Question 1023101
Let the numbers be a,b,c and d.
Average of first 3 numbers is double the 4th
(a+b+c)/3 = 2d => (a+b+c) = 6d -----------eq.1

Avg. of all 4 numbers is

(a+b+c+d)/4 = 28 => /(a+b+c+d) = 28*4 = 112 ----eq.2
Put eq.1 in eq.2

6d + d = 112
7d = 112
d = 112/7

d = 16
So, the 4th number is 16