CAT 2005 | Question: 05

Question

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in $45$ games both the players were girls, and in $190$ games both were boys. The number of games in which one player was a boy and other was a girl is 

• A. $200$
• B. $216$
• C. $235$
• D. $256$

Answer 1

Let there be x girls and y boys.

Number of games between two girls = $\binom{x}{2}$

Number of games between two boys = $\binom{y}{2}$

$\frac{x(x-1)}{2}$ = 45

$x^{2} - x$ – 90 = 0 

(x– 10)(x + 9) = 0

x = 10

Similarly,

$\frac{y(y-1)}{2}$ = 190

$x^{2} - x$ – 380 = 0

 (y + 19)(y – 20) = 0

 y = 20

Total number of games = $\binom{x+y}{2}$ = $\binom{30}{2}$ = 435

Number of games in which one player is a boy and the other is a girl = 435 – 45 – 190 = 200

 

Hence, option (1)200 is the correct choice.