Math, asked by anita167, 10 months ago

A tap A can fill a cistern in 4 hours and the tap B can empty the full cistern in 6 hours. If both the taps are opened together in the empty cistern, in how much time will the cistern be filled up?

Answers

Answered by Anonymous
35

Answer:

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your answer is here !

Step-by-step explanation:

Time taken by tap A to fill the cistern = 4 hours.

Work done by tap A in 1 hour = 1/4ᵗʰ

Time taken by tap B to empty the full cistern = 6 hours.

Work done by tap B in 1 hour = -1/6ᵗʰ (since, tap B empties the cistern).

Work done by (A + B) in 1 hour (¹/₄ - ¹/₆) = (3 - 2)/12 = 1/12th part of the tank is filled.

Therefore, the tank will fill the cistern = 12 hours.

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Answered by sonabrainly
23

Answer:

Step-by-step explanation:

A cistern can be filled by a tap in 4 hrs

A cistern can be filled by a tap in 1 hr = 1/4

A cistern emptied by an outlet pipe in 6 hrs

A cistern emptied by an outlet pipe in 1 hr =1/6

Time taken to fill the cistern if both taps are opened together in 1 hr = 1/4-1/6=1/12

Total time taken to fill the cistern = 12 hrs.

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