1) y <0. Thus y can not be 0 and it will be a negative integer

Lets say x >0

Then x^2y+ 9 > 0
x^2y > -9
x^2 < -9/y

we know that y is negative ....the greatest value -9/y will have will be when y = -1

x^2 < -9/-1
x^2 < 9
-3 < x < 3 but x >0 so 0<x<3 is the greatest value of x

Now lets look at x <0

-x^2y+ 9 > 0
-x^2y > -9
x^2y < 9
x^2 < 9/y

Now this is not possible because y is negative so x can not be negative

Thus the only way this can be true is when 0<x<3 ...so x <6 ...Hence sufficient

2) |y| < 1

-1 < y < 1

We know that y is an integer. so the only possibility in this case is y = 0(Ok first of all this is not a GMAT question because GMAT questions choices never contradict each other whereas this does)

putting y =0

|x|×.0 + 9 > 0

This just cancels out the x's so wont help us in anyways ...Thus Insufficient

So IMO the answer should be A

Sowree but the OA is E

Source MGMAT CAT

neoreaves wrote:1) y <0. Thus y can not be 0 and it will be a negative integer

Lets say x >0

Then x^2y+ 9 > 0
x^2y > -9
x^2 < -9/y

we know that y is negative ....the greatest value -9/y will have will be when y = -1

x^2 < -9/-1
x^2 < 9
-3 < x < 3 but x >0 so 0<x<3 is the greatest value of x

Now lets look at x <0

-x^2y+ 9 > 0
-x^2y > -9
x^2y < 9
x^2 < 9/y

Now this is not possible because y is negative so x can not be negative

Thus the only way this can be true is when 0<x<3 ...so x <6 ...Hence sufficient

2) |y| < 1

-1 < y < 1

We know that y is an integer. so the only possibility in this case is y = 0(Ok first of all this is not a GMAT question because GMAT questions choices never contradict each other whereas this does)

putting y =0

|x|×.0 + 9 > 0

This just cancels out the x's so wont help us in anyways ...Thus Insufficient

So IMO the answer should be A

If |x| * y + 9 = 0
|x|*y=-9

Since |x| cannot be negative y is negative
Hence y <0

Statement 1 ----- Tells us what we already know Insuff

Statement 2 ------ Does not tell us anything about x so insuff

Both Combined - Imples y is between 0 and 1 but that again leads to a variety of numbers

Hence E