If x ≠ 0 and the average (arithmetic mean) of set A is twice the avera

There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}

This means the average of set A is 2.5x
And the average of set B is 3x²

If the average (arithmetic mean) of set A is twice the average of set B, we can write: 2.5x = 2(3x²)
Simplify: 2.5x = 6x²
To make things easier on ourselves, let's multiply both sides of the equation by 2 to get: 5x = 12x²
Subtract 5x from both sides: 0 = 12x² - 5x
Factor: 0 = x(12x - 5)
So, either x = 0 or 12x - 5 = 0

Since we are told x ≠ 0, it must be the case that 12x - 5 = 0
Add 5 to both sides: 12x = 5
Divide both sides by 12 to get: x = 5/12

Answer: B