A pipe when work with total efficiency can fill a tank in 2 hours.However, in first two hours that pipe uses only twelth part of efficieny and in next two hours uses 9 th part of efficiency and in next two hours uses 6 th part and in next uses 1/4 th part and in next two hours uses only 1/3 rd part.now same procedure done by another pipe that has been filled the tank in 4 hours. a drain pipe is connected to that tank i.e. empty the tank with definite rate. if all three pipe can open together tank filled in 10 hours. if both first and second pipe is closed then in how much time drain pipe can empty the tank?
Answers
Answer:
24hrs
Step-by-step explanation:
let total capacity of tank = 360units
pipe1 - normal eff = 360/2 = 180/hr
working :: 2*180/12 + 2*180/9 + 2*180/6 + 2*180/4 + 2*180/3
= 30 +40 +60 +90+120 = 340units (in10hrs)
now pipe 2- normal eff = 360/4 =90/hr
working :: 2*90/12 + 2*90/9 + 2*90/6 + 2*90/4 + 2*90/3
= 15+20+30+45+60 = 170 (10hrs)
now a pipe 3 - drain pipe
p1+p2 in 10 hrs would have been filled = 340+170 = 510 units
but due to pipe 3 they filled 510-360 = 150 less in 10 hours
so pipe 3 drained 150units in 10 hours so in 1hr 15units
time taken to empty whole tank = 360/15 = 24hrs
Answer:
26 hrs ( approx)
Step-by-step explanation:
% efficiency of each A & B used when all 3 pipes worked together
= 100% ( 1/12 + 1/9 +1/6 + 1/4 + 1/3)
= 1700/18 %
in comparison to A's efficiency actual time used of A while working together
= 1700/18% of 2 = 3400/1800
= 17/9 hrs ,
of B = 1700/18 % of 4 = 68/18 = 34/9 hrs
work done by all three pipes per hour
= 1/ 10 units
1/17/9 + 1/34/9 - 1/x = 1/10
9/17 + 9/34 - 1/x = 1/10
1/x = 27/34 - 1/10
x =340/236
capacity of tank
= L.C.M ( 17/9, 34/9 , 340/236 )
= 340/9 units ( approx) .
drain pipe can empty the tank in
= 340/9/ 340/236 = 236/9 hr
= 26 hrs (approx),