Question

An empty cistern can be filled by tab in 10 hours tab in 12 hours and tap C in 15 hours respectively. All taps being together but tap A stopped working after 2 hours and tap B stopped working 3 hours before the cistern is filled. How long did tap C worked to fill the empty cistern.​

Answer 1

Answer:

From the information, in 1 hour;

Tap A, running alone, would fill 1/10 of the tank.

Tap B, running alone, would fill 1/15 of the tank.

Running together, both taps would fill

1/10+1/15= 1/6 of the tank in 1 hour I.e it would take 6 hours to fill the tank with both taps running.

If Tap A alone runs for 8 hours it would fill 1/10 x 8 hours = 4/5 of the tank.

This means both taps had already filled 1/5 when they were running together.

Since 1/6 of the tank is filled per hour when both tanks are running, how much time would both take to fill 1/5 of the tank. DIVIDE

=1/5 ÷ 1/6

=1.2 hours = 1 hour 12 minutes

Both taps were running for that time before tap B was close