two pipes can fill the tank in 4 hours and 8 hours respectively. while the third pipe empties the full tank in 12 hours. if all three pipes are simultaneously operated how long will it take to for tank to be full
Answers
Answer: I hope it will help u...
Step-by-step explanation:
Let two filling pipes be A and B, draining pipe is C.
Given that pipe A can fill a tank in 10 hours
=> In one hour it fills 1/10th part of the tank ( by cross multiplication method)
And pipe B can fill a tank in 12 hours
=> In one hour it fills 1/12th part of the tank.
And also given pipe C empties the tank in 20 hours => in one hour it empties 1/20th part.
If all three pipes operate simultaneously work done by both pipes A and B in one hour will be added and C work is subtracted.
Therefore work done by three pipes in one hour = 1/10 + 1/12 -1/20 = 8/60
In 1 hour 8/60th part of a tank can be filled.
Then complete tank can be filled in 60/8 = 7.5hrs
( by cross multiplication)
OR
Consider x, y and z are the velocity in three different pipes and v as volume of he tank. Then,
10x = 12y = 20z = v
Hence
y = x/2 x and z = 5x/6
Considering all the three pipes are working,
No of hours * (x+y-z) = v
No of hours *(x + 1/2x + 5/6x) = v
No of hours*4x/3 = v
With velocity of x, the tank is filled in 10 hrs.
Then with velocity of 4x/3, time taken will be
3*10/4 = 7.5 hrs
When all the three pipes are working simultaneously, time taken to fill the tank is 7.5 hours