Difference between revisions of "1985 AJHSME Problem 19"

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== Solution ==
 
== Solution ==
Let the length and width of the original rectangle be <math>L</math> and <math>W,</math> respectively. The area of the original rectangle is <math>LW.</math> If we apply the changes as described in the problem, the area of the new rectangle is <math>(\frac{11L}{10})(\frac{11W}{10}) = \frac{121LW}{100}.</math> This is an area increase of 21%, so the answer is <math>\text{(D)}.</math>
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Let the length and width of the original rectangle be <math>L</math> and <math>W,</math> respectively. The perimeter of the original rectangle is <math>2L+2W.</math> If we apply the changes as described in the problem, the perimeter of the new rectangle is <math>2 \cdot (\frac{11L}{10}) + 2 \cdot (\frac{11W}{10}) = \frac{11}{10} (2L+2W).</math> This is an increase of 10%, so the answer is <math>\text{(B)}.</math>

Revision as of 10:54, 19 September 2021

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 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 $\text{(B)}.$