"The Two Trains"

When the trains are traveling in opposite directions, we can imagine that train B is still and train A has a speed of $(54 + 36)$ km/hr $= 90$ km/hr $= 25$ m/s. Now in this scenario, for train A to fully pass train B it will need to travel a distance of $(a + b)$ m relative to train B, (i.e., at the time the fronts of the two trains coincide, the back end of train A will be a distance of $(a + b)$ m from the back end of train B). As this event takes $10$ s, train A will have traveled $25*10 = 250$ m relative to train B, and so $a + b = 250$ m.

Now when they are traveling in the same direction, we can again imagine that train B is still, but this time train A will have a speed of $54 - 36 = 18$ km/hr = $5$ m/s. Now for a person in train A to fully cross train B they will have to have covered a distance $b$ m at $5$ m/s relative to train B, and since this event takes $30$ s we have that $b = 5*30 = 150$ m.

Thus $a = 250 - b = 100$ m, and so $\frac{a}{b} = \frac{100}{150} = \boxed{0.67}$ to $2$ decimal places.

Difference in speed = 18 km/hr = 18000m/hr = $\frac{18000}{3600}$ m $s^{-1}$ = 5m $s^{-1}$ . Using distance = time $\times$ speed (1), the length of the slow train = 5 $\times$ 30 = 150m. Now their combined speed is 90km\hr = 25m $s^{-1}$ .

Lets denote the length of the faster train as ${x}$ . Now the scenario of two trains coming past each other at a combined speed of 25m $s^{-1}$ is mathematically equivalent to the faster train going past the other (stationary) train at 25m $s^{-1}$ . This means that the distance covered in 10 seconds is 150 + ${x}$ $\Rightarrow$ 10 $\times$ 25 = 250 = 150 + ${x}$ (from (1) ) so ${x}$ = 100. $\therefore\frac{a}{b}\ = \frac{100}{150}\ = \frac{2}{3}\ = 0.67 (2 d.p)$