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# Is x > 0 ?

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Director
Joined: 07 Jun 2004
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Location: PA
Is x > 0 ?  [#permalink]

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21 Sep 2010, 08:14
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85% (hard)

Question Stats:

50% (01:26) correct 50% (01:01) wrong based on 393 sessions

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Is x > 0 ?

(1) x < x^2
(2) x < x^3

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

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Joined: 18 Jul 2010
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21 Sep 2010, 08:35
(1) x < x^2 so x can be less than -1 and greater than 1 !
insufficient!
(2) x < x^3 so x is greater than -1 so it can be negative or positive
insufficent!

(1) and (2) only said than x is greater than -1 so x can be positive or negative....
insufficient!

ANS:. E
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Joined: 02 Sep 2009
Posts: 49417

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21 Sep 2010, 08:36
3
rxs0005 wrote:
Is x > 0

x < x^2

x < x^3

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Question: is $$x>0$$?

When considering statements together we can see that YES answer is easy to get: $$x=2>0$$, so answer YES --> $$(x=2)<(x^2=4)<(x^3=8)$$;

Now, for NO answer we should pick negative $$x$$ --> ANY negative $$x$$ would be less than $$x^2$$ as $$x^2$$ would be positive, so satisfying (1) is easy. Next, in order negative $$x$$ to be less than $$x^3$$, it shoul be a fraction in the range (-1,0) --> $$x=-\frac{1}{2}<0$$, so answer NO --> $$(x=-\frac{1}{2})<(x^3=-\frac{1}{8})<(x^2=\frac{1}{4})$$.

We have two different answers, not sufficient.

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21 Sep 2010, 09:22
Bunuel

cheers

R
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21 Sep 2010, 13:33
Can't I write first equation

x < x2 as 1 < x.
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Joined: 30 May 2010
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21 Sep 2010, 13:45
onedayill wrote:
Can't I write first equation

x < x2 as 1 < x.

I'm assuming you are trying to divide both sides by x. The problem with an inequality is that you don't know if x is positive or negative. If it is negative, you need to change the sign to greater than in this case.
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21 Sep 2010, 14:32
rxs0005 wrote:
Is x > 0

x < x^2

x < x^3

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

Can I deduce as following :
1) $$x < x^2$$
holds true for x > 1 and x < 0
Thus, not sufficient

2) $$x < x^3$$
holds true for x > 0 and 0 < x < 1
Not sufficient

No unique value is determined when combining 1 & 2.
Thus, E.

Please correct me if I am incorrect in the above logic.
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Joined: 30 Apr 2008
Posts: 1840
Location: Oklahoma City
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21 Sep 2010, 15:16
Orange08 wrote:
rxs0005 wrote:
Is x > 0

x < x^2

x < x^3

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

Can I deduce as following :
1) $$x < x^2$$
holds true for x > 1 and x < 0
Thus, not sufficient

2) $$x < x^3$$
holds true for x > 0 and 0 < x < 1
Not sufficient

No unique value is determined when combining 1 & 2.
Thus, E.

Please correct me if I am incorrect in the above logic.

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Math Expert Joined: 02 Sep 2009 Posts: 49417 Re: DS Algebra [#permalink] ### Show Tags 21 Sep 2010, 23:43 1 onedayill wrote: Can't I write first equation x < x2 as 1 < x. Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So you can not divide both parts of inequality $$x<x^2$$ by $$x$$ as you don't know the sign of this unknown: if $$x>0$$ you should write $$1<x$$ BUT if $$x<0$$ you should write $$1>x$$. Hope it helps. _________________ Math Expert Joined: 02 Sep 2009 Posts: 49417 Re: DS Algebra [#permalink] ### Show Tags 21 Sep 2010, 23:46 1 jallenmorris wrote: Orange08 wrote: rxs0005 wrote: Is x > 0 x < x^2 x < x^3 Can some one explain how do we eliminate choice "C" for this DS question using plug in values thanks Can I deduce as following : 1) $$x < x^2$$ holds true for x > 1 and x < 0 Thus, not sufficient 2) $$x < x^3$$ holds true for x > 0 and 0 < x < 1 Not sufficient No unique value is determined when combining 1 & 2. Thus, E. Please correct me if I am incorrect in the above logic. You are correct. Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself. Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question. No your ranges for (2) are not correct, it should be: $$-1<x<0$$ or $$x>1$$. Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution: Is $$x>0$$? (1) $$x<x^2$$ --> $$x*(x-1)>0$$ --> $$x<0$$ or $$x>1$$ --> ---------0----1----. Not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO. (2) $$x<x^3$$ --> $$x(x-1)(x+1)>0$$ --> $$-1<x<0$$ or $$x>1$$ --> ----(-1)----(0)----(1)----. Not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO. (1)+(2) Intersection of the ranges from (1) and (2) is: ----(-1)----(0)----(1)---- Again not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO. Answer: E. Hope it helps. _________________ SVP Joined: 30 Apr 2008 Posts: 1840 Location: Oklahoma City Schools: Hard Knocks Re: DS Algebra [#permalink] ### Show Tags 22 Sep 2010, 06:26 You're right. I hadn't thought of negative fractions. -1/2 ^3 = -1/8, which is greater than -1/2. Thanks. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Joined: 19 Sep 2010
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16 Oct 2010, 08:39
+1 for E...took more than 3 mins though..
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Joined: 26 Jul 2012
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01 Nov 2012, 13:49
Bunuel wrote:

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.

No your ranges for (2) are not correct, it should be: $$-1<x<0$$ or $$x>1$$. Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution:

Is $$x>0$$?

(1) $$x<x^2$$ --> $$x*(1-x)>0$$ --> $$x<0$$ or $$x>1$$ --> ---------0----1----. Not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO.

(2) $$x<x^3$$ --> $$x(x-1)(x+1)>0$$ --> $$-1<x<0$$ or $$x>1$$ --> ----(-1)----(0)----(1)----. Not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO.

(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)----
Again not sufficient as if $$x$$ is in the green zone answer is YES but if $$x$$ is in the red zone answer is NO.

Hope it helps.[/quote]

Hi Bunuel,

Could you please explain why you're considering only the negative factors while considering $$x<x^2$$ --> $$x*(1-x)>0$$ --> $$x<0$$ or $$x>1$$.. From $$x*(1-x)>0$$ can't we consider both x and (x-1) to be positive also ?

Similarly for $$x(x-1)(x+1)>0$$ --> $$-1<x<0$$ or $$x>1$$, I seem to be missing your logic.

Thanks
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Joined: 02 Sep 2009
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01 Nov 2012, 14:09
kheba wrote:
Hi Bunuel,

Could you please explain why you're considering only the negative factors while considering $$x<x^2$$ --> $$x*(1-x)>0$$ --> $$x<0$$ or $$x>1$$.. From $$x*(1-x)>0$$ can't we consider both x and (x-1) to be positive also ?

Similarly for $$x(x-1)(x+1)>0$$ --> $$-1<x<0$$ or $$x>1$$, I seem to be missing your logic.

Thanks

Actually, both cases were considered.

$$x<x^2$$ --> $$x*(x-1)>0$$
x>0 and x-1>0 (x>1) --> x>1;
x<0 and x-1<0 (x<1) --> x<0.

Thus, $$x<x^2$$ holds true for $$x<0$$ and $$x>1$$.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html
xy-plane-71492.html

Hope it helps.
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01 Nov 2012, 14:49
Bunuel wrote:
onedayill wrote:
Can't I write first equation

x < x2 as 1 < x.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality $$x<x^2$$ by $$x$$ as you don't know the sign of this unknown: if $$x>0$$ you should write $$1<x$$ BUT if $$x<0$$ you should write $$1>x$$.

Hope it helps.

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52
TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y?
1. X<Y and XY = 100
2. X^2 <100< Y^2

In the explanation part
2nd stem is directly reduced to X<10<Y

Pls help in getting my concept right.
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Joined: 02 Sep 2009
Posts: 49417

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01 Nov 2012, 14:55
1
Archit143 wrote:
Bunuel wrote:
onedayill wrote:
Can't I write first equation

x < x2 as 1 < x.

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.

So you can not divide both parts of inequality $$x<x^2$$ by $$x$$ as you don't know the sign of this unknown: if $$x>0$$ you should write $$1<x$$ BUT if $$x<0$$ you should write $$1>x$$.

Hope it helps.

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52
TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y?
1. X<Y and XY = 100
2. X^2 <100< Y^2

In the explanation part
2nd stem is directly reduced to X<10<Y

Pls help in getting my concept right.

First of all, they are not reducing, they are taking square root. Also, notice that the stem says that x and y are positive numbers.

P.S. This question is discussed here: if-x-and-y-are-positive-is-x-10-y-139873.html
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Re: Is x > 0 ?  [#permalink]

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01 Nov 2012, 15:07
Thanks BUnnel
really missed the statement.Sorry

But is it ok to take square root,considering signs are unknown in above scenario.
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Posts: 49417
Re: Is x > 0 ?  [#permalink]

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01 Nov 2012, 15:12
Archit143 wrote:
Thanks BUnnel
really missed the statement.Sorry

But is it ok to take square root,considering signs are unknown in above scenario.

If we were not told that x and y are positive numbers, then from x^2<100<y^2, we would have |x|<10<|y| (since $$\sqrt{x^2}=|x|$$).

Generally:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.
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Re: Is x > 0 ?  [#permalink]

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13 Jan 2013, 21:50
I don't think that we need plug-in here. It has very easy algebraical solution "solvable" within 80 sec
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Re: Is x > 0 ?  [#permalink]

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15 Jan 2013, 07:22
rxs0005 wrote:
Is x > 0 ?
(1) x < x^2
(2) x < x^3

1. x < x^2
Test x = 2: 2 < 4 OK!
Test x = -1: -1 < 1 OK!
Test x = -1/4: -1/4 < -1/16 OK!
INSUFFICIENT!

2. x < x^3
Test x = 2: 2 < 8 OK!
Test x = -1: -1 = -1 NOT OK!
Test x = -1/4: -1/4 < -1/64 OK!
INSUFFICIENT!

Together: When x is negative fraction or x is greater than 2, both inequalities work!

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Re: Is x > 0 ? &nbs [#permalink] 15 Jan 2013, 07:22

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