Trains Meeting!

One train travels from $A$ towards $B$ till $10 : 00~am$ and then another train starts from $B$ towards $A$ .

$\color{#20A900}\text{Till 10 : 00 am}$ :

Only one train is traveling at a constant speed of $40~km/hr$ . So, we can directly find the distance in the two hours that is from $8~am$ to $10~am$ .

$Distance = 40 \times 2 = 80~km$

$\color{#20A900}\text{From 10 : 00 am}$ :

Separation between the two trains at $10~am = 110 - 80 = 30~km$ . Now another train starts from station $B$ . Let us assume that both the trains will meet after time $t$ . Also, if the first train travels for a distance $x$ then the second train will travel for a distance $(30 - x)$ .

$\implies x = 40t \quad \quad \quad and \quad \quad \quad 30 - x = 50t$

$\implies 30 - (40t) = 50t$

$\implies 90t = 30 \implies t = \dfrac{1}{3}hours = 20~minutes$

$\therefore \color{#3D99F6}\text{Both the trains will meet at 10 : 20 am .}$

At 10:00 AM, train from A would have traveled for 2 hours and a distance of $40 \times 2 = 80$ km and the remaining distance between the two trains is $110-80 = 30$ km. The remaining distance of 30 km is covered by the two train with a combined speed of $40+50=90$ km/h. So the time taken for them to cover the distance after 10:00 AM is $\frac {30}{90} = \frac 13 \times 60= 20$ minutes. Therefore, the trains meet at $\boxed {\text{10:20 AM}}$ .

Train A travels 80 km in 2 hours, so when train B starts at 1000, the trains are 30 km apart. They are approaching each other at a relative speed of 90 km per hour. Therefore they will meet in 1/3 hours = 20 minutes, or 1020.