(1) 1/x > - 1
1/x + 1 > 0
1 + x > 0
x > -1

x could be -0.5 or 5. NSF..

(2) 1/x^5 > 1/x^3
1/x^5 - 1/x^3 > 0
(1 - x^2)/x^5 > 0
1 > x^2

If x^2 is smaller than 1, x is always smaller than 1. SUFF.
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GT

oh what a silly question. I retract it. In case anyone had the same question, the answer is in the math basics intro. It goes into details of multiplying/dividing by the variable in an inequality and instances where it can be done and in what ways.

hiii everyone...

i solved the problem this way... any mistake, plz point it out...

1) 1/x>-1 ====> 1>-x ====> x>-1.
so x can be 0 or any +ve number. so insuff.

2) [1/(x^5)]>[1/(x^3)] ====> x^2<1 ====> -1<x<1.
so x cannot be greater than 1.
hence answer B.
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Deserve before you Desire

The above approach is not right. First of all I'd suggest to state in the stem that x does not equal to zero, as x is in denominator and GMAT almost always states when providing the fraction that the variable in denominator is not zero.

Now this approach is not correct because when we multiply by variable we don't know its sign and thus don't know whether we should flip sign of inequality or not.

Algebraic solution for this question would be:

Question is x>1?

(1) \frac{1}{x}>- 1 --> \frac{1+x}{x}>0, two cases:

A. x>0 and 1+x>0, x>-1 --> x>0;

B. x<0 and 1+x<0, x<-1 --> x<-1.

We got that given inequality holds true in two ranges: x>0 and x<-1, thus x may or may not be greater than one. Not sufficient.

(2) \frac{1}{x^5}> \frac{1}{x^3} --> \frac{1-x^2}{x^5}>0, two cases:

A. x>0 (it's the same as x^5>0) and 1-x^2>0, -1<x<1 --> 0<x<1;

B. x<0 and 1-x^2<0, x<-1 or x>1 --> x<-1;

We got that given inequality holds true in two ranges: 0<x<1 and x<-1, ANY x from this ranges will be less than 1. So the answer to our original question is NO. Sufficient.

Answer: B.

As we can see the ranges for (1) and (2) obtained by this solution differ from the ranges obtained by the solution provided by dpgxxx in his/her post.
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I got this right. Furthermore, from the solution provided by "bb" for (2)
I noted that 1/x^5 > 1/x^3 can be re-written as: x^3 > x^5.

Usually, i reduced the equation to 1/x^3 > 1/x .......multiplied both sides by x^2
I now realized that the same equation is retained when u multiplied both sides by
(x^5)(x^3); after all, -ve * -ve = +ve.
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KUDOS me if you feel my contribution has helped you.

You might get a clue from reading the explanations i gave above. Else, simply multiply both sides by X^5 AND X^3.
This way, you have canceled the likelihood of any change in sign: After all, negative x negative = positive.
Do you get my explanations?
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KUDOS me if you feel my contribution has helped you.

Thanks Bunuel for such a good explanation.

Hi,

Regarding Q No 9. Why are u considering x=-2 for I option as already stated in question that x is greater than 1. Considering that I option will be TRUE always. And for 2nd it is false. So A is the answer. Please post if i am wrong.


Regards,
martin

Is x greater than 1?

(1) 1/x > -1

(2) 1/x^5 > 1/x^3

Stat 1:

Multiply both sides of the inequality by x^2, a positive number, to get;

        x > -(x^2)

        x^2 + x > 0

        x(x + 1) > 0

(i) any positive number, including those less than 1, satisfies 

     x(x + 1) > 0

(ii) negative numbers less than -1 are also consistent with the ineq.

Since x can be greater than or less than 1, Stat 1 is insufficient.       

Stat 2:

Using the rule that a smaller bottom of a fraction results in a greater number:

       1/x^5 > 1/x^3 =

        x^5 < x^3

(i) exclude values of x > 1, which increase when raised to a power that is greater than 1.  Sufficient.

(ii) the domain:

    *(iia) 1/x^5 > 1/x^3 =

            x > x^3 

      (iib) x - x^3 > 0

**  (iic)  x(1 - x^2) > 0

            For x > 0, x^2 < 1
            
            x^2 - 1 < 0
            
            (x + 1)(x - 1) <0

           -1 < x < 1 
            0 < x < 1

            For x < 0, x^2 > 1

            x^2 - 1 > 0

            (x + 1)(x - 1) > 0 

            x > 1, x < - 1
            x < -1

      (iid) x < - 1, 0 < x < 1
    
             x < 1

*(iv) multiply by x^6, a positive  
number, to clear the bottom of the fraction.

**(v) x(1 - x^2) can be greater than 0 
if the products are (positive)(positive) or (negative)(negative).

A suggestion:
Since in both statements x appears in the denominator, it is clear that x is non-zero. Then the inequality in (1) can be multiplied by x^2 which is definitely positive and inequality in statement (2) can be multiplied by x^6, which is also positive. In each case, we obtain an inequality without fractions, much easier to deal with. The inequality in (1) becomes x+x^2=x(1+x)>0, and that in (2) becomes x-x^3=x(1-x)(1+x)>0. In this way, it is much easier to see that the correct answer is (B).
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