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# D01-10

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Intern
Joined: 18 Jun 2017
Posts: 13
D01-10  [#permalink]

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14 Feb 2018, 22:57
can we take x not equal to zero for granted in such questions? If yes, then here is the simple way of solving it

1/x >-1; x>-1 {A alone is not sufficient}
1/x^5 >1/x^3; X^3 > X^5 which is possible only is X<1. { so B alone is sufficient ..Simple}
Intern
Joined: 08 Aug 2017
Posts: 27
Re: D01-10  [#permalink]

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04 Mar 2018, 11:56
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.
Thanks in advance.
Math Expert
Joined: 02 Sep 2009
Posts: 52429
Re: D01-10  [#permalink]

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04 Mar 2018, 12:13
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.
Thanks in advance.

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.
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Manager
Status: EAT SLEEP GMAT REPEAT!
Joined: 28 Sep 2016
Posts: 167
Location: India
Re: D01-10  [#permalink]

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30 May 2018, 09:32
What sort of values shall we use in such type of questions?

Can you suggest Bunuel

Bunuel wrote:
Is $$x$$ greater than 1?

(1) $$\frac{1}{x} \gt -1$$

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

_________________

Regards,
Adi

Intern
Joined: 28 Jul 2018
Posts: 6
Re: D01-10  [#permalink]

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08 Sep 2018, 16:15
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
Intern
Joined: 19 Mar 2018
Posts: 1
Re: D01-10  [#permalink]

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19 Sep 2018, 03:20
This is probably elementary in nature, but why do you have to take cases A and B? x > 0 and x < 0?
Manager
Joined: 09 Jun 2014
Posts: 218
Location: India
Concentration: General Management, Operations
Schools: Tuck '19
D01-10  [#permalink]

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29 Oct 2018, 01:26
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
Joined: 02 Feb 2018
Posts: 36
D01-10  [#permalink]

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31 Oct 2018, 03:10
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
D01-10 &nbs [#permalink] 31 Oct 2018, 03:10

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# D01-10

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