1/8 - 1/x = 1/10
=> x = 40 hrs
Now, it is the turn of A and B (3/20) part is filled by A and B in 1 hour.
Therefore, Total time taken to fill the tank =(6+1)hrs= 7 hrs
Part filled in 2 hours = 2/6=1/3
Remaining part =
(A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }
=1/6-2/21 = 1/14
C alone can fill the tank in 14 hours.
Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour =4*1/6 =2/3
Remaining part =
=> x =
So, total time taken = 3 hrs. 45 mins.
If the total area of pump=1 part
The pumop take 2 hrs to fill 1 part
The pumop take1 hour to fill 1/2 portion
Due to lickage
The pumop take 7/3 hrs to fill 1 part
The pumop take1 hour to fill 3/7 portion
Now the difference of area = (1/2-3/7)=1/14
This 1/14 part of water drains in 1 hour
Total area=1 part of water drains in (1x14/1)hours= 14 hours
So the leak can drain all the water of the tank in 14 hours.
Leak will empty the tank in 14 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 : 8/15 :: 1:x
x = (8/15*1*20) =
The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec
Clearly,pipe B is faster than pipe A and so,the tank will be emptied.
part to be emptied = 2/5
part emptied by (A+B) in 1 minute=
so, the tank will be emptied in 6 min
Let the filling capacity of the pump be x /min.
Then, emptying capacity of the pump=(x+10) /min.
so,
1/10 + 1/15 = 1/6 x 3 = 1/2
1 - 1/2 = 1/2
1/10 + 1/15 - 1/x = 1/2
x = 8.57 min
Work done by the waste pipe in 1 minute = [-ve sign means emptying]
Volume of part = 3 gallons
Volume of whole = (3 x 40) gallons = 120 gallons.
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