A train covers a distance in 1 h 40 min, if it runs at a speed of 96 km/h on an average. Find the speed at which the train must run reduce the time of journey to 1 h 20 min.

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

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

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

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