Read the following data carefully and answer the questions given below. The data shows the three different people (A, B & C) win or lose multiple games.

Total games win by B is 50 and the ratio of game win by A and games lose by B is 3:2 respectively. The ratio of total games loses by A to total games win by A is 4:3. The average of number of games lose by A and win by B is 105. Total games win by C is 20% more than total games loss of B and the ratio of total games win to lose by C is 4 : 5.

Find the average of total games win by A & B and total games lose by C.

The correct option is 96.67.

Let us analyze the given information step-by-step to find the required values for people A, B, and C:

1. Data for Person B:

Total games won by B = 50.

We are given that the average of the number of games lost by A and games won by B is 105.

$\frac{\text{Games lost by A}+\text{Games won by B}}{2}=105$

Games lost by A + 50 = 210
Games lost by A = 210 - 50 = 160.

2. Data for Person A:

The ratio of total games lost by A to total games won by A is 4 : 3.

Since games lost by A = 160:

$\frac{4}{3}=\frac{160}{\text{Games won by A}}$

Games won by A = $\frac{160\times 3}{4}=120$

3. Data for Person B (Games lost):

The ratio of games won by A and games lost by B is 3 : 2.

Since games won by A = 120:

$\frac{3}{2}=\frac{120}{\text{Games lost by B}}$

Games lost by B = $\frac{120\times 2}{3}=80$

4. Data for Person C:

Total games won by C is 20% more than the total games lost by B.

Games won by C = 80 + (20% of 80) = 80 + 16 = 96.

The ratio of total games won to lost by C is 4 : 5.

$\frac{4}{5}=\frac{96}{\text{Games lost by C}}$

Games lost by C = $\frac{96\times 5}{4}=120$

5. Finding the Required Average:

We need to find the average of total games won by A & B and total games lost by C.

Games won by A = 120
Games won by B = 50
Games lost by C = 120

$\text{Required Average}=\frac{120+50+120}{3}$

$\text{Required Average}=\frac{290}{3}\approx 96.67$

Hence, the average of total games won by A & B and total games lost by C is 96.67.