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Tough DS question ...explanations please

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Source: — Data Sufficiency |
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by luckylucky » Fri Jan 09, 2009 5:40 am
source for the question is Manhattan study guide
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by maihuna » Fri Jan 09, 2009 6:32 am
take it this way:

if u is between 0 and 1, u^3 will be even lesser and u^1/3 more, for example: if u = 1/8 , u^3 = 1/24, and u^1/3 = 1/2

But if u>1, always u^3>u and u^1/3<u, like if u = 8, u^3= 512, u^1/3 = 2

Putting this info in perspective:

I) If u^3/v<1 we know u^3<v but u<v, we cant say so, as seen above ex: u = 1/2, v = 1/4, so u^3/v = 4/8 = 1/2 so u^3<v but u>v
u = 2, v = 12, u^3/v<1, u<v so Insuff

2) similarly 2 could be proved insuff

On combination: u^3/v <1 so u^3<v,
u^1/3/v<1 so u^1/3<v

If both are less than v it means u is not between 0 and 1, for such scenario u<v ans so C
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Junior | Next Rank: 30 Posts
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by luckylucky » Sat Jan 10, 2009 4:34 am
thanks dude ..OA is C
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