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m1033512
Apt0810
Taking option B as true, we can clearly say that the expression in the problem would be positive, which gives a definite answer

Posted from my mobile device

from statement 2

X can be 0 also , then

we have 0<0 which is not possible .


Could you help on this whether this reasoning is correct .

The question asks whether x^2/y < 0.

(2) says that y is positive. Thus, x^2/y is 0, if x = 0, OR positive, if x ≠ 0. In both cases x^2/y is NOT less than 0. So, we have a definite NO answer to the question. That's why (2) is sufficient.

Hope it helps.
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Bunuel
m1033512
Apt0810
Taking option B as true, we can clearly say that the expression in the problem would be positive, which gives a definite answer

Posted from my mobile device

from statement 2

X can be 0 also , then

we have 0<0 which is not possible .


Could you help on this whether this reasoning is correct .

The question asks whether x^2/y < 0.

(2) says that y is positive. Thus, x^2/y is 0, if x = 0, OR positive, if x ≠ 0. In both cases x^2/y is NOT less than 0. So, we have a definite NO answer to the question. That's why (2) is sufficient.

Hope it helps.


Got it , Thanks :)

even if it is 0, we can say it is not less than 0
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saurabh9gupta
Is x^2/y < 0 ?

(1) -2 < x < 3

(2) 1 < y < 3

\(x^2\) will always be non negative OR

\(x^2 >= 0\) always.

So now everything comes down to the value of "y"

Statement 1 : \(2 < x < 3\)

No info about "y".

If "y" is negative, answer to the question is YES i.e, \(\frac{x^2}{y} < 0\)

but if "y" is positive, answer to the question is NO i.e, \(x^/y\) not less than "0"

NOT SUFFICIENT


Statement 2 : \(1 < y < 3\)

This shows that \(y > 0\) always.

Therefore, answer to the question is definite NO. SUFFICIENT

\(\frac{x^2}{y} < 0\)
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Is x^2/y < 0 ?

(1) -2 < x < 3

(2) 1 < y < 3

#1
x would be +ve ; y is not clear
insufficient
#2
y wil be +ve and x whether - pr + no difference it makes
IMO B
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