A cistern can be filled by one of two pipes in 30 minutes and by the other in 36 minutes. Both pipes are opened together for a certain time but being particularly clogged only 5 / 6 of the full quantity of water flows through the former and only 9 / 10 through the later. The obstructions, however, being suddenly removed the cistern is filled in 15 ¹/ ₂ minutes from that moment. How long was it before the full flow of water began?

T = (10 x 15 x 20) / (10 x 15 + 10 x 20 + 15 x 20) = 60/13 minutes
Now, applying the given rule, we have [60/13 x y]/ [y - 60/13 + 3] = 20
or y = 21/10 = 2 min. 6 seconds.

Part filled by 1st pipe in 1 h = 1/14
Part filled by 2nd pipe in 1 h = 1/16
Part filled by the two pipes in 1 h
= ( 1/14 + 1/16 ) = (8 + 7)/112 = 15/112
? Time taken by these two pipes to fill the cistern = 112/15 h = 7 h 28 min
Due to leakage, the time taken 7 h 28 min + 92 min = 9 h
&thee4; Work done by ( two pipes + leak ) in 1 h = 1/9
Work done by the leak in 1 h = 1/9 - 15/112 = (112 - 135)/1008
= - 23/1008
? Time taken by leak to empty the full cistern
= 1008/23 = 43¹⁹/₂₃ h

Time taken to fill the tank by the tap having 2 cm diameter = 61 min

? Time taken to fill the tank by the tap having 1 cm diameter
= 61 x (2/1)² = 244 min

Similarly, time taken to fill the tank by the tap having 4/3 cm diameter
= 61 x [(2 x 3)/4]² = 61 x 9/4 = 549/4 min.

? Part of the tank filled by all the three pipes in 1 min
= 1/61 + 1/244 + 1/(549/4)
= (36 + 9 + 16)/2196 = 61/2196 = 1/36
Hence, required time taken = 36 min

Let the cistern gets emptied in m h after 3:pm
Work done by A in m h, by B in (m - 1) h and by c in (m - 2) h= 0
? m/3 + (m - 1)/4 - (m - 2) = 0
? 4m + 3(m - 1) - 12(m - 2) = 0
? 5m = 21
? m = 21/5 = 4.2
? m = 4 h 12 min
? Required time = 7 : 12 pm.

Part of the tank filled by both taps in 5 min = 5 x (1/20 + 1/60)
= 5 x (6 +2 )/120 = 8/24 = 1/3
? Remaining part = (1 - 1/3) = 2/3
? 1/60 Part is now filled in 1 min.
? 2/3 Part is now filled in 60 x 2/3 = 40 min.

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

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

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

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