1967 AMC 12 Problem 19

Attempt Problem 19 of the 1967 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1967 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

The area of a rectangle remains unchanged when it is made 2122\dfrac12 inches longer and 23\dfrac23 inch narrower, or when it is made 2122\dfrac12 inches shorter and 43\dfrac43 inch wider. Its area, in square inches, is:
3030
803\dfrac{80}{3}
2424
452\dfrac{45}{2}
2020
Answer: E
Concepts:rectanglesystem of equationsarea
Small hint:
Let the original length and width be ll and ww
Big hint:
Expand both equal-area equations; the lwlw terms cancel
Solution:
The two conditions are lw=(l+52)(w−23),lw=(l−52)(w+43). \begin{aligned} lw&=\left(l+\frac52\right)\left(w-\frac23\right),\\ lw&=\left(l-\frac52\right)\left(w+\frac43\right). \end{aligned} After expanding, these give −4l+15w=10-4l+15w=10 and 8l−15w=20.8l-15w=20. Hence l=152,l=\frac{15}{2}, w=83,w=\frac{8}{3}, and lw=20.lw=20.
Therefore, the correct answer is E.
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