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# If M = \sqrt4 + \sqrt 4 + \sqrt 4 then M is A) less than 3

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Joined: 11 Apr 2008
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If M = \sqrt4 + \sqrt 4 + \sqrt 4 then M is A) less than 3 [#permalink]

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18 Aug 2008, 14:07
If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is

A) less than 3
B) equal to 3
C) between 3 and 4
D) equal to 4
E) greater than 4

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Re: roots of 4 [#permalink]

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18 Aug 2008, 14:13
brokerbevo wrote:
If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is

A) less than 3
B) equal to 3
C) between 3 and 4
D) equal to 4
E) greater than 4

E.

This is going to be $$2 + \sqrt[3]4 + \sqrt{2}$$

We know that $$\sqrt{2}$$ is approx 1.4, so 2 + 1.4 = 3.4 and the cube root of 4 should be between the two other values, so we have 2 + 1.4 + [some value between 2 and 1.4] therefore, it must be greater than 4.
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings SVP Joined: 07 Nov 2007 Posts: 1733 Location: New York Re: roots of 4 [#permalink] ### Show Tags 18 Aug 2008, 14:14 brokerbevo wrote: If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is A) less than 3 B) equal to 3 C) between 3 and 4 D) equal to 4 E) greater than 4 sqrt(2)~1.41 2 + 2* sqrt(2) + sqrt(2) --> 2+ 2*1.4 +1.4 > 4 E. _________________ Your attitude determines your altitude Smiling wins more friends than frowning SVP Joined: 30 Apr 2008 Posts: 1841 Location: Oklahoma City Schools: Hard Knocks Re: roots of 4 [#permalink] ### Show Tags 18 Aug 2008, 14:16 $$\sqrt[3]4$$≠$$2\sqrt{2}$$ The cube root cannot be greater than the square root. x2suresh wrote: brokerbevo wrote: If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is A) less than 3 B) equal to 3 C) between 3 and 4 D) equal to 4 E) greater than 4 sqrt(2)~1.41 2 + 2* sqrt(2) + sqrt(2) --> 2+ 2*1.4 +1.4 > 4 E. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Joined: 07 Nov 2007
Posts: 1733
Location: New York
Re: roots of 4 [#permalink]

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18 Aug 2008, 14:19
jallenmorris wrote:
$$\sqrt[3]4$$≠$$2\sqrt{2}$$
x2suresh wrote:
brokerbevo wrote:
If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is

A) less than 3
B) equal to 3
C) between 3 and 4
D) equal to 4
E) greater than 4

sqrt(2)~1.41
2 + 2* sqrt(2) + sqrt(2) --> 2+ 2*1.4 +1.4 > 4

E.

you are right.. thanks for pointing out.

it should be Power 2/3 I took it as power 3/2
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Smiling wins more friends than frowning

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Joined: 30 Apr 2008
Posts: 1841
Location: Oklahoma City
Schools: Hard Knocks
Re: roots of 4 [#permalink]

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18 Aug 2008, 14:20
It's a small point in this question. Just hope it helps you in the future for another question. You still had the right answer!
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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 12 Jul 2008 Posts: 511 Schools: Wharton Re: roots of 4 [#permalink] ### Show Tags 18 Aug 2008, 15:02 brokerbevo wrote: If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is A) less than 3 B) equal to 3 C) between 3 and 4 D) equal to 4 E) greater than 4 the nth root of any number greater than 1 is always greater than 1. Therefore, E SVP Joined: 30 Apr 2008 Posts: 1841 Location: Oklahoma City Schools: Hard Knocks Re: roots of 4 [#permalink] ### Show Tags 18 Aug 2008, 15:04 That rule eliminates A & B, but doesn't really help you decide between C, D, and E. zoinnk wrote: brokerbevo wrote: If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is A) less than 3 B) equal to 3 C) between 3 and 4 D) equal to 4 E) greater than 4 the nth root of any number greater than 1 is always greater than 1. Therefore, E _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Intern
Joined: 17 Aug 2008
Posts: 19
Re: roots of 4 [#permalink]

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18 Aug 2008, 15:16
The numbers can be simplified to 2 + sqrt(2) + (2^2/3); I went the long way...sqrt(2) ~ 1.414; (2^2/3) ~ 1.6; Sum of all these terms is > 4; So, E.
Director
Joined: 12 Jul 2008
Posts: 511
Schools: Wharton
Re: roots of 4 [#permalink]

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18 Aug 2008, 20:39
jallenmorris wrote:
That rule eliminates A & B, but doesn't really help you decide between C, D, and E.

zoinnk wrote:
brokerbevo wrote:
If M = $$\sqrt4 + \sqrt[3]4 + \sqrt[4]4$$ then M is

A) less than 3
B) equal to 3
C) between 3 and 4
D) equal to 4
E) greater than 4

the nth root of any number greater than 1 is always greater than 1.

Therefore, E

I assumed that anyone would realize sqrt(4) is 2...
SVP
Joined: 30 Apr 2008
Posts: 1841
Location: Oklahoma City
Schools: Hard Knocks
Re: roots of 4 [#permalink]

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19 Aug 2008, 04:49
Sometimes people that read it might not see exactly what you're saying, and for the sake of learning, it's often best to just spell it out even when it appears obvious. For those people who do not see it as obvious aren't likely to speak out and say they don't understand. By stating what might appear to be obvious on how the rule brings us to the answer step-by-step, it helps people learn. Some users don't care to hit a 750+ GMAT score. They just want to go from a 550 to a 620 maybe and they'll be happy. We want to help those individuals as well.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Re: roots of 4   [#permalink] 19 Aug 2008, 04:49
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# If M = \sqrt4 + \sqrt 4 + \sqrt 4 then M is A) less than 3

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