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Intern  B
Joined: 28 Sep 2016
Posts: 4

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Hey, is this approach right:

Just for a:

1/x > -1 ,

if x > 0 ,

a) multiply both sides by x

x(1/x) > x(-1)

1 > -x

b) Divide by -1 and flip signs,

-1 < x

if x < 0
a) multiply both sides by x and flip signs cos x is negative

1 < -x

b) divide by -1 and flip signs

-1 > x

Same answer as with your process. Not sure if it is right though
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

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1
jjacosta37 wrote:
Hey, is this approach right:

Just for a:

1/x > -1 ,

if x > 0 ,

a) multiply both sides by x

x(1/x) > x(-1)

1 > -x

b) Divide by -1 and flip signs,

-1 < x

if x < 0
a) multiply both sides by x and flip signs cos x is negative

1 < -x

b) divide by -1 and flip signs

-1 > x

Same answer as with your process. Not sure if it is right though

1/x > -1:

1. If x > 0, then x > -1 but since we are considering the range when x > 0, then x > 0.

2. If x < 0, then x < -1.
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Intern  B
Joined: 09 Sep 2017
Posts: 1

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I think this is a poor-quality question and I don't agree with the explanation. In the case of option one. whithout solving it, by just considering the inequality, by taking inverse cant we write that x<-1?
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

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HarshineV wrote:
I think this is a poor-quality question and I don't agree with the explanation. In the case of option one. whithout solving it, by just considering the inequality, by taking inverse cant we write that x<-1?

The solutions HERE clearly show TWO possible cases for (1). Then again HERE it's explained why your way is NOT correct. This is further elaborated HERE and HERE and yet again HERE. And finally, the links to the theory which explains inequalities in detail is given HERE.

Still more:

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Manager  B
Joined: 02 Jan 2017
Posts: 71
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34 GPA: 3.41

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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I think its extremely important that explanations are student friendly, the explanations by expert Bunnel are amazing but often it feels likes those explanations assume that students have all the knowledge. We dont. The jargon of variables without some explanation can leave big gaps to fill. For this question, i felt explanation could have been MUCH MUCH easier.
Take it as constructive critique.
Manager  B
Joined: 02 Jan 2017
Posts: 71
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 650 Q47 V34 GPA: 3.41

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Statement 1)
$$\frac{1}{x}$$$$>-1$$

$$\frac{1}{x + 1}$$ $$>0$$

$$\frac{(1+x)}{x}$$ $$>0$$

Now if $$x=2$$ , then statement 2 becomes $$\frac{(1 + 2)}{2}$$ $$= 3/2$$which is greater 1
But if$$x=1/2$$ then statement 2 becomes $$\frac{1.5}{.5}$$$$>0$$ ( which is true) but less than 1

Therefore statement 1 is insufficient.

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

$$\frac{1}{x^5}$$ - $$\frac{1}{x^3}$$ >0

$$\frac{(x^3 -x^5)}{x^5x^3}$$ $$>0$$

$$\frac{x^3[1-x^2]}{x^5x^3}$$ $$>0$$

$$\frac{(1 - x^2)}{x^5}$$$$>0$$

Now Only way expression above is $$>0$$ when $$x^2$$is <1 therefore $$-1< x < 1$$
Which answers the question $$x>1$$ ? as NO. Hence Statement 2 is sufficient.
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

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mtk10 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I think its extremely important that explanations are student friendly, the explanations by expert Bunnel are amazing but often it feels likes those explanations assume that students have all the knowledge. We dont. The jargon of variables without some explanation can leave big gaps to fill. For this question, i felt explanation could have been MUCH MUCH easier.
Take it as constructive critique.

Dear friend,

It's assumed that a student is not attempting questions, especially harder ones, and especially GMAT Club's hard questions, if the fundamentals are not clear. Solution for this particular questions includes all necessary steps and cannot possibly include underline theory. Still, if the solution is not clear you can always ask specific questions (will be glad to assist) and/or read the whole thread, often there you can find missing gaps.

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

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Intern  B
Joined: 31 Mar 2017
Posts: 4

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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hi,
How did we go from (x+1)/x>0 to x+1>0 because we cant multiply both sides by x as we dont know whether x is positive or negative.
Best Regards.
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

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Furqan90 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hi,
How did we go from (x+1)/x>0 to x+1>0 because we cant multiply both sides by x as we dont know whether x is positive or negative.
Best Regards.

For $$\frac{1+x}{x} \gt 0$$ we should consider two cases:

Case A - both numerator and denominator are positive: $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

Case B - both numerator and denominator are negative: $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

Can you please tell me which step is unclear?
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Intern  B
Joined: 08 Aug 2017
Posts: 31

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Hi Bunuel

I have a doubt in Statement 1 for the case x<0.

I got this,
for x<0, 1/(-x)>-1, so when we multiply by -1 and flip signs,we get 1/x<1, so x>1.

Can you please explain how you got 1+x<0 for x<0.
Please correct me. I am doing something wrong here. Is there any concept file to know more about inequalities? I was not able to find any chapter in GMATClub Maths book.
Math Expert V
Joined: 02 Sep 2009
Posts: 56307

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1
pantera07 wrote:
Hi Bunuel

I have a doubt in Statement 1 for the case x<0.

I got this,
for x<0, 1/(-x)>-1, so when we multiply by -1 and flip signs,we get 1/x<1, so x>1.

Can you please explain how you got 1+x<0 for x<0.
Please correct me. I am doing something wrong here. Is there any concept file to know more about inequalities? I was not able to find any chapter in GMATClub Maths book.

If x is negative, then x denotes negative quantity, no need to replace x with -x.

9. Inequalities

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________
Intern  B
Joined: 28 Jul 2018
Posts: 6

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I didn't see this approach for Statement 2 posted yet:
- multiply both sides by x^4, which is possible because any number to an even power is positive.
- this results in: (1/x) > (x)
- plug in numbers between inf, -1, 0, 1, and inf -- I always start on the positive side, so as soon as I tried >1, I got the answer
Senior Manager  P
Joined: 09 Jun 2014
Posts: 341
Location: India
Concentration: General Management, Operations

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banmai8x96 wrote:
I think this is a high-quality question and I agree with explanation.

Yes.I think too its a high quality question.

However another approach that I found useful was to to test it with numbers.

STAT 1:

Check at X=1/2 for which value of expression X>1= FALSE
Check at X=2 for which value of expression X>1= TRUE

so insuffcient!!

STAT2:

The denominator of LHS is of higher power than greater tan the deno at right side..

so for number greater than 1 it will always be false ..That range will not be apart here only.

and will be definite NO.

Press Kudos if you like the post Intern  B
Joined: 02 Feb 2018
Posts: 34

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dharan wrote:

Solving this way consumes more time. We need to pick right numbers. And if we miss any of it, we dont get correct values.

Using inequality wavy curve method you can solve much early in a generic way [ w/o picking the numbers ]

Please search for wavy curve method in this forum or in google.

Hi there,

I understand the algebraic approach by Bunuel but I can't find my mistake using the wavy line method to test statement (2). When I draw $$\frac{(1−x^2)}{x^5} > 0$$ or $$\frac{(1+x)(1-x)}{x^5} > 0$$ I get the zero points -1, 1, and 0 (has to be excluded from final solution set) and as a result $$x>1$$ and $$-1<x<0$$.

I just discovered this method yesterday through the excellent post by EgmatQuantExpert so I'm sorry if this comes across as a silly question Thanks!

Edit: I found my mistake: Make sure that the factors are of the form (ax - b), not (b - ax) -> $$\frac{(1−x^2)}{x^5} > 0$$ (=) $$\frac{( x^2-1 )}{x^5} < 0$$ and then I get the right answer
Manager  G
Joined: 22 Jun 2017
Posts: 178
Location: Argentina
Schools: HBS, Stanford, Wharton
GMAT 1: 630 Q43 V34 ### Show Tags

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Is there any other way to analyze the second statement?
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The HARDER you work, the LUCKIER you get.
Manager  S
Joined: 12 Jul 2017
Posts: 188
GMAT 1: 570 Q43 V26 GMAT 2: 660 Q48 V34 ### Show Tags

Hi Bunuel,
Please let me know whether my understanding for deriving range for option 2 is proper.
Given:
(1/x^5) > (1/x^3)
if x < 0
x^5 < x^3 ( as we are taking reciprocal, signs will be reversed)

x^5 - x^3 < 0
x^3 (x^2-1) < 0
this gives us range x < -1

and if x > 0
given (1/x^5) > (1/x^3)
x^5 < x^3
here again x^3 (x^2-1) <0 gives us range 0 <x < 1

so our final range is x < -1 or 0 < x < 1.

Is this proper?

Regards,
Rishav
Manager  S
Joined: 26 Nov 2018
Posts: 99
Concentration: Technology, Entrepreneurship
GPA: 3.3
WE: Manufacturing and Production (Manufacturing)

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Bunuel wrote:
Official Solution:

(1) $$\frac{1}{x} \gt - 1$$. Rewrite as $$\frac{1+x}{x} \gt 0$$, two cases:

A. $$x \gt 0$$ and $$1+x \gt 0$$, ($$x \gt -1$$). So, $$x \gt 0$$;

B. $$x \lt 0$$ and $$1+x \lt 0$$, ($$x \lt -1$$). So, $$x \lt -1$$.

So, the given inequality holds true in two ranges: $$x \gt 0$$ and $$x \lt -1$$, thus $$x$$ may or may not be greater than 1. Not sufficient.

(2) $$\frac{1}{x^5} \gt \frac{1}{x^3}$$. Rewrite as $$\frac{1-x^2}{x^5} \gt 0$$, two cases:

A. $$x \gt 0$$ (it's the same as $$x^5 \gt 0$$) and $$1-x^2 \gt 0$$, ($$-1 \lt x \lt 1$$). So, $$0 \lt x \lt 1$$;

B. $$x \lt 0$$ and $$1-x^2 \lt 0$$, ($$x \lt -1$$ or $$x \gt 1$$). So, $$x \lt -1$$;

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

Hallo Bunuel,

At a glance, statement 1 satisfy.
considering statement 1, 1/X > -1.

While reversing, it will be X< -1. Considering this statement 1 satisfy.

I have understood your explanation. But when will I understand that it should be 1+X/X > 0 ?
Intern  B
Joined: 02 Jul 2018
Posts: 24

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Hi Bunuel, VeritasKarishma or anyone: I tried and tried and tried but I am not able to comprehend how is B Sufficient. I have tried to solve it via Wavy Curve Method, can you please help me understand what I am doing wrong
>> !!!

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Happy to receive feedback on my post. Please give Kudos if you find my content interesting. Veritas Prep GMAT Instructor D
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Posts: 9449
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1
amanrsingh wrote:
Hi Bunuel, VeritasKarishma or anyone: I tried and tried and tried but I am not able to comprehend how is B Sufficient. I have tried to solve it via Wavy Curve Method, can you please help me understand what I am doing wrong

Your stmnt 2 analysis is not correct.

$$1/x^5 > 1/x^3$$

$$1/x^5 - 1/x^3 > 0$$

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

$$(x^2 - 1)/x^5 < 0$$ (Multiply both sides by -1)

$$(x + 1)(x - 1)/x^5 < 0$$ (Note that to use the wavy line method as discussed, we must have factors in the form (ax + b) or (ax - b), not (b - ax))

Transition points, -1, 0, 1

The expression is negative when 0 < x< 1 or x < -1

So in any case, x is less than 1.
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Veritas Prep GMAT Instructor

Intern  B
Joined: 02 Jul 2018
Posts: 24

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amanrsingh wrote:
Hi Bunuel, VeritasKarishma or anyone: I tried and tried and tried but I am not able to comprehend how is B Sufficient. I have tried to solve it via Wavy Curve Method, can you please help me understand what I am doing wrong

Your stmnt 2 analysis is not correct.

$$1/x^5 > 1/x^3$$

$$1/x^5 - 1/x^3 > 0$$

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

$$(x^2 - 1)/x^5 < 0$$ (Multiply both sides by -1)

$$(x + 1)(x - 1)/x^5 < 0$$ (Note that to use the wavy line method as discussed, we must have factors in the form (ax + b) or (ax - b), not (b - ax))

Transition points, -1, 0, 1

The expression is negative when 0 < x< 1 or x < -1

So in any case, x is less than 1.

Thank You so much, this was really helpful, will keep this point in mind _________________
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Happy to receive feedback on my post. Please give Kudos if you find my content interesting.  Re: D01-10   [#permalink] 07 Jul 2019, 11:26

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# D01-10

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