Stats question- Distributions

Need a bit of explainign how this works
continuous random variable Y
mean=100
variance=256

Find K such that P(100-k more than /equal to Y less than eual to 100+k)
= 0.516

ans :k=11.2

Mark scheme says
p(Y less than/equal to 100 + k )= 0.516 + 1/2(1-0.516)
=0.758
P(Z less than/eual to k/16) = 0.758
and so on

i can't see how they arrived to these 2 conclusions

It also says some candidates stated
(100 +k -100) root 256
can be simplified to k/16 also

I'm not following this :s
from June 2001 paper