The width of a rectangle is doubled and the length is halved. This produces a square with a perimeter of $P$. What is the perimeter of the original rectangle?
(A) $P$
(B) $2 P$
(C) $\frac{1}{2} P$
(D) $\frac{5}{4} P$
(E) $\frac{5}{2} P$
Since the perimeter of the square is $P$ and the 4 sides of a square are equal in length, then each side of the square has length $\frac{1}{4} P$. We now work backward to determine the width and length of the rectangle.

The width of the rectangle was doubled to produced the side of the square with length $\frac{1}{4} P$. Therefore, the width of the rectangle is half of the side length of the square, or $\frac{1}{2} \times \frac{1}{4} P=\frac{1}{8} P$. The length of the rectangle was halved to produce the side of the square with length $\frac{1}{4} P$. Therefore, the length of the rectangle is twice the side length of the square, or $2 \times \frac{1}{4} P=\frac{1}{2} P$.

Finally, we determine the perimeter of the rectangle having width $\frac{1}{8} P$ and length $\frac{1}{2} P$, obtaining $2 \times\left(\frac{1}{8} P+\frac{1}{2} P\right)=2 \times\left(\frac{1}{8} P+\frac{4}{8} P\right)=2 \times\left(\frac{5}{8} P\right)=\frac{5}{4} P$.

ANSWER: (D)

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