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If the side length of a square is reduced by p percent, what

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If the side length of a square is reduced by p percent, what [#permalink] New post 03 Aug 2014, 12:44
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A
B
C
D
E

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Question Stats:

58% (02:09) correct 42% (00:38) wrong based on 71 sessions
If the side length of a square is reduced by p percent, what is the resulting percent reduction in the area of the square?

(A) \(\frac{p^2}{100}%\)
(B) \([1-\frac{p^2}{100}]^2%\)
(C) \([2p-\frac{p^2}{100}]%\)
(D) \(\frac{(100-p)^2}{100} %\)
(E) \(\frac{p^2-2p}{100}%\)
[Reveal] Spoiler: OA
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Re: If the side length of a square is reduced by p percent, what [#permalink] New post 03 Aug 2014, 13:48
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oss198 wrote:
If the side length of a square is reduced by p percent, what is the resulting percent reduction in the area of the square?

(A) \(\frac{p^2}{100}%\)
(B) \([1-\frac{p^2}{100}]^2%\)
(C) \([2p-\frac{p^2}{100}]%\)
(D) \(\frac{(100-p)^2}{100} %\)
(E) \(\frac{p^2-2p}{100}%\)



I set the side length of a square is 1 ( just to simplify the calculation). Set x = p/100 ( since p is in percent, for instance, p = 50% , x = 0.5) Area = 1^2 = 1

New side length = 1*( 1-x) = 1-x , Area after the reduction = (1-x)^2

the resulting percent reduction = (1- (1-x)^2)/1 *100% = [1- ( 1- 2x+ x^2)] * 100% =( 2x - x^2 ) *100%

replace p for x: \([2p-\frac{p^2}{100}]%\)

C is the answer
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Re: If the side length of a square is reduced by p percent, what [#permalink] New post 03 Aug 2014, 14:10
oss198 wrote:
If the side length of a square is reduced by p percent, what is the resulting percent reduction in the area of the square?

(A) \(\frac{p^2}{100}%\)
(B) \([1-\frac{p^2}{100}]^2%\)
(C) \([2p-\frac{p^2}{100}]%\)
(D) \(\frac{(100-p)^2}{100} %\)
(E) \(\frac{p^2-2p}{100}%\)


Two ways to solve this problem. My preferred option here is to determine the pattern (I always like to do that over memorizing complex formulas or rules).

I drew 3 squares with sides of 10, 9 and 8. The areas are 100, 81, 64.

Based on that, I was able to easily figure out the answer choice C fits that mold just by plugging in. It all took less than 90 seconds.

The second way to do this is algebraically.

Side of a square = s^2.
Side reduced by p% = s(1-(p/100))
Area after side reduced by p% = (s(1-(p/100)))^2
Difference = ((s^2) - (s(1-(p/100)))^2)/(s^2)

Simplify (and it's not easy to simplify this thing!), and you end up with answer choice C.
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Re: If the side length of a square is reduced by p percent, what [#permalink] New post 04 Aug 2014, 00:05
Lets say side = x;

Original Area \(= x^2\) ........ (1)

Reduction \(= p% = \frac{xp}{100}\)

New side dimension \(= x - \frac{xp}{100}\)

New Area \(= x^2(1-\frac{p}{100})^2\)

\(= x^2 (1 - \frac{2p}{100} + \frac{p^2}{100^2})\).............. (2)

Reduction = (1) - (2)

\(x^2 (1 - 1 + \frac{2p}{100} - \frac{p^2}{100^2})\)

Percentage reduction

\(= \frac{x^2 (1 - 1 + \frac{2p}{100} - \frac{p^2}{100^2})}{x^2} * 100\)

\(= 2p - \frac{p^2}{100}\)

Answer = C
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Re: If the side length of a square is reduced by p percent, what [#permalink] New post 10 Aug 2014, 11:25
The most effective way to solve this problem is to use smart numbers.
Ex: s=10, p=50
A = 100
A_reduced = 25
% reduction = (100-25)/100*100% = 75%
Only C gives the correct answer of 75.
Re: If the side length of a square is reduced by p percent, what   [#permalink] 10 Aug 2014, 11:25
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