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In an ABC Apartment, a carroms tournament is conducted. In that tournament, both boys and girls
played. Every kid in that apartment has to play exactly one game with another kid. In 45 matches, both
the players are girls. In 190 matches, both the players are boys. The number of games in which one
player was a boy and the other was a girl is?​

Answer 1

Answer:

200

Step-by-step explanation:

Answer 2

Given :- Every kid in that apartment has to play exactly one game with another kid. In 45 matches, both the players are girls. In 190 matches, both the players are boys.

To Find :- The number of games in which one player was a boy and the other was a girl is ?

Answer :-

we know that, total matches played between n teams with each other is given by,

• n(n - 1)/2 .

so, given that, In 45 matches, both the players are girls.

then,

→ n(n - 1)/2 = 45

→ n(n - 1) = 45 * 2

→ n(n - 1) = 90

→ 10 * (10 - 1) = 90

→ n = 10 = Total Number of girls in the apartment .

similarly, given that, In 190 matches, both the players are boys.

then,

→ n(n - 1)/2 = 190

→ n(n - 1) = 190 * 2

→ n(n - 1) = 380

→ 20 * (20 - 1) = 380

→ n = 20 = Total Number of boys in the apartment .

therefore,

→ The number of games in which one player was a boy and the other was a girl is = Total Number of girls * Total Number of boys = 10 * 20 = 200 (Ans.)

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