A, B, C and D were the members of a team. The average runs

Solution:

Let us assume the average of A, B, C and D be M. Thus we can say \(\frac{A+B+C+D}{4}=M⇒A+B+C+D=4M.....(i)\)

We are said that this average decreased when E with a score of 45 joins, thus we can say \(\frac{A+B+C+D+E}{5}=\frac{A+B+C+D+45}{5}=M-2\)
\(⇒A+B+C+D+45=5M-10\)\(\)
\(⇒4M+45=5M-10\)
\(⇒M=55\)
\(⇒A+B+C+D=4M=220.....(ii)\)

We are told that A and B scored in a ratio \(13:12\). So, let us assume \(A=13x\) and \(B=12x\)

We also know that \(45<A<65\) or \(45<13x<65\) or \(3.4<x<5\) which means \(x= 4\)

and \(A=13x=52\) and \(B=12x=48\)

We are asked the ratio of B to average of C and D

Average of C and D \(= \frac{C+D}{2}=\frac{220-A-B}{2}\) (from (ii))
\(=\frac{220-52-48}{2}\)
\(=\frac{120}{2}\)
\(=60\)

Thus, the required ratio \(=\frac{B}{60}=\frac{48}{60}=\frac{4}{5}\)

Hence the right answer is Option A