A carnot engine operating between temperatures `T_(1)` and `T_(2)` has efficiency. When `T_(2)` is lowered by 62K, its efficience increases to `(1)/(3)`. Then `T_(1)` and `T_(2)` are respectively:

Explanation

`eta=1-(T_(2))/(T_(1))rArr(1)/(6)=1-(T_(2))/(T_(1))rArr(T_(2))/(T_(1))=(5)/(6)`.......(1)
`eta_(2)=1-(T_(2)-62)/(T_(1))rArr(1)/(3)=1-(T_(2)-62)/(T_(1))` ....(2)
On solving Equation (1) and (2)
`T_(1)=372K` and `T_(2)=310K`