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# M18-16

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Math Expert
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M18-16  [#permalink]

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16 Sep 2014, 01:03
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85% (hard)

Question Stats:

38% (00:58) correct 62% (01:16) wrong based on 282 sessions

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Is integer $$x$$ positive?

(1) $$x \gt x^3$$

(2) $$x \lt x^2$$

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M18-16  [#permalink]

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16 Sep 2014, 01:03
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Official Solution:

Is integer x positive?

(1) $$x \gt x^3$$.

$$x^3-x \lt 0$$;

$$x(x-1)(x+1) \lt 0$$;

$$x \lt -1$$ or $$0 \lt x \lt 1$$.

Since we are told that $$x$$ is an integer and there is no integer in the range $$0 \lt x \lt 1$$ then $$x \lt -1$$, so $$x$$ is a negative integer. Sufficient.

(2) $$x \lt x^2$$.

$$x^2-x \gt 0$$;

$$x(x-1) \gt 0$$;

$$x \lt 0$$ or $$x \gt 1$$, so integer $$x$$ could be negative as well as positive. Not sufficient.

Answer: A
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M18-16  [#permalink]

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09 Nov 2014, 22:44
Hi, Bunuel
I understand everything you said except x<0 in st-2 and x>0 in st-1.
Please help me to understand.

Actually, I think integer 0 is not possible for both. I need either +ve or -ve integers.

Now in st1 - no positive integer will satisfy it. So we are sure that integers here are negative.
in st2 whatever we put negative or positive we can satisfy the condition. so not sufficient.

My confusion is in 2 if I think it other way.
We can express x^2> x in mod x> x. Now this condition is sufficient only if x is negative.
I think I am making mistake in thinking in the 2nd way.
Where am I making mistake in my 2nd thought?
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Re: M18-16  [#permalink]

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10 Nov 2014, 02:56
1
Raihanuddin wrote:
Hi, Bunuel
I understand everything you said except x<0 in st-2 and x>0 in st-1.
Please help me to understand.

Actually, I think integer 0 is not possible for both. I need either +ve or -ve integers.

Now in st1 - no positive integer will satisfy it. So we are sure that integers here are negative.
in st2 whatever we put negative or positive we can satisfy the condition. so not sufficient.

My confusion is in 2 if I think it other way.
We can express x^2> x in mod x> x. Now this condition is sufficient only if x is negative.
I think I am making mistake in thinking in the 2nd way.
Where am I making mistake in my 2nd thought?

x^2 > x cannot be written as |x| > x.
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Re: M18-16  [#permalink]

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30 Nov 2014, 09:27
Hi Bunuel,

I am having problems with this question and I would like to get your opinion.

When I see this type of questions in which exponents with no coefficients are involved and I am dealing with a DS question then I use "the zones" method. That is, I fix my attention on -1, 0 and 1 but not evaluating those exact numbers, rather I randomly pick a number less than -1, a number betwen -1 and 0, a number between 0 and 1 and a number greater than one:
The above would look something like this:
If x=-2, x^2= 4; x^3= -8 --> summary x^3<x<x^2
If x= -1/2, x^2= 1/4 ; x^3= -1/8 --> summary x<x^3<x^2
If x= 1/2; x^2 = 1/4; x^3= 1/8 --> summary x^3<x^2<x
If x=2; x^2=4; x^3=8 --> summary x<x^2<x^3

Once I have this I can evaluate statements. S-1 that x>x^3 is true for x=-2 and x= 1/2. Not sufficient
S-2 that x<x^2 is true for x=-2, x=-1/2, x=2. Not sufficient
If we combine the information, we have sufficiency
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Re: M18-16  [#permalink]

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01 Dec 2014, 05:41
marquinho wrote:
Hi Bunuel,

I am having problems with this question and I would like to get your opinion.

When I see this type of questions in which exponents with no coefficients are involved and I am dealing with a DS question then I use "the zones" method. That is, I fix my attention on -1, 0 and 1 but not evaluating those exact numbers, rather I randomly pick a number less than -1, a number betwen -1 and 0, a number between 0 and 1 and a number greater than one:
The above would look something like this:
If x=-2, x^2= 4; x^3= -8 --> summary x^3<x<x^2
If x= -1/2, x^2= 1/4 ; x^3= -1/8 --> summary x<x^3<x^2
If x= 1/2; x^2 = 1/4; x^3= 1/8 --> summary x^3<x^2<x
If x=2; x^2=4; x^3=8 --> summary x<x^2<x^3

Once I have this I can evaluate statements. S-1 that x>x^3 is true for x=-2 and x= 1/2. Not sufficient
S-2 that x<x^2 is true for x=-2, x=-1/2, x=2. Not sufficient
If we combine the information, we have sufficiency

Notice that the stem specifies that x is an integer.
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Re: M18-16  [#permalink]

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16 Jan 2015, 01:52
Bunuel wrote:
Official Solution:

Is integer x positive?

(1) $$x \gt x^3$$.

$$x^3-x \lt 0$$;

$$x(x-1)(x+1) \lt 0$$;

$$x \lt -1$$ or $$0 \lt x \lt 1$$.

Since we are told that $$x$$ is an integer and there is no integer in the range $$0 \lt x \lt 1$$ then $$x \lt -1$$, so $$x$$ is a negative integer. Sufficient.

(2) $$x \lt x^2$$.

$$x^2-x \gt 0$$;

$$x(x-1) \gt 0$$;

$$x \lt 0$$ or $$x \gt 1$$, so integer $$x$$ could be negative as well as positive. Not sufficient.

Answer: A

I understood all but the following,
How $$x \lt 0$$ or $$x \gt 1$$ can be derived from $$x(x-1) \gt 0$$ ?
Wouldn't it be $$x(x-1) \gt 0$$ ---> $$x \gt 0$$ OR $$(x-1) \gt 0$$ ?
Would be helpful if someone clarifies this.........

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Re: M18-16  [#permalink]

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16 Jan 2015, 02:47
1
3
Mahmud6 wrote:
Bunuel wrote:
Official Solution:

Is integer x positive?

(1) $$x \gt x^3$$.

$$x^3-x \lt 0$$;

$$x(x-1)(x+1) \lt 0$$;

$$x \lt -1$$ or $$0 \lt x \lt 1$$.

Since we are told that $$x$$ is an integer and there is no integer in the range $$0 \lt x \lt 1$$ then $$x \lt -1$$, so $$x$$ is a negative integer. Sufficient.

(2) $$x \lt x^2$$.

$$x^2-x \gt 0$$;

$$x(x-1) \gt 0$$;

$$x \lt 0$$ or $$x \gt 1$$, so integer $$x$$ could be negative as well as positive. Not sufficient.

Answer: A

I understood all but the following,
How $$x \lt 0$$ or $$x \gt 1$$ can be derived from $$x(x-1) \gt 0$$ ?
Wouldn't it be $$x(x-1) \gt 0$$ ---> $$x \gt 0$$ OR $$(x-1) \gt 0$$ ?
Would be helpful if someone clarifies this.........

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M18-16  [#permalink]

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18 Jul 2016, 08:12
Bunuel wrote:
Is integer $$x$$ positive?

(1) $$x \gt x^3$$

(2) $$x \lt x^2$$

Alternative approach: Plug in numbers

1) When I see $$x \gt x^3$$, I think in fractions and negative values to satisfy the condition.

X= 1/2.......... $$1/2 \gt (1/2)^3$$.. Answer to question is Yes.

X= -2 ............ $$-2 \gt -2^3$$....... Answer to question is Yes.

However, it mentioned that X is integer so eliminate any fraction. Only negative integers satisfy. so Sufficient.

2) $$x \lt x^2$$

X = 2 & X =-2 .... both integers satisfy the equation, yielding Yes & No to the question. Insufficient

Answer: A
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Re: M18-16  [#permalink]

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23 Jul 2016, 10:04
Mo2men wrote:
Bunuel wrote:
Is integer $$x$$ positive?

(1) $$x \gt x^3$$

(2) $$x \lt x^2$$

Just one doubt:
xsq - x > 0
should not be
x>0, x>1?
plz help in this...
thanks
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Posts: 49231
Re: M18-16  [#permalink]

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23 Jul 2016, 10:09
Celestial09 wrote:
Mo2men wrote:
Bunuel wrote:
Is integer $$x$$ positive?

(1) $$x \gt x^3$$

(2) $$x \lt x^2$$

Just one doubt:
xsq - x > 0
should not be
x>0, x>1?
plz help in this...
thanks

x > 0 or x > 1 does not make any sense. What dies it mean can x be in this case say 1/2? Follow the links given above for solving inequalities.

Also, please format properly: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
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Re: M18-16  [#permalink]

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14 Dec 2016, 09:46
In Equation 1:

If I put value of x=1 then 1>1^3. Then, this equation fails to prove x=1(which is positive). So, I discarded first option.
Can you please help me with correct understanding.

Thank you.
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Posts: 49231
Re: M18-16  [#permalink]

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15 Dec 2016, 04:58
VaibhavSeth wrote:
In Equation 1:

If I put value of x=1 then 1>1^3. Then, this equation fails to prove x=1(which is positive). So, I discarded first option.
Can you please help me with correct understanding.

Thank you.

Really not sure what are you trying to say there but 1 = 1^3.
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Re: M18-16  [#permalink]

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16 Feb 2017, 23:33
4
Solution provided by Bunuel is self-explanatory. But I saw few posters were having issue in understanding how to get to $$x<−1$$ or $$0<x<1$$ from $$x(x−1)(x+1)<0$$

To break any equality of this type - First, we will find the trigger point. Trigger points are the point on which equality will give output = 0 .

So in this equation, we have 3 i.e. $$x=0$$ , $$x=1$$ and $$x=-1$$

Now we will mark these points on number line.
___-1____0_____1____

Here we are having 4 regions
1. $$x <-1$$
2. $$-1<x<0$$
3. $$0<x<1$$
4. $$x>1$$

Second step , test one value from each region and see if you get the value < 0

$$x(x−1)(x+1)<0$$

1. $$x <-1$$ ---> Put x = -2 ; -6 < 0 ; So this range is valid .
2. -$$1<x<0$$ ---> Put x = -1/2 ; Some +ve < 0 ;So this range is not valid.
3. $$0<x<1$$ ---> Put x = 1/2 ; Some -ve value < 0 ; So this range is valid .
4.$$x>1$$ ---> Put x =2 ; SOme +ve value < 0 ; So this range is not valid.

So from here we get 2 ranges $$x <-1$$ or $$0<x<1$$

Same way try for Statement 2 .

Please +1 kudos If this post helps.
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Re: M18-16  [#permalink]

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22 Jun 2017, 06:01
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Re: M18-16  [#permalink]

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22 Jun 2017, 07:11
The answer has to be option A. My take:

This question asks us if integer x is positive or not. In questions of this type, we need to evaluate each option.Substitute values for each option and check.

Statement 1 - Put both positive and negative values. We can ascertain if x is positive or not
Statement 2 - Put both positive and negative values. We can't ascertain if x is positive or not

Hence answer is option A.
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28 Sep 2017, 05:39
I think this the explanation isn't clear enough, please elaborate. Did not understand how 1<x<0 for Statement 1. Please elaborate.
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28 Sep 2017, 07:38
Re: M18-16 &nbs [#permalink] 28 Sep 2017, 07:38
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