So we can solve my bonus question using algebra. If the distance from 4 is double the distance from 16, x can be either smaller than both, between both, or bigger than both. We already solved the between option. The smaller than both option doesn't make sense because if it's smaller than both, it must be closer to 4, so how can it be twice as far from 4 as from 16? But that would work if x is bigger than both. So we can set up an equation as follows:
x  4 = 2(x  16)
Expand:
x  4 = 2x  32
Collect terms:
28 = x
This makes sense since 28 is 24 greater than 4, and 12 greater than 16.
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