Solving a Pool Leak: Calculating Inches Lost Per Hour

[xowe]

Hello, again. I have a question problem, fairly simple, that I have most likely done wrong. It is about a pool leak.

When the pools water pump is on, from 3:50pm to 10:10am the pool lost 2.25in.

When the pump was turned off from 10:10am, to 1:50pm, it lost 7/8 in. They want to know how many inches are lost every hour. Here's how I did it.

(When water to the pools pumpe is on, from 3:50pm to 10:10am the pool lost 2.25in.)
3:50-10:10= 7hr. 2.25 divided by 7 right? And the same with the other?...xowe

 

[HallsofIvy]

It seems to me a lot easier than you make it.
First, from 3:50 pm to 10:10 am is NOT 7 hours: it is 18 hours and 20 minutes.

From 3:50pm to 10:10am the pump is on (Am I to assume that the pump is actually pumping water into the pool and not just recirculating it?- otherwise the pump is really irrelevant.) so there are two reasons for the level of water to change: change= I- O where "I" is the water pumped in (in "inches per hour") and O is the water flowing out because of the leak (also in "inches per hour")- you have
I- O= 2.25in/18.33 hrs= 0.147 in/hr. From 10:10 am to 1:50 pm, a period of 3 hours and 40 min (= 3.67 hrs) the pump was turned off so the only reason for the change in height, 7/8 in, is the leak itself:
O= (7/8 in)/(3.67 hr)= .24 in/hr.
"They want to know how many inches are lost every hour." I assume those two numbers are the answers you want.

Another possibility is that they want the average over the whole time period. To do that you would not that there was a total of
2.25+ 7/8= 2.25+ 0.875= 3.125 inches in a total of 22 hours. The average rate of drop is 3.125/22= 0.142 in/hour.