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The length of rectangle S is 20 percent longer than the length of

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Bunuel
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The length of rectangle S is 20 percent longer than the length of [#permalink] New post   28 Sep 2017, 23:46
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The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is

(A) 20% greater than the area of rectangle R
(B) 4% greater than the area of rectangle R
(C) equal to the area of rectangle R
(D) 4% less than the area of rectangle R
(E) 20% less than the area of rectangle R
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Re: The length of rectangle S is 20 percent longer than the length of [#permalink] New post   29 Sep 2017, 02:12
Let the length of rectangle R= l
width of rectangle R = w
area = lw

then length of rectangle S = 1.2l
width of rectangle S = 0.8w
area = 0.96lw

so difference in area = lw-0.96lw = 0.04
4% less than the area of rectangle R

Answer should be D
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The length of rectangle S is 20 percent longer than the length of [#permalink] New post   29 Sep 2017, 15:28
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Bunuel wrote:
The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is

(A) 20% greater than the area of rectangle R
(B) 4% greater than the area of rectangle R
(C) equal to the area of rectangle R
(D) 4% less than the area of rectangle R
(E) 20% less than the area of rectangle R

Let R's length = 10
Let R's width = 5
R's area = (10 * 5) = 50

S's length is 20 percent longer than R's length
(10)(1.2) = 12

S's width is 20 percent shorter than R's width
(5)(.8) = 4

S's area = (12 * 4) = 48

The area of rectangle S is what percent greater or less than rectangle R?

Percent change =

\((\frac{(new-old)}{old}\) * 100) = \((\frac{S - R}{R}\) * 100) =

\((\frac{48 - 52}{100})\) = \((\frac{-4}{100})\) =

-.04 * 100 = -4%

The area of a rectangle S is 4% less than the area of rectangle R

Answer D
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Re: The length of rectangle S is 20 percent longer than the length of [#permalink] New post   03 Oct 2017, 16:33
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Bunuel wrote:
The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is

(A) 20% greater than the area of rectangle R
(B) 4% greater than the area of rectangle R
(C) equal to the area of rectangle R
(D) 4% less than the area of rectangle R
(E) 20% less than the area of rectangle R


We can let the length of rectangle R = L and the width of rectangle R = W.

Thus, the area of rectangle R = LW and the area of rectangle S = 1.2L x 0.8W = 0.96LW.

We see that the area of rectangle S is 4% less than the area of rectangle R.

Answer: D
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Re: The length of rectangle S is 20 percent longer than the length of [#permalink]
03 Oct 2017, 16:33
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