A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = 1200/(5/3) km/hr.     [We can write 1⅔ hours as 5/3 hours]

∴ Required speed =	❨	1200 x	3	❩km/hr	= 720 km/hr.
			5

Let speed of the car be x kmph.

Then, speed of the train =	150	x	=	❨	3	x	❩kmph.
	100				2

∴	75	-	75	=	125
	x		(3/2)x		10 x 60

⟹	75	-	50	=	5
	x		x		24

⟹ x =	❨	25 x24	❩	= 120 kmph.
		5

Time taken = 1 hr 40 min 48 sec = 1 hr 40	4	min = 1	51	hrs =	126	hrs.
	5		75		75

Let the actual speed be x km/hr.

Then,	5	x x	126	= 42
	7		75

⟹ x =	❨	42 x 7 x 75	❩	= 35 km/hr.
		5 x 126

Let the duration of the flight be x hours.

Then,	600	-	600	= 200
	x		x + (1/2)

⟹	600	-	1200	= 200
	x		2x + 1

⟹ x(2x + 1) = 3

⟹ 2x² + x - 3 = 0

⟹ (2x + 3)(x - 1) = 0

⟹ x = 1 hr.      [neglecting the -ve value of x]

Let the actual distance travelled be x km.

Then,	x	=	x + 20
	10		14

⟹ 14x = 10x + 200

⟹ 4x = 200

⟹ x = 50 km.

Speed =	❨	600	❩m/sec.
		5 x 60

   = 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

=	❨	2 x	18	❩	km/hr
			5

   = 7.2 km/hr.

Total time taken =	❨	160	+	160	❩hrs.	=	9	hrs.
		64		80			2

∴ Average speed =	❨	320 x	2	❩km/hr	= 71.11 km/hr.
			9

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km =	❨	9	x 60	❩min	= 10 min.
		54

Let Abhay's speed be x km/hr.

Then,	30	-	30	= 3
	x		2x

⟹ 6x = 30

⟹ x = 5 km/hr.

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then,	120	+	480	= 8       ⟹	1	+	4	=	1	....(i)
	x		y		x		y		15

And,	200	+	400	=	25	⟹	1	+	2	=	1	....(ii)
	x		y		3		x		y		24

Solving (i) and (ii), we get: x = 60 and y = 80.

∴ Ratio of speeds = 60 : 80 = 3 : 4.