In a meter bridge a $30\Omega$ resistance is connected in the left gap and a pair of resistances P and Q in the right gap. Measured from the left, the balance point is$37.5cm$ , when P and Q are in series and $71.4cm$ when they the parallel. The values of P and Q $\left(in\Omega \right)$ are :

Resistance $R=30\Omega$

Resistance $S=PandQ$

Balancing length $l=37.5cm$

Balancing length $l^1=71.4cm$

1^st case :

when P and Q are in series

$\frac{R}{S}=\frac{l}{\left(100-l\right)}$

$\frac{30}{P+Q}=\frac{37.5}{\left(100-37.5\right)}$

$\frac{30}{P+Q}=\frac{37.5}{62.5}$

$P+Q=\frac{30\times 62.5}{37.5}$

$P+Q=50$                         …(i)

IInd case :

when P and Q are in parallel

$\frac{R}{S}=\frac{l^1}{\left(100-l^1\right)}$

$\frac{30}{\frac{PQ}{P+Q}}=\frac{71.4}{(100-71.4)}$

$\frac{30\left(P+Q\right)}{PQ}=\frac{71.4}{28.6}$

$\frac{30\times 50}{PQ}=\frac{71.4}{28.6}$

$PQ=\frac{30\times 50\times 28.6}{71.4}$

$PQ=600$                           …(ii)

So, from Eqs. (i) and (ii)

$P+Q=50$

$PQ=600$

$Q=50-P$

$P(50-P)=600$

$50P-P^2=600$

$P^2-50P+600=0$

$P^2-30P-20P+600=0$

$P(P-30)-20(P-30)=0$

$P=20,P=30$

$\therefore P=30\Omega$

$\therefore Q=20\Omega$