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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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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.

Therefore answer is 55 i.e option C
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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nomomuffins wrote:
*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




Can you plz check how you got 10 after solving, I believe it should be 11
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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pranab01 wrote:
nomomuffins wrote:
*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




Can you plz check how you got 10 after solving, I believe it should be 11


Total 11 teams , each team plays only one match with other , therefore each team plays a total of 10 matches

And thus 11*10=110 matches in all
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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nomomuffins wrote:

Total 11 teams , each team plays only one match with other , therefore each team plays a total of 10 matches

And thus 11*10=110 matches in all


Right, just got skipped :)
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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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!
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TRICKY! There are n teams playing in a basketball tournam [#permalink]
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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.

Thus, the problem is impossible
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
pranab223 wrote:
Case 3 : each of the remaining teams won all of its games
So the no. of games won = N * (N-1)(N - 9)

I think you made a small mistake in the explanation here. The number of games won should just be = (N-1)(N-9) correct?
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
Let x be number of teams, therefore number of games


x(x - 1)/2


Now according to the question


4 teams = 5 losses

5 teams = 3 wins

Since each game has exactly 1 winner and 1 loser, number of wins must equal number of losses

So there are 2 teams left


Number of teams = 4 + 5 + 2 = 11


11x 10/2 = 55

Answer C

Adewale Fasipe, quant instructor from Lagos Nigeria.
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Re: TRICKY! There are n teams playing in a basketball tournam [#permalink]
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