Xavier Aptitude Test 2011 Paper

(d) No. of teams in the bottom group = 10
No. of matches played among the bottom group teams = ${}^{10}$${\mathit{C}}_2$= 45
$\therefore$ No. of points that bottom group gets
playing amongst themselves = 45 (1 pt per match).
It is given in the question that the bottom teams get 45 points from their matches against the top group , i.e.45 out 10 n points where n is the no. of teams in the top group.
No. of points that the top group gets from the matches played among themselves = ${}_{}^{\mathit{n}}{\mathit{C}}_2^{}$
No. of points that the top group gets against the bottom group = 10 n – 45
${}_{}^{\mathit{n}}{\mathit{C}}_2^{}=10\mathit{n}-45$
$\mathit{n}\left(\mathit{n}+1\right)=20\mathit{n}-90$
${\mathit{n}}^2-21\mathit{n}+90=0$
$\left(\mathit{n}-6\right)\left(\mathit{n}-15\right)=0$
If $\mathit{n}=6,$ the top group would get ${}_{}^{\mathit{n}}{\mathit{C}}_2^{}+10\mathit{n}-45$
If $\mathit{n}=6,$ the top group would get ${}_{}^{\mathit{n}}{\mathit{C}}_2^{}+10\mathit{n}-45$
${=}^{\mathit{n}}{\mathit{C}}_2+10\left(6\right)-45=30$ points or an average of 5 points per team the bottom group would get $\frac{\left(45+45\right)}{10}$ or an avg of $9.$ This is not possible as points of the top group should be higher than that of the bottom group.
$\therefore \mathrm{}\mathit{n}=15.$
$\therefore$ Total no. of teams $=25$