Data Sufficiency

x is not 1, (1-x^5)/(1-x) < 1/1-x ?
(1) x>0
(2) x<1
using (1-x^5)/(1-x) < 1/1-x)
we can arrange the equation to be (1/1-x)- (x^5/1-x) <1/1-x
then rearrange again and get -x^5/1-x <0
multiplying each side by (1-x)^2 since its positive
=> -x^5(1-x) which is also x^5*(x-1)<0
dividing each side by x^4, we get x(x-1)<0, or 0<x<1 so the answer must be c

but what if from -x^5/(1-x)<0, instead of making it into x^5*(x-1)<0,
can u just take -x^5/(1-x)<0 and switch it to x^5/(1-x)>0 (taking out the negative sign?
for some reason if i do it that way my answer comes out differently..
(the answer is C btw)

-x^5/(1-x)<0, so is it legal to do this?
==> x^5/(1-x)>0
so is there a difference between x^5/(1-x)>0
and x^5*(x-1)<0 ?

my answer comes out different when i multiply each side by a negative (x^5/(1-x)>0 )

x is not 1, (1-x^5)/(1-x) < 1/1-x ?
(1) x>0
(2) x<1

cross multiply = (1-x^5)* (1-x) < 1-x

1-x-x^5 +x^6 < 1-x

Is x^5+x^6<0 ?

(1) x>0
Sufficient

2) x<1
Not Sufficient
x could be -1, where result = 0
x could be -2, where result >0

Did you say the answer was (C)?

If you back solve the answer it might be easier

1). says x>0, so plug in 1 to the equation
(1-1^5)/1-1 < 1/(1-1)
= 0 < 1/0 insuffient

2). says x<1, so plug in 0
(1-0^5)/1-0 < 1/(1-0)
= 1<1 insuffient

Together they say that x is between 0 and 1
If you plug in 1/2 into the equation it is sufficient.