DIRECTIONS for the question: Solve the following question and mark the best possible option.

In a cricket match ,five batsmen A, B, C, D and E. scored an average of 36 runs. D scored 5 more than E; E scored 8 fewer than A; B scored as many as D and E combined; and B and C scored 107 between them. How many runs did E score?

Let us assume that B scored B runs. 
Runs of B and C = 107, so C scored 107 - B runs.
B = D + E where D = E + 5. So, B = 2E + 5 and E = (B-5)/2.
D = (B+5)/2.
E = A-8, so A = E + 8 ==> (B+11)/2.
sum of runs of A, B, C, D and E = $$\frac{B+11}{2}+B+\left(107-B\right)+\frac{B+5}{2}+\frac{B-5}{2}=\frac{3B}{2}+112.5$$.
As their average is 36, the sum of their runs is also 5*36 = 180.
So, $$180=\frac{3B}{2}+112.5$$ ==> B = 45.
Runs scored by E = (B-5)2 = 20.