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1) Sufficient. This has to be an integer (fractions would really change up the answer to this one). So with a negative number, when you cube it, you come up with a smaller (to the left on the number line) negative number. x = -2. x^3 = -8. -2 > -8. What if x is positive? x = 2, x^3 = 8. 2 !> 8 ( ! = not in programming). Since we are to take the statements as true, a value that does not conform to the statement is not possible. The only values that are possible to make #1 true are negative numbers. So the answer to the question posed is "No, x is not positive." and the data presented is sufficient.

2) Insufficient. we have x < x^2. x = -2...so x^2 = 4. -2 < 4 => TRUE. x = 2, x = 4 2 < 4 =>TRUE. We have one negative number that works for statement 2 and one positive number for statement 2. This is not enough to give a difinitive answer as to whether x is positive.

aaron22197 wrote:

Is integer x positive?

1. \(x \gt x^3\) 2. \(x \lt x^2\)

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

x^3-x <0 ---> x(x^2-1) <0 we have two solutions x<0 and x^2-1>0 --(I) x>0 and x^2-1<0 --(II) II - this one not the solutions because x is interger.. and x>0 --> x^2-1<0

So we have only one solution i.e x <0 and x^2>1 x can be any -ve integers < -1 i.e -2,-3......................

2) x<x^2 x^2-x>0 -- x(x-1)>0

two solutions x>0 and x-1>0 (III) x<0 and x-1<0 (IV) both are possible solutions (can be +ve or -ve) Insuffcient

A
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

i dont get how A is sufficient ... i rearranged to get x(1-x^2) > 0 ....

this gives me solutions like x>0 or -1<x<1 or x>1 or x<-1 ...

x(1-x^2) > 0 for the product tobe >0 both (x and (1-x^2) should have same sign) two solutions are: x >0 and (1-x^2)>0 (but this solution is not possible because x is integer and >0 (1-x^2) never >0. so we can strike out this solution. or x <0 and (1-x^2)<0

Got it???
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

i dont get how A is sufficient ... i rearranged to get x(1-x^2) > 0 ....

this gives me solutions like x>0 or -1<x<1 or x>1 or x<-1 ...

x(1-x^2) > 0 for the product tobe >0 both (x and (1-x^2) should have same sign) two solutions are: x >0 and (1-x^2)>0 (but this solution is not possible because x is integer and >0 (1-x^2) never >0. so we can strike out this solution. or x <0 and (1-x^2)<0

Got it???

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

sometime plugging in is easy way to solve these kind of problems.

the question says x is an integer. this statement gives half solution.

1: since x is an integer, x cannot be +ve. so it must be -ve. so A is suff. 2: since x is an integer, x can be either -ve or +ve. so not suff.

jallenmorris wrote:

A

1) Sufficient. This has to be an integer (fractions would really change up the answer to this one). So with a negative number, when you cube it, you come up with a smaller (to the left on the number line) negative number. x = -2. x^3 = -8. -2 > -8. What if x is positive? x = 2, x^3 = 8. 2 !> 8 ( ! = not in programming). Since we are to take the statements as true, a value that does not conform to the statement is not possible. The only values that are possible to make #1 true are negative numbers. So the answer to the question posed is "No, x is not positive." and the data presented is sufficient.

2) Insufficient. we have x < x^2. x = -2...so x^2 = 4. -2 < 4 => TRUE. x = 2, x = 4 2 < 4 =>TRUE. We have one negative number that works for statement 2 and one positive number for statement 2. This is not enough to give a difinitive answer as to whether x is positive.

Pick numbers that satisfy the equation S1. x x^3 x>x^3 -1 -1 No (does not satisfy the equation so you can cancel it) -2 -8 Yes 2 8 NO (does not satisfy the equation so you can cancel it) For all postivie numbers you will not satisfy S1. So you get all -ve numbers only that satisfy the equation Hence BCE out keep AD

S2. x x^2 x<x^2 -1 1 Yes 2 4 Yes You get a Maybe case so D is out

I can't figure out why - but I'm not doing this right. Can tell me why my logic is wrong?

a) x > x^3

Then I simplified to: 0 > x^3 - x 0 > x(x^2-1) 0 > x(x+1)(x-1)

This gives me solution: 0>x, -1>x, 1>x - what this tells me is that x < -1. Ok, I'm good here.

Then I have to test the opposite side of the inequality: 0 < x(x+1)(x-1)

This gives me the solution: 0 < x, x<-1, 1<x - the solution doesn't make sense here. Does it mean it's not valid, therefore the only solution is the one above?

b) x<x^2

I applied the same logic as above, testing both sides of the inequalities.

Both of the solutions were not consistent with each other, so this to me was not sufficient.