Question

Pool A starts with 300 gallons of water. It has a leak and is losing water at a rate of 6 gallons of water per minute. At the same time, Pool B starts with 400 gallons of water and also has a leak. It is losing water at a rate of 10 gallons per minute. The variable t represents the time in minutes. After how many minutes will the two pools have the same amount of water? How much water will be in the pools at that time?

Answer 1

If time be x and volume be y

The equations are

• y=300-6t
• y=400-10t

Equate and solve

• 300-6t=400-10t
• 4t=100
• t=25min

Now

• y=400-10(25)
• y=400-250
• y=150gallons

Answer 2

25 minutes

150 gallons

Explanation:

Definition of variables

Let t = time in minutes

Let V = volume in gallons

Create two equations with the given information:

Pool A

• Starting volume = 330 gallons
• Leak rate = 6 gallons/min

⇒ V = 300 - 6t

Pool B

• Starting volume = 400 gallons
• Leak rate = 10 gallons/min

⇒ V = 400 - 10t

To find after how many minutes the two pools will have the same amount of water, substitute the equation for pool A into the equation for pool B and solve for t:

⇒ 300 - 6t = 400 - 10t

⇒ 300 - 6t + 10t = 400 - 10t + 10t

⇒ 300 + 4t = 400

⇒ 300 + 4t - 300 = 400 - 300

⇒ 4t = 100

⇒ 4t ÷ 4 = 100 ÷ 4

⇒ t = 25

Therefore, the two pools will have the same amount of water after 25 minutes.

To find how much water will be in the pools at that time, substitute the found value of t into one of the equations and solve for V:

⇒ V = 400 - 10t

⇒ V = 400 - 10(25)

⇒ V = 400 - 250

⇒ V = 150

Therefore, the amount of water in the pools at the time when they have the same amount of water is 150 gallons.