Two pipes P and Qbcan fill a tank in 8 hours and 24 hours respectively. There is a leak which can empty the full tank in 12 hours. If leak is actually at 2/3 of the height from the base, then how many hours will the empty tank will be filled if all pipes are opened simultaneously?
Answers
Answered by
3
Step-by-step explanation:
Therefore, n2 = 9 x 6.25 => n = 3 x 2.5 = 7.5 Thus, working together, pipes A and B require 7.5 hours. Three pipes A, B and C were opened to fill a cistern. Working alone, A, B and C require 12, 15 and 20 minutes respectively.
Please mark as brainliest
Answered by
2
Answer:
Step-by-step explanation:
P(Filling pipe) - 8 hrs
Q(Filling pipe) - 24 hrs
R(Leak pipe) - 12 hrs
Taking LCM we get 24 which is total wotk
So, Efficiencies will be-
P - 3
Q - 1
R (-2)
Now till 2/3 height there is no use of leak pipe R so p+q will do the 2/3 work i.e ( 16 unit) in 16/4 = 4 hrs
Now Work left is 8 unit
This will be done by all three taps
So, 8 unit work will be done by p+q+r with efficiency (3+1-2) in
8/2= 4 hrs.
Total time required is 4+4=8 hrs
✔️
Similar questions