**Correct option: (D) 500 K and 300 K**

**Explanation:**

Initial efficiency = η = 0.4

When T_{2} is lowered by 50k, η' = 0.5

η = 1 – (T_{2} / T_{1})……. efficiency of cannot engine

Hence 0.4 = 1 – (T_{2}/T_{1}) (1)

0.5 = 1 – [(T_{2} – 50) / T_{1}] (2)

From (2)

0.5 = 1 – (T_{2} / T_{1}) + (50 / T_{1})

From (1),

0.5 = 0.4 + (50 / T_{1})

∴ 0.1 = (50 / T_{1})

∴ T_{1} = 500K

From (1),

1 – 0.4 = [T_{2} / (500)] hence T_{2} = 500 × 0.6 = 300K