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If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:00
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Re: If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:21
Bunuel wrote: If x^3 < x^2, what is the value of the integer x?
(1) x < 2 (2) x^3 = x \(x^3 < x^2.......x^2x^3>0.....x^2(1x)>0\) therefore 1x>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 \(x^3=x.....x^3x=0......x(x^21)=0\) 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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Re: If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:31
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^3x^2<0 x^2(x1)<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^3x=0 x(x^21)=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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If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:36
Thanks chetan2uI have not counted that x not equal to 0 so the answer i got was completely wrong
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Re: If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:39
SonGoku wrote: Thanks chetan2uI 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?
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31 Aug 2018, 23:40
chetan2u wrote: Bunuel wrote: If x^3 < x^2, what is the value of the integer x?
(1) x < 2 (2) x^3 = x \(x^3 < x^2.......x^2x^3>0.....x^2(1x)>0\) therefore 1x>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 \(x^3=x.....x^3x=0......x(x^21)=0\) 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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Re: If x^3 < x^2, what is the value of the integer x?
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31 Aug 2018, 23:44
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 \(x^3 < x^2.......x^2x^3>0.....x^2(1x)>0\) therefore 1x>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 \(x^3=x.....x^3x=0......x(x^21)=0\) 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 TRUE1^3 < 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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If x^3 < x^2, what is the value of the integer x?
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01 Sep 2018, 20:28
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, \(x^3x^2<0\) Or, \(x^2(x1)<0\) Or,\(x:\:\left(\infty \:,\:0\right)\cup \left(0,\:1\right)\)(a) St1: x < 2Or, 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, \(x^3x=0\) Or, \(x(x^21)=0\) Or, \(x(x+1)(x1)=0\) 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?
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01 Sep 2018, 22:35
Bunuel wrote: If x^3 < x^2, what is the value of the integer x?
(1) x < 2 (2) x^3 = x chetan2ui 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?
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02 Sep 2018, 05:44
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 Answer: D



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If x^3 < x^2, what is the value of the integer x?
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Updated on: 02 Sep 2018, 05:57
saurabh9gupta wrote: Bunuel wrote: If x^3 < x^2, what is the value of the integer x?
(1) x < 2 (2) x^3 = x chetan2ui 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 > 0and 1 > x^2 x < 1 or 1 < x < 1 is this the correct way to break the stem? saurabh9guptaThe 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?
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02 Sep 2018, 05:56
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, \(x^2(1x) > 0\).. 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 \(x^2(1x) > 0\). Hence, x = 1Sufficient.
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