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Given that the length of the side of a square is L and that

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Manager
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Given that the length of the side of a square is L and that [#permalink]

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New post 14 Jul 2009, 13:11
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Given that the length of the side of a square is L and that the length of the side is increased by x%. State whether the area of the square is increased by more than 10%.

(1) x < 10
(2) x > 5

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Re: DS- square [#permalink]

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New post 14 Jul 2009, 13:56
Before the increase, the area is \(L^2\). If we choose L=10, then A=100.

After the increase, now area of L = \([(1 + x/100)*L]^2\)

1. x<10, so let's choose 9 and plug into our new area formula.
A = (1+(9/100)*10)^2 = 10.9^2 = 118.81
So if we use 9, then 118.81 is definitely more than 10%
Let's pick 1 now:
A = (1+(1/100)*10)^2 = 10.1^2 = 102.01
So if we use 1, the increase is less than 10%.
INSUFFICIENT

2. x>5, so let's choose 5 and plug in:
A = (1+(5/100)*10)^2 = 10.5^2 = 110.25
If x>5, then the new area must be greater than 110.25, so this statement alone is SUFFICIENT.

B

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Re: DS- square [#permalink]

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New post 14 Jul 2009, 18:17
OA is B. Thanks for the explanation....

Thought the problem is easy, I am just worried with getting the calculations all correct under 2 minutes.

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Re: DS- square   [#permalink] 14 Jul 2009, 18:17
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Given that the length of the side of a square is L and that

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