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GRE Math Challenge #6-area of a rectangle whose length
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06 Sep 2014, 12:53
06 Sep 2014, 12:53
2
Expert Reply
00:00
Question Stats:
90% (00:54) correct
10% (00:36) wrong based on 20 sessions
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What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?
A 1
B 6
C 2/3
D 4/3
E 8/9
A 1
B 6
C 2/3
D 4/3
E 8/9
Re: GRE Math Challenge #6-area of a rectangle whose length
[#permalink]
10 Apr 2020, 15:56
10 Apr 2020, 15:56
Expert Reply
Length = L, Width = W, Area = A, Perimeter = P
L = 2W
A = L*W = 2W*W = 2W^2
P = 2L + 2W
P = 6W
Perimeter of of a square whose area is 1
= S^2 = 1 (S= Side of a square)
= S = 1
Then, P = 4*S = 4
Now,
4 = 6W
W = 4/6
A = 2*W^2 = 2*(4/6)^2
A = 2*(16/36)
A = 8/9
E is the answer
L = 2W
A = L*W = 2W*W = 2W^2
P = 2L + 2W
P = 6W
Perimeter of of a square whose area is 1
= S^2 = 1 (S= Side of a square)
= S = 1
Then, P = 4*S = 4
Now,
4 = 6W
W = 4/6
A = 2*W^2 = 2*(4/6)^2
A = 2*(16/36)
A = 8/9
E is the answer
Re: GRE Math Challenge #6-area of a rectangle whose length
[#permalink]
16 Aug 2021, 03:04
16 Aug 2021, 03:04
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