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Re: 10 teams (A, B, C, D, E, F, G, H, I and J) participate [#permalink]
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GreenlightTestPrep wrote:
10 teams (A, B, C, D, E, F, G, H, I and J) participate in a soccer tournament in which each team plays every other team once.
For each game played, points are awarded to the teams as follows:
    0 points for losing the game
    1 point each or tying the game
    2 points for winning the game

After all the games are played, we learn that:
- teams A, B, C, D and E each won 5 games and tied 3 games.
- teams F, G, H and I each tied 2 games and lost 5 games.

How many games did team J lose?
A) 0
B) 2
C) 4
D) 6
E) 8


Hi shad1968

Method would be based on the logic that the games won should be equal to games lost..

So

- teams A, B, C, D and E each won 5 games and tied 3 games.
They played 9 matches each so each lost 9-(5+3)=1 game
Games won 5*5=25
Games lost 1*5=5
Games drawn 3*5=15

- teams F, G, H and I each tied 2 games and lost 5 games.
They played 9 matches each so each won 9-(2+5)=2 games
Games won 2*4=8
Games lost 4*5=20
Games drawn 2*4=8

Total games won = 25+8=33 and total games lost = 5+20=25
But games lost=won ..33=25+?
So ? Is the number of games lost by the 10th team that is J
?=33-25=8

If you want to see the effect of 9th game, it will be draw because total draws are 15+8=23 but draw has to be an even number as two teams have one draw..
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10 teams (A, B, C, D, E, F, G, H, I and J) participate [#permalink]
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Shad1968 wrote:
GreenlightTestPrep wrote:
10 teams (A, B, C, D, E, F, G, H, I and J) participate in a soccer tournament in which each team plays every other team once.
For each game played, points are awarded to the teams as follows:
    0 points for losing the game
    1 point each or tying the game
    2 points for winning the game

After all the games are played, we learn that:
- teams A, B, C, D and E each won 5 games and tied 3 games.
- teams F, G, H and I each tied 2 games and lost 5 games.

How many games did team J lose?
A) 0
B) 2
C) 4
D) 6
E) 8


Method ?


A simpler method.

There are 45 games in total.
10 teams each played 9 opponents. Divide by 2 to remove the duplicate counts. ---> \((10 * 9) / 2\)

A, B, C, D, E all won 5 and tied 3. This gives record of only 8 games for these five teams. This means that the 9th game for each of the remaining team was lost by each team. Therefore,

5 teams each won 5 games (5*5 = 25 games)
5 teams each tied 3 games (5*3 = 15 games)
5 teams each lost 1 game (5*1 = 5 games)


Teams A-E
W- 25
T - 15
L - 5



Now for teams F,G,H,I.
4 teams each tied 2 games (4*2 = 8)
4 teams each lost 5 games (4*5 = 20)
5 teams each won 2 games (5*2 = 10)


Teams F-I
W-10
T - 8
L - 20


Combining data for teams A-I.
W - 33 (25+10)
T - 23 (15+8)
L - 25 (5+20)


Note: each game had two outcomes. (W or L) / T
If 23 were tied then the rest of the 22 (45 - 23) games resulted in W / L. (The number of won and lost game would always be the same. A game won by one team is lost by another)
Since according to the combined data, 33 were won and only 25 were lost, so 8 games were also lost. The only team not mentioned above is team J.

So J lost 8 games.

Answer E.
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10 teams (A, B, C, D, E, F, G, H, I and J) participate [#permalink]
1
Team----- A B C D E F G H I J
Wins----- 5 5 5 5 5 2 2 2 2
Draws---- 3 3 3 3 3 2 2 2 2
Losses--- 1 1 1 1 1 5 5 5 5

Okay so we were given information of 8 (A,B,C,D,E) & 7 (F,G,H,I) games and each team played 9 games. So, A-E won 5 drew 3 means they lost 1, while F-I drew 2 and lost 5 each they must have won 2 each. Since, 9 games are played all teams from A-E must have lost their final game which adds 1 into the table above. While, teams F-I played 7 games with 2 games unaccounted for which must have been wins, hence 2s are added in the table above. Now it should be obvious that the number of wins and losses should be equal. Meaning if 20 matches were won by teams, 20 matches must have been lost by teams. It cannot be that the total wins is greater or lesser than the matches lost. Adding the row of wins we get = 33 wins. While the loss row adds up to 25. Wins - Loss = 33-25 = 8 losses are unaccounted for. Since, the only team we have no data on is J it is safe to assume that the remaining 8 matches lost were team J which is the answer to this question: E

P.S. Just because the question is difficult/complex does not mean the answer will be the same. Took me less than 2 min to come up with this.
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10 teams (A, B, C, D, E, F, G, H, I and J) participate [#permalink]
1
Easy way

Number of wins must equal number of losses.
Also Total game per team =9 i.e. 10 - 1

First 5teams
Won 5*5= 25games
Lost 5*(9-5-3)= 5*1= 5games

Second set of 4 teams
Won 4*(9-2-5)=4*2=8games
Lost 4*5=20games.

Team J can only lose to other teams, so it will make up for the difference in losses not already accounted for i.e. Wins by other teams - losses accounted

= (25+8)-(20+5)=8games it must lose to balance for wins by the other 9 teams.

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