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by bblast » Sun Jun 05, 2011 10:21 am
Is 1/x > y/(y^2+3)?

(1) y > 0
(2) x = y


C
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Source: — Data Sufficiency |
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by Frankenstein » Sun Jun 05, 2011 10:25 am
Hi,
From(1): we know nothign about x
Insufficient
From(2) : (1/x)- x/(x^2+3)= 3/x.(x^2+3). This is greater than 0 if x> 0 else less than 0.
Insufficient
Both (!) and (2):
y>0 =>x>0 as x=0
So, (1/x)- x/(x^2+3)= 3/x.(x^2+3)>0
Sufficient

Hence, C
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by cans » Sun Jun 05, 2011 10:28 am
1/x > y/(y^2+3)??
a)y>0
1>r.h.s>0
l.h.s. can be +ve or -ve depending on value of x
Insufficient.
b)x=y
1/y > y/(y^2+3) or 1/y - y/(y^2+3) > 0
3/[(y^2+3)*y] >0 ->y>0
Insufficient.
a&b together) y>0 and x=y and thus x>0
cross multiplying, y^2 + 3 > y^2 -> 3>0 true always
Sufficient
IMO C
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by jainnikhil02 » Sun Jun 05, 2011 12:17 pm
C is the right answer.
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