Pipe A takes of the time required by pipe B to fill an empty tank individually. When an outlet pipe C is also opened simultaneously with pipes A and B, it takes more time to fill the empty tank than it takes when only pipe A and pipe B are opened together. If it takes 40 hours to fill the tank when all three pipes are opened simultaneously, at what times ( in hours) will pipe C empty the full tank, operating alone?
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 50
Let the time pipe A, pipe B, and pipe C fill the tank be x, y, and z respectively.
Pipe A takes 4/5 of the time taken by pipe B.
So, x = (4/5)y
⇒ y = (5/4)x
Pipe A and B fill in one hour = (1/x) + (1/y) = (1/x) + (4/5x) = 9/5x of the tank
Time taken by A and B together to fill the tank = 5x/9
When A, B, and C are open, it takes 4/5 more time than A and B together.
Time taken by A, B and C together = (1 + 4/5)(5x/9) = (9/5)(5x/9) = x
It takes 40 hours to fill the tank when all three pipes are open.
So, x = 40
Time taken by A and B together = 5(40/9) = 200/9 hours
Pipe A, B, and C fill in one hour = (1/x) + (1/y) - (1/z)
= (1/x) + (4/5x) - (1/z)
= (9/5x) - (1/z) of the tank
Now, (9/5x - 1/z) × 40 = 1
⇒ (9/200 - 1/z) × 40 = 1
⇒ 9/5 - 40/z = 1
⇒ 40/z = 4/5
⇒ z = 50
Therefore, Pipe C will empty the full tank in 50 hours.