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

1^st Method:
Let length of train be x m and speed of train is s kmph.
Speed, s = $$\frac{{x + 800}}{{100}}$$ . . . . . (i)
Speed, s = $$\frac{{x + 400}}{{60}}$$ . . . . . (ii)
Equating equation (i) and (ii), we get,
Or, $$\frac{{x + 800}}{{100}}$$  = $$\frac{{x + 400}}{{60}}$$
Or, 5x + 200 = 3x + 2400
Or, 2x = 400
Or, x = 200m

2^nd Method:
As in both cases, speed of train is constant, and n we have;
Time α distance
$$\eqalign{ & \frac{{100}}{{60}} = \frac{{x + 800}}{{x + 400}} \cr & {\text{or,}}\,\,x = 200{\text{m}} \cr} $$

Distance covered in 1 min = 50 m
Distance covered in 2 min = 90 m
Similarly, 1^st min, 2^nd min, 3^rd min . . . . . 15^th
Distance → 50 m + 90 m + 130 m + . . . . . . . .
By using A.P.
a = 50 m
d = (90 - 50) = 40 m
Tn = a + (n - 1)d
     = 50 + (15 - 1) × 40
     = 50 + 560
     = 610 minutes

Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ $$\frac{1}{{{\text{Time}}}},$$   When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.

Then,
4x - 3x = $$\frac{{20}}{{60}}$$
Or, x = $$\frac{{1}}{{3}}$$
Hence, time taken by A = 4x
hours = 4 × $$\frac{1}{3}$$ = $$1\frac{1}{3}$$ hours.

Ratio of speed of A, B and C = 6 : 3 : 1
Then, ratio of time taken;
= $$\frac{{1}}{{6}}$$ : $$\frac{{1}}{{3}}$$ : 1 = 1 : 2 : 6
Hence, time taken by A = $$\frac{{72}}{{6}}$$ = 12 minutes.

Let speed of Ravi = x km/hr
Then Ajay's speed will be = (x + 4) km/hr
Total distance, covered by Ajay :
= (60 + 12) km
= 72 km
Total distance, covered by Ravi :
= (60 - 12) km
= 48 km
According to question,
They run at same time
$$\eqalign{ & \Rightarrow \frac{{72}}{{(x + 4)}} = \frac{{48}}{x} \cr & \Rightarrow 72x = 48x + 192 \cr & \Rightarrow 24x = 192 \cr & \Rightarrow x = 8{\text{ km/hr}} \cr} $$
Therefore, Ravi's speed = 8 km/hr

$$\eqalign{ & {{\text{S}}_{{\text{average}}}} = \frac{{2ab}}{{a + b}} \cr & = \frac{{2 \times 25 \times 4}}{{25 + 4}} \cr & = \frac{{200}}{{29}}{\text{ km/hr}} \cr & {\text{Now, 2D = }}\frac{{200}}{{29}} \times \left( {5 + \frac{4}{5}} \right) \cr & = \frac{{200}}{{29}} \times \frac{{29}}{5} \cr & = 40{\text{ km}} \cr & {\text{ = D = 20 km}} \cr} $$

Speed = 10/sec
Speed = 10 × $$\frac{18}{5}$$ km/hr
Speed = 36 km/hr

1 km/hr = $$\frac{5}{18}$$
∴ 90 km/hr
= 90 × $$\frac{5}{18}$$ = 25 m/s

When A runs 1000m, B runs 900m.
Hence, when A runs 500m, B runs 450m.
Again, when B runs 400m, C runs 360m.
And, when B runs 450m, C runs = 360 × $$\frac{{450}}{{400}}$$ = 405m.
Required distance = 500 - 405 = 95 meter.
That means when A runs 500 meter n B can run 450m n C runs 405m.