A bus covers a certain distance in 16 h. It covers half the distance at 40 km/h and the rest at 60 km/h. Find the length of the journey.?

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

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 th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h

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 they meet after T h.
then, according to the question, 3T + 4T = 17.5
? 7T = 17.5
? T = 17.5/7 = 2.5 h
So, they meet 2.5 h after 8 : 00 am.
It means they meet at 10 : 30 am.

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.