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1985 AJHSME Problems/Problem 19

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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 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}}

See Also

1985 AJHSME (Problems • Resources)
Preceded by
Problem 18 		Followed by
Problem 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
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