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The length of each side of square A is increased by 100 perc
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21 Apr 2019, 11:21
21 Apr 2019, 11:21
Expert Reply
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Question Stats:
65% (00:56) correct
34% (00:54) wrong based on 26 sessions
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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%
A. 75%
B. 80%
C. 100%
D. 150%
E. 180%
Re: The length of each side of square A is increased by 100 perc
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16 Apr 2020, 14:48
16 Apr 2020, 14:48
Expert Reply
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
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
gmatclubot
Re: The length of each side of square A is increased by 100 perc [#permalink]
16 Apr 2020, 14:48
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