1985 AJHSME Problem 19


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

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


Let the length and width of the original rectangle be $L$ and $W,$ respectively. The perimeter of the original rectangle is $2L+2W.$ If we apply the changes as described in the problem, the perimeter of the new rectangle is $2 \cdot (\frac{11L}{10}) + 2 \cdot (\frac{11W}{10}) = \frac{11}{10} (2L+2W).$ This is an increase of 10%, so the answer is $\text{(B)}.$

Solution 2 (by mathmax12) proceed by coordbash