UGC NET Paper 1 GA 2013 June Solved Paper

Let number of teams participated in the tournament are n then, number of game played by,
first team = (n – 1)
Second team = (n – 2)
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(n – 1)th team = 1
According to the questions,
(n – 1) + (n – 2) + (n – 3) +.....+ 1 = 45
This is equivalent to, selection of two ‘teams’ out of n teams ${}_{}^{\mathit{n}}{\mathit{C}}_2^{}$
∴ ${}_{}^{\mathit{n}}{\mathit{C}}_2^{}$ = 45
$\Rightarrow \frac{\mathit{n}\left(\mathit{n}-1\right)}{2}=45$
or $\mathit{n}\left(\mathit{n}-1\right)=90$
$\Rightarrow {\mathit{n}}^2-\mathit{n}-90=0$
$\Rightarrow \left(\mathit{n}-10\right)\left(\mathit{n}+9\right)=0$
$\Rightarrow \mathit{n}=10$