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TarunKumar1234
If x & y are integers and y = \(x^2 + x^3\), is y < 0?

Stat1: x < 0
y = \(x^2 (x+1)\); if -1 < x <0 then, y >0. And, if x <-1 then, y <0. Not Sufficient.

Stat2: y < 1; between 0 and 1, y >0 and for rest of the condition , y<0 . Not Sufficient.

Combining both, x <0 and y <1, still we have -1 < x <0 condition valid, in which y >0 and for others, y<0. Not Sufficient.

So, I think E. :)

TarunKumar1234, thank you for the soln. If I may point out, just a small oversight, I think, \(x\) and \( y\) are integers.

Please note : \(x \) and \(y\) are integers:

Lets simplify the stem :

\(y = x^2(1+x)\) is \(y < 0\)

I) \(x <0 \)

if \(x=-1\) then \(y =0\) ans. is NO.

if \(x= -2\) then \(y=-4\) ans is YES

INSUFF.

II) \(y <1\)

if \(x=-1\) then \(y =0\) Ans. is NO

if \(x=-2\) then \(y = -4\) Ans. is Yes.

INSUFF.

1 + 2

\(x= -1\) then \(y=0\) Ans is NO
\(x=-2\) then \(y=-4\) Ans is yes.

Ans- E

Hope it helps.
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TarunKumar1234
If x & y are integers and y = \(x^2 + x^3\), is y < 0?

Stat1: x < 0
y = \(x^2 (x+1)\); if -1 < x <0 then, y >0. And, if x <-1 then, y <0. Not Sufficient.

Stat2: y < 1; between 0 and 1, y >0 and for rest of the condition , y<0 . Not Sufficient.

Combining both, x <0 and y <1, still we have -1 < x <0 condition valid, in which y >0 and for others, y<0. Not Sufficient.

So, I think E. :)

stne! Thanks for pointing it out. Appreciate it. :)
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