Question

A cistern has three pipes, A, B and C. The pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are" opened in order at 1, 2 and 3 A.M. when will the cistern be empty​

Answer 1

Answer:

7 : 12 pm

Step-by-step explanation:

Pipe A is opened at 3 p.m.,

pipe B at 4 p.m. and the pipe C at 5 p.m.

Part of the tank, filled by pipe

A in 2 hours =23

Part of the tank filled by pipe

B in 1 hour =14

Part of the tank filled till 5 p.m.

=23+14=8+312=1112

Remaining part =1−1112=112

Net part emptied when A, B and C are opened

=13+14−1=4+3−1212=−512

∴512Partis emptied in1hour

∴1112is emptiedin

=125×1112=115hours

=2hours12minutes

Hence the tank will be emptied at 7:12p.m