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Joined: 10 Apr 2015
Posts: 6218
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TRICKY! There are n teams playing in a basketball tournam
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05 Oct 2018, 08:32
05 Oct 2018, 08:32
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Question Stats:
32% (02:00) correct
67% (02:29) wrong based on 49 sessions
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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?
A) 36
B) 45
C) 55
D) 66
E) 78
A) 36
B) 45
C) 55
D) 66
E) 78
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: TRICKY! There are n teams playing in a basketball tournam
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07 Oct 2018, 07:36
07 Oct 2018, 07:36
6
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
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
Answer: C
RELATED VIDEO FROM OUR COURSE
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Re: TRICKY! There are n teams playing in a basketball tournam
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05 Oct 2018, 22:03
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
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 =\(\frac{{N(N-1)}}{2} = \frac{{11 * 10}}{2} = 55\)
