IMO A

we basically need to find if X is positive

statement 1

x^2<x
this is possible only if x is positive

ex- if x=1/2; then 1/4<1/2

thus sufficient

Statement 2


x^3<x

possible both when x is negative and postive (fractions)

not sufficient


Thus IMO A

put couple of fractions between (-1,1)

for x = -1/2

stmt 1 ain't true it is true only when x>0

stmt 2 can be true for any regular fraction both + and - hence Insuff.

A.

I get E.
Pls let me know if I have missed something here.


-1 < x < 1 is x > 0 ?

stmt 1: x^2 < x

ie. x^2 -x < 0

x(x -1) < 0

ie x < 0 or x < 1

so not sufficient.


stmt 2:
x^3 -x < 0

x(x^2 -1) < 0

ie x < 0 or x < 1 or x < -1.
so insufficient.


Even combining we are left with x < 0 or x < 1.
ie x < 0 or x > 0

so E

Stmt I

If x (x-1) <0 then there are 2 situations we can think of

Case 1: x>0 and x-1<0
x>0 x<1

Case 2: x<0 x-1>0
x<0 x>1

IMPOSSIBLE

Only case 1 possible which tells us x>0 and x<1

SUFF

Stmt II

x(x^2-1) < 0

Case1: x<0 and x^2-1>0 i.e x^2>1

Case 2: x>0 and x^2-1 <0 x^2<1

In case 1 x has to be outside the range -1<x<1 so not possible
In case 2 x has to be between 0 and 1

SUFF


I am going to go with D and add more to the confusion since we already have 2 A's and an E

Regards,
CR

By taking values that satisfies statement 1 and 2, I can surely conclude the OA should be 'D'.

Statement 1 - x^2 < x

x = 1/2 -> (1/2) ^ 2 < 1/2 - true
x = -1/2 -> (-1/2)^2 < 1/2 - false

Statement 2 - X^3 < x

x = 1/2 -> 1/8 < 1/2
x = -1/2 -> -1/8 < 1/2 - false, therefore x has to be a +ve fraction.

Therefore, statement 2 is also sufficient.

cramya wrote:Stmt I

If x (x-1) <0 then there are 2 situations we can think of

Case 1: x>0 and x-1<0
x>0 x<1

Case 2: x<0 x-1>0
x<0 x>1

IMPOSSIBLE

Only case 1 possible which tells us x>0 and x<1

SUFF

Stmt II

x(x^2-1) < 0

Case1: x<0 and x^2-1>0 i.e x^2>1

Case 2: x>0 and x^2-1 <0 x^2<1

In case 1 x has to be outside the range -1<x<1 so not possible
In case 2 x has to be between 0 and 1

SUFF


I am going to go with D and add more to the confusion since we already have 2 A's and an E

Regards,
CR

I got D as well

1
try using a +ve and -ve fraction within the terms

x= 1/2 x^2=1/4

x=-1/2 x^2 = 1/4

x has to be +ve for the inequality to hold

Suff

2
x=1/2 x^3=1/8

x=-1/2 x^3=-1/8

since -1/8 > =-1/2

x has to be a +ve fraction to work

Suff

for ineqality one thumb rule is put back the value we get from solving into original inequality.

do that for stmt 2 and see why x^3<x gives us two different answers.

HTH

for -1<x<1

1. x^2<x for x = -0.9 as well as 0.9 so no help here

2. x^3<x for x = -0.9 x^3 = -.729 > -0.9
so for positive no only x^3<x

Other way to look at it: x(x^2-1) <0
if x<0 => x2-1 >0 or x >1 or x<-1 so it doesnt hold given |x| =1
if x>0 => x^2-1<0 or x^2<1 simply |x| <1 as given premise

So B is the answer.

maihuna

1. x^2<x for x = -0.9 as well as 0.9 so no help here

When x = -0.9 x^2 = .81 > -.9 will not be less

Pakaswa,
So far we have A,B,D,E only C left. I am sure we r not helping u or your friend with an OA.

I still think it should be D

Regards,
CR

Definitely D

From the stem we know that x is a fraction or x=0. If you raise a positive fraction to an exponent you get a smaller number. If you raise a negative fraction to an exponent you always get a larger number because it will either be positive (even exponent) or less negative. For proof draw a number line and start with -1/2, soon you will see that all the numbers you fill in will be to the right of -1/2. Therefore (-1/2)^3 can never be less than -1/2 and x must be greater than 0.

Hi CR and all,
Thanks for your help. It's trickier than it looks eh?

CR, my approach is similar to yours, yet I got different answer.

From stmt 1, we can get x(x-1)<0
so there are two solutions:
1. x<0, x>1, or
2. x>0, x<1 --> 0<x<1
The question stem told us that -1<x<1, so 1st solution is out. 2nd solution stands, so stmt 1 SUFF.

From stmt 2, we can get x(x+1)(x-1)<0
Then there are 3 possiblities:
1. x>0, x+1>0, x-1<0 --> 0<x<1
2. x>0, x+1<0, x-1>0 --> x>1 (out)
3. x<0, x+1>0, x-1>0 --> no solution (out)
So from stmt 2 we know x>0. SUFF.

Choice is D.

This is an edited answer after CR pointed out my error. Hopefully it's correct.

Hi Pakasawa,

It's trickier than it looks eh

Agree!!!

Stmt II

3. x<0, x+1>0, x-1<0 --> -1<x<0

Dont Agree.

Pick any x <0 and >-1

x (x-1) (x+1) will be positive > 0

x=-.9

-.9 (-.9-1) (-.9+1) > 0 not < 0

x=-.1

-.1 ( -.1-1) (-.1+1) >0 and not less than 0


This is the case that made stmt II INSUFF in your analysis.



The key to this problem I think is keeping -1<x<1 in mind at all times.Hope this helps and I dint miss something!


Good luck.

Regards,
CR

You are right CR. I made a mistake with Stmt 2.

From stmt 2, we can get x(x+1)(x-1)<0
Then there are 3 possiblities:
1. x>0, x+1>0, x-1<0 --> 0<x<1
2. x>0, x+1<0, x-1>0 --> x>1 (out)
3. x<0, x+1>0, x-1>0 --> no solution
So from stmt 2 we know that x>0. SUFF!

Choice is D!
I'll edit on my previous posting so that it won't mislead readers. Thanks CR.

cramya wrote:Hi Pakasawa,

It's trickier than it looks eh

Agree!!!

Stmt II

3. x<0, x+1>0, x-1<0 --> -1<x<0

Dont Agree.

Pick any x <0 and >-1

x (x-1) (x+1) will be positive > 0

x=-.9

-.9 (-.9-1) (-.9+1) > 0 not < 0

x=-.1

-.1 ( -.1-1) (-.1+1) >0 and not less than 0


This is the case that made stmt II INSUFF in your analysis.



The key to this problem I think is keeping -1<x<1 in mind at all times.Hope this helps and I dint miss something!


Good luck.

Regards,
CR

Pick a no is not a good strategy, all the mod questions are solvable step by step, I have recently paid huge price to such ad-hoc methods when I ended up scoring a mere 47 in quant and so ruining up my chance to apply for an mba this year. Oppurtunity cost is too much, in practice look for a method that is guaranteed to work and dont waste time tricking on other questions.....

as for as part b is concerned, dont factorize x^2-1 to durther cases just consider it this way: for 0<x<1 x^2<1 for other values x^2>1 it will help minimizing number of cases to be considered.