M18-06

Question

Teams A, B, C, and D are participating in a tournament where each team plays against every other team once. A victory earns a team 3 points, a draw earns 1 point, and a defeat earns 0 points. The current standings are as follows:
 
Team B: 7 points;
 
Team C: 4 points;
 
Team A: 0 points;
 
Team D: 0 points;
 
How many games are yet to be played in the tournament?

A. 0
B. 1
C. 2
D. 3
E. it cannot be determined from the information given

Bunuel · Answer

Official Solution:

Teams A, B, C, and D are participating in a tournament where each team plays against every other team once. A victory earns a team 3 points, a draw earns 1 point, and a defeat earns 0 points. The current standings are as follows:
 
Team B: 7 points;
 
Team C: 4 points;
 
Team A: 0 points;
 
Team D: 0 points;
 
How many games are yet to be played in the tournament?

A. 0
B. 1
C. 2
D. 3
E. it cannot be determined from the information given

With four teams in the tournament, each team must play three games, resulting in a total of 6 games (4C2 = 6).
 
Let's examine each team's points.
 
Team B must have played all its three games, as it is not possible to accumulate 7 points in fewer than 3 games. Thus, Team B played three games: 2 wins and 1 draw. Since Teams A and D have 0 points each, Team B must have drawn with Team C, with each team earning 1 point from the draw.
 
To accumulate 4 points, Team C must have played 2 games: 1 win (against either A or D) and 1 draw (against B). (Note that Team C cannot have played 3 games - 1 win, 1 draw, and 1 loss - because it cannot lose to Team B since they drew, and it cannot lose to Team A or Team D since these teams have 0 points).
 
Consequently, there are two games yet to be played: Team C against either Team A or Team D (whichever it has not yet played) and the match between Team A and Team D.

Answer: C

shasadou · Answer

I usually try drawing a competition box. Please see the pic. The black box means teams do not play with themselves. If there were 2 games" one home and one away then the fields on both sides of the black row would represent the 12 different games. However we are given that they play only once with each other then we have only 6 games and hence regard only on side of the box - the other just a mirror reflection.

So in total there 2 games left: A-D and either C-D or C-A.

Attachments

abcd.jpg [ 28.33 KiB | Viewed 31753 times ]

So in total there 2 games left: A-D and either C-D or C-A.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

A_ · Answer

I like the solution - it’s helpful.

Harsh1110 · Answer

I like the solution - it’s helpful. nice question

jipandeyji · Answer

Total Games - 4C2 = 6 Matches

A(0) B(7) C(4) D(0) 
0------3-----3 ------ 0
------- 3-----1 
--------1-----0

This gives us - 3 Wins / 3 Zeros and 1 Draw = Total 4 matches

Hence, 2 Matches remaining = Option C

Hope this helps!

Shaurs · Answer

really great technique to solve the question.

Aashimabhatia · Answer

For every +3, there should be a corresponding 0 and vice versa. Similarly, for every +1, there should also be another +1.
B= +3, +3, +1
C= +3, +1, 0
A & D= 0, 0, 0
As there are 3 3's and 2 1's, it means 4 games were played in total. Hence, 4C2 - 4= 6 - 4 = 2 remaining games

Paattaa · Answer

I like the solution - it’s helpful.

Poojakhatri241 · Answer

I like the solution - it’s helpful.

M18-06

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