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# Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1

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Intern
Joined: 24 May 2018
Posts: 6
Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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10 Sep 2018, 18:53
00:00

Difficulty:

75% (hard)

Question Stats:

46% (01:44) correct 54% (01:59) wrong based on 71 sessions

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Is x > 0?

(1) (x^2 - 1)(x^3) > 0

(2) x^2 < 1
Math Expert
Joined: 02 Aug 2009
Posts: 7756
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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10 Sep 2018, 19:14
jennysussna wrote:
Is x>0?

1. (X^2 - 1) (x^3) > 0
2. X^2 < 1

Posted from my mobile device

1) $$(x^2-1)(x^3)>0$$
If x>0.....x^3>0 and x^2-1>0 or x^2>1...X>1
If x<0...x^3<0 and x^2-1<0 or x^2<1.... So x>-1
Both cases possible
Insufficient

2) x^2<1
-1<X<1
Insufficient

Combined
If x>0, x>1 but statement II says otherwise it says X is <1, so X can not be greater than 0
So x<0
Sufficient

C
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Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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10 Sep 2018, 19:35
chetan2u wrote:
jennysussna wrote:
Is x>0?

1. (X^2 - 1) (x^3) > 0
2. X^2 < 1

Posted from my mobile device

1) $$(x^2-1)(x^3)>0$$
If x>0.....x^3>0 and x^2-1>0 or x^2<1...0<X<1
If x<0...x^3<0 and x^2-1<0 or x^2>1.... So x<-1
Both cases possible
Insufficient

2) x^2<1
-1<X<1
Insufficient

Combined
If x <0, X<-1 but statement II says otherwise..
So x>0

For statement one I combined to have X^5-X^3>0 as I thought that was easier to test than solve out the inequality.

Can you clarify on the last part? What do you mean statement II means otherwise? How do you go from having 0<X<1 and x<-1 from the first statement and then -1<X<1 from the second to having combined 'If x <0, X<-1'? I am still confused how this makes x>0?

I tested -(1/2) and it was true for both statements, so it seems that x is negative.
Math Expert
Joined: 02 Aug 2009
Posts: 7756
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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10 Sep 2018, 19:57
jennysussna wrote:
chetan2u wrote:
jennysussna wrote:
Is x>0?

1. (X^2 - 1) (x^3) > 0
2. X^2 < 1

Posted from my mobile device

1) $$(x^2-1)(x^3)>0$$
If x>0.....x^3>0 and x^2-1>0 or x^2<1...0<X<1
If x<0...x^3<0 and x^2-1<0 or x^2>1.... So x<-1
Both cases possible
Insufficient

2) x^2<1
-1<X<1
Insufficient

Combined
If x <0, X<-1 but statement II says otherwise..
So x>0

For statement one I combined to have X^5-X^3>0 as I thought that was easier to test than solve out the inequality.

Can you clarify on the last part? What do you mean statement II means otherwise? How do you go from having 0<X<1 and x<-1 from the first statement and then -1<X<1 from the second to having combined 'If x <0, X<-1'? I am still confused how this makes x>0?

I tested -(1/2) and it was true for both statements, so it seems that x is negative.

Hi..
I wrote the opposite signs by mistake..
So (x^2-1)(x^3)>0...
So two case
1) both are positive
x^3>0, x^2>1..X>1
2) both negative
x^3<0, x^2<1...-1<X<0
Insufficient

But statement II says x^2<1..
Individually it is insufficient
But when you add it to the info of statement I..
x^2<1 means x^3 is also <0 or X<0
Sufficient
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Joined: 07 Aug 2018
Posts: 110
Location: United States (MA)
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GMAT 2: 670 Q48 V34
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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12 Sep 2018, 08:57
chetan2u

Just to be sure i fully understand it:

S(1) + S(2): results in the following range for x -->

-1<x<0

Which implies that x is not x>0, am I correct???
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Math Expert
Joined: 02 Aug 2009
Posts: 7756
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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12 Sep 2018, 09:06
1
T1101 wrote:
chetan2u

Just to be sure i fully understand it:

S(1) + S(2): results in the following range for x -->

-1<x<0

Which implies that x is not x>0, am I correct???

Yes combined it tells us -1<X<0
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Joined: 22 Feb 2018
Posts: 428
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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13 Sep 2018, 11:46
jennysussna wrote:
Is x > 0?

(1) (x^2 - 1)(x^3) > 0

(2) x^2 < 1

OA:C

$$(1) \quad (x^2 - 1)(x^3) > 0$$

Case 1
$$(x^2-1)>0$$ AND $$x^3>0$$
This leads to $$x>1$$

OR

Case 2
$$(x^2-1)<0$$ AND $$x^3<0$$
This leads to $$-1<x<0$$

So $$x$$ can be either $$x>1$$ OR $$-1<x<0$$
Statement $$1$$ alone is insufficient

$$(2) \quad x^2 < 1$$
This leads to $$-1<x<1$$
Statement $$2$$ alone is insufficient

Combining $$(1)$$ and $$(2)$$, we get $$-1<x<0$$.
After combining $$(1)$$ and $$(2)$$, there is a definite answer to the question: Is x > 0? No
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Joined: 12 Feb 2015
Posts: 862
Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1  [#permalink]

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14 Sep 2018, 09:31
This is a yes/no Q.

Start with statement 2 as it is easier to evaluate. According to statement 2 x could be a positive fraction such that x lies between 0 & 1 or a negative fraction which lies between -1 and 0. Clearly we have two answers (a Yes & a No) to the same questions hence statement 2 alone is not sufficient to answer whether x>0? Strike out option B & D.

Statement 1 alone would give the same result as in statement 2; if x is a negative fraction which lies between -1 and 0 then we have a no but if x >1 then we have a yes. Hence insufficient.

Combining 1 and 2 we get only one answer that is x is a negative fraction which lies between -1 and 0. And we have a No to the Q. Ans C is correct.
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Re: Is x > 0? (1) (x^2 - 1)(x^3) > 0 (2) x^2 < 1   [#permalink] 14 Sep 2018, 09:31
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