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# M09-19

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Joined: 02 Sep 2009
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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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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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Joined: 15 Sep 2016
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Schools: CBS '20
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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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14 Nov 2017, 03:03
Senior Manager
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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.

Hi Bunuel

if $$x^3 < 27$$

what will be the value of x?
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If you wish to spend wisely on your gmat prep material, check my post titled: How to Spend Money On GMAT Material Wisely, link: https://gmatclub.com/forum/how-to-buy-gmat-material-wisely-tag-free-gmat-resources-236174.html

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

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19 Aug 2018, 23:40
1
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.

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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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?
_________________
If you find my post useful, please give me a kudos.

Thank you.
Regards,
ENEM

If you wish to spend wisely on your gmat prep material, check my post titled: How to Spend Money On GMAT Material Wisely, link: https://gmatclub.com/forum/how-to-buy-gmat-material-wisely-tag-free-gmat-resources-236174.html

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

simple template for more vs greater and fewer vs less: https://gmatclub.com/forum/simple-template-for-more-vs-greater-and-fewer-vs-less-242216.html
Math Expert
Joined: 02 Sep 2009
Posts: 58325

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20 Aug 2018, 03:55
1
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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# M09-19

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