Question

Two taps can fill a cistern in 6 min. and 7 min. respectively. if these taps are opened alternatively for a minute, in what time will the cistern be filled

Answer 1

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First tap can fill 1/6 part of cistern in 1 min

Second tap can fill 1/7 part of cistern in 1 min

Let both can fill it in x min(together)
Both can fill 1/x part in 1 min
1/x=1/6+1/7
1/x=13/42
x=42/13
Both together can fill the cistern in 42/13 minutes.

Answer 2

Cistern will be filled in 6 3/7 min.

Given:

Two taps can fill a cistern in 6 min. and 7 min. respectively.

To find:

If these taps are opened alternatively for a minute, in what time will the cistern be filled.

Solution:

As the pipes are operating alternatively, thus their 2 minutes job is = 1/6 + 1/7 = 13/42

In the every 2 minutes the pipes can fill another 13/42 part of cistern.

Taking an estimate of 6 minutes to fill the cistern.

Therefore, In 6 minutes the two pipes which are operating alternatively will fill

13/42 + 13/42 + 13/42 = 13/14 of the cistern

Remaining part = 1 - 13/14 = 1/14

As both pipes have filled alternatively, now is the turn of pipe A to fill th cistern.

Pipe A can fill 1/6 of the cistern in 1 minute

Pipe A can fill 1/14 of the cistern in = 6/14 = 3/7 min

Therefore, Total time taken to fill the Cistern 6 + 3/7 min.

Hence, our answer is 6 3/7 min.

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