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Is x^2 greater than x ?

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Is x^2 greater than x ?  [#permalink]

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New post 02 Jul 2017, 07:44
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Is x^2 greater than x ?

(1) x^2 is greater than 2x.
(2) 2x^2 is greater than x.

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Is x^2 greater than x ?  [#permalink]

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New post 02 Jul 2017, 09:38
sananoor wrote:
Is x^2 greater than x ?

(1) x^2 is greater than 2x.
(2) 2x^2 is greater than x.



hi

so we are looking for \(x^2>x.....x^2-x>0......x(x-1)>0\)
two cases..
a) x is positive and x-1>0 0r x>1..
b) x is negative and x-1<0 or x<1... or x<0..

lets see the statements..

1) \(x^2>2x...x^2-2x>0...x(x-2)>0\)
so we have two ranges.. x>2 or x<0..
it suffices to answer as YES
suff

2) \(2x^2>x....x(2x-1)>0\)..
here we get \(x>\frac{1}{2}\) and \(x<0\)..
but x can be between \(\frac{1}{2}\) and 1..
so insufficient

A
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is x^2 greater than x ?  [#permalink]

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New post 13 Jan 2018, 08:42
sananoor wrote:
Is x^2 greater than x ?

(1) x^2 is greater than 2x.
(2) 2x^2 is greater than x.


To Find is \(x^2 > x\) ------ equ (1)
=> on simplifying we have
=> \((x^2-x)>0\)
=> OR \(x(x-1)>0\)
=> Above hold true in 2 case- both x & (x-1) are +ve or both -ve
=> case 1 both x & (x-1) are +ve implies x>0 & x-1>0 or x>1
=> case 2 both x & (x-1) are -ve implies x<0 & x-1<0 or x<1
Thus equ (1) hold true if x<0 OR x>1


Statement 1 \(x^2>2x\)
=> OR \(x^2-2x>0\)
=> OR \(x(x-2)>0\) ------ equ (2)
=> Above hold true in 2 case- both x & (x-2) are +ve or both -ve
=> case 1 both x & (x-2) are +ve implies x>0 & x-2>0 or x>2
=> case 2 both x & (x-2) are -ve implies x<0 & x-2<0 or x<2
Thus equ (2) hold true if x<0 OR x>2
BUT these values x<0 OR x>2 hold TRUE for equ (1) too. THEREFORE SUFFICIENT

Statement 2 \(2x^2>x\)
=> OR \(2x^2-x>0\)
=> OR \(x(2x-1)>0\) ------ equ (3)
=> Above holds true in 2 case- both x & (2x-1) are +ve or both -ve
=> case 1 both x & (2x-1) are +ve implies x>0 & 2x-1>0 or \(x>\frac{1}{2}\)
=> case 2 both x & (2x-1) are -ve implies x<0 & 2x-1<0 or \(x<\frac{1}{2}\)
Thus equ (3) hold true if x<0 OR \(x>\frac{1}{2}\)
Now when x<0 OR x>1 equ (1) is ALSO TRUE
BUT for \(\frac{1}{2}<x<1\) equ (1) FAILs.
Since no UNIQUE answer THEREFORE IN-SUFFICIENT

Hence 'A'

Thanks
Dinesh
Re: Is x^2 greater than x ? &nbs [#permalink] 13 Jan 2018, 08:42
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