If a train runs at 5/6 of its original speed, then it reaches the station 10 min late. Then find out the usual time taken by train to cover the distance.

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

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

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

New speed = 6/7 of usual speed
Now, time taken = 7/6 of usual time
(7/6 of the usual time) - (usual time) = 10 min
? 1/6 of the usual time = 10 min
? Usual time = 60 min

Speed in cm/minute = (Speed in km/hr x 1000 x100 )/60
= 47.52x(50/3)
=79200 cm/min

And Circumference of circle = 2?r
=2 x ( 22 / 7 ) x 21
=132

No. of revolutions = ( Speed in cm/minute ) / circumference of circle in cm
=79200 / 132
=600 rpm

speed = 3 x ( 5/18 ) m/sec.
= 5 / 6 m/sec.

? Distance covered in 2 min. = ( 5 / 6) x 2 x 60 m
= 100 m.

? Length of diagonal = 100 m

Area of the field = 1/2 x diagonal²
=1/2 x 100 x 100 m²
= 5000 m²
= 50 acres.

Let x km . be covered in y hrs.
then, 1st speed = (x / y) km/hr.

2nd speed = [(x/2) / 2y)] km/hr.
= (x/4y) km/hr.

? Ratio of speed = x/y : x/4y = 1 :1/4 = 4: 1

Let C' s speed = x km/hr.
Then, B's speed = 3x km/hr.
and A's speed = 6x km/hr.

? Ratio of speed of A, B, C
= 6x : 3x : x = 6 : 3 : 1

Ratio of times taken = 1/6 : 1/3 : 1 or 1 : 2 : 6

? 6 : 1 : : 42 : t
? 6t = 42
? t = 7 min.

Let the distance be x km.
Then, x/3 - x/4 = 30/60
? (4x - 3x ) / 12 = 1/2
? x = 6 km.