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

Ajay → (x + 4) kmph.
A ________ 60 km _________ B
Ravi → x kmph.

Let speed of Ravi be x kmph;
Hence, Ajay's speed = (x + 4) kmph;
Distance covered by Ajay = 60 + 12 = 72 km;
Distance covered by Ravi = 60 - 12 = 48 km.

According to question,
$$\eqalign{ & \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & {\text{or,}}\,\frac{3}{{x + 4}} = \frac{2}{x} \cr & {\text{or,}}3x = 2x + 8 \cr & {\text{or,}}x = 8\,{\text{kmph}} \cr} $$

Let speed of truck is = x km/hr
Their relative speed in same direction = (45 - x) km/hr
[Here (45 - x) has been written because bus crosses truck which is running 150 metres ahead of it. i.e., Truck speed will be lower than that of bus]
According to question,
$$\eqalign{ & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \Rightarrow \frac{{150}}{{\left( {45 - x} \right) \times \frac{5}{{18}}}} = 30 \cr & \Rightarrow \frac{{150 \times 18}}{{\left( {45 - x} \right) \times 5}} = 30 \cr & \Rightarrow x = 27{\text{ km/hr}} \cr} $$

1^st method:
Let speed of train be S. And length of train be x.
When a train crosses a man, its travels its own distance.
$$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$

Total distance
= $$4 \times 3\frac{3}{4}$$
= 15 km
Time taken on cycle :
= $$\frac{{15}}{{16.5}} \times 60$$
= 54.55 minutes

Let Speed of man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or, $$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph.

Detailed Explanation:
A _____________M_______________B
A = Bus Terminal.

B = Let meeting point of first bus and man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well.

Distance Covered by Bus in 10 min = AB = $$\frac{{20}}{{60}} \times 10$$   = $$\frac{{10}}{3}$$ km.

Now, M is Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes.
Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each or. Let Speed of man = x kmph.

Relative speed = 20 + x

To meet at Point M, bus and Man has covered distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

1 hour 45 minutes = $$1 + \frac{{45}}{{60}}$$  = $$\frac{7}{4}$$ hours.
Speed of motorboat up-stream,
$$\eqalign{ & = \frac{{{\text{Distance}}}}{{{\text{Time}}\,\,{\text{Taken}}}} \cr & = \frac{{56\,{\text{km}}}}{{\frac{7}{4}{\text{hours}}}} \cr & = \frac{{56 \times 4}}{7} \cr & = 32\,{\text{kmph}} \cr} $$
Let speed of current = x kmph
Hence, 36 - x = 32
Or, x = 36 - 32 = 4 kmph
Speed of boat down stream = 36 + 4 = 40 kmph.
∴ Time taken to cover 56 km at 40 kmph = $$\frac{{56}}{{40}}$$ = $$\frac{7}{5}$$ hours
or 1 hours 24 minutes.

According to question,


5 units → 12 hr
1 unit → $$\frac{12}{5}$$ hr and,
4 units → $$\frac{12}{5}$$ × 4
             = $$\frac{48}{5}$$
Distance travelled on foot :
= $$\frac{48}{5}$$ × 5
= 48 km

Let length of train be $$l$$ metre
According to question,
Time = $$\frac{{{\text{Distance}}}}{{{\text{Speed}}}}$$
⇒ 100 =$$\frac{{{\text{500 + }}l}}{{{\text{Speed}}}}$$
⇒ Speed = $$\frac{{500 + l}}{{100}}$$    ..... (i)
Again,
60 = $$\frac{{{\text{250 + }}l}}{{{\text{Speed of Train}}}}$$
Speed = $$\frac{{{\text{250 + }}l}}{{{\text{60}}}}$$    ..... (ii)
Equation (i) and (ii)
$$\eqalign{ & \Rightarrow \frac{{500 + l}}{{100}} = \frac{{250}}{{60}} \cr & \Rightarrow 1500 + 3l = 1250 + 5l \cr & \Rightarrow 2l = 250 \cr & \refore {\text{Length of train = 125 metres}} \cr} $$

$$\eqalign{ & {\text{Average speed :}} \cr & {\text{ = }}\frac{{2xy}}{{x + y}} \cr & = \frac{{2 \times 16 \times y}}{{16 + y}} \cr & \Rightarrow 6.4 = \frac{{32y}}{{16 + y}} \cr & \Rightarrow 102.4 + 6.4 = 32y \cr & \Rightarrow 102.4 = 25.6y \cr & \Rightarrow y = \frac{{102.4}}{{25.6}} \cr & \Rightarrow y = 4{\text{ km/hr}} \cr} $$