Difference between revisions of "1985 AJHSME Problems/Problem 19"

Revision as of 21:05, 1 June 2009

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 $\begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}$

$\boxed{\text{B}}$

1985 AJHSME (Problems • Answer Key • 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
All AJHSME/AMC 8 Problems and Solutions

@@ Line 7: / Line 7: @@
 ==Solution==
-Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+1)</math>.
+Let the width be <math>w</math> and the length be <math>l</math>. Then, the original perimeter is <math>2(w+l)</math>.
 After the increase, the new width and new length are <math>1.1w</math> and <math>1.1l</math>, so the new perimeter is <math>2(1.1w+1.1l)=2.2(w+l)</math>.

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Difference between revisions of "1985 AJHSME Problems/Problem 19"

Revision as of 21:05, 1 June 2009

Problem

Solution

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