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The length of each side of square A is increased by 100 percent to

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The length of each side of square A is increased by 100 percent to [#permalink]

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The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?

A. 75%
B. 80%
C. 100%
D. 150%
E. 180%
[Reveal] Spoiler: OA

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Re: The length of each side of square A is increased by 100 percent to [#permalink]

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New post 03 Dec 2015, 05:39
Let length of each side of square A be 10
Area of A = 10^2 = 100

Since , length of each side of square A is increased by 100 percent to make square B
length of each side of square B = 2*10 = 20
Area of B = 20^2 = 400

Since , length of the side of square B is increased by 50 percent to make square C
length of each side of square C= 1.5*20 = 30
Area of C= 30^2 = 900

Difference in areas of C and cummulative areas of A and B = 900 -(400+100) = 400
percent is the area of square C greater than the sum of the areas of squares A and B = (400/500) * 100 % = 80%
Answer B
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Re: The length of each side of square A is increased by 100 percent to [#permalink]

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New post 03 Dec 2015, 05:56
Let the side of square A be a
So area = a*a

Side of square B =2a

Area =2a*2a

Side of square C=3a

Area =3a*3a

percent is the area of square C greater than the sum of the areas of squares A and B =(3a*3a-(2a*2a+a*a))/(2a*2a+a*a)
=80%
Hence B
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Re: The length of each side of square A is increased by 100 percent to [#permalink]

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New post 03 Dec 2015, 11:33
Bunuel wrote:
The length of each side of square A is increased by 100 percent to make square B.
Let side of square A be 1 , so area of A is 1

Side of square B is 2 , so area of B is 4
Bunuel wrote:
length of the side of square B is increased by 50 percent to make square C,
Side of square C is 3 , so area of B is 9
Bunuel wrote:
by what percent is the area of square C greater than the sum of the areas of squares A and B?
Area of C is 9 and Area of A & B is 5

So, The required percentage is \(\frac{(9-5)}{5}*100\)% = \(\frac{4}{5}*100\) =>80%

Hence answer is definitely (B) 80%
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Re: The length of each side of square A is increased by 100 percent to [#permalink]

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New post 02 Feb 2018, 11:24
Bunuel wrote:
The length of each side of square A is increased by 100 percent to make square B. If the length of the side of square B is increased by 50 percent to make square C, by what percent is the area of square C greater than the sum of the areas of squares A and B?

A. 75%
B. 80%
C. 100%
D. 150%
E. 180%


We can let the length of each side of square A = 2, so the length of each side of B is 4, and the length of each side of square C is 1.5 x 4 = 6.

The sum of the areas of squares A and B is 4 + 16 = 20, and the area of C is 36. Let’s now use the percent change formula:

(36 - 20)/20 = 16/20 = 4/5 = 0.80 = 80%

Answer: B
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Re: The length of each side of square A is increased by 100 percent to   [#permalink] 02 Feb 2018, 11:24
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