Let downstream speed be D and Upstream speed be U. Then

(⁴⁸/_D + ⁷²/_U) = 12 -------- (1)

and

(⁴⁸/_D + ⁷²/_U) = 13 ------------ (2) On solving the eqn (1) and (2), we get

D = 12 kmph and

U = 8 kmph

Thus, speed of the current = ^(D-U)/₂ = ⁴/₂ = 2 kmph.

It will come 11 times together.

A .....B .....C

Let both the trains start from point A.

At point B third train overtook the first train.

To overtake the first train by third train, Third train needs to cover,

Distance covered by first train in 1(¹/₂) h = distance covered by third train in 1 h.

[as third train has stared 30 minutes later]

In this situation distance is constant,

then from,

S*T = D; we get, S ∝ ¹/_T

Now, ^(t+1/2)/_t = ^S/₄₀

(^2t+1/_t) = ^S/₄₀ . . . . . . (1)

From equation (1),

it is clear that time is in the ratio 3:2 then speed will be in 2:3 ratios.

Hence, Speed of the Third train will be 60 km/h.

Ratio of the speed of A, B and C = 6 : 3 : 1

Then, ratio of time taken;

= ¹/₆ : ¹/₃ : 1 = 1 : 2 : 6;

Hence, time taken by A = ⁷²/₆ = 12 minutes.

Let the distance of the station from the house of the person = x km \begin{align} & {\text{Difference}}\, {\text{of}}\, {\text{time}} \cr & = 8 + 7 \cr & = 15\, {\text{minutes}} \cr & = \frac{1}{4}\, hr \cr & {\text{Since}}, \, \cr & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \therefore \frac{x}{6} - \frac{x}{{10}} = \frac{1}{4} \cr & \Rightarrow \frac{{10x - 6x}}{{60}} = \frac{1}{4} \cr & \Rightarrow \frac{{2x}}{{30}} = \frac{1}{4} \cr & \Rightarrow x = \frac{{15}}{4} = 3\frac{3}{4}km \cr\end{align}