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M09-19

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Re: M09-19  [#permalink]

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New post 14 Nov 2017, 02:01
KungFuGmat wrote:
Hi Bunuel,

Many thanks for this great question. I got the solution, the way you showed it. Only today, while I was attempting another question from the GMAT Club tests, I stumbled upon the sign method for solving inequalities. I was just wondering if you or someone else could kindly explain how to solve THIS question using the sign method. Many thanks!


Can you please tell me what "sign method" are you referring to? Thank you.
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New post 14 Nov 2017, 02:38
Bunuel,

Apologies if I got the name wrong. I just learnt about it today. Don't have strong handle on it.

The discussion is here:
https://gmatclub.com/forum/inequalities ... 91482.html
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New post 14 Nov 2017, 02:57
I think VeritasPrepKarishma gave a detailed explanation on the process. Sorry, I am just learning it, so it's really rusty for me.
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New post 14 Nov 2017, 03:03
KungFuGmat wrote:
I think VeritasPrepKarishma gave a detailed explanation on the process. Sorry, I am just learning it, so it's really rusty for me.


Check for more below links:

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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M09-19  [#permalink]

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New post 19 Aug 2018, 23:12
Bunuel wrote:
Official Solution:


Is \(|x - 5| \gt 4\)?

Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \lt 0\) and thus \(x^2 \lt 4\) therefore \(|x| \lt 2\) and we can deduce that \(-2 < x < 2\), so we can have an YES answer as well as a NO answer (consider \(x=0\) and \(x=1.5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B


Hi Bunuel

if \(x^3 < 27\)

what will be the value of x?
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Re: M09-19  [#permalink]

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New post 19 Aug 2018, 23:40
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ENEM wrote:
Bunuel wrote:
Official Solution:


Is \(|x - 5| \gt 4\)?

Is \(|x-5| \gt 4\)? Opening the absolute value we get the question: is \(x \lt 1\) or \(x \gt 9\)?

(1) \(x^2-4 \lt 0\) and thus \(x^2 \lt 4\) therefore \(|x| \lt 2\) and we can deduce that \(-2 < x < 2\), so we can have an YES answer as well as a NO answer (consider \(x=0\) and \(x=1.5\)). Not sufficient.

(2) \(x^2-1 \lt 0\) thus \(x^2 \lt 1\) therefore \(|x| \lt 1\) and finally, \(-1 \lt x \lt 1\). Sufficient.


Answer: B


Hi Bunuel

if \(x^3 < 27\)

what will be the value of x?


Recall that we can always raise both parts of an inequality to an odd power and take an odd root of both sides of an inequality. So, by taking the third root from \(x^3 < 27\) we'll get x < 3.
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Re: M09-19  [#permalink]

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New post 20 Aug 2018, 03:24
[/quote]if \(x^3 < 27\)

what will be the value of x?[/quote]

Recall that we can always raise both parts of an inequality to an odd power and take an odd root of both sides of an inequality. So, by taking the third root from \(x^3 < 27\) we'll get x < 3.[/quote]

thank you Bunuel

and if \(x^3 < -27\)

then \(x < -3\)

is that correct?
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Simple and handy template for CR: https://gmatclub.com/forum/simple-and-handy-template-for-cr-242255.html

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Re: M09-19  [#permalink]

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New post 20 Aug 2018, 03:55
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ENEM wrote:
if \(x^3 < 27\)

what will be the value of x?[/quote]

Recall that we can always raise both parts of an inequality to an odd power and take an odd root of both sides of an inequality. So, by taking the third root from \(x^3 < 27\) we'll get x < 3.[/quote]

thank you Bunuel

and if \(x^3 < -27\)

then \(x < -3\)

is that correct?[/quote]

Absolutely: recall that odd roots have the same sign as the base of the root, for example: \(\sqrt[3]{125} = 5\) and \(\sqrt[3]{-64} = -4\).
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Re: M09-19   [#permalink] 20 Aug 2018, 03:55

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