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Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

(1) x^2 is less than x --> \(x^2<x\) --> \(x(x-1)<0\):

Multiples must have opposite signs: \(x<0\) and \(x-1>0\), or \(x>1\) --> no solution (\(x\) cannot be simultaneously less than zero and more than 1); \(x>0\) and \(x-1<0\), or \(x<1\) --> \(0<x<1\);

So \(x(x-1)<0\) holds true when \(0<x<1\). Sufficient.

St1: x>x^2-----> x(1-x)>0 So either x>0 and 1-x>0 or 1>x and thus 0<x<1 or x<0 or 1-x<0 or x>1...this is not possible cause we said x<0 so st1 alone is sufficent

St2: x^3 is positive...well there are many options like x=2 and x^3=8 or x=1/2 and x^3=1/8

So ans is A

Difficulty level 600 is okay

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(1) x^2 is less than x -> only possible for values between 0 and 1. (if x < 0, x^2 always > x; if x > 1, x^2 always > x). (2) x^3 is positive -> example: x = 2 or 1/2, Not sufficient.

(1) x^2 is less than x --> \(x^2<x\) --> \(x(x-1)<0\):

Multiples must have opposite signs: \(x<0\) and \(x-1>0\), or \(x>1\) --> no solution (\(x\) cannot be simultaneously less than zero and more than 1); \(x>0\) and \(x-1<0\), or \(x<1\) --> \(0<x<1\);

So \(x(x-1)<0\) holds true when \(0<x<1\). Sufficient.

No need to talk why B is wrong. A) If X^2 < X it means that 0<X<1, X can not be negative -> -2^2 isalways > -2, so X is positive. And a positive X^2 can ONLY be smaller than X if its range is between 0 and 1.

It's actually a property of a normal fraction that X^2 < X (see MGMAT)
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