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

Expert replies
Source: — Data Sufficiency |

u and v

by GMATGuruNY » Mon Mar 26, 2018 2:47 am
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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If u and v are positive real numbers, is u>v?

by Mo2men » Mon Mar 26, 2018 3:27 am
GMATGuruNY wrote:
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 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.
Thanks Mitch for you support,

I have general question to understand better the method that you used to solve the problem.

1- is not it incorrect to jump to the conclusion based on testing one condition ( v < u)?

2-You have test that v < u , what if the result of your test proved that v < u? Should I test if v > u to cover all situations? or can I conclude directly that u >v?

Thanks
Join the discussion

by GMATGuruNY » Mon Mar 26, 2018 4:43 am
Mo2men wrote:Thanks Mitch for you support,

I have general question to understand better the method that you used to solve the problem.

1- is not it incorrect to jump to the conclusion based on testing one condition ( v < u)?
The question stem asks whether v < u.
As shown in my solution above, v < u would require that u simultaneously be both a fraction and a value greater than 1.
Since it is not possible for u to be simultaneously both a fraction and a value greater than 1, we can conclude that v is NOT less than u.
Thus, the answer to the question stem is NO.
2-You have test that v < u , what if the result of your test proved that v < u? Should I test if v > u to cover all situations? or can I conclude directly that u >v?
If we were able to show that v < u is possible, we would then test whether u < v is also possible.
If both cases were possible, then the correct answer would be E.
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I unlock the best way for YOU to solve problems.

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