Saga Internationals have conducted a chess competition between young boys and girls, wherein every individual has to play exactly one game with every other individual. It was found that in 66 games, both the players were girls, and in 210 games, both were boys. Find the number of games in which one player was a boy and the other was a girl.

Question

Arun selected an integer \( x \) between 2 and 40, both inclusive. He noticed that the greatest common divisor of the selected integer \( x \) and any other integer between 2 and 40, both inclusive, is 1. How many different choices for such an \( x \) are possible?

Answer 1

The correct answer is option (C):

252

Let's denote the number of girls as 'g' and the number of boys as 'b'.

The number of games played between two girls is given by the combination formula, which is gC2 or g(g-1)/2. We are given that this number is 66. So, we have:

g(g-1)/2 = 66
g(g-1) = 132

By trying a few values or recognizing that 132 = 12 * 11, we find that g = 12. There are 12 girls.

Similarly, the number of games played between two boys is bC2 or b(b-1)/2. We are given that this is 210. So:

b(b-1)/2 = 210
b(b-1) = 420

By trying values, or recognizing that 420 = 21 * 20, we find that b = 21. There are 21 boys.

Now, we want to find the number of games played between a boy and a girl. This means each boy plays against each girl, so the number of such games is the product of the number of boys and the number of girls: b * g.

So, the number of games with one boy and one girl is 21 * 12 = 252.

Therefore, the correct answer is 252.

The Correct Option is C

Solution and Explanation

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