In a metre bridge [Fig], the gaps are closed by two resistances P and Q and the balance point is obtained 

at 40 cm. When Q is shunted by resistance of 10 $\mathrm{\Omega }$, the balance point shifts to 50 cm. The values of P and Q are

For the balance of metre bridge
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}}=\frac{40}{60}=\frac{2}{3}$
or $\mathrm{P}=\frac{2}{3}\mathrm{Q}$……….(1)
When Q is shunted by a resistance of 10 $\mathrm{\Omega }$ then net resistance in this gap should be a parallel combination of Q and 10 $\mathrm{\Omega }$ resistance i.e., 10Q /10 +Q).
Now the balance shifts to 50 cm i.e., R = 50cm and S = 50 cm. In this case
$\frac{\mathrm{P}}{[10\mathrm{Q}/(10+\mathrm{Q}\left)\right]}=\frac{50}{50}=1$……..(2)
Substituting the value of  P from eq. (1) in eq. (2), we get
$\frac{(2/3)\mathrm{Q}}{[10\mathrm{Q}/(10+\mathrm{Q}\left)\right]}=1$
Solving we get $\mathrm{Q}=5\mathrm{\Omega }$
From eq. (1),  $\mathrm{P}=\frac{2}{3}\times 5=\frac{10}{3}\mathrm{\Omega }$