This task illustrates that different but equivalent expressions can be used in a function rule to provide valuable information about the context of the problem.
Sandra and Tim have a circular above-ground pool in their backyard that is 5 meters in diameter. They decide to have the water drained from the pool before the winter season. They hire Paul’s Pool Service to pump the water from the pool.
The water is pumped from the swimming pool at a constant rate. Below is one representation of the amount of water (in liters) remaining in the pool after n minutes have passed since the pump started emptying the full pool: f(n) = 20,100 − 100n
Paul from Paul's Pool Service uses the function f(n) = 100(201 – n) to represent the same quantity—that is, the amount of water (in liters) remaining in the pool after n minutes have passed since the pool was full.
This task illustrates that sometimes when we ask students to “simplify” their answers, we lose valuable information about the context. It also provides an opportunity to illustrate the value of using the distributive property to show equivalent expressions.
The value 20,100 represents the number of liters of water in the pool before any water has been removed. The value 100 represents the rate at which water is being pumped from the pool—100 liters per minute.
Solution methods may vary.
Substituting 10 for n in the function,
f(n) = 20,100 – 100n = 20,100 – 100(10) = 19,100
After 10 minutes there were 19,100 liters of water remaining in the pool.
Use the table of values from the graphing calculator to find the number of liters when n = 25.

After 25 minutes there were 17,600 liters of water remaining in the pool.
When f(n) = 0, the pool will be empty. Therefore, set the function f(n) = 20,100 – 100n equal to 0 and solve for n.
0 = 20,100 – 100n
100n = 20,100
n = 201
After 201 minutes or 3 hours and 21 minutes, the pool will be empty.
The two functions are equivalent. When the functions are graphed simultaneously, only one line appears on the screen, indicating that the two functions represent the same line. Additionally, the table of values shows that both functions generate the same output (Y1 and Y2) for the same input (x).


The distributive property can be used to show the two functions are equivalent.
f(n) = 100(201 – n)
= 100(201) – 100(n) = 20,100 – 100n
Writing the function as f(n) = 100(201 – n) will allow Paul to quickly see how many minutes it will take to empty all the water from the pool. This may help Paul figure out how much time to schedule for an appointment.
f(n) = 38500 – 100n
f(n) = 100(385 – n)
g(n) = 20100 – 120n
g(n) = 120(167.5 – n)