Magical Mathematics[Interesting Approach]> in a chess tournament, where the particip...

in a chess tournament, where the participants are to play one game with one another two player fell ill, having played only 3 games each .If the total numberof games played in the tournament is equal to 84 then find the number of participants in the beginning.please tell me how it is 15

riya , 8 Years ago

Grade 11

3 Answers

Riddhish Bhalodia

Last Activity: 8 Years ago

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

Approved

Ankit Jaiswal

Last Activity: 8 Years ago

x=15whic results in $x^2-5x -150 =0$ this results in a quadratic as $\frac{(x-2)(x-3)}{2} + 6 = 84$ hilet there are x players from the startthus x-2 non ill playersTHe 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

manmath

Last Activity: 8 Years ago

x=15 which results in $x^2-5x -150 =0$ this results in a quadratic as $\frac{(x-2)(x-3)}{2} + 6 = 84$ hilet there are x players from the startthus x-2 non ill playersTHe 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

2 days ago