two stations A and B are 144 miles apart. A fast train starts from B at 9 AM another fast train travelling at the same rate starts from A at 10 AM. A slow train starts from B at 10:20 AM. The fast train from A meets the other fast train from B at 11:30 AM, and the slow train at 12:30 PM. When does the slow train reach A ? i know its a question of relative speed but i want answer.

Dear Student,

Please find below the solution to the asked query:

Let speed of fast train start from A and B  =  x  mile/ hr and Speed of slow train start from B = y mile/ hr

Let , Total distance cover by fast train from A to meet fast train from B =  z , So Total distance cover by fast train from B to meet fast train from A = 144 - z 

We know :  Time = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$

From first condition we get
$$\begin{gathered} \frac{z}{x} = \frac{144 - z}{x} \\ \Rightarrow z x = 144 x- z x \\ \Rightarrow 2 z x = 144 x \\ \mathbf{\Rightarrow }\mathit{z}\mathbf{=}\mathbf{ }\mathbf{72} \end{gathered}$$
So,
Total distance cover by fast train from A to meet fast train from B = 72 mile

The fast train from A meets the other fast train from B at 11:30 AM , So time taken by train A to cover 72 mile = 1 hr 30 minute

So,

$$\begin{gathered} 1+\frac{30}{60} = \frac{72}{x} \\ \Rightarrow 1+\frac{1}{2} = \frac{72}{x} \\ \Rightarrow \frac{3}{2} = \frac{72}{x} \\ \Rightarrow x = 72 \times \frac{2}{3} \\ \Rightarrow x = 24 \times 2 \\ \mathbf{\Rightarrow }\mathit{x}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{48} \end{gathered}$$
So,

Speed of fast train start from A and B  =  48 mile/ hr

Also given : The fast train from A meets the slow train at 12:30 PM. So distance travel by fast train from A to meet slow train in time 2 hr 30 minutes :

distance travel by fast train from A to meet slow train in time = $\left(2+\frac{30}{60} \right)\times 48 = \left(2+\frac{1}{2} \right)\times 48= \frac{5}{2}\times 48 = 5 \times 24$ = 120 mile

Then , Distance travel by slow train =  144 - 120 =  24 mile in ( 12:30 AM - 10:20 AM ) 2 hr 10 minutes 

So,

Speed of slow train from B  = $\frac{24}{2+{\displaystyle \frac{10}{60}}} = \frac{{\displaystyle 24}}{{\displaystyle 2+\frac{1}{6}}}= \frac{{\displaystyle 24}}{{\displaystyle \frac{13}{6}}} = \frac{\mathbf{144}}{\mathbf{13}}$ mile / hr

So, Time taken by slow train to reach from B to A = $\frac{144}{{\displaystyle \frac{144}{13}}} = 144 \times \frac{13}{144}$ = 13 hr

Therefore,

The slow train reach A  = ( 10 : 20 AM  + 13 hr ) =  11 : 20 PM                                ( Ans )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards