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Bunuel
If A = 2B, is A^4 > B^4?

(1) A^2 = 4B^2.
(2) 2A + B < A/2 + B.

Hi..
\(A=2B.......A^4=16B^4\)...
is \(A^4 > B^4\)?
if A=B=0 ans is NO, otherwise YES

lets see the statements

(1) \(A^2 = 4B^2\).
Nothing New .. already given from A=2B
insuff

(2) 2A + B < A/2 + B
\(\frac{3A}{2}<0\)..
so A is NOT equal to 0
suff

B
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A = 2B, then the question is if 16B^4 > B^4?
B^4 is always positive and if B^4 ≠ 0, we can affirm that 16B^4 > B^4. But as B could be equal to Zero, we need to evaluate the statements

Statement 1:
A^2 = 4B^2
Well, the question stem gave the information that A = 2B so A^2 is 4B^2. We know that from the question stem and this statement doesn't help at all.

Statement 2:
2A + B < A/2 + B (rule out the Bs)
2A < A/2
2A - A/2 < 0
3A/2 < 0

So A is a negative number and ≠ than 0.
Statement 2 is sufficient to answer the question and affirm that 16B^4 > B^4!
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