Mark partially drained his pool to clean it. At 1:00 P.M., he started to refill it. At 5:00 P.M., the pool had 10,400 gallons of water. At 9:00 P.M., the pool had 13,800 gallons of water. If filling the pool is modeled by a linear function, how many gallons of water still remained in the pool when he first started to refill it?

Question

Responses
A 6,1506,150
B 6,5006,500
C 7,0007,000
D 7,150

Answer 1

To determine how many gallons of water were in the pool when Mark first started refilling it, we can model the situation with a linear function.

1. Identify the points:
   We have the following two data points based on the times and volumes of water:

   • At 5:00 P.M. (4 hours after starting), the pool has 10,400 gallons (let's denote this point as \((4, 10400)\))
   • At 9:00 P.M. (8 hours after starting), the pool has 13,800 gallons (denote this point as \((8, 13800)\))

2. Calculate the slope (rate of filling):
   We can find the slope (m) of the line representing the water volume over time using the formula:
   \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Where:

   • \((x_1, y_1) = (4, 10400)\)
   • \((x_2, y_2) = (8, 13800)\)

   Plugging in the values: \[ m = \frac{13800 - 10400}{8 - 4} = \frac{3400}{4} = 850 \text{ gallons per hour} \]

3. Equation of the function:
   We can express the function in point-slope form. Using the point \((4, 10400)\): \[ y - 10400 = 850(x - 4) \] Simplifying this: \[ y - 10400 = 850x - 3400 \] \[ y = 850x + 7000 \]

4. Finding the initial volume (at \(x = 0\)):
   To find how many gallons were in the pool when Mark started refilling at 1:00 P.M. (when \(x = 0\)): \[ y = 850(0) + 7000 = 7000 \]

So, the number of gallons of water that were in the pool when he first started refilling it is 7,000 gallons.

Thus, the answer is: C 7,000.