Since u and v are POSITIVE, the inequalities in the statements can be simplified by multiplying each side by v.Mo2men wrote:If u and v are positive real numbers, is u>v?
1. u^3/v < 1
2. (u^1/3) /v < 1
Statement 1, rephrased: u³ < v
Case 1: u=1 and v=2
In this case, u < v, so the answer to the question stem is NO.
Case 2: u=1/2 and v=1/3
In this case, u > v, so the answer to the question stem is YES.
INSUFFICIENT.
Statement 2, rephrased: ∛u < v
Case 1: u=1 and v=2
In this case, u < v, so the answer to the question stem is NO.
Case 3: u=8 and y=7
In this case, u > v, so the answer to the question stem is YES.
INSUFFICIENT.
Statements combined:
Test whether it's possible for both statements to be satisfied if v < u.
Adding v < u to Statement 1 yields the following:
u³ + v < u + v
u³ < u.
Here, u must be a FRACTION.
Adding v < u to Statement 2 yields the following:
∛u + v < u + v
∛u < u.
Here, u must be GREATER THAN 1.
Since it is not possible for u simultaneously to be both a fraction and a value greater than 1, the two statements cannot both be satisfied if v < u.
Implication:
For both statements to be satisfied, u must NOT be greater than v.
Thus. the answer to the question stem is NO.
SUFFICIENT.
The correct answer is C.
