A certain distance is covered at a certain speed. If half of this distance is covered in double the time, the ratio of the two speed is?

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.

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

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

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

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.

Distance covered by thief in (1/2 ) hour = 20 km.
Now , 20 km is compensated by the owner at a relative speed of 10 km/hr in 2 hours so, he overtake the thief at 4 p.m.

Suppose they meet after y hours .
Then, 21y - 16y = 60
? y = 12

?required distance = (16 x 12 + 21 x 12) km.
= 444 km.

Let the original speed be x km/hr.
Then, [ 840/x - 840/( x + 10) ] = 2
? 840 (x + 10) - 840x = 2x (x+10)
? x² + 10x - 4200 = 0
? (x + 70) (x - 60 ) = 0
? x = 60 km/hr.