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

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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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36% (03:12) correct 63% (01:18) wrong based on 1 sessions
If u and v are positive real numbers, is u>v?

1. u^3/v < 1

2. (u^1/3) /v < 1
[Reveal] Spoiler: OA
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Re: If u and v are positive real numbers, is u>v [#permalink] New post 16 Jul 2011, 07:50
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 [#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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Manager
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Re: If u and v are positive real numbers, is u>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 [#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   [#permalink] 31 Jul 2011, 10:46
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