grockit DS

[bblast]

Is 1/x > y/(y^2+3)?

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


C

[Frankenstein]

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

[cans]

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

[jainnikhil02]

C is the right answer.