# Difference between revisions of "1985 AJHSME Problems/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 #1

Let the width be $w$ and the length be $l$. Then, the original perimeter is $2(w+l)$.

After the increase, the new width and new length are $1.1w$ and $1.1l$, so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$.

Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}

$\boxed{\text{B}}$

## Solution #2 (Quick Fakesolve)

Assume WLOG that the rectangle is a square with length $10$ and width $10$. Thus, the square has a perimeter of $40$. Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is $44$, and because $44$ is $110\%$ of $40$, our answer is $\boxed{\text{(B) } 10\%}$.

## See Also

 1985 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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