Smita04 wrote:

There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?

A 10

B 20

C 40

D 60

E 120

There are different ways of approaching this question.

Method 1:

Take the first amateur. He plays a game with each of the other four i.e. 4 games.

Now take the second one. He has already played a game with the first one. He plays 3 games with the rest of the 3 amateurs i.e. 3 more games are played.

Now take the third amateur. He has already played a game each with the first and the second amateur. Now he plays 2 games with the remaining 2 amateurs so 2 more games are played.

Now go on to the fourth amateur. He has already played 3 games with the first 3 amateurs. He just needs to play a game with the last one i.e. 1 more game is played.

The last amateur has already played 4 games.

Total no of games = 4+3+2+1 = 10

Method 2:

Each person is one participant of 4 games. So there are in all 4*5 = 20 instances of one participant games. But each game has 2 participants so total number of games = 20/2 = 10

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[b]Karishma

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