Jeff O. answered • 3d

Knowledgeable Ivy League Math and Science Tutor

A) Since the water is draining out of the swimming pool at a constant rate of 780 gallons/hour, the amount of water remaining in the pool after h hours can be expressed by the following equation:

V_{remain}(h) = 45000 - 780*h

B) As the chemical additive must be added to the pool when it has more than 15000 gallons of water remaining, we can translate this statement directly into an inequality to solve for the least number of hours before the chemical need to be added:

45000 - 780*h > 15000

-780*h > -30000

h < 38.46 (hours)

Of course, this is the most number of hours to add chemical to the pool before pool has less than 15000 gallons of water left. Since the problem asks for the least number of hours before the chemical need to be added, the answer is 0 hour, right when the pool is drained. After all, according to the problem, the chemical must be added when there are more than 15000 gallons of water in the pool (45000 gal > 15000 gal)

C) No, based on answer from B), it will take at most 38.46 hours before chemical need to be added. This is still less than 2 days (48 hours).