Moving 6/7 of its usual speed a train is 10 min late. Find its usual time to cover the journey.

Let L km distance be covered in T h. So, speed in first case = L/T km/h.
And speed in second case = (L/2)/2T = L/4T km/h
? Required ratio = L/T : L/4T =1 : 1/4 = 4 : 1

Total distance = 30 + 40 = 70 km
Total time taken = (30/6) + 5 = 10 h
? Required average speed = 70/10 = 7 km/h

We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2

We know that, speed = Distance / Time
Speed = 12/90 km/min

? Distance covered in 3 h = Speed x Time
= (12/90) x (3 x 60) = 24km

We know that for relative speed, the subtraction of speeds takes place, if the motions of the two objects are in the same direction.
? Required relative speed = 9 - 5 = 4 km/h

John's speed : Vinod's speed = 75 : ( 125 - 75 )
= 75 : 50 = 3 : 2

Average speed = 2AB/(A + B)
= 2 x 5 x 3/(5 + 3)
= 30/8
= 3.75 km/h

Distance= Speed x Time = 96 x (100/60) = 160 km
New time = 80/60 h = 4/3 h
New speed = 160 x 3/4 = 40 x 3 =120 km/h

Required average speed = 75 x {(100 - 40)/100} x {(100 + 50)/100}
= 75 x 3/5 x 3/2
= 67.5 km/h

Let the actual speed of train be x and actual time taken be y

Then new speed of train = 5x/6

Therefore, new time taken = 6y/5 (as distance is same in both case)

Given, 6y/5 - y = 1/6 hr , therefore actual time = 50 min