$$\eqalign{ & {\text{Average speed :}} \cr & = \frac{{{\text{Total distance covered }}}}{{{\text{Total time taken}}}} \cr & = \left( {\frac{{4 \times 9 \times 400}}{{88 + 96 + 89 + 87}}} \right){\text{m/min}} \cr & = \left( {\frac{{14400}}{{360}}} \right){\text{m/min}} \cr & = 40{\text{ m/min}} \cr} $$

In first situation,
⇒ Total distance covered by train :
= 80 × $$4\frac{1}{2}$$
= 360 kms
⇒ Therefore,
The speed of train to cover same distance 360 km in 4 hours is :
$$\eqalign{ & = \frac{{360}}{4}\left\{ {{\text{Speed }} = \frac{{{\text{Distance }}}}{{{\text{Time}}}}} \right\} \cr & = 90{\text{ km/h}} \cr} $$

Rani goes to school from her house = 30 minutes
Raja goes to school from his house = 45 minutes
Required ratio :
= 30 : 45
= 2 : 3

Let distance be x km
According to question,
$$\eqalign{ & \frac{x}{{12}} + \frac{x}{9} = 2 + \frac{{20}}{{60}} \cr & \frac{{3x + 4x}}{{36}} = \frac{7}{3} \cr & \frac{{7x}}{{36}} = \frac{7}{3} \cr & \boxed{x = 12{\text{ km}}} \cr} $$

At time of meeting, let distance travelled by first train be x km.
Then, discount covered by second train = (x + 120) km
$$\eqalign{ & \refore \frac{x}{{50}} = \frac{{x + 120}}{{60}} \cr & \Leftrightarrow 60x = 50x + 6000 \cr & \Leftrightarrow 10x = 6000 \cr & \Leftrightarrow x = 600 \cr} $$
So, distance between A and B
= (x + x + 12) km
= 1320 km

$$\eqalign{ & {\text{Due}}\,{\text{to}}\,{\text{stoppages,}}\,{\text{it}}\,{\text{covers}}\,{\text{9}}\,{\text{km}}\,{\text{less}} \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{cover}}\,{\text{9}}\,{\text{km}} \cr & = \left( {\frac{9}{{54}} \times 60} \right)\min \cr & = 10\ \text{minutes} \cr} $$

Suppose thief is overtaken x hrs after 2.30 pm
The, distance covered by thief in x hrs = Distance covered by owner in $$\left( {x - \frac{1}{2}} \right)$$  hrs
$$\eqalign{ & \refore 60x = 75\left( {x - \frac{1}{2}} \right) \cr & \Leftrightarrow 15x = \frac{{75}}{2} \cr & \Leftrightarrow x = \frac{5}{2}{\text{ hrs}} \cr} $$
So, if is overtaken at 5 pm

Actual time taken for journey :
= 4 hrs 15 min - (10 + 2 × 5 + 3) min
= 4 hrs 15 min - 23 min
= 3 hrs 52 min
= $$3\frac{26}{30}$$ hrs
= $$\frac{116}{30}$$ hrs
∴ Average speed :
$$\eqalign{ & = \left( {\frac{{580}}{3} \times \frac{{30}}{{116}}} \right){\text{ km/hr}} \cr & = 50{\text{ km/hr}} \cr} $$

Average speed :
$$\eqalign{ & = \frac{{\left( {2 \times \frac{5}{2} \times \frac{{13}}{4}} \right)}}{{\left( {\frac{5}{2} + \frac{{13}}{4}} \right)}}{\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{65}}{4} \times \frac{4}{{23}}} \right){\text{ km/hr}} \cr & = \left( {\frac{{65}}{{23}}} \right){\text{ km/hr}} \cr} $$
Total time taken :
= 4 hr 36 min
$$= 4\frac{36}{60}\,\, hr$$
$$= 4\frac{3}{5}\,\, hr$$
$$= \frac{23}{5}\,\, hr$$
Total distance covered uphill and downhill :
$$\eqalign{ & = \left( {\frac{{65}}{{23}} \times \frac{{23}}{5}} \right){\text{ km}} \cr & = 13{\text{ km}} \cr} $$
∴ Distance walked uphill :
$$\eqalign{ & = \left( {\frac{{13}}{2}} \right){\text{ km}} \cr & = 6\frac{1}{2}{\text{ km}} \cr} $$

In final minute of collision, two boats travel $$\frac{{5}}{{60}}$$ km and $$\frac{{10}}{{60}}$$ km.
As y moves towards each or relative distance covered in 1 minute,
$$\eqalign{ & = \frac{1}{{12}} + \frac{1}{6} \cr & = \frac{1}{4}\,\,{\text{km}} \cr} $$