In a round-robin tournament with 6 teams, each team plays one game aga

Question

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end on the tournament?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Archit3110 · Answer

not able to completely solve question 
total games played ; 6c2; 15
let each winning point ; 1 ; so total points if a team wins all games; 15
and out of 6 teams maximum teams which can be tied for most wins can be 5 ; if all wining 3 games at most and 1 team loosing all games.
IMO D

LevanKhukhunashvili · Answer

There will be 15 games. if we assume that the winner gets 1 point, there will be total of 15 points.

15 points can be distributed equally with 5 teams, each earning 3 points (that will be maximum points earned), the sixth team losing every game

I will bet on ans  D

arorni · Answer

Can someone please help with this?

neha283 · Answer

There are 6 teams and each team plays 5 matches, now think like this
Team A - played 5, won 5. 
Team B - played 5, won 4 (lost 1 to A)
Team C - played 5, won 3 (lost 2 to A and B)
Team D - played 5, won 2 (lost 3 to A, B and C)
Team E - played 5, won 1 (lost 4 to A, B, C and D)
Team F - played 5, won 0 (lost 5 to all teams)

Thus, maximum number of teams to win - 5.

Hope it helps.

ScottTargetTestPrep · Answer

Since 6C2 = 15, the 6 teams will play 15 games in total. Since each game results 1 team winning and 1 team team losing, the sum of wins and loses is 15 + 15 = 30. For example, if no teams have the same number of wins and loses, then the 6 teams have W-L records as 5-0, 4-1, 3-2, 2-3, 1-4 and 0-5. As you can see, if you add up all these numbers, the sum is 30.

We see that 30/6 = 5, however, we can’t have all 6 teams to have the same W-L records. That is, they can’t have all have 5-0, 4-1, or 3-2 W-L records. On the other hand, we can let the team with the worst W-L record be 0-5 and for each of the 5 remaining teams to have W-L records as 3-2.

Answer: D

Vivek1707 · Answer

One intuitive way of solving this question in under 2 mins | Otherwise Building cases would take a long time
 
***   Key thing to remember --> Since there are no ties, the number of  win results  have to be equal to  loss results .

There are a total of 6 teams ---> Each team plays 5 games

6 teams, there will be total 6 x 5 i.e.  30 game results  i.e. wins and losses and  total of 15 games  i.e. (6 x 5) /2 games played.

Now we want a situation in which the maximum number of teams can be drawn tied for the most wins

Say the max wins a team can have is 5. This only 1 team can have at max, cause this would imply that all the other teams have lost at least 1 match

Now, if the  max number of wins  for a team is  4 . We ll have the following situations

2 Teams with 4 wins, total 8 win results, hence 22 loss results. This is not possible since win results and loss results have to be equal
3 Team with 4 wins each ---> win results ---> 12, loss results 18, not possible
4 Teams with 4 wins each ---> win results ---> 16, loss results 14, not possible
5 Teams with 4 wins each ---> win results ---> 20, loss results 10, not possible

Try max number of wins as 3

5 team with 3 wins each---> win results ---> 15, loss results 15,  Possible  Hence max 5 teams can be tied for the most wins.

Regor60 · Answer

15 games played so 15 wins and 15 losses to be allocated among the six players.

The goal is to maximize the number of players who have the same winning score.

Obviously all six players can't have three wins each because that would be 18 total wins.

Can six players have two wins each ? That would equal 12 wins leaving 3 more wins to be allocated back among the six meaning that three players would now have three wins and three other players two wins.

Can we do better than having three players with three wins ?

If we allocate three wins to five players that's a total of 15 wins matching the required total.

That means one player has zero victories and five losses.

That leaves 10 more losses to be allocated among the other five players who have 3 victories each.

So, overall, five teams have a 3 and 2 record and one team has a 0 and 5 record.

In a round-robin tournament with 6 teams, each team plays one game aga

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In a round-robin tournament with 6 teams, each team plays one game aga

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards