A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

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.

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So,	x	+	(61 -x)	= 9
	4		9

⟹ 9x + 4(61 -x) = 9 x 36

⟹ 5x = 80

⟹ x = 16 km.

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

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.

Let distance = x km and usual rate = y kmph.

Then,	x	-	x	=	40	⟹   2y(y + 3) = 9x ....(i)
	y		y + 3		60

And,	x	-	x	=	40	⟹   y(y - 2) = 3x ....(ii)
	y -2		y		60

On dividing (i) by (ii), we get: x = 40.

Let the distance travelled by x km.

Then,	x	-	x	= 2
	10		15

⟹ 3x - 2x = 60

⟹ x = 60 km.

Time taken to travel 60 km at 10 km/hr =	❨	60	❩hrs	= 6 hrs.
		10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

∴ Required speed =	❨	60	❩kmph.	= 12 kmph.
		5