Is x^2/y > 0 ? (1) -4 < x < 1 (2) 1 < y < 4

No takers?

Target question: Is \(\frac{x^2}{y} > 0\)

[b] Statement 1: \(-4 < x < 1\)
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = -1 and y = 2. In this case, the answer to the target question is YES, \(\frac{x^2}{y}\) is greater than 0
Case b: x = 0 and y = 2. In this case, the answer to the target question is NO, \(\frac{x^2}{y}\) is not greater than 0
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: \(1 < y < 4\)
This statement also seems insufficient.
TIP: Rather than come up with new counterexamples, check whether we can repurpose the pairs of values we used for statement 1.
Since the counter-examples we used in statement 1 also satisfy statement 2, let's reuse them.
Case a: x = -1 and y = 2. In this case, the answer to the target question is YES, \(\frac{x^2}{y}\) is greater than 0
Case b: x = 0 and y = 2. In this case, the answer to the target question is NO, \(\frac{x^2}{y}\) is not greater than 0
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: x = -1 and y = 2. In this case, the answer to the target question is YES, \(\frac{x^2}{y}\) is greater than 0
Case b: x = 0 and y = 2. In this case, the answer to the target question is NO, \(\frac{x^2}{y}\) is not greater than 0
Since we can’t answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

BrentGMATPrepNow ; well I had posted a solution to this yesterday via mobile app, not sure how it got deleted

Is x^2/y>0

(1) −4<x<1
(2) 1<y<4

given target would be true when y is +ve and x is not =0
#1
−4<x<1
no info about y and also x can be 0 so insufficient
#2
1<y<4
since y is +ve but value of x is not known as it could be 0 insufficient
from 1 &2
y is +ve but x can be either 0 or number from −4<x<1
therefore insufficient
option E is correct