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Hi, I can't see why you think E is the correct answer here. All three values that you used for \(x\) are less than 1. If you plug in \(x=-\frac{1}{2}\), the inequality doesn't hold true, you're right here. However, it doesn't prove that \(x\) can be both greater than and less than 1. Plugging in \(-\frac{1}{2}\) only shows that \(x=-\frac{1}{2}\) is not in the range of \(x\) that hold the inequality true.

In other words, while we consider S2, we have to use only those \(x\) that hold the S2 true. Your example in iii) conflicts with S2 and can't be used to prove anything.

I'll have to modify the OE to use a more algebraic way of solving.

I hope my answer helps and doesn't confuse you.

sunny4frenz wrote:

Hi...

I feel answer to this question must by E .

Statement (1) is obvious why it is not sufficient to find the answer

I thought you weren't allowed to multiple or divide by the variable because until you know whether the variable is positive, you can't know whether to change the inequality sign or not.

1st post so I'll just applaud this incredible forum!

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.

\(\frac{1}{x^5} > \frac{1}{x^3}\) is equivalent to \(x^5 < x^3\). because if two fractions have equal numerators, the fraction with a smaller denominator is the bigger one.

...I simply went ahead and pluggeed-in "2" and "-2" for the the second expression (just like the first). In this way, I concluded a "NO" answer for both 1/32 > 1/8 and -1/32 > -1/8

I solved this one pretty quickly because I somewhat remembered the rule. I didn't remember it exactly but I had a sense of what concept the problem was testing and approached the STMT's accordingly. I just plugged in a positive and negative for STMT1. I only plugged in negatives for STMT2 because I believed that 1/x^5 needed to be negative for it to be greater than 1/x^3 since fractions decrease when multiplied.

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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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.

Hi James,

Can you please provide a link to this, I still don't get why we can cross multiply, x can be negative and inequality sign can change.

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.

Hi James,

Can you please provide a link to this, I still don't get why we can cross multiply, x can be negative and inequality sign can change.

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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We should probably revise the difficulty level of the question. I like your approach. +1. Can't see anything wrong with it.

dpgxxx wrote:

Here is my way, which worries me by its simplicity in comparison to a "750" level problem. Please let me know if there are any errors in the following approach. I am not the best with these question types and am not sure if plugging in numbers, like others did, considers an element my approach does not.

S1) 1/X > - 1 ---> 1>-X ---> (multiply by -1/turn sign)---> -1<X ..... This means X can be -.5, 0, 1, 2..insufficient

S2) 1/X^5 > 1/X^3 --->(multiply both sides by X^3)---> 1/x^2 > 1 ---> 1>X2 ---> +-1 > X ...if X is less than 1 or -1, we know its not greater than 1; sufficient.

Answer: B

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.

Bunuel's approach definitely helped me, but I was wondering why we can't assume the two situations (x>0 and x<0) right off the bat.

I have been trying to do this, but unfortunately I always yield the same (wrong) answer: 1<x. I would really appreciate if someone could explain to me where I have made an unwarranted assumption. Here is my work:

Scenario A: Assume x > 0. (1/x) > -1 1 > -x 1 < x

Scenario B: Assume x < 0 (1/x) > -1 1 < x

In both cases, it yields 1 < x, but obviously it is the wrong answer since x can be any value [-infinity, -1) and (1, infinity]

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.

Substitute x=2 in second option, you get 0.125 < 0.03, which can only be true if x is -ve, As for the first option, multiplying -1 on both sides give 1/x > -1, which doesn't infer anything.

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient

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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Last edited by EvaJager on 17 Jun 2012, 12:37, edited 1 time in total.