X and Y start walking toward each other at 8 : 00 am at the speed of 3 km/h and 4 km/h. respectively. They were initially 17.5 km apart, at what time do they meet?

Let usual speed of a man is V and time is t h.
By the formula, V₁t₁ = V₂t₂
Vt = 1¹/₂ V(t - 20/60)
? t = 3/2(t - 1/3)
? 3/2 t - t = 1/2
? t = 1 h
Hence, the time taken by the man to reach his office at his usual speed is 1 h.

Let the total distance covered = L = Length of journey
According to the question,
(L/2 x 1/40) + (L/2 x 1/60) = 16
? L/80 + L/120 = 16
? (3L + 2L)/240 = 16
? 5L = 16 x 240
? L = (16 x 240)/5 = 16 x 48 = 768 km

Due to stoppages, bus covers 9 km less per hour.
Time taken to cover 9 km =(9/54) x 60 = 10 min
Hence, the train stops on an average 10 min per hour.

Due to stoppages, it covers 10 km less per hour.
Time taken to cover 10 km = (10/50) x 60 = 12 min
Hence, the train stops on an average 12 min per hour.

Let Brijesh Catches Amit after x h.
Then, distance travelled by Amit in (x + 4) h = Distance travelled by Brijesh in x h.

According to the question,
4(x + 4) = 20 x
? 4x + 16 = 20x
? 16x = 16
? x = 16/16 = 1

Distance travelled by Brijesh in1h = 1 x 20 = 20 km

Required average speed = Total distance covered/Total time taken
= (3 x 39)/[39/26 + 39/39 +39/52]
= 3 x 13/(1/2 + 1/3 + 1/4)
= 3 x 13 x 12/13 = 36 km/h

Relative speed = 21 - 15 = 6 m per minute.
Relative distance = 114 m
Time taken to catche = 144/6 min = 19 min

Let the certain distance be d and time t.
Now, by given condition,
d/3 = ( t + 15 ) min ( t +15 )/60 h
? 20d = t + 15
? t = 20d - 15 ......(i)

And from another condition,
d/4 = (t - 15) min = ( t - 15 )/60 h
? 15d = t - 15
? t = 15d + 15 ...(ii)

From Eqs. (i) and (ii), we get
20d - 15 = 15d + 15
? 5d = 30
? d = 6 km.

Let distance and original speed of the man be d km and s km/h.
Then,
d/s - d/s + 3 = 2/3
? [d(s + 3 - s)]/[s(s + 3)] = 2/3
? 9d = 2s(s + 3) ...(i)

And d/(s - 2) - d/s = 2/3
? [d(s - s + 2)]/[s(s - 2)] = 2/3
? 3d = s(s - 2) ...(ii)

From Eqs.(i) and (ii), we get
3s(s - 2) = 2s(s + 3)
? 3s² - 6s = 2s² + 6s
? s² = 12s
? s = 12

From Eq.(ii), we get
3d = 12(12 - 2)
? d = 40 km

Here, a = 10 km/h
b = ( 10 + 2 ) = 12 km/h,
t₁ = 6 min = 1/10 h
and t₂= 1/10 h.
? Required distance = ab( t₁ + t₂ )/(b - a)
= 10 x 12 x ( 1/10 + 1/10 ) /(12 - 10) = (10 x 12 x 1)/(5 x 2) = 12km