If x and y are nonzero integers, is ...

answer should not be A

If you solve the inequality, you will get:

Is xy / x + y > 1/xy

statement 1 -> x = 2y

2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2

y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2

Answer should be C.

Abdulla wrote:OA is A

mehravikas wrote:answer should not be A

If you solve the inequality, you will get:

Is xy / x + y > 1/xy

statement 1 -> x = 2y

2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2

y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2

Answer should be C.

Abdulla wrote:OA is A

No dude, you made some mistakes..

When you solve the inequality you will get

Is xy / x + y > xy ?

go further...

Now in statement one..
x =2y ... plug it ..

2y^2 / 3y = 2y^2 and since y^2 will never be negative then we can cross them out without flipping the sign.

Is 1/3y >1 ? Sufficient.

The answer is A

My mistake I read the right part as - (x^1 y ^1)^-1

But it is [(x-1)(y-1)]-1

In that case the answer will be A

Hi Abdullah,

when we come to following equation:

1/3Y > 1 ??

how can u assume that this is true. This will only be true, if
0<3Y<1 ==> 0<Y<1/3

but the Question said X and Y are non zero integer.

I think answer should be E.

Hi guys,

I think the OA should indeed be A.

Unless i am wrong, (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 is the same as X+Y>XY

So, if you pick values, and consider that the integers are not 0, you will see that statement 1 is sufficient while statement 2 is not.

sunnyjohn wrote:Hi Abdullah,

when we come to following equation:

1/3Y > 1 ??

how can u assume that this is true. This will only be true, if
0<3Y<1 ==> 0<Y<1/3

but the Question said X and Y are non zero integer.

I think answer should be E.

Hi Sunnyjohn,

We know that x and y is NON zero INTEGERS, so when you plug any negative or positive INTEGER, the expression won't be greater than 1..
1/3(2) > 1 No
1/3(-2) > 1 No

The answer should be A because the result (NO.)