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If u and v are positive real numbers, is u>v? 1. u^3/v
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16 Jul 2011, 06:17
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If u and v are positive real numbers, is u>v? 1. u^3/v < 1 2. (u^(1/3))/v < 1
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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16 Jul 2011, 07:01
(1) u^3/v < 1 => u^3 < v (Because v is positive, we can multiply both sides by v) If u = 2 and v = 9, then the condition holds and u < v If u = 1/2 and v = 1/3, then the condition holds and u > v Insufficient (2) (u)^1/3 < v If u = 8 and v = 3 then the condition holds and u > v If u = 8 and v = 9, then the condition holds and u < v Insufficient (1) + (2) u^3 < v and u < v^3 (By cubing both sides) This is only possible only when u < v Sufficient Answer  C
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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16 Jul 2011, 06:50
vivgmat wrote: If u and v are positive real numbers, is u>v?
1. u^3/v < 1
2. (u^1/3) /v < 1 First Statement can be written as u^3 < v, sufficient; A cube of one number (u) can be lesser than another number (v) only if u<v Second statement can be written as u^1/3 < v, insufficient; can be tested with ( u=9, v=5), (u=9, v=10) I would go with A
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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16 Jul 2011, 07:13
Yaa now I get it you are right. In the first statement, I missed the condition of (U<1, V<1) +1 subhashghosh
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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31 Jul 2011, 09:46
Guys, whats the most efficient way to conclude that: u^3 < v and u < v^3 (By cubing both sides) This is only possible only when u < v thanks.
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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24 Apr 2016, 18:45
what about if v is negative



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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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24 Apr 2016, 21:41



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Re: If u and v are positive real numbers, is u>v? 1. u^3/v
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26 Aug 2018, 01:08
144144 wrote: Guys, whats the most efficient way to conclude that: u^3 < v and u < v^3 (By cubing both sides)
This is only possible only when u < v
thanks. A > B > 0 and P > Q > 0. Then, as with addition, we can multiply inequalities with the same direction: A*P > B*Q must be true. And, as with subtraction, we can divide inequalities with the opposite direction: A/Q > B/P. Again, remember the caveat: everything must be positive for these patterns to work. If anything can be negative, things get much more complicated, so complicated that the GMAT won’t ask about them. >>>from Magooshin this case, you can conclude u^4<v^4Square Root Property Taking a square root will not change the inequality (but only when both a and b are greater than or equal to zero). If a ≤ b then √a ≤ √b (for a,b ≥ 0) in this case,you can say u<v.




Re: If u and v are positive real numbers, is u>v? 1. u^3/v &nbs
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26 Aug 2018, 01:08






