Three men A, B, C play cards. If one loses the game he has to give Rs. 3. If he wins the game he will gain Rs. 3 each from the other two losers. If A has won 3 games, B loses Rs. 3. C wins Rs. 12 , then the total number of games played is
Let the games won by \(B\) and \(C\) be \(x\) and \(y\) respectively. So losses of A will be \(\mathrm{x}+\mathrm{y}\) and of B will be \(y+3\) and of \(C\) will be \(x+3\). For B, \(6 x-3(y+3)=-3\) For \(\mathrm{C}, 6 \mathrm{y}-3(\mathrm{x}+3)=12\) By adding these two equations we get \(x+y=9\). Hence total number of games will be \(\mathrm{x}+\mathrm{y}+3=12\).\[\]Choice (A)\[\]
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