1985 AJHSME Problem 19

Problem

If the length and width of a rectangle are each increased by $10\%$, then the perimeter of the rectangle is increased by

$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$

Solution

Let the length and width of the original rectangle be $L$ and $W,$ respectively. The area of the original rectangle is $LW.$ If we apply the changes as described in the problem, the area of the new rectangle is $(\frac{11L}{10})(\frac{11W}{10}) = \frac{121LW}{100}.$ This is an area increase of 21%, so the answer is $\text{(D)}.$

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