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statement 1 says, x > 0 . This is not sufficient to know whether x^3 >x^2. suppose x =1/2, then x^2 = 1/4 and x^3 = 1/8 , so as 1/4 is greater than 1/8, x^2 > x ^3. but if x =2. then x^3 > X^2. so this statement is insufficient.

statement 2 says x ^ 2 > x. this means either x is a negative number or x is positive number more than 1. Essentially x can have any value except 0 to +1. so this is not enough to tell whether x ^3 is greater than x ^2 or not.

combining both statement we know that x is positive number and x is more than 1. so it sufficiently answers that x^3 > x^2. Any number more than 1 will have its value raised to power of 3 more than its value raised to power of 2.

Regarding your second doubt . your wrote " same applies for 2 ... sincex^2>x x^3>x^2"

No, X^2 > X does not necessarily means x^3 will be greater than X^2 and so on...suppose x= -2 (minus 2) then x^2 = +4 which means x^2 >x but x^3 will be -8 which is not more than +4. so you have to consider +- and fraction value while solving such questions.

Regarding your second doubt . your wrote " same applies for 2 ... sincex^2>x x^3>x^2"

No, X^2 > X does not necessarily means x^3 will be greater than X^2 and so on...suppose x= -2 (minus 2) then x^2 = +4 which means x^2 >x but x^3 will be -8 which is not more than +4. so you have to consider +- and fraction value while solving such questions.

Can we solve this algebrically Question says\(x^3 > x^2\) ?

Thus we can reduce this to \(x^3 - x^2\) > 0?

\(X^2 (x-1)> 0?\)------ (1)

Now this equation 1 of of the form xy >0 so either x and y are both positive or both negative ...

since x^2 is always +ve this x-1 is also positive

so equation 1 become x> 0 or x> 1

thus the question becomes is x >1 ??

Now lets come to the statements

1) x>0

This just tells us x> 0 but we dont know if x > 1 ---- Insufficient

2) \(x^2 > x\)

\(X^2 - x > 0\)

thus, x (x-1) >0

x and (x-1) must have same sign either both positive or both negative

Case 1 : x and x-1 both positive

x > 0 or x-1 >0

x> 0 or x > 1

thus from case 1 we can conclude x>1

Case 2: x and x-1 both negative

x<0 or x-1<0

x<0 or x<1

thus x < 0 from case 2

combining case 1 and 2 we get :

i dont know how to proceed ahead after this... can someone help???

+1 , you have solved it correctly using algebra , can't imagine how u got it wrong first time itself.

from what you solved through equation 2. you get x >1 or x <0, so its it. on a number line x is either right of +1 or left of 0. from this we cant conclude anything. if x is more than 1, x^3 > x^2 but if x is less than 0, then x^2 > x^3. so this statement is insufficient.

But once we combine statement 1 and 2 we get to know x is positive and more than 1. positive value mentioned in statement 1 rules out one of the possibility of value of x we get from statement 2. so only 1 value remains that is x >1. sufficient.

I searched for this but couldnt find it... Is \(x^3 > x^2?\) (1) x > 0 (2) \(x^2 >x\)

There are many different ways to approach these questions, some of which have been discussed above. I'll just point out one useful simplification that you can often use in these questions. When we are asked if x^3 > x^2, we can rephrase this question easily. We cannot divide by x on both sides (without breaking the problem down into awkward cases), since x might be negative (and if we divide by a negative, we need to reverse the inequality). But we *can* divide by x^2 on both sides, since x^2 cannot be negative. So we can instantly rephrase the question as "Is x > 1?".

Then we can immediately see that Statement 1 is not sufficient. Further, since Statement 2 is true for any negative value of x, but is also true when x > 1, it cannot be sufficient alone either. Combining the statements, if x > 0 from Statement 1, then we can safely divide by x on both sides in Statement 2, so we get x > 1, which is what we want. So the answer is C.
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[quote="siddhans"]I searched for this but couldnt find it...

Is \(x^3 > x^2?\) (1) x > 0 (2) \(x^2 >x\)

I got a C in 20 seconds and I used a simple thought process:

If x^3 has to be greater than x^2, x has to be a positive integer greater than zero. In all other cases it will not hold true. Now we go to the statements and see whether we can get a YES or NO.

Statement 1 tells us that x is positive and greater than zero but tells nothing about it being an integer. Hence INSUFFICIENT

statement 2 tells us that x is an integer but says nothing about the sign of x. Hence INSUFFICIENT.

Combining both statement, we get a YES to the question. Hence SUFFICIENT
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I hope yo know the basic structure of DS questions? You need to first check the suffecieny of statement 1...then for statement 2....and if neither is insuffecient on its own...then test for suffeciency together etc...

St1: X>0 his means x could be a number like 1/2 (which between 0 and 1) or could be a number like 2 (>1)... for x=1/2, \(x^3<x^2\) as 1/8<1/4 for x=2, \(X^3>X^2\) as 8>4. As both cases are possible, statement 1 alone is not suffecient to say if \(x^3>x^2\) or not

St2: this says x^2>x => x(x-1)>0 =>x>1 or x<0 =>x could be numbers like -1/2,-2, 2 for x =-1/2, x^3<x^2 for x= -2, x^3<x^2 for x= 2, X^3>x^2.... hence not suffecient to say if\(x^3\) is greater than\(X^2\) or not

St1 and St2 together: X>0 AND (X>1 or X<0)....only numbers satisfying both cases (or in other words only common area for both cases on number line) is for x>1....for all x greater than 1. x^3>x^2...so suffecient to say YES to the question asked.

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Re: Is x^3 > x^2? (1) x > 0 (2) x^2 > x [#permalink]

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31 Aug 2017, 01:58

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Re: Is x^3 > x^2? (1) x > 0 (2) x^2 > x [#permalink]

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31 Aug 2017, 03:45

We have to find if x^3 > x^2

Generally we know if x= +ve : x^3 > x^2 or else if x is -ve x^3< x^2 But here we have 2 more conditions for x=+ve if x=1 x^3 =x^2 and if x is +ve but less than 1 (ie fraction less than 1).. then x>x^2> x^3 . eg 1/2 > 1/4> 1/8.

So we have to check all the above conditions to answer the question.

1) x> 0. We know x is +ve . But we dont knw if x is greater than or less than 1. So not sufficient

2) x^2 > x This is True for all + and -ve numbers. This equation just tell that x is not between -1 and 1. So it omits the fraction part but we don't know if x is +ve or -ve. So Not Sufficient

1+2 Tells that x is positive and x is greater than 1. so we can conclude x^3 > x^2