Required answer = (12 x 16) /(12 + 16) = 48 / 7 = 66/7 minutes.
Work done by leak in 1 hour = (1/8 - 1/10) = 1/40
? The leak will empty the cistern in 40 hours.
We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1h = 1/6 part
Part filled A in 1 hr= (1/45)
Part filled B in 1 hr= (1/36)
Part filled by (A+B) together in 1 hr=(1/45)+(1/36)=1/20
So, The tank will be full in 20 hr.
Let the work lasted for N days,
Then, Rashmi's 4 day's work + Ravina (N - 3) day's + Gitika's N day's work = 1
? (4/16) + (N - 3) / (124/5) + N/32 = 1
? 5(N - 3)/64 + N/32 = 1 - 1/4
? [5(N - 3) + 2N] / 64 = 3/4
? 7N - 15 = 48
? N= (48 + 15)/7 = 63/7 = 9 days
Let the work be finished in N days.
Then, A's N day's work + B's (N - 1) days + work + C's (N - 2) day's work = 1
? N/16 + (N - 1/32) + (N - 2/48) = 1
? N/1 + N - 1/2 + N - 2/3 = 16
? (6N + 3N - 3 + 2N - 4)/6 = 16
? 11N - 7 = 96
? 11N = 103
? N = 103/11 = 94/11 days
Here x = 8 hrs. and y = 8 + 2 = 10 hrs.
Now, applying the given rule, we have the
Required answer = (8 x 10) /(10 - 8) = 40 hrs.
We know that, when a pipe fills 1/m part of a tank in 1 h, then the pipe takes m h to fill the tank.
Here, 1/m = 1/8 ? m = 8
? Required time to fill the tank = 8 h
Time taken by the pipe to empty the cistern = 3 h
Then, time taken by the pipe to empty the 2/3 part = 3 x 2/3 = 2 h
Let the leak empties the full tank in x h, then
Part emptied in 1 by leak = 1/x
Part filled by inlet in 1 h = 1/20
According to the question,
1/20 + 1/x = 1/40
? 1/x = 1/40 - 1/20 = 1- 2/40 = -1/40 [-ve sign Indicates emptying.]
Clearly, leak will empty the full tank in 40 h.
We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1 h = 1/6 part
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