Carnot Engine Temperature Change

Using the framework of NCERT Class 11, Chapter 12 (Thermodynamics), the efficiency of a Carnot engine is governed solely by the absolute temperatures through $\eta =1-\frac{T_C}{T_H}$, and in this problem the sink temperature $T_C$ is held fixed at 300 K while the source temperature is varied. First find the original source temperature from the 40 percent condition: $0.40=1-\frac{300}{T_H}$ gives $\frac{300}{T_H}=0.60$, so $T_H=\frac{300}{0.60}=500$ K. Now impose the new requirement of 50 percent efficiency: $0.50=1-\frac{300}{T_H^{\prime }}$ gives $\frac{300}{T_H^{\prime }}=0.50$, so $T_H^{\prime }=\frac{300}{0.50}=600$ K. Thus the source must be raised from 500 K to 600 K. The option 500 K is wrong because that is the original source temperature, which only yields 40 percent efficiency. The option 450 K is wrong because lowering the source would reduce efficiency below 40 percent, the wrong direction entirely. The option 550 K is wrong because $1-300/550=0.455$, which is not 50 percent. A plausibility check confirms the logic: with a fixed sink, raising the source temperature must raise efficiency, and indeed 600 K > 500 K corresponds to 50 percent > 40 percent, consistent in both direction and magnitude.