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# Is x |y| > y^2? (1) x > y (2) y > 0

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Manager
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Is x |y| > y^2? (1) x > y (2) y > 0 [#permalink]

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08 Jun 2009, 05:37
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Is x·|y| > y^2?

(1) x > y

(2) y > 0
Director
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08 Jun 2009, 10:12
prinits wrote:
Is x·|y| > y^2?

(1) x > y

(2) y > 0

E, I think

x-lyl-y^2>0?

1)x>y

say x=-1, y=-2

then -1-2-4> 0? no

say x=5, y=-1

then 5-1-1>0? yes

insuff.

2) y>0

will yeild the same results because we are taking the absolute value of y and the square of y, so y ends up being greater than 0 regardless

together

same results, insuff
Manager
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08 Jun 2009, 16:06
prinits wrote:
Is x·|y| > y^2?

(1) x > y

(2) y > 0

Question:$$(x|y|>y^{2}?)$$

Question:$$(x>\frac{y^{2}}{|y|}?)$$

Question:$$(x>\frac{y}{|y|}y?)$$

Case 1: y>0 $$\longrightarrow (x>y?)$$
Case 2: y<0 $$\longrightarrow (x>-y?) \longrightarrow x+y>0?$$

(1) $$x>y$$ INSUFFICIENT

(2) $$y>0$$ INSUFFICIENT

Together sufficient as they're both positive
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08 Jun 2009, 16:40
prinits wrote:
Is x·|y| > y^2?

(1) x > y

(2) y > 0

is this x minus lyl or x times lyl ?

i thought it was minus
Manager
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08 Jun 2009, 17:28
We agree that alone none is suff.

Lets combine....due to stmt. b we can say that both X and Y are +ve.
From stmt a we can say that X = A + Y.
So the question becomes...
Is (A+Y) * Y > Y^2
Is (AY + Y^2) > Y^2.........and this is clearly true.

Hence C.
Senior Manager
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08 Jun 2009, 19:02
Is x·|y| > y^2?

stmt 1 : x > y

when x and y are +ve x = 4 and y = 3. 12 > 9
when x and y are -ve y = -3 and x = 4 12 >9
y = -3 and x = -2 6 < 9
Insufficient.

stmt 2 : y >0 . Not sure of x insuff

togther yes x >y and y >0 tell x >0 and hence sufficient.
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10 Jun 2009, 17:37
prinits wrote:
Is x·|y| > y^2?

(1) x > y

(2) y > 0

Misplaced Post - user warning issued
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Intern
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11 Jun 2009, 00:00
C is the correct answer, for the reasons given above.

How hard is this question by the way? At what scorelevel can you expect to see such questions?
Manager
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11 Jun 2009, 00:07
Not terribly, I'd say a mid 40 Q

You can always get the answer by plugging in some values
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Re: Tough DS   [#permalink] 11 Jun 2009, 00:07
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