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I didn't understand. Pls can anyone give any easier solution? Why x-2 is coming everything time?
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which part didn't you understand?
Let's say we have 3 teams A, B, C. Each team plays each other once so we have 3 games: AB, AC, BC
In case of 4 teams A, B, C, D, we have 6 games : AB, AC, AD, BC, BC, CD
You can see the pattern here. When we have m teams:
The 1st team plays with other (m-1) teams -> we have (m-1) games
The 2nd team plays with (m-2) teams (as the 2nd team can't play game with itself and the game between 1st team and 2nd team is already counted in number of games that 1st team plays) --> we have (m-2) games
...
So when there are m teams, each team plays each other once, the number of games = (m-1) + (m-2) + ... +2+1 = \(\frac{[(m-1)+1]*(m-1)}{2}\) = \(\frac{m*(m-1)}{2}\)
In this question:
when there are x teams, the number of games = \(\frac{x*(x-1)}{2}\)
when there are (x-1) teams, the number of games = \(\frac{(x-1)*[(x-1)-1]}{2}\) = \(\frac{(x-1)*(x-2)}{2}\)
Now we have \(\frac{(x-1)*(x-2)}{2}\) = \(\frac{3}{4}\) * \(\frac{x*(x-1)}{2}\)
Solve this equation we have the result x = 8.
Answer: B.
Hope it helps.