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# M18-06

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Math Expert
Joined: 02 Sep 2009
Posts: 58312

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16 Sep 2014, 01:03
3
8
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Difficulty:

95% (hard)

Question Stats:

50% (02:10) correct 50% (02:22) wrong based on 220 sessions

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Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

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16 Sep 2014, 01:03
2
Official Solution:

Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

Team B has played all of its three matches - it could not have accumulated 7 points in 2 games. Also, because team A and team D have no points, team B tied with team C. Therefore team C beat either team A or team D. In either case, two games are yet to be played: A-D and D-C (or A-C).

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26 Jan 2016, 09:14
2
1
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.
>> !!!

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21 Jul 2016, 04:44
1
Bunuel

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23 Sep 2016, 15:02
I think this the explanation isn't clear enough, please elaborate. Can someone point out the flaw in the below scenario-

Given that, each team plays three

TEAM B - Won, Won, D(draw with C)
TEAM C- Won, D(draw with B), Not played
TEAM A- L (lost to B), L (lost to C), Not played
TEAM D- L (lost to B), Not played, Not played

Hence four are yet to be played?!
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23 Sep 2016, 19:22
2
KarishmaParmar wrote:
I think this the explanation isn't clear enough, please elaborate. Can someone point out the flaw in the below scenario-

Given that, each team plays three

TEAM B - Won, Won, D(draw with C)
TEAM C- Won, D(draw with B), Not played
TEAM A- L (lost to B), L (lost to C), Not played
TEAM D- L (lost to B), Not played, Not played

Hence four are yet to be played?!

Hi,
Four teams will play,BUT two teams will play in ONE match..
So the moment ONE match is played TWO of 'not played' will cancel out and remaining two 'not played' will cancel out by second match..
So matches yet to be played= (not played)/2= 4/2=2
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17 Jun 2017, 01:11
1
Bunuel wrote:
Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

=========================

B = 2 games Win (3*2) + 1 draw (1) = 6+1 = 7
C = 1 game win (3*1) + 1 draw (1) = 3+1 =4
A and D both 0 means 0 wins.
Total matches should be 6 as (4*3)/2=6 matches, 4 matches have already played so only option left is 2. No need to go for win and loose in deep.

Kudos if this helps.
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13 Mar 2018, 02:21
Hi Bunuel,

I still don't get @ chetan2u's explanation as to how it's 2 games left to be played as opposed to 4.
How do you know to divide by 2?

Best,

Tosin
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01 Sep 2018, 08:35
Hmm..
There are 4 teams :
Each has to play with the other three teams.Therefore the [b]total matches that must be played is 12.[/b]

Team B: 7 points; Lets see B has 2 wins and 1 draw;Here B has played all the three matches and some other teams(opponents) have played 3 matches therefore total six matches were played

Team C: 4 points; C has played 2 matches 1 win and 1 draw; C has played 2 matches and some other teams(opponents) have played 2 matches therefore total 4 matches were played.

Team A: 0 points;opponents

Team D: 0 points;opponents

THEREFORE
Total 6+4=10 matches were played ;
we have the total match count of 12;
Left over matches =12-10 =2;
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14 Aug 2019, 23:03
Bunuel wrote:
Official Solution:

Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

Team B has played all of its three matches - it could not have accumulated 7 points in 2 games. Also, because team A and team D have no points, team B tied with team C. Therefore team C beat either team A or team D. In either case, two games are yet to be played: A-D and D-C (or A-C).

Hi Bunuel,

Please guide. Here's my approach. Total games possible 4C2.. meaning 6 games.
If B has 7 points; then it has 2 victories and 1 draw
If C has 4 point; then it has 1 victory and 1 draw.

Since some of these games might have been between B&C, max 5 games have been played; 2+1+1+1 ---> implies only 1 game is remaining

What's wrong with this approach ?

Thanks and Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 58312

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14 Aug 2019, 23:36
1
azhrhasan wrote:
Bunuel wrote:
Official Solution:

Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

Team B has played all of its three matches - it could not have accumulated 7 points in 2 games. Also, because team A and team D have no points, team B tied with team C. Therefore team C beat either team A or team D. In either case, two games are yet to be played: A-D and D-C (or A-C).

Hi Bunuel,

Please guide. Here's my approach. Total games possible 4C2.. meaning 6 games.
If B has 7 points; then it has 2 victories and 1 draw
If C has 4 point; then it has 1 victory and 1 draw.

Since some of these games might have been between B&C, max 5 games have been played; 2+1+1+1 ---> implies only 1 game is remaining

What's wrong with this approach ?

Thanks and Regards,

There are the following cases possible:
B beat A --> 3 point for B
B beat D --> 3 point for B
B tied with C --> 1 point for B, and 1 point for C
C beat A (or D) --> 3 point for C.

B = 7
C = 4
A = 0
D = 0.

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14 Aug 2019, 23:49
Bunuel wrote:
azhrhasan wrote:
Bunuel wrote:
Official Solution:

Team A, team B, team C, and team D take part in a tournament in which every team plays once with each of its opponents. 3 points are awarded for victory, 1 for a draw, and 0 for defeat. The scoring table currently is 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

Team B has played all of its three matches - it could not have accumulated 7 points in 2 games. Also, because team A and team D have no points, team B tied with team C. Therefore team C beat either team A or team D. In either case, two games are yet to be played: A-D and D-C (or A-C).

Hi Bunuel,

Please guide. Here's my approach. Total games possible 4C2.. meaning 6 games.
If B has 7 points; then it has 2 victories and 1 draw
If C has 4 point; then it has 1 victory and 1 draw.

Since some of these games might have been between B&C, max 5 games have been played; 2+1+1+1 ---> implies only 1 game is remaining

What's wrong with this approach ?

Thanks and Regards,

There are the following cases possible:
B beat A --> 3 point for B
B beat D --> 3 point for B
B tied with C --> 1 point for B, and 1 point for C
C beat A (or D) --> 3 point for C.

B = 7
C = 4
A = 0
D = 0.

Ok. Got it. I am counting the tied game between B&C twice (once with B and once with C). So, i need to reduce 1 game. Therefore, Max possible games 4.

Thanks for the guidance
Re: M18-06   [#permalink] 14 Aug 2019, 23:49
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# M18-06

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