Following data shows the number of family members living in different bungalows of a locality :

Number of Members	$\displaystyle 0$-$\displaystyle 2$	$\displaystyle 2$-$\displaystyle 4$	$\displaystyle 4$-$\displaystyle 6$	$\displaystyle 6$-$\displaystyle 8$	$\displaystyle 8$-$\displaystyle 10$	Total
Number of Bungalows	$\displaystyle 10$	p	$\displaystyle 60$	q	$\displaystyle 5$	$\displaystyle 120$

If the median number of members is found to be $\displaystyle 5$, find the values of p and q.

Number of members	Number of bungalows	Cumulative frequency
$\displaystyle 0$-$\displaystyle 2$	$\displaystyle 10$	$\displaystyle 10$
$\displaystyle 2$-$\displaystyle 4$	p	\(\displaystyle 10+\mathrm{p}\)
$\displaystyle 4$-$\displaystyle 6$	$\displaystyle 60$	\(\displaystyle 70+\mathrm{p}\)
$\displaystyle 6$-$\displaystyle 8$	q	\(\displaystyle 70+\mathrm{p}+\mathrm{q}\)
$\displaystyle 8$-$\displaystyle 10$	$\displaystyle 5$	\(\displaystyle 75+\mathrm{p}+\mathrm{q}\)
	$\displaystyle 120$

\[\begin{aligned} & 75+p+q=120 \\ & p+q=45 \end{aligned} \] Median is $\displaystyle 5$ ∴ Median class is $\displaystyle 4$-$\displaystyle 6$ \[5=4+\left[\frac{\dfrac{120}{2}-(10+\mathrm{p})}{60}\right] \times 2 \] On solving, we get \(\displaystyle \mathrm{p}=20\) and \(\displaystyle \mathrm{y}=25\) $\displaystyle 120$ Let distance of gate at P from point B is \(\displaystyle x\) m Then distance of gate at P from point A is \(\displaystyle (35+x) \mathrm{m}\) In right \(\displaystyle \Delta \mathrm{APB}\) \[(x+35)^{2}+x^{2}=(65)^{2} \] \(\displaystyle x^{2}+35 x-1500=0\) \(\displaystyle (x+60)(x-25)=0\) \(\displaystyle x=25\) Hence, \(\displaystyle x+35=60\) Distance of P from A = $\displaystyle 60$ m Distance of P from \(\displaystyle \mathrm{B}=25 \mathrm{~m}\)For real roots, \(\displaystyle \mathrm{D} \geq 0\) \[\begin{array}{r} {[-2(\mathrm{p}+1)]^{2}-4 \mathrm{p}^{2} \geq 0} \\ \Rightarrow \mathrm{p} \geq-\frac{1}{2} \\ \therefore \text { smallest value of } \mathrm{p}=-\frac{1}{2} \end{array} \] At \(\displaystyle \mathrm{p}=-\frac{1}{2}\) given equation becomes \[\begin{aligned} & x^{2}-2\left(\frac{-1}{2}+1\right) x+\left(\frac{-1}{2}\right)^{2}=0 \\ & x^{2}-x+\frac{1}{4}=0 \text { or } 4 x^{2}-4 x+1=0 \\ & \quad(2 x-1)(2 x-1)=0 \\ & \therefore \text { roots are } \frac{1}{2}, \frac{1}{2} \end{aligned} \]