Which team played the highest number of matches?
Answer: A
One of these 5 teams played exactly two matches and it scored the same number of runs in both the matches.
⇒ Runs scored should be an even number. Hence, E is only possible team who could have scored same number of runs both its matches i.e., 157 in each match.
One of these 5 teams played exactly two matches and both the teams that played against it scored the same number of runs.
⇒ Runs against should be an even number. Hence, D is the only possible team who could have conceded same number of runs in both its matches i.e., 157 in each match.
Now, there is a possibility that E played against D but we can’t be sure yet. One of D’s matches will definitely be with some other team.
One of teams played only one match. It cannot be E and D. Since maximum runs scored in a match can be 300, it cannot be A and B as well. Hence, C played only 1 match.
We can make the following table with the information so far.
Now, C could have played against A or B only.
Case 1: C played against A.
⇒ A scored 283 in M4.
Now A has to score (435 – 283 =) 152 more runs and it has to concede (437 – 287 =) 150 more runs.
Since a team scores at least 100 runs in any match it means A would play one more match and B would have played 3 matches.
A cannot play against D or E since A has to score 152 runs and concede 150 runs. Hence, A plays against B in M5 and D plays against E. B will play against D and E.
In M2 B scores (427 – 157 - 150 =) 120 runs.
In M2 B conceded (457 – 157 – 152 =) 148 runs.
In M1 D scores (297 – 148 =) 149 runs.
We can make the following table with the information that we have.
The team playing 3 matches i.e., B has won 1 out of 3 matches, hence it’s win rate is less than 50%. This violates the last condition and hence this case is rejected.
Case 2: C played against B.
⇒ B scored 283 in M4.
Now B has to score (427 – 283 =) 144 more runs and it has to concede (456 – 287 =) 170 more runs.
Since a team scores at least 100 runs in any match it means B would play one more match and A would have played 3 matches.
B cannot play against D or E since B has to score 144 runs and concede 170 runs. Hence, B plays against A in M5 and D plays against E. A will play against D and E.
In M2 A scores (435 - 157 – 170 =) 108 runs.
In M1 E conceded (269 – 108 =) 161 runs.
In M3 D scores (297 – 161 =) 136 runs.
We can make the following table with the information that we have.
The team playing 3 matches i.e., A has won 2 out of 3 matches, hence it’s win rate is more than 50%.
All the conditions are satisfied in this case and hence we will accept it.
∴ A played highest number of matches i.e., 3.
Hence, option (a).