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Bunuel
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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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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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chetan2u
Bunuel
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 what`s wrong with my approach ? :?

thanks and have a great day :)
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chetan2u
Bunuel
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 what`s 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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Bunuel
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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Bunuel
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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Bunuel
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

Answer: D
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saurabh9gupta
Bunuel
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
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Bunuel
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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Bunuel
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.

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

Answer (D)
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Bunuel
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 \(x^2\). Since x<1 and x cannot be 0, the question becomes "If x < 0 what is the value of integer x?"

Statement 1:
-2 < x < 2, combined with x < 0 we get that the range of x is -2 < x < 0. x=-1. Sufficient.

Statement 2:
Divide by x. \(x^2 = 1\) since x<0, x= -1 only. Sufficient.

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

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

If I divide x^3 by x^2 in the stem, it would give me x<1, then from (1), x = -1 and 0, making (1) insufficient, although without dividing the stem, (A) is sufficient.
Is it that we are dividing the stem because of 0, but in other questions we are getting the correct value with this approach?
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