Question

Kamal has $x$ children by his first wife. Ritu has $\left(\mathrm{x}+1\right)$ children by her first husband. They marry and have children of their own. The whole family has $24$ children. Assuming the two children of the same parents do not fight, then the maximum possible number of fights that can take place is :

• A. $190$
• B. $191$
• C. $200$
• D. $52$

Answer 1

The correct option is B

$191$

Explanation for the correct option :

Step 1: Defining an equation for $\mathrm{Z}$

Kamal has $x$children by his first wife and Ritu has $\left(\mathrm{x}+1\right)$ children by her first husband.

We assume that Kamal and Ritu marry and they have $y$ children together.

Then, total number of children $=24$

Therefore
$$\begin{gathered} \Rightarrow \mathrm{x}+\mathrm{x}+1+\mathrm{y}=24 \\ \Rightarrow 2\mathrm{x}+\mathrm{y}=23 \\ \Rightarrow \mathrm{y}=23-2\mathrm{x}-\left(1\right) \end{gathered}$$

Let the number of fights be equal to $\mathrm{Z}$.

It is given that a child from $\mathrm{X}$ will fight with $\left(\mathrm{x}+1\right)$ or from $\mathrm{Y}$which

$\Rightarrow {}^{\mathrm{x}}\mathrm{C}_1\times {}^{\mathrm{x}+1}\mathrm{C}_1+{}^{\mathrm{x}}\mathrm{C}_1\times {}^{\mathrm{y}}\mathrm{C}_1$

Similarly $\left(\mathrm{x}+1\right)$will fight with $\mathrm{X}$ or $\mathrm{Y}$and $\mathrm{Y}$will fight with $\mathrm{X}$or $\left(\mathrm{x}+1\right)$.

Therefore total number of fights is equal to

$$\begin{gathered} \Rightarrow \mathrm{Z}={}^x{C}_1{\mathrm{x}}^{\mathrm{x}+1}{\mathrm{C}}_1+{}^x{C}_1{\times }^{\mathrm{y}}{C}_1+{}^{\mathrm{x}+1}C_1\times {}^y{C}_1 \\ \Rightarrow \mathrm{Z}=\mathrm{x}(\mathrm{x}+1)+\mathrm{xy}+(\mathrm{x}+1)\mathrm{y} \end{gathered}$$

Step 2: Equating the equation for finding the value of $\mathrm{Z}$

Substitute Equation (1) here we get,
$$\begin{gathered} \Rightarrow \mathrm{Z}=\mathrm{x}(\mathrm{x}+1)+\mathrm{x}(23-2\mathrm{x})+(\mathrm{x}+1)(23-2\mathrm{x}) \\ \Rightarrow \mathrm{Z}=-3{\mathrm{x}}^2+45\mathrm{x}+23 \\ \Rightarrow \left.-3{\mathrm{x}}^2+45\mathrm{x}+(23-Z\right)=0 \\ \Rightarrow 3{\mathrm{x}}^2-45\mathrm{x}-(23-Z)=0 \end{gathered}$$

Now $x$ is a real number, therefore discriminant is greater then or equal to zero

Therefore

$$\begin{gathered} \Rightarrow \mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}\ge 0 \\ \Rightarrow \mathrm{D}=(-45{)}^2-4\times 3\times (Z-23)\ge 0 \\ \Rightarrow (45{)}^2+12\times 23\ge 12Z \\ \Rightarrow 12Z\le 2301 \\ \Rightarrow \mathrm{Z}\le 191.75 \end{gathered}$$

But $\mathrm{Z}$ should be a whole number, hence $\mathrm{Z}=191$

Therefore maximum possible number of fights that can take place is equal to $191$.

Hence, option B is the correct answer.