Mo2men wrote:If u and v are positive real numbers, is u>v?
1. u^3/v < 1
2. (u^1/3) /v < 1
Since u and v are POSITIVE, the inequalities in the statements can be simplified by multiplying each side by v.
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.
Last edited by
GMATGuruNY on Mon Mar 26, 2018 12:05 pm, edited 1 time in total.
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