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.
Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120
For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120
So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120
So remaining tank = 13/120
Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280.
So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes
Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.
The time taken by the leak to empty the tank =
Therefore, the leak empties the tank in 40 hours.
Let pipe A takes p min to fill
Then,
pipe B takes 3p min to fill
=> 3p - p = 32
=> p = 16 min => 3p = 48 min
Required, both pipes to fill = (48 x 16)/(48 + 16) min = 12 min.
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the taps in 1 hour = 4 x 1/6 = 2/3
Remaining part = 1 - 1/2 = 1/2
2/3 : 1/2 :: 1 : p
p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min
So, total time taken = 3 hrs 45 min.
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.
Let the leak will empty the tank in x hrs
Then, 1/9 - 1/x = 1/13
x = 29.25 hrs
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.
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.
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