in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ---- 10 / 15 / 12 / 14 ?

Question

Riddhish Bhalodia · Answer

OK say there are x players from the start
thus x-2 non ill players
THe total matches they played with one another is
(x-2)(x-3)/2 …..........(because each player will play will every other player by (x-2)(x-3), but this will be double of the total answer so we half it)
in addition to this there are 6 more games played by ill players
$\frac{(x-2)(x-3)}{2} + 6 = 84$
this results in a quadratic as
$x^2-5x -150 =0$
whic results in
x=15

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in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ---- 10 / 15 / 12 / 14 ?

in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ----

10 / 15 / 12 / 14 ?

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in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ---- 10 / 15 / 12 / 14 ?

in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ---- 10 / 15 / 12 / 14 ?

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in a chess tournament, all should play one game with each other. 2 players fell ill after playin 3games each. if total no. of games played is 84, then total no.of participants in the starting is ----

10 / 15 / 12 / 14 ?