A liquid A of mass 100g at 100 ° C is added to 50g of a liquid B at...

Heat loss will always be equal to heat gain.

As per ${1}^{st}$ condition,

100 \times {S}_{A} \times ( 100 - 90 ) = 50 \times {S}_{B} \times ( 90 - 75 ) ___ ( i )

As per ${2}^{nd}$ condition,

100 \times {S}_{A} ( 100 - \theta ) = 50 \times {S}_{B} ( \theta - 50 ) ___ ( ii )

Dividing $( ii )$ by $( i )$, we get
$\dfrac{100 - \theta}{100 - 90} = \dfrac{\theta - 50}{90 - 75}$

$300 - 3\theta = 2\theta - 100$

$\theta = 80 ° C$