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TRICKY! There are n teams playing in a basketball tournam
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05 Oct 2018, 08:32
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Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?
Re: TRICKY! There are n teams playing in a basketball tournam
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07 Oct 2018, 07:36
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GreenlightTestPrep wrote:
Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?
A) 36 B) 45 C) 55 D) 66 E) 78
Let x = the total number of teams in the tournament.
There are a few ways to determine the total number of GAMES played. One approach is to recognize that the total number of games played = the total number of ways to select 2 teams (which will then play against each other) Since the order in which we select the 2 teams does not matter, we can use COMBINATIONS We can select 2 teams from x teams in xC2 ways, which equals (x)(x-1)/2 (see video below to see how we derived this) So, the total number of GAMES PLAYED = (x)(x-1)/2
Since each of the (x)(x-1)/2 games results in 1 winner, we can also say that.... ..the total number of WINS = (x)(x-1)/2
From this point on, I'll keep track of the WINS only
IMPORTANT: If there x teams, then each team plays x-1 games (since each team plays every team, other than itself)
4 teams lost exactly 5 games Each team plays x-1 games So, if a team LOST 5 games, then it must have WON the remaining games. In other words, the number of games that each team WON = (x - 1 - 5) games So, the total number of WINS from these 4 teams = (4)(x - 1 - 5)
5 teams won exactly 3 games Perfect. We're keeping track of WINS only. So, the total number of WINS from these 5 teams = (5)(3) = 15
Each of the remaining teams won all of its games We started with x teams, and we have dealt with 9 teams so far. So, the number of teams remaining = (x - 9) Each team plays x-1, so if each of the (x - 9) teams won ALL of their games, then . . . The total number of WINS from these (x - 9) teams = (x - 9)(x - 1) = x² - 10x + 9
We're now ready for our big equation! We know that: total number of wins = (x)(x-1)/2 So, we can write: (4)(x-1 - 5) + 15 + x² - 10x + 9 = (x)(x-1)/2 Expand both sides to get: 4x - 24 + 15 + x² - 10x + 9 = (x² - x)/2 Simplify left side: x² - 6x = (x² - x)/2 Multiply both sides by 2 to get: 2x² - 12x = x² - x Subtract x² from both sides: side: x² - 12x = -x Add x to both sides: side: x² - 11x = 0 Factor to get: x(x - 11) = 0 So, EITHER x = 0 or x = 11
Since it x = 0 makes NO SENSE, we can be certain that x = 11
What is the total number of games played during the tournament? We already determined that the total number of games played = (x)(x - 1)/2 So, plug in x = 11 We get: total number of games played =(11)(11 - 1)/2 = 55
Re: TRICKY! There are n teams playing in a basketball tournam
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05 Oct 2018, 22:03
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GreenlightTestPrep wrote:
Several teams are competing in a basketball tournament, and each team plays every other team once. Each game has exactly 1 winner and 1 loser (no ties). If 4 teams lost exactly 5 games, 5 teams won exactly 3 games, and each of the remaining teams won all of its games, what is the total number of games played during the tournament?
A) 36 B) 45 C) 55 D) 66 E) 78
Let the Total number of teams be = N
1st case: 4 teams lost exactly 5 games
So the no. of games Lost = \(4 * 5 = 20\)
and the no. of game won = \(4* ([N- 1] - 5) = 4 * ( N -6)\) (Here remaining team = N-1)
Case 2:: 5 teams won exactly 3 games
So the no. of games Lost = \(5 *([N-1] - 3)= 5 * ( N - 4)\)
and the no. of games won = \(5 * 3 = 15\)
Case 3 : each of the remaining teams won all of its games
So the no. of games won = \(N * (N-1)(N - 9)\)
There are no matches lost for the remaining teams.
Therefore,
\(20 + 5 * ( N - 4) = 4 * ( N -6) + 15 + N * (N-1)(N - 9)\)
or \(5N = N^2 - 6N\)
or N^2 - 11N =0
so N cannot be = 0 , so N = 11
Therefore the total no. of teams =\(\frac{{N(N-1)}}{2} = \frac{{11 * 10}}{2} = 55\)
Re: TRICKY! There are n teams playing in a basketball tournam
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06 Oct 2018, 01:52
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I’m no expert , this method I feel is not optimised
Trick lies in the statement exactly 1 win or loss.
Therefore , no of losses = no of wins ..............(1)
*n(n - 1)/2 = no of matches played , solving options we get C equals 110 , therefore each team will play a total of 10 matches
Since it is given that 4 teams lost exactly 5 games , thus, we have (4*5) 20 losses and to compensate these losses we have 20 wins
Similarly, 5 teams won exactly 3 games , we have (5*3) 15 wins and (5*7) 35 losses
No of losses = 55 No of wins = 35
Now, rest of the teams won all the games and not a single game was lost so total losses equals 55 . Therefore , 20 must be added to no of wins to make both the above quantities equal.
Re: TRICKY! There are n teams playing in a basketball tournam
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30 Oct 2022, 07:50
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I think there is a flaw in the argument of those reasoning for no of teams = 11. There are two important statements given: (1) "each team plays every other team once" (2) "each of the remaining teams won all of its games" Then, it is logically necessary that we add only one team to the 9 teams already mentioned. Because if we added two new teams, they would also have to play against each other, and one of them would lose. Hence, if each of the remaining teams won all of their games, there can only be one remaining team. Only one team can win all of their games!
TRICKY! There are n teams playing in a basketball tournam
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25 Mar 2023, 14:19
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This question is impossible as posed.
Say N is the total number of teams, A refers to the 5 '3 wins' teams, and B refers to the 4 '5 losses' teams. Also note # of wins must equal # of losses, bc each win comes with a loss, and there can only ever be 1 remaining team if they are to win all their games, bc 2+ teams can't play all the other teams and win them all.
Case 1 - N <= 8: This is impossible because if A and B overlap (some A teams have also lost 5), the total games they played = 8, but there aren't 8 other teams to play. if N <= 8 (Games played per team = N - 1)
Case 2 - N = 9: Assume no remaining teams, so set A and B don't overlap. Games played by each team = N-1 = 8, so Set A would have lost 5 of their 8 if they won 3, but that means more than 4 lost 5 games. Assume 1 remaining team, the same problem above results.
Case 3: - N = 10: Note there can only be 1 remaining team, so this is the last case. The table looks like this #: 1 2 3 4 5 6 7 8 9 10 L: 5 5 5 5 x x x x x 0 W: y y y y 3 3 3 3 3 9
x = 6, and y = 4. bc they each played 9 games. Total wins = 4(4)+5(3) +9 = 40, and total losses = 4(5) + 5(6) + 0 = 50. Doesn't work.