Bunuel wrote:

Is a^2 > 3a – b^4?

(1) 3a – b^4 = -5

(2) a > 5 and b > 0

Kudos for a correct solution.

Question:

Is \(a^2 > 3a - b^4?\)

Solution:

Is \(a^2 > 3a - b^4\)?

→ question becomes: Is \(a^2 + b^4 -3a > 0\)?

1) \(3a – b^4 = -5 → b^4 – 3a = 5.\) Substituting in inequality above, we obtain:

Is \(a^2 + 5 > 0?\) Since \(a^2\) is always positive and 5 is a positive number, their sum will always be greater than zero. Hence, statement I is sufficient.

2) a>5 and b>0.

Let a = 5. →\(a^2 = 25.\)

Let \(b=1 → b^4 = 1 → 3a = 15\)

→ Is 25 > 15-1 ? → Is 25 > 14? Answer is yes. This is an extreme case as a is always greater than 5. Integer a could be 5.001 for example.

With any value of a and b chosen that satisfy above condition in statement 2, inequality always hold. Hence, statement 2 is sufficient.

Answer: D.