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

Question

If u and v are positive real numbers, is u>v?

1. u^3/v < 1

2. (u^(1/3))/v < 1

subhashghosh · Accepted Answer

(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

anordinaryguy · Answer

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

anordinaryguy · Answer

Yaa now I get it you are right. In the first statement, I missed the condition of (U<1, V<1)

+1 subhashghosh

144144 · Answer

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.

PK32 · Answer

what about if v is negative

Bunuel · Answer

v cannot be negative since we are told in the stem that "u and v are  positive  real numbers".

shmba · Answer

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

Saupayan · Answer

How do we prove that highlighted part?

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

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