Question

A sludge pool is filled by two inlet pipes. One pipe can fill the pool in 10 days and the other pipe can fill it in 25 days. However, if no sewage is added, waste removal will empty the pool in 31 days. How long (in days) will it take the two inlet pipes to fill an empty pool?

Answer 1

9.28 days are taken to completely fill the pool

It is given that one pipe can fill the pool in 10 days

the other pipe can fill it in 25 days.

if no sewage is added, waste removal will empty the pool in 31 days.

The amount filled by pipe one in 1 day=1/10

The amount of pool filled by second pipe in 1 day=1/25

The waste removal pipe's work in 1 day= 1/31

Therefore, the work done together by all pipes in one day=1/10+1/25-1/31

= (775+310-250)/31*25*10=835/7750

Therefore, number of days taken to empty the pool= 7750/835=9.28 days

As half day cannot be counted, hence 9.28 days are taken to completely fill the pool

Disclaimer: The waste removal pipe has also been taken into consideration

For further reference:

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Answer 2

Main Answer: 8.91 days

Concept and definitions should be there:

The smallest positive number that is a multiple of two or more numbers. The Least common Multiple we said LCM

Given data

A sludge pool is filled by two inlet pipes. One pipe can fill the pool in 10 days and the other pipe can fill it in 25 days. However, if no sewage is added, waste removal will empty the pool in 31 days.

Solving part

Let t = time in days to fill the pool

let the competed job (a full pool) = 1

Each will do a fraction of the job, the 3 fractions will add up to 1

waste removal will be negative.

[(t/10)+(t/25)-(1/36)] = 1

multiply equation by LCM, 1800, to get rid of the denominators

180t + 72t - 50t = 1800

202t = 1800

t = 1800/202

t = 8.91 days

Final Answer:8.91 days

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