CAT Exam Model Paper 2 with solutions for free online practice

Number of common children of Mr. John and Ms. Bashu = 10 − ( x + x + 1) = 9 − 2 x
Let N = The number of fights between children of different parents = (Total number of fights that can take place among all the children) − (The number of fights among the children of the same parents)
${=}^{10}{\mathit{C}}_2-{[}^{\mathit{x}}{\mathit{C}}_2{+}^{\mathit{x}+1}{\mathit{C}}_2{+}^{9-2\mathit{x}}{\mathit{C}}_2]$...(i)
$=45-\left[\frac{\mathit{x}\left(\mathit{x}-1\right)}{2}+\frac{\left(\mathit{x}+1\right)\left(\mathit{x}\right)}{2}+\frac{\left(9-2\mathit{x}\right)\left(8-2\mathit{x}\right)}{2}\right]$
$=45-\frac{1}{2}\left[{\mathit{x}}^2-\mathit{x}+{\mathit{x}}^2+\mathit{x}+72-34\mathit{x}+4{\mathit{x}}^2\right]$
$=\frac{397}{12}-3{\left(\mathit{x}-\frac{17}{6}\right)}^2$
For N to be maximum, x must be $\frac{17}{6}$ As x can not be in fraction,
we take x = 3 $\left(\approx \frac{17}{6}\right)$
Thus, maximum value of N = 33 which is attained at x = 3.