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# square root, cube root etc..

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Joined: 20 Jul 2009
Posts: 193
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Schools: Kellogg; Ross (\$\$); Tuck
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Kudos [?]: 26 [0], given: 6

square root, cube root etc.. [#permalink]  08 Sep 2009, 03:47
00:00

Difficulty:

5% (low)

Question Stats:

50% (01:51) correct 50% (00:24) wrong based on 6 sessions
Hi peple,

Just wanted to share this with you, I got it wrong on a test prep and i find it interesting.
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square.JPG [ 23.42 KiB | Viewed 2192 times ]
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Current Student
Affiliations: ?
Joined: 20 Jul 2009
Posts: 193
Location: Africa/Europe
Schools: Kellogg; Ross (\$\$); Tuck
Followers: 2

Kudos [?]: 26 [1] , given: 6

Re: square root, cube root etc.. [#permalink]  08 Sep 2009, 03:49
1
KUDOS
The way i found to solve this is that

sqrt(4) =2

the other terms are all grather than 1, so the summ is greather than 2+1+1 = 4.

any otther shorcut?
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Joined: 28 Jul 2009
Posts: 128
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Followers: 6

Kudos [?]: 46 [2] , given: 12

Re: square root, cube root etc.. [#permalink]  08 Sep 2009, 03:58
2
KUDOS
The way i found to solve this is that

sqrt(4) =2

the other terms are all grather than 1, so the summ is greather than 2+1+1 = 4.

any otther shorcut?

I think that is the only way here.

\/4 = 2
4\/4 = 1.41
1.41 < 3\/4 < 2

Therefore, 2+1.41+1.xx > 4.

Not sure if there is any other shortcut to it!
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Kudos [?]: 110 [0], given: 10

Re: square root, cube root etc.. [#permalink]  02 May 2011, 00:01
both 4^(1/3) and 4^(1/4) are greater than 1.
hence M = 2 + 1 +1

thus greater than 4.
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Re: square root, cube root etc..   [#permalink] 02 May 2011, 00:01
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