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# In a certain play, each of four players must play with each

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Manager
Joined: 04 Sep 2006
Posts: 113
In a certain play, each of four players must play with each  [#permalink]

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04 Jun 2009, 11:38
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In a certain play, each of four players must play with each of other players exactly once. If no game ended in a tie, did someone win one and lose two games?
1). No one won all the games.
2). No one lost all the games.

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Senior Manager
Joined: 16 Jan 2009
Posts: 338
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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04 Jun 2009, 14:35
Number of games =6

1). No one won all the games.
Suppose A won all the games.
AB-A
AC-A
BC
BD
CD
In any case the required condition “did someone win one and lose two games” will not be met.
So (1) is SUFFICIENT.

2). No one lost all the games.
EX1.
Suppose A lost all the games.
AB-B
AC-C
BC-B
BD
CD-D
So C won 1 game and lost 2
EX2.
Suppose A lost all the games.
AB-B
AC-C
BC-C
BD
CD-D
So C won 2 game and lost 1

So (1) is INSUFFICIENT.

IMO A
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Lahoosaher

Manager
Joined: 14 May 2009
Posts: 187

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04 Jun 2009, 17:07
vcbabu wrote:
In a certain play, each of four players must play with each of other players exactly once. If no game ended in a tie, did someone win one and lose two games?
1). No one won all the games.
2). No one lost all the games.

There are 4x3 games, then dividing by 2 to eliminate repeats ie (player 1 w/ 2 is the same as 2 w/ 1).

So 6.

In total we have 6 wins, 6 losses, and each person gets 3.
Nobody won all the games --> everyone gets >=1 loss. So we have 2 losses left.

A LLL
B LWW
C LWW
D LWW

2

Sufficient

After playing around with it

Will post a more detailed solution
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Senior Manager
Joined: 16 Jan 2009
Posts: 338
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)

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04 Jun 2009, 17:10
BTW whats OA?
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Lahoosaher

Manager
Joined: 14 May 2009
Posts: 187

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04 Jun 2009, 23:16
It's a network flow.

I set it up, very basic. I tried to solve it mentally but no dice, there is no configuration that works. Let me explain it another way

We're going to have 6 matches, so 6 wins & 6 losses. Let's try and make a case where nobody Wins 1 game & Loses 2 games. If nobody can have 3 losses nor 2 losses (from condition in (2), and from additional condition we impose), then someone can have at most 1 loss. But we need to distribute 6 losses total...so let's give everyone 1 loss, and we have 2 losses left. But where do we put those? If we put another loss onto 2 different people, we violate our own condition (no 2 losses). And if we try to stick the 2 losses onto one player, we violate the condition in (2), which states that no player can lose all 3. Hence there will always be at least 1 player who wins 1 but loses 2.

B
Attachments

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Manager
Joined: 14 May 2009
Posts: 187

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05 Jun 2009, 00:31
More graphically:

We're going to have $$6$$ wins and $$6$$ losses.

Let's represent the players via $$A/B/C/D$$.

$$L$$ = loss
$$W$$ = win

Condition in (2): No player can have 3 losses[/m].

Let's look at the question: [b]Can a player have 2 losses?

The YES case means a player CAN have 2 losses.

Let's see if we can make a NO case, that is no player has 2 losses. If it's impossible to make a NO case,
then the answer is always YES

So we have 6 losses: $$LLLLLL$$
& 4 players:

$$A ???$$
$$B ???$$
$$C ???$$
$$D ???$$

So since nobody can have 3 losses, or 2 losses, let's give everyone 1 loss:

$$A L??$$
$$B L??$$
$$C L??$$
$$D L??$$

We have 2 L's left. Where do we put them? If we put the 2 losses onto 1 player, we have 3 losses, which violates our condition
that no player has 3 losses. But our only other choice is to put one loss onto 2 players-- which violates our assumption for the NO
case that no player has 2 losses.

Hence the NO case is impossible.

Sufficient
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Manager
Joined: 11 Aug 2008
Posts: 132

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09 Jun 2009, 02:09
From Hade's approach, it clearly that
We have 4 people A,B,C,D and 6 games as follow:
AB
AC
BC
BD
DC
See that we have total 6 Lose and 6 Win, each person play 3 games with other people.
1) If no one wills all, at most a person has 1 lose. So remains 2 lose. These 2 loses can be distributed to one A,B,C or D to make a person who has 3 all loses and each other has 1 lose 2 wins; or they can be distributed to two other people to make 2 people who have 2 loses 1 win. So insuf
2) Because no one lose all 3 games, let's distribute 6 loses into 3 people have 2 loses each; or 2 people have 2 loses each and 2 people have 1 lose each. So suf
Hence B

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: play &nbs [#permalink] 09 Jun 2009, 02:09
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