Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15
Remaining part = 1 - 7/15 = 8/15
Part filled by B in 1 minute = 1/20
1/20 : 8/15 :: 1 : k
k = (8/15 )x 1 x 20 = 10( 2/3) min = 10 min 40 sec.
The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.
Let the leak will empty the tank in x hrs
Then, 1/9 - 1/x = 1/13
x = 29.25 hrs
Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by K = 3 liters.
Capacity of tank filled in 1 minute by L = 15 liters.
Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.
Given that the waste tap can empty the filled tank in 32 min.
Now, the rate at which the waste tap can empty the tank = (40 + 60)8/32 = 100/4 = 25 lit/min.
1/k - 1/20 = -1/24
k = 120
120 x 4 = 480
Therefore, the capacity of the cistern is 480 liters.
Work done by the third pipe in 1 min = 1/50 - (1/60 + 1/75) = - 1/100.
[-ve sign means emptying]
The third pipe alone can empty the cistern in 100 min.
Given that the diameters of the three pipes are 2 cm, 3 cm and 4 cm
From the given data,
Amount of water from three pipes is 4 units, 9 units and 16 units.
Let the capacity of cistern be 'p' units.
? p/58 = 16
? p = 928 units.
In 1 minute, quantity to be filled by 3 pipes = 29 units
? Total time required = 928/29 = 32 minutes.
Let the slower pipe alone fill the tank in x min.
Then, faster pipe will fill it in x/3 min.
1/x + 3/x = 1/36
2 pipes open for 2 hours will fill = (2/3) x (2/3) x 3 = 4/3 buckets
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.