Question

pipe a could fill an empty cistern in 18 hr while pipe b can drain a filled a cistern in 30 hrs when the cistern empty pipe a is turned on for an hour and then turned off now pipe is allowed to drain out water from the cistern for an hour and then turned off the pipe were alternately left open for an hour each time till the cistern was full how much time did it take for the cistern too be full​

Answer 1

Answer:

90 hours

Step-by-step explanation:

We analyze the question as follows :

In one hour the amounts emptied and filled are as follows :

a = 1/18

b = 1/30

These taps are opened alternatively.

In 1 hr the amount in the tank will be = 1/18

The amount emptied in 1 hr = 1/30

The amount left in the tank after 2 hours is:

1/18 - 1/30 = 1/45

In 2 hours the amount that will be in the tank will be = 1/45

It takes 2hrs to fill 1/45 of the tank, how long will it take to fill the tank?

2 hrs — 1/45

? — 1

1 × 45/1 × 2 = 90 hours

It takes 90 hours to fill the tank if the taps are opened alternatively.

Answer 2

Answer:

86 hrs 48 minutes

Step-by-step explanation:

A--->18 5

90

B--->30 3

1st hr=5 part filled by pipe A

2nd hr=3 part drained by Pipe B

i,e., in 2 hour 2 part is filled only.

so, in 86 hour 86 part will be filled.

Reaming 4 part will be filled by A with 5 eff. so time required=⅘hr =48 minutes.

Total time=86 hr 48 minutes.