A train left point A at 12 noon. Two hours later, another train started from point A in the same direction. It overtook the first train at 8 PM. It is known that the sum of the speeds of the two trains is 140 km/hr. Then, at what time would the second train overtake the first train, if instead the second train had started from point A in the same direction 5 hours after the first train? Assume that both the trains travel at constant speeds.

A train left point A at 12 noon. Let its speed be T1. Two hours later, another train started from point A in the same direction. Let the speed of the second train be T2.

The second train overtook the first train at 8 PM. This means that the distance covered by the first train in 8 hours is covered by the second train in 6 hours

=> $$8T_1=6T_2$$ => $$T_1=\dfrac{3T_2}{4}$$

It is known that the sum of the speeds of the two trains is 140 km/hr.

=> $$T_1+T_2=140$$

=> $$\dfrac{3T_2}{4}+T_2=140$$

=> $$T_2=80$$ km/hr, and $$T_1=60$$ km/hr

Now, if the second train had started from point A in the same direction 5 hours after the first train, then the distance travelled by the first train will be $$60\times5=300$$ km.

Now, at 5 PM, the second train will leave. We need to find out at what time the second train will overtake the first train. 

Thus, we will apply the relative velocity concept here. We will assume the first train to be at rest; thus, the second train's speed will be (80-60) = 20 km/hr, and the second train has to cover 300 km. 

The time taken to meet = $$\dfrac{300}{20}=15$$ hours

Now, at 5 PM, the second train starts. And, after 15 hours, they will meet. Therefore, the two trains will meet at 5 PM + 15 hours = 8 AM next day.