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
Is X>0?
Source: Beat The GMAT — Data Sufficiency |
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blackarrow
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cubicle_bound_misfit
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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.
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.
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vittalgmat
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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
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
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cramya
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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
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
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mehravikas
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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.
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
Last edited by mehravikas on Thu Apr 16, 2009 3:24 am, edited 1 time in total.
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Sher1
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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
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
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cubicle_bound_misfit
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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
do that for stmt 2 and see why x^3<x gives us two different answers.
HTH
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maihuna
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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
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
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cramya
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When x = -0.9 x^2 = .81 > -.9 will not be less1. x^2<x for x = -0.9 as well as 0.9 so no help here
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
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rossmj
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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.
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.
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pakaskwa
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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.
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.
Last edited by pakaskwa on Thu Apr 16, 2009 3:04 pm, edited 2 times in total.
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cramya
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Hi Pakasawa,
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
It's trickier than it looks eh
Agree!!!
Dont Agree.Stmt II
3. x<0, x+1>0, x-1<0 --> -1<x<0
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
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pakaskwa
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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.
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.
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maihuna
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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.....cramya wrote:Hi Pakasawa,
It's trickier than it looks eh
Agree!!!
Dont Agree.Stmt II
3. x<0, x+1>0, x-1<0 --> -1<x<0
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
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.
