Question 284124
The vent fills the pool in 10 hours, so it fills at a rate of 1/10 of the job per hr.
The drain empties the pool in 20 hours , so it drains at a rate of -1/20 per hr.
We can imagine the pool being full as = 1.
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x(1/10) -x(1/20) = 1 job
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Multiply through by 20.
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2x - x = 20
x = 20
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Checking, we can find how much water was put in vs. how much water was pulled out.
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20 hr * 1/10 per hr = 2  :: So in 20 hrs the pool can be filled twice.
20 hr * 1/20 per hr = 1 :: So in 20 hrs the pool can be drained once.
2-1 = 1, which suggests the pool is full at that instant.
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But I suspect the pool cannot hold twice it's normal volume. That is, once you have it full, adding more water just flows over the rim and floods the building.  So, perhaps 20 hrs is too long.
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Perhaps a better model is to suggest the pool is filling and draining simultaneously.  That means it fills at a net rate of 1/10 - 1/20 per hr.
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x(1/10-1/20) = 1
x(1/20) = 1
x = 20
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Hmmm...20 again.
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So, the answer is 20 hrs.
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But, after 20 hrs the pool will be flooding the building since it will still be adding water at the rate of 1/10 the volume of the pool per hr minus 1/20 of the volume per hr or a net rate of 1/20 of the volume per hr.
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Done.