filling a pool without rate of starting amount

Question

We are filling our backyard pool which holds 8000 gallons of water.  It is known to have 1200 gallons of water after 90 minutes and there are 1800 gallons of water after 3.5 hours.  

(a)  Find an equation for the volume of water in the pool as a function of time in minutes where   

(b)  Based on your findings, how much water was in the pool when we began filling it?

(c)  How much time is needed to fill the pool to maximum capacity?

Answer 1

Hello Ella, I hope this helps.

I started by making a table of what I do know:

time (minutes) | Volume (gallons)

--------------------|-----------------------

0 | x (original amount of water in the pool)

90 | 1200

210 | 1800

? | 8000 (maximum)

*please note the 210 minutes comes from converting 3.5 hours to minutes:

3.5 hrs 60 minutes 210 minutes

--------- X --------------- = ------------------ = 210 minutes

1 1 hr 1

a) To write a linear equation we first need to decide what form we want to write it in. I am going to move forward with using point-slope (y=mx+b). m is the slope (or in this case, the rate of change of the volume of water) and b is the y-intercept (or in this case the amount of water that was in the pool to begin with).

(the change in y) (1800 - 1200) 600

m= ------------------------- = -------------------- = ---------- = 5

(the change in x) (210 - 90) 120

We can now rewrite our formula with m:

y=5x+b

To solve for b, we just plug in one of the known points for x & y and solve for b:

1200 = 5(90)+b *multiply 5*90

1200 = 450+b *subtract 450 from both sides

750 = b

Our final equation: y=5x+750

**since we are supposed to express the formula in terms of Volume and Time we get: V(t) = 5t + 750

b) Since we already calculated the y-intercept, we know how much water was in the pool to begin with: 750 gallons

c) Plug the maximum gallons the pool can take into our equation from a).

8000 = 5t + 750 *subtract 750 from both sides

7250 = 5t *divide both sides by 5 to isolate x

1450 = t

1450 minutes (divide by 60 if you want to convert back to hours)

Answer 2

Will have to assume a constant rate of filling which would be the slope(m) of what would be a linear equation. We are given 2 points ( 90min, 1200gal) and (210min, 1800 gal)

We can write 2 equations in y = mx +b form

1800 = 210m + b

1200 = 90m +b

Subtract equation 2 from 1 yields 600 = 120m m=5gal/minute

Can now solve for b using either equation. I will use the one with smaller number

1200 = (5)(90) +b

1200 = 450 +b b= 750 So our equation is y= 5x + 750

750 gallons were in the pool when filling began which would be at x=0

To solve for the time to completely fill will need to solve the equation 8000= 5x +750. I will let you solve this