Net filling in 1 min. = (1/32 + 1/36 - 1/20) = 13/1440
? Time taken to fill the tank = (1440/13) min.
Time taken to fill half of the tank = (1440/13) x 2 min. = (720/13) min.
= 555/13 min.
Work done by waste pipe in 1 min. = (1/10) + (1/15) - (1/18)
= (1/6) - (1/18) = 1/9
? Waste pipe can empty the cistern in 9 min.
Work done by waste pipe in 1 min. = (1/12 + 1/15) - 1/20
= (3/20 - 1/20) = 1/10
? Waste pipe can empty the cistern in 10 min.
Work done by both the taps in 5 min
= 5 (1/20 +1/25) = (5 x 9)/100 = 9/20
Remaining part = (1 - 9/20) = 11/20
Now, 1/20 part is filled in 1 min.
So, 11/20 part will be filled in 11 min.
Hence, the tank will be full in 11 more min.
Let the number of minutes taken to empty the cistern be T min.
According to the questions,
T/6 - (T + 5)/12 - (T + 5)/15 = 0
? T/6 - T/12 - 5/12 - T/12 - T/12 - T/15 = 0
? T/6 - T/12 - T/15 = 5/12 + 5/15
? (10T - 5T - 4T)/ 60 = (25 + 20)/60
? T/60 = 45/60
? T = 45 min.
Here, A = 24 min, B = 32, T = 9 min
? Required time = B(1 - T/A )
= 32 ( 1 - 9/24 )
= 32 x 15/24 = 20 min.
Part filled in 2 hours = 2 x 1/6 = 1/3
Remaining part = (1 - 1/3) = 2/3
(A + B)'s 7 hour's work = 2/3
? (A + B)'s 1 hour's work = (2/3 x 1/7) = 2/21
(A + B + C)'s 1 hour's work = 1/6
C's 1 hours work = (1/6 - 2/21) = 1/14
Hence, C alone can fill the cistern in 14 hours.
Suppose that one pipe takes N hours to fill the reservoir. Then another pipe takes (N - 10) hours.
? 1/N + 1/(N - 10) = 1/12
? 12(N - 10 + N) = N(N - 10)
or N2 - 34N + 120 = 0
or (N - 30)(N - 4) = 0
? N = 30 or N = 4
So, the faster pipe takes 30 hours to fill the reservoir.
Part filled in 10 hours = 10 x (1/15 + 1/20 - 1/25) = 23/30
Remaining part = (1 - 23/30) = 7/30
Now, (1/15 + 1/20) part is filled by A and B in 1 hr.
7/30 part be filled them in (60/7) x (7/30) = 2 hrs.
? Total time taken to fill the tank = (10 + 2) hrs. = 12 hrs.
Let the leak empty the full cistern in N hours
Now, applying the given rule (9 x N) / (N - 9) = 9 + 1
or N = 90 hours.
? 7/24 of the cistern will be filled up in 1/2 hr.
? The whole of the cistern will be filled up in (1/2) x (24/7) = 12/7 hrs.
Let the pipe C be empty the whole cistern in h hours
Now, applying the given rule we have,
(2 x 3 x h) / (3 x h + 2 x h - 2 x 3) = 12/7
or 42h = 60h - 72
? h = 4 hours.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.