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
Net filling in last 151/2 minutes
= 31/2 (1/30 + 1/36) = 341/360
Now, suppose they remained clogged for x minutes.
Net filling in these x minutes
= (x/30 x 5/6 + x/36 x 9/10) = 19 x/360
Remaining part = (1 - 19x/360) = (360 - 19x/360)
360 - 19x/360 = 341/360 or x = 1.
Hence, the pipes remained clogged for 1 minutes.
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
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
? Time left = (1/3 X 45/60) hr.
= 1/4 hr.
Distance left = 3km
? speed required = [3 / (1/4) ] km/hr.
= 3 x 4
=12km/hr .
Total Distance = 48 x (50/60) km.
= 40 km
Required speed = [40 / (40/60)] km.
= (40 x 60 )/ 40 km/hr.
= 60 km/hr.
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