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If the side length of a square is reduced by p percent, what  [#permalink]

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Question Stats: 60% (02:07) correct 40% (02:15) wrong based on 252 sessions

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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) $$\frac{p^2-2p}{100}%$$
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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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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: $$[2p-\frac{p^2}{100}]%$$

##### General Discussion
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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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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) $$\frac{p^2-2p}{100}%$$

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]

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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

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

$$= 2p - \frac{p^2}{100}$$

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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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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.
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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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I used p=10% and neither of the answer choices gives the correct answer of 19%.
Answer choice C wrongly gives 20% reduction for p=10%
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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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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) $$\frac{p^2-2p}{100}%$$

We can let the original side of the square = 4 and let p = 50. That is, if the side is reduced by 50%, the new side of the square is 4 x 0.5 = 2.

Thus, the original area is 4 x 4 = 16 and the new area is 2 x 2 = 4.

The percentage reduction is (4 - 16)/16 x 100 = -12/16 x 100 = -3/4 x 100 = -75%.

Let’s now test our answer choices.

A) 50^2/100

50^2/100 = 2500/100 = 25

We see that this is not 75 percent.

B) [1 - 50^2/100]^2

[1 - 50^2/100]^2 = [1 - (2500/100)]^2 = [1 - 25]^2

We see that this will not equal 75 percent.

C) 2(50) - 50^2/100

2(50) - 50^2/100 = 100 - 2500/100 = 100 - 25 = 75

We see that this is 75 percent.

Alternate Solution:

Let’s let the side of the square be 100 units. After a p percent reduction, the side of the square becomes (100 - p) units. The area of the original square was 10,000 square units and the area of the reduced square is (100 - p)^2. Thus, the reduction is 10000 - (100 - p)^2 = 10000 - (10000 -200p + p)^2 = 200p - p^2 square units. Compared to the original area, this corresponds to a [(200p - p^2)/10000]x100 = (200p - p^2)/100 = 2p - (p^2/100) percent reduction.

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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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_Alex_ wrote:
I used p=10% and neither of the answer choices gives the correct answer of 19%.
Answer choice C wrongly gives 20% reduction for p=10%

Could you elaborate?
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Re: If the side length of a square is reduced by p percent, what  [#permalink]

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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) $$\frac{p^2-2p}{100}%$$

$$p + p -\frac{(p*p)}{100}$$

= $$2p -\frac{p^2}{100}$$, Answer must hence be (C)
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