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02 Dec 2014, 09:11
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
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59% (00:57) correct
41%
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based on 470
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Tough and Tricky questions: Geometry.
If the sides of a square are doubled in length, the area of the original square is now how many times as large as the area of the resultant square?
A) 25%
B) 50%
C) 100%
D) 200%
E) 400%
Kudos for a correct solution.
Source: Chili Hot GMAT
intheend14
GMAT 1: 740 Q49 V41
Posts: 125
02 Dec 2014, 10:24
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Let the original square have sides of length L and the new square have sides of length 2L.
The resulting areas are L^2 for the original square and 4L^2 for the new square (as the length of the side is square to get the area).
The original square's area is L^2/4L^2 or 25% of the new square's area. Choice A
The resulting areas are L^2 for the original square and 4L^2 for the new square (as the length of the side is square to get the area).
The original square's area is L^2/4L^2 or 25% of the new square's area. Choice A
02 Dec 2014, 19:50
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Answer = A) 25%
Say side of original square = 1
Area\(= 1^2 = 1\)
New square side is doubled = 2
Area \(= 2^2 = 4\)
Area of original square (=1) is \(\frac{1}{4}\) times area of new square (=4)
Percentage \(= \frac{1}{4} * 100 = 25%\)
Say side of original square = 1
Area\(= 1^2 = 1\)
New square side is doubled = 2
Area \(= 2^2 = 4\)
Area of original square (=1) is \(\frac{1}{4}\) times area of new square (=4)
Percentage \(= \frac{1}{4} * 100 = 25%\)
