Walking with 3/4 of his usual speed, Sachin covers a certain distance in 2 h more than the time he takes to cover the distance at his usual speed. Find the time taken by him to cover this distance with usual speed.

Let required distance = L
According to the question,
L/16 - L/20 = 15/20
? (5L - 4L)/80 = 1/4
? L = (1/4) x 80 = 20 km

Let B takes x h to walk D km.
Then, A takes ( x + 4 ) h to walk D km.
With doube of the speed,
A will take ( x+ 4)/2 h.

According to the question,
x - (x + 4)/2 = 2
? 2x - (x + 4) = 4
? 2x - x - 4 = 4
? x = 4 + 4 = 8 h

Let the speed in return journey = x
According to the question,
6( x + 2 ) = 9x
? 6x + 12 = 9x
? 9x - 6x = 12
? 3x = 12
? x = 12/3 = 4km/h

We know that speed and time are inversely proportional.
? Ratio of time taken = 1/2 : 1/3 : 1/5
= 30/2 : 30/3 : 30/5 [ LCM of 2, 3, 5 = 30 ]
= 15 : 10 : 6

Let speed of 1st bus = 5k
and speed of 2nd bus = 3k
According to the question,
5k = 400/8 = 50
? k = 50/5 = 10
? Speed of 2nd bus = 3k = 3 x 10 = 30 km/h

Speed of car = Distance/Time = 200/2⁴⁰/₆₀
= 600/8 = 75 km/h
Speed of jeep = 200/2 = 100 km/h
? Required ratio = 75/100 = 3/4 3 : 4

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

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.

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.

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