Mark is draining his pool to clean it. At 12:00 P. M., he started to drain the pool. At 4:00 P. M., the pool had 11,600 gallons of water. At 7:00 P. M., the pool had 6,800 gallons of water. If draining the pool is modeled by a linear function, how many gallons of water were initially in the pool? A) 15,600 B) 18,000 C) 19,400 D) 21,200

Question

Mathematics

Aspen Schmidt

14 May, 01:25

Mark is draining his pool to clean it. At 12:00 P. M., he started to drain the pool. At 4:00 P. M., the pool had 11,600 gallons of water. At 7:00 P. M., the pool had 6,800 gallons of water. If draining the pool is modeled by a linear function, how many gallons of water were initially in the pool? A) 15,600 B) 18,000 C) 19,400 D) 21,200

+4

Answers (1)

Know the Answer?

Ryder Nixon · Answer 1

12:00 P. M. to 7:00 P. M. is a total of 7 hours.

4:00 P. M. to 7:00 P. M. is a total of 3 hours.

This means that from 4:00 p. m to 7:00 P. M. this represents 3/7 of what was lost from 12:00 P. M. to 7:00 P. M.

After knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

We then divide 4,800 by 3 which gets 1,600.

Because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200. After adding 11,200 to how much water was left at 7:00 P. M. which was 6,800. 11,200 + 6,800 = 18,000

This means that the pool originally had 18,000 gallons in the pool.

You can also check your work by doing 18,000 - 4 (1600), which becomes 1800 - 6400 which gets you 11,600.