Inequalities

[fruti_yum]

Is xy > x/y?
(1) xy > 0
(2) y < 0



Need help on approach!
OA on further discussion

[Gdieterling]

IMO : B

You can rewrite the inequality by dividing both sides of the equation by x :

y > 1/y

1 tells you that: both x and y are positive, or both x & y are negatives. INSUFFICIENT


2 tells you that y < 0. SUFFICIENT

Let's try for dfferent values of y :

y = -2
Is -2 > -1/2 ? No


y = -10000
Is - 10000 > - 1/10000 ? No

B is sufficient, as y will always be smaller than 1/y.

[fruti_yum]

IMO : B

You can rewrite the inequality by dividing both sides of the equation by x :

y > 1/y

1 tells you that: both x and y are positive, or both x & y are negatives. INSUFFICIENT


2 tells you that y < 0. SUFFICIENT

Let's try for dfferent values of y :

y = -2
Is -2 > -1/2 ? No


y = -10000
Is - 10000 > - 1/10000 ? No

B is sufficient, as y will always be smaller than 1/y.

OA is not B

[tom4lax]

IMO answer is E. Take your first equation one step further so that y^2 > 1

condition II: if y is -2 then 4 > 1, yes. But if y is -1/2 then 1/4 < 1, no. Insufficient.

[wccotton]

IMO answer should be E

if you take 1) x = 1 and y =0.1 x*y>0, but x*y<x/y (0.1 < 10). If you make x = 2 and y = 2, x*y > x/y (4>1)

if you take 2) x = -1 and y = -0.1 satisfies y<0 but again x*y < x/y and if x=-2 and y=-2, x*y >x/y

The two problems together don't give you any additional information.

[Gdieterling]

Hmm, by reading your post I agree : OA should be E, forgot about the fractions in my example !

[blr_gmat_prep]

xy>x/y

can be expressed as

xy-x/y >0
=> (xy^2 -x)/y >0
=>(x(y^2 -1))/y ----(1)

In 1 we can see that y^2 will be positive always.
the term (y^2 -1) will be negative only if y=0.


From (a) xy> 0 we know that x and y have same sign (+ve or -ve). We can also say that neither x nor y = 0.

for all cases 1 will give +ve value except for y=1 where it becomes 0.
Insuff.

From (b),
since we dont know x we cant say Insuff.

Combining we still cant say so answer = E.