Question

pipe a can fill up a cistern in 6hrs and pipe b can fill in 8 hrs. both pipes are opened and after 2 hrs pipe a is closed. how much time will b take to fill the remaining part of the tank?

Answer 1

Pipe A can fill the cistern in 6 hours

1 hour = 1/6 of the cistern

Pipe B can find the cistern in 8 hours

1 hour = 1/8 of the cistern

If both worked together

1 hour = 1/6 + 1/8 = 7/24 of the cistern

2 hours = 7/24 x 2 = 7/12 of the cistern

Find amount of work left after 2 hours

work left = 1 - 7/12 = 5/12 of the cistern

Find the amount of time needed for B to finish the rest of the work:

1 hour = 1/8 of the cistern

Number of hours needed = 5/12 ÷ 1/8 = 3 1/3 hours

3 1/3 hours =  3 hours 20 mins

Answer: It will take 3 hours 20 mins for Pipe B to fill the remaining of the tank.

Answer 2

Hey mate ^_^

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Answer:
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Part of the tank filled by pipe A and B in 2 hours

$= 2( \frac{1}{6} + \frac{1}{8} )$

$=2( \frac{4 + 3} {24})$

$= \frac{7}{12}$

Remaining part is

$= 1 - \frac{7}{12}$

$= \frac{5}{12}$

Now,

This part is filled by Pipe B.

Therefore,

The required time is

$= \frac{5}{12} \times 8$

$= \frac{10}{3}$

$= 3 \frac{1}{3}hours.$

So,

'B' will take $3 \frac{1}{3}$hours to fill the remaining part of the tank.

#Be Brainly❤️