The temperature of equal masses of three different liquids  $x,y\text{ and }z$  are  $10°C,20°C$ and $30°C$  respectively. The temperature of mixture when x is mixed with y is $16°C$ and that when y is mixed with z is $26°C$ . The temperature of mixture when x and z are mixed will be 

Given masses of three liquids are equal, i.e $m_x=m_y=m_z=m$ 
Temperature of mixture of liquids x and $y_1$ , $T_1=16°C$
Temperature of mixture liquids  y and z ,  $T_2=26°C$
While mixing the liquids x and y , the heat energy lost by liquid y will be equal to the heat energy absorbed by liquid x 
Consider the specific heat capacity of liquid x is $S_x$ , of liquid y is $S_y$ and of liquid z  is $S_z$

Now, for liquid x and y , using principle of calorimetry, 
Heat gained by liquid  x = Heat lost by liquid y 
 $\Rightarrow m{S}_x\left(T_1-10\right)=m{S}_y\left(20-T_1\right)\Rightarrow S_x\left(16-10\right)=S_y\left(20-16\right)\Rightarrow \frac{S_x}{S_y}=\frac{2}{3}\mathrm{......}\left(i\right)$
Similarly, for liquid y and z using principle of calorimetry, 
Heat gained by liquid y = Heat lost by liquid z
 $m{S}_y\left(T_2-20\right)=m{S}_z\left(30-T_2\right)\Rightarrow S_y\left(26-20\right)=S_z\left(30-26\right)\Rightarrow \frac{S_y}{S_z}=\frac{2}{3}\mathrm{.......}\left(ii\right)$
Multiply Eq (i) by (ii) , we get $\frac{S_x}{S_z}=\frac{4}{9}$  
Consider the mixing of liquids x and z , let the temperature of mixture be $T_3$ .
Using principle of calorimetry for liquids x  and  z
   $m{S}_x\left(T_3-10\right)=m{S}_z\left(30-T_3\right)$$\Rightarrow S_x\left(T_3-10\right)=S_z\left(30-T_3\right)$

$\Rightarrow \frac{S_x}{S_z}=\frac{\left(30-T_3\right)}{\left(T_3-10\right)}$   $\Rightarrow \frac{4}{9}=\frac{\left(30-T_3\right)}{\left(T_3-10\right)}$
    
 $\Rightarrow 4T_3-40=270-9T_3\Rightarrow T_3=23.84°C$