Question

the efficiency of a carnot heat engine has increased from 40% to 50% which the temperature of source and sink are reduced by 100 degree centigrade find the source and sink temperature

Answer 1

When the temperatures of source and sink are $T_{1}$ and $T_{2}$ respectively, the efficiency of Carnot engine is given by, $\eta=1-\frac{T_{2}}{T_{1}}$.
From the information provided, we get two equations,
$\frac{T_{2}}{T_{1}}=\frac{3}{5}$ → 1
$\frac{T_{2}-100}{T_{1}-100}=\frac{1}{2}$ → 2
Solving them, you get $T_{1}=500 K$ and $T_{2}=300 K$.