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GMAT Diagnostic Test Question 10

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Re: GMAT Diagnostic Test Question 9  [#permalink]

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New post 14 Oct 2009, 12:58
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.

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New post 15 Oct 2009, 09:21
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.
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New post 26 Nov 2009, 12:28
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!
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New post 26 Nov 2009, 20:20
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.
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New post 19 Dec 2009, 12:37
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.
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New post 24 Dec 2009, 09:54
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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New post 14 Jan 2010, 20:01
stilite wrote:
Since I didn't know about this rule:

\(\frac{1}{x^5} > \frac{1}{x^3}\) 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.
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New post 24 Jan 2010, 07:11
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.

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.
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New post 25 Jan 2010, 04:21
I guess I didn't give it a thorough look like Bunuel did. I will improve. Thanks, Bunuel! +1.
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New post 02 Apr 2010, 08:47
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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New post 13 May 2010, 12:55
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.
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New post 14 May 2010, 09:17
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?
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New post 08 Jul 2010, 08:15
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
if possible please explain
Thank you
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New post 09 Oct 2010, 02:45
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.

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.


Thanks Bunuel for such a good explanation.
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New post 17 May 2011, 09:46
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]
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New post 21 Jun 2011, 21:47
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,
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New post 05 Jul 2011, 07:36
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
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New post 16 Jul 2011, 11:41
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).
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New post 17 Apr 2012, 00:24
Thx dzyubam.

i solved it in a similar fashion... but when x < -1... i forgot that x>0.
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Re: GMAT Diagnostic Test Question 9  [#permalink]

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New post Updated on: 17 Jun 2012, 11:37
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.
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Re: GMAT Diagnostic Test Question 9   [#permalink] 17 Jun 2012, 10:54

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