IPMAT 2021 - Quantitative Aptitude Question 26
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IPMAT 2021 - Quantitative Aptitude
If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined
Each Draw is reported twice as in eg:-
If E vs F → Draw then both E & F get 1 point in draw
Number of matches resulted in draw = 12/6 = 6 matches