Question

Pipes a,b,c empty a tank in 12, 15, 20 hrs respectively such that pipe a is opened always. Pipes b and c are opened alternatively. In how many hours tank will be emptied?

Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

 Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 minutes

Answer 2

[HeY Mate]

Answer:

Part filled by pipe A in 1 hour

=1/12

Part filled by pipe B in 1 hour

=1/15

Part filled by pipe C in 1 hour

=1/20

In first hour, A and B are open.

In second hour, A and C are open.

then this pattern goes on till the tank is completely filled.

Part filled by pipe A and B in 1 hour

=1/12 + 1/15

=9/60

=3/20

Part filled by pipe A and C in 1 hour

=1/12 + 1/20

=8/60

=2/15

Part filled in 2 hour

=3/20 + 2/15

=17/60

Remaining part

=(1−17/20)

=3/20

Now, 6 hours are over and only

3/20 part needed to be filled.

In 7th hour, A and B are open and fill this remaining 3/20 part. Thus, tank is totally filled.

Therefore, total time taken

=7 hour

I Hope It Helps You✌️