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Math Expert V
Joined: 02 Sep 2009
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If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Question Stats: 38% (01:34) correct 62% (01:37) wrong based on 106 sessions

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If x^3 < x^2, what is the value of the integer x?

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

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Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

$$x^3 < x^2.......x^2-x^3>0.....x^2(1-x)>0$$
therefore 1-x>0 or x<1 but $$x\neq{0}$$

(1) |x| < 2
so -2<x<2...
possible values = -1,0,1
but x<1 and not equal to 1, so only value left -1
sufficient

(2) x^3 = x
$$x^3=x.....x^3-x=0......x(x^2-1)=0$$
so x=o or x^2=1 that is -1 and 1
but as seen above only -1 is possible out of -1,0 and 1
sufficient

D
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GMAT 1: 460 Q42 V14 Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Quote:
If x^3 < x^2, what is the value of the integer x?

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

Please correct me if I am wrong:

Given x^3<x^2
x^3-x^2<0
x^2(x-1)<0
x<0 or x<1
Quote:

STATEMENT 1:

|x|<2:
-2<x<2
means x can be -1,0,1:
If we take x<0 then we will have x=-1 so suff
If we take x<1 then x can be 0 or -1 so two solutions not possible
Hence statement 1 is not sufficient.

Quote:
STATEMENT 2:

x^3=x
x^3-x=0
x(x^2-1)=0
x=0 or |x|=1
we have x= -1,0,1;
Same as statement 1 ; so not sufficient .

combining 1+2

even after clubbing both the statements we have the same result. so the answer must be E.
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GMAT 1: 460 Q42 V14 If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Thanks
chetan2u
I have not counted that x not equal to 0 so the answer i got was completely wrong
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Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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SonGoku wrote:
Thanks
chetan2u
I have not counted that x not equal to 0 so the answer i got was completely wrong

yes, the question is tricky in that sense, one can easily miss out on this aspect.

therefore look at each question for the hidden meaning too
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If x^3 < x^2, what is the value of the integer x?  [#permalink]

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chetan2u wrote:
Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

$$x^3 < x^2.......x^2-x^3>0.....x^2(1-x)>0$$
therefore 1-x>0 or x<1 but $$x\neq{0}$$

(1) |x| < 2
so -2<x<2...
possible values = -1,0,1
but x<1 and not equal to 1, so only value left -1
sufficient

(2) x^3 = x
$$x^3=x.....x^3-x=0......x(x^2-1)=0$$
so x=o or x^2=1 that is -1 and 1
but as seen above only -1 is possible out of -1,0 and 1
sufficient

D

hi there chetan2u

regarding STATEMENT ONE, i dont get how it can be sufficient.

This is how i understand it

|x| < 2

When X is positive x < 2 so x can be 1 , 0 , -1 etc

now plug in initital question

-1^3 < -1^2, ----> -1<1 TRUE

1^3 < 1^2 ---- > 1<1 NOT TRUE

When x is negative -|x| < 2

-x < 2 (divide by -1)

x > -2 so X can be -1, 0, 1 etc

plug in into initial question

-1^3 < -1^2 ----- > -1 < 1 TRUE

1^3 < 1^2 ----- > 1 < 1 NOT TRUE

So whats wrong with my approach ? thanks and have a great day Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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dave13 wrote:
chetan2u wrote:
Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

$$x^3 < x^2.......x^2-x^3>0.....x^2(1-x)>0$$
therefore 1-x>0 or x<1 but $$x\neq{0}$$

(1) |x| < 2
so -2<x<2...
possible values = -1,0,1
but x<1 and not equal to 1, so only value left -1
sufficient

(2) x^3 = x
$$x^3=x.....x^3-x=0......x(x^2-1)=0$$
so x=o or x^2=1 that is -1 and 1
but as seen above only -1 is possible out of -1,0 and 1
sufficient

D

hi there chetan2u

regarding STATEMENT ONE, i dont get how it can be sufficient.

This is how i understand it

|x| < 2

When X is positive x < 2 so x can be 1 , 0 , -1 etc

now plug in initital question

-1^3 < -1^2, ----> -1<1 TRUE

1^3 < 1^2 ---- > 1<1 NOT TRUE

When x is negative -|x| < 2

-x < 2 (divide by -1)

x > -2 so X can be -1, 0, 1 etc

plug in into initial question

-1^3 < -1^2 ----- > -1 < 1 TRUE

1^3 < 1^2 ----- > 1 < 1 NOT TRUE

So whats wrong with my approach ? thanks and have a great day look at the colored portion RED and BLUE
in red you took x as positive but substituted -1 in the equation
and similarly in blue portion, you took x as negative and substituted x as 1

hope you understood
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If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

(1) |x| < 2
(2) $$x^3=x$$

Question stem:- x=?

Given, x is an integer and
$$x^3 < x^2$$
Or, $$x^3-x^2<0$$
Or, $$x^2(x-1)<0$$
Or,$$x:\:\left(-\infty \:,\:0\right)\cup \left(0,\:1\right)$$-------------(a)

St1:- |x| < 2
Or, -2 < x < 2
Or, x: (-2, 2)
Or, x={-1,0,1}------------(b)
From (a)& (b), the only possible integer value of x=-1 (we have an unique value of x)
Sufficient.

St2:- $$x^3=x$$
Or, $$x^3-x=0$$
Or, $$x(x^2-1)=0$$
Or, $$x(x+1)(x-1)=0$$
x={-1,0,1}-----------------------(c)

From (a)& (c), the only possible integer value of x=-1 (we have an unique value of x)
Sufficient

Ans. (D)
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Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

chetan2u

i have a doubt

in the question stem ---> x^3 - x^2 < 0;
then x^2 - x^3 > 0
x^2(1 - x^2) > 0

so since the product is greater than zero, therefore x can never be Zero (right?)
secondly,
x^2 > 0;
x > 0

and

1 > x^2
|x| < 1 or -1 < x < 1

is this the correct way to break the stem?
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Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

x^3 < x^2 means 2 things:

1) x does not equal to 0
2) x^2 is always positive,

Therefore, we can divide both sides safely by x^2 WITHOUT flipping the sign.

It will be if x < 1, Integer x value?

(1) |x| < 2

-2< |x| < 2.........3 intgere values 1, 0 ,-1

0 & 1 are invalid values.........So the only integers left is -1........We have unique value

Sufficient

(2) x^3 = x

Refer to note 1 above( zero is not considered), we can divide by x...Hence

x^2 =1........Only value is -1 (It can't be 1 as stem says x<1)

Sufficient

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If x^3 < x^2, what is the value of the integer x?  [#permalink]

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saurabh9gupta wrote:
Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

chetan2u

i have a doubt

in the question stem ---> x^3 - x^2 < 0;
then x^2 - x^3 > 0
x^2(1 - x^2) > 0

so since the product is greater than zero, therefore x can never be Zero (right?)
secondly,
x^2 > 0;
x > 0

and

1 > x^2
|x| < 1 or -1 < x < 1

is this the correct way to break the stem?

saurabh9gupta

The part in redis incorrect.

x^2(1 - x) > 0...It should be x, after factoring out x^2.

Also the highlighted part is incorrect

x^2 >0 is ok but it is incorrect to conclude that only x > 0

2 > 0 ......2^2 >0

-2 <0 ......but also -2^2 >0

So x could take negative or positive value

If I continue your way above, I would do the following:

x^2 is +ve and does not equal to zero...So divide both sides safely ........So

1- x <.....1<x

I hope it helps

Originally posted by Mo2men on 02 Sep 2018, 05:49.
Last edited by Mo2men on 02 Sep 2018, 05:57, edited 1 time in total.
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Re: If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

Statemnt I:
Given, $$x^2(1-x) > 0$$.. So, x can be -1,-2,-3.. etc.
But as $$|x| <2$$, x can be only -1.............Sufficient.

Statement II:
From, this statement - $$x = 0, x = -1$$
But x can't be 0 as $$x^2(1-x) > 0$$. Hence, x = -1

Sufficient.
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If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

$$x^3<x^2…x=integer…x<0$$

(1) |x| < 2: sufic.

$$|x|<2…-2<x<2…(x<0)…-2<x<0…x=-1$$

(2) x^3 = x: sufic.

$$x^3=x…x(x^2-1)=0…x(x+1)(x-1)=0…x=(-1,0,1)…(x<0)…x=-1$$

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GMAT 1: 780 Q51 V45 GRE 1: Q170 V167 If x^3 < x^2, what is the value of the integer x?  [#permalink]

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Bunuel wrote:
If x^3 < x^2, what is the value of the integer x?

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

Analyzing the question:
x cannot be 0, so $$x^2$$ must be positive and we are allowed to divide both sides by $$x^2$$. Then the question becomes "If x < 1 what is the value of integer x?" We can continue to list integers that are viable, x = 0 is not in the list so we start from x = -1, -2, -3 etc.

Statement 1:
This gives -2 < x < 2, combining with x < 1 we get the range of x is -2 < x < 1 and because x is an integer, either x = -1 or x = 0. We also should recall x cannot be 0 so only x = -1 is viable. Sufficient.

Statement 2:
Move all terms to the left side. $$x^3 - x = 0$$ can have three solutions at most. Factoring gives $$x*(x+1)(x-1) = 0$$, so x = 0, x = -1, and x = 1. We can take x= -1 only. Sufficient.

Ans: D
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