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Can we really work like this with inequalities?schumi_gmat wrote:1) x^4 - x^3 > 0
i.e. x^3 (x-1) >0
The above will be true if bot x^3 >0 and x-1>0 or x^3 < 0 and x-1 <0
Two answers ....Insuff
2) x^5 - x^3 > 0
x^3 (x-1) (x+1) >0
using same concept above, Again multiple answers and hence insuff.
Combining, we cannot get unique answer
Hence E
Bro..neoreaves wrote:IMO B
2) x^3 < x^5
This is only true for x>1 ....we can test other values
x = fraction ....this won't be true for fractions so discard
x = negative integers ....wont be true either so the only way this can be true is x > 1
I disagree with the OA = E ....unless someone can give a counter example to prove this is not sufficient.
ok i see your point ...though your example of x = 0.5 or 0.25 wont work ....gmatmachoman wrote:Bro..neoreaves wrote:IMO B
2) x^3 < x^5
This is only true for x>1 ....we can test other values
x = fraction ....this won't be true for fractions so discard
x = negative integers ....wont be true either so the only way this can be true is x > 1
I disagree with the OA = E ....unless someone can give a counter example to prove this is not sufficient.
But again going by ur wrds that X >1, then u shuld have opted D. Still it is WRONG!!
here is the catch!
Did u try for X =0.5,0.25 etc??
Logic shall fail!!
hahahhaa..nice catch...Yeah it should have been x=-0.5,-0.25...Seems sunday midnite blues has struck me!neoreaves wrote:ok i see your point ...though your example of x = 0.5 or 0.25 wont work ....gmatmachoman wrote:Bro..neoreaves wrote:IMO B
2) x^3 < x^5
This is only true for x>1 ....we can test other values
x = fraction ....this won't be true for fractions so discard
x = negative integers ....wont be true either so the only way this can be true is x > 1
I disagree with the OA = E ....unless someone can give a counter example to prove this is not sufficient.
But again going by ur wrds that X >1, then u shuld have opted D. Still it is WRONG!!
here is the catch!
Did u try for X =0.5,0.25 etc??
Logic shall fail!!
but x = -0.5 will ...so yeah it should be E
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