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Q.
A Carnot engine has efficiency of $50 \%$. If the temperature of sink is reduced by $40^{\circ} C$, its efficiency increases by $30 \%$. The temperature of the source will be :
$ \eta=1-\frac{ T _{ L }}{ T _{ H }} $
$ \frac{1}{2}=1-\frac{ T _{ L }}{ T _{ H }} $
$ \frac{1}{2}(1 \cdot 3)=1-\left(\frac{ T _{ L }-40}{ T _{ H }}\right) $
$ \frac{1}{2}(1 \cdot 3)=\frac{1}{2}+\frac{40}{ T _{ H }}$
$ T _{ H }=266.7 \,K $