - \frac{2}{3}x + \frac{5}{2}y - \frac{5}{3} = xy + \frac{4}{3}x - \frac{5}{2}y - \frac{10}{3}$

Taking the last two parts and moving everything to the left gives: $-2x + 5y + \frac{5}{3} = 0$

Taking the first two parts and multiplying by $3$ gives $-2x + \frac{15}{2}y - 5 = 0$

Solving both equations for $-2x$ and setting them equal to each other gives $5y + \frac{5}{3} = \frac{15}{2}y - 5$, which leads to $y = \frac{8}{3}$

Plugging that in to $-2x + 5y + \frac{5}{3} = 0$ gives $x = \frac{15}{2}$.

The area of the rectangle is $xy = 20$, or $\fbox{E}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 