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 $\boxed{\text{B}}$ .

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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1985 AJHSME Problems/Problem 19

Contents

Problem

Solution #1

Solution #2 (Quick Fakesolve)

See Also