Question

A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours respectively, and C can empty it in 1 hour. If the pipes are opened at 3 p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be empty at​

Answer 1

Solution :-

Let the cistern be emptied at x p.m.

We know that x - 3/3 + x – 4/4 = x – 5/1

=> 4x – 12 + 3x – 12/12 = x – 5/1

=> 7x – 24 = 12x – 60

=> 5x = 36

We know x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.