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.

Answer: C

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