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If u and v are positive real numbers, is u>v? 1. u^3/v

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vivgmat
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If u and v are positive real numbers, is u>v? 1. u^3/v [#permalink] New post  16 Jul 2011, 07:17
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42% (02:29) correct 58% (02:12) wrong based on 432 sessions
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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 [#permalink] New post  16 Jul 2011, 08:01
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(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 [#permalink] New post  16 Jul 2011, 07:50
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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 [#permalink] New post  16 Jul 2011, 08:13
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 [#permalink] New post  31 Jul 2011, 10:46
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 [#permalink] New post  24 Apr 2016, 19:45
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 [#permalink] New post  24 Apr 2016, 22:41
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prabhanshukumar32 wrote:
what about if v is negative


v cannot be negative since we are told in the stem that "u and v are positive real numbers".
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v [#permalink] New post  26 Aug 2018, 02:08
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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 Magoosh
in this case, you can conclude u^4<v^4

Square 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.
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v [#permalink] New post  02 Sep 2023, 09:34
subhashghosh wrote:
(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



How do we prove that highlighted part?
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Re: If u and v are positive real numbers, is u>v? 1. u^3/v [#permalink]
02 Sep 2023, 09:34
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