Three rods of material X and three rods of material Y are connected as shown in figure. All are identical in length and cross-sectional area. If end A is maintained at ${60}^{0}C,$ end E at ${10}^{0}C,$ thermal conductivity of X is $0.92ca{l}^{-1}c{m}^{-1}{}^{0}C^{-1}$ and that of Y is  $0.46ca{l}^{-1}c{m}^{-1}{}^{0}C^{-1}$, then sum of the temperatures of junctions B,C,D is______  $°C$

Let L be the length of each rod. 
Temperature of A = $60°$ C temperature of E= $10°$ C 
Let ${\theta }_1,{\theta }_2,{\theta }_3$  be respective temperture of  B, C, D
If $Q_1,Q_{2,}{Q}_3,Q_4,Q_5,Q_6$ are the amounts of heat 
following/sec respectively from A to B;Bto C ; Bto D; C to D;D
to E and C to E, then using figure.

$$\begin{gathered} Q_1=\frac{0.46A(60-{\theta }_1)}{L},Q_2=\frac{0.92A({\theta }_1-{\theta }_2)}{L} \\ {Q}_3=\frac{0.46A({\theta }_1-{\theta }_2)}{L},Q_4=\frac{0.92A({\theta }_2-{\theta }_3)}{L} \\ {Q}_5=\frac{0.46A({\theta }_3-10)}{L},Q_6=\frac{0.92A({\theta }_2-10)}{L} \\ Now, Q_1=Q_2+Q_3 \\ \frac{0.46A(60-{\theta }_1)}{L}=\frac{0.92A({\theta }_1-{\theta }_2)}{L}+\frac{0.46A({\theta }_1-{\theta }_3)}{L} \\ 60-{\theta }_1=2({\theta }_1-{\theta }_2)+{\theta }_1-{\theta }_3 \end{gathered}$$

Again,  $Q_2=Q_4+Q_6$ gives  ${\theta }_1-3{\theta }_2-{\theta }_3=10°$ 
Solving  Eqs. (i) , (ii) and (iii) we get   ${\theta }_1+30°C,{\theta }_2=20°C,{\theta }_3=20°C$