Sump Pump (whatever that is) Question


New member
Oct 9, 2004
A sump pump in my basement kicks on automatically when the water level reaches a certain depth and turns off automatically when the depth is down to 0. One rainy night, water began to accumulate in the basement at 12:00 midnight, rising at a constant rate. The sump pump turned on at 7:30 AM and began pumping at a constant rate. At 8:00 AM the water depth was 2 inches, and water stopped coming into the basement. By 10:30 AM all the water had been pumped out.

a. Using r in./h for the rate of water accumulation and s in./h for the rate at which it is pumped out, write an equation modeling the situation at 8:00 AM

b. Using the rest of the info, find the s

c. Find the rate r at which water had been accumulating

HELP!! I can't even figure out a, let alone the rest of them... :oops:


A sump pump is a small pump that is used for removing water from a low level pit(drainage area).

Now for your problem

look at what is going on for the 8 hours, from start of rain untill 8.00 am.
rain in = 8 x r
water pumped out = 1/2 x s

Depth of water(d) = in - out
d = 8r - (1/2)s

by 10.30 no water, ie d=0, but the variables above change

d = t x r - (t-7.5) x s where t= time in hours from midnight

This is the best way to write the equation as it allows any condition to be solved.