#### The temperature of equal masses of three ifferent liquids x,y and z are respectively.The temperature of mixture when x is mixed with y is and that when y is mixed with z is , The temperature of mixture when x and z are mixed will be :Option: 1Option: 2Option: 3Option: 4

$T_{mix}= \frac{n_{1}T_{1}+n_{2}T_{2}}{n_{1}+n_{2}}$
$T_{x}= 10^{\circ}C= 283\, k$
$T_{y}= 20^{\circ}C= 293\, k$
$T_{z}= 30^{\circ}C= 303\, k$
For x & y mixture ,
$T_{1}= \frac{n_{x}T_{x}+n_{y}T_{y}}{n_{x}+n_{y}}$
$289= \frac{\frac{m_{x}T_{x}}{M_{x}}+\frac{m_{y}T_{y}}{M_{y}}}{\frac{m_{x}}{M_{x}}+\frac{m_{y}}{M_{y}}}$
$289= \frac{\frac{283}{M_{x}}+\frac{293}{M_{y}}}{\frac{1}{M_{x}}+\frac{1}{M_y}}\rightarrow \left ( 1 \right )$
Mixture of y & Z
$299= \frac{\frac{293}{M_{y}}+\frac{303}{M_{z}}}{\frac{1}{M_{y}}+\frac{1}{M_z}}\rightarrow \left ( 2 \right )$
Mixture of x & z
$T_{3}= \frac{\frac{283}{M_{x}}+\frac{303}{M_{z}}}{\frac{1}{M_{x}}+\frac{1}{M_z}}\rightarrow \left ( 3 \right )$
From eqn (1)
$\frac{289}{M_{x}}+\frac{289}{M_{y}}= \frac{283}{M_{x}}+\frac{213}{M_{y}}$
$6M_{y} = 4M_{x}$
$3M_{y} = 2M_{x}\rightarrow \left ( 4 \right )$
$6M_{z} = 4M_{y}$
$3M_{z} = 2M_{y}\rightarrow \left ( 5 \right )$
$3\times \frac{3M_{z}}{2} = 2M_{x}$
$9 M_{z} = 4M_{x}\rightarrow \left (6 \right )$
$T_{3}= \frac{\frac{283}{\frac{9}{4}M_{z}}+\frac{303}{M_{z}}}{\frac{1}{\frac{9}{4}M_{z}}+\frac{1}{M_{z}}}= \frac{\frac{4\times 283}{9}+\frac{303\times 9}{9}}{\frac{4}{9}+\frac{9}{9}}$
$= \frac{4\times283+303\times9}{13}= 296\cdot 84 k$
$T_{3}= 23\cdot 84\degree C$
The correct option is (4)