Laxman has to cover a distance of 6 km in 45 minutes . If he cover one half of the distance in 2/3 ^rd time . what should be his speed to cover the remaining distance in the remaining time?

Let the side of the square field be x and the average speed of plane be y

x/200 + x/400 + x/600 + x/800 = 4x/y
? 25x/2400 = 4x/y
? y =384

? Average speed is 384 km/hr

Rohit's speed = 10 Km/hr = (10 x 5/18) m/sec = 50/18 m/sec

Time taken to cover 800 m = (800 / (50/18) ) sec
= (800 x 18/50) sec = 288 sec = 4 min 48 sec

Speed = (1500/300 ) m/sec
= 5 m/sec
= (5 x 18/5 ) km/hr
= 18 km/hr

Suppose they will meet after T hours.
Distance = Speed x Time
Sum of distance traveled by them after T hours
6T + 4T = 20 km
T = 2 hours.
So they will meet at 7:00 AM + 2 hours = 9:00 AM

Let the leak empties it in T hours
From the given rule, we have (T x 30) / (T - 30) = 40
? T = 120 minutes = 2 hours.

Now, from the question, applying the rule, we have time taken by B to fill the tank when crack in the bottom develops
= (120 x 40) / (120 - 40) = 60 minutes = 1 hour

Total Distance = 48 x (50/60) km.
= 40 km

Required speed = [40 / (40/60)] km.
= (40 x 60 )/ 40 km/hr.
= 60 km/hr.

Due to stoppages, it covers 9 km. less per hour

Required time = Time taken to cover 9 km.
= 9 / ( 54 /60) min
= 10 min

Let the distance be y km.
From question
y/3 - y/3.75 = 1/2
? (3.75y - 3y) / (3 x 3.75) = 1/2
? 1.5y= 3 x 3.75
? y = ( 3 x 3.75 ) / 1.5
= 7.5 km.

Total time taken = (3/10 + 3/20 + 3/30 + 3/60) hrs.
= 3/5 hrs.

? Average speed = {12 / (3 / 5)} km/hr.
= (12 x 5 ) / 3 km/hr.
= 20 km/hr.

Let the length of total journey be y km.

From question
( y/2 x 1/40 ) + ( y/2 x 1/60) = 8
? y/80 + y/120 = 8
? 3y + 2y = 1920
? y = 384 km.