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Intern  Joined: 03 Aug 2009
Posts: 7
Schools: LBS
Re: GMAT Diagnostic Test Question 9  [#permalink]

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on S2: I used the same approach as dpgxxx,

but

x^2 < 1 holds only true if -1<x<1. IMO answer B remains correct.

Just my 2 cents.

Jochen
SVP  Joined: 29 Aug 2007
Posts: 1711
Re: GMAT Diagnostic Test Question 9  [#permalink]

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bb wrote:
Is x greater than 1?

(1) $$\frac{1}{x} > -1$$
(2) $$\frac{1}{x^5} > \frac{1}{x^3}$$

(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.
Intern  Joined: 25 Nov 2009
Posts: 14
Location: San Francisco
Schools: Wharton West eMBA, Haas EW, Haas eMBA
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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!
Intern  Joined: 25 Nov 2009
Posts: 14
Location: San Francisco
Schools: Wharton West eMBA, Haas EW, Haas eMBA
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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.
Manager  Joined: 08 Jul 2009
Posts: 147
Re: GMAT Diagnostic Test Question 9  [#permalink]

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I would use the number line approach. Test 1 number in each section :small than -1, between -1 and 0, between 0 and 1, and larger than 1.
Intern  Joined: 22 Dec 2009
Posts: 12
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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.
Manager  Joined: 24 Jul 2009
Posts: 162
Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: GMAT Diagnostic Test Question 9  [#permalink]

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stilite wrote:

$$\frac{1}{x^5} > \frac{1}{x^3}$$ 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.

This was a fair problem.
Math Expert V
Joined: 02 Sep 2009
Posts: 65829
Re: GMAT Diagnostic Test Question 9  [#permalink]

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3
1
dzyubam wrote:
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.

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.

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.
_________________
CIO  Joined: 02 Oct 2007
Posts: 1179
Re: GMAT Diagnostic Test Question 9  [#permalink]

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I guess I didn't give it a thorough look like Bunuel did. I will improve. Thanks, Bunuel! +1.
Senior Manager  Joined: 21 Dec 2009
Posts: 442
Concentration: Entrepreneurship, Finance
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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.
Manager  Joined: 21 Mar 2010
Posts: 67
Re: GMAT Diagnostic Test Question 9  [#permalink]

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james12345 wrote:
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.
Senior Manager  Joined: 21 Dec 2009
Posts: 442
Concentration: Entrepreneurship, Finance
Re: GMAT Diagnostic Test Question 9  [#permalink]

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tryinghard wrote:
james12345 wrote:
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?
Intern  Joined: 31 Jul 2009
Posts: 2
Schools: Texas A&M,University of North Florida,University of Texas
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Dear Bunuel,

In statement 1-\frac{1+x}{x}>0,and
statement 2- \frac{1-x^2}{x^5}>0,
How you derive those two, i cant understand
Thank you
Manager  Joined: 15 Apr 2010
Posts: 93
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Bunuel wrote:
dzyubam wrote:
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.

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.

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.

Thanks Bunuel for such a good explanation.
Intern  Joined: 02 May 2011
Posts: 5
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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]
Intern  Joined: 21 Jun 2011
Posts: 1
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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
Intern  Joined: 27 May 2011
Posts: 2
Re: GMAT Diagnostic Test Question 9  [#permalink]

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I just used old substitution method,

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.

So the solution is B
Intern  Joined: 14 Sep 2010
Posts: 10
Re: GMAT Diagnostic Test Question 9  [#permalink]

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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).
Intern  Joined: 21 Jan 2012
Posts: 9
Re: GMAT Diagnostic Test Question 9  [#permalink]

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Thx dzyubam.

i solved it in a similar fashion... but when x < -1... i forgot that x>0.
Director  Joined: 22 Mar 2011
Posts: 576
WE: Science (Education)
Re: GMAT Diagnostic Test Question 9  [#permalink]

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1
bb wrote:
GMAT Diagnostic Test Question 9
Field: number properties, fractions
Difficulty: 750
 Rating:

Is x greater than 1?

(1) $$\frac{1}{x} > -1$$
(2) $$\frac{1}{x^5} > \frac{1}{x^3}$$

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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Originally posted by EvaJager on 17 Jun 2012, 10:54.
Last edited by EvaJager on 17 Jun 2012, 11:37, edited 1 time in total. Re: GMAT Diagnostic Test Question 9   [#permalink] 17 Jun 2012, 10:54

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# GMAT Diagnostic Test Question 10

Moderator: Bunuel  