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Bunuel
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If x & y are integers and y = x^2 + x^3, is y < 0? 21 Apr 2021, 04:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

61% (01:38) correct 39% (01:22) wrong based on 92 sessions

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

(1) x < 0
(2) y < 1
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E
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If x & y are integers and y = x^2 + x^3, is y < 0? 21 Apr 2021, 04:56
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I think the answer is A

statement 1:
If X<0
That gives \(X^3 \)= negative number
and \(X^2\) will be positive giving y to be negative

2nd statement : Y can be 0 also
Hence not sufficient.
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If x & y are integers and y = x^2 + x^3, is y < 0? Updated on: 03 May 2021, 10:32
Originally posted by TarunKumar1234 on 21 Apr 2021, 04:59.
Last edited by TarunKumar1234 on 03 May 2021, 10:32, edited 1 time in total.
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If x & y are integers and y = \(x^2 + x^3\), is y < 0?

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

Stat2: y < 1; then, y <=0. Not Sufficient.

Combining both, x <0 and y <1, still we have y<=0 condition valid. So, we can't be sure, that y is negative always. Not Sufficient.

So, I think E. :)
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If x & y are integers and y = x^2 + x^3, is y < 0? 03 May 2021, 07:39
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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. :)
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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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If x & y are integers and y = x^2 + x^3, is y < 0? 03 May 2021, 10:34
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stne
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. :)
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stne! Thanks for pointing it out. Appreciate it. :)
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If x & y are integers and y = x^2 + x^3, is y < 0? 15 Jan 2024, 13:17
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