Directions: For the given problem, write an algebraic equation using the given variable. Then, show each step necessary to solve the equation in order to answer the question. Container A and Container B have leaks. Container A has 1200 oz of water, and is leaking 6 oz per minute. Container B has 1800 oz of water, and is leaking 10 oz per minute. How many minutes, m, will it take for the two containers to have the same amount of water? (Bonus Question: How many ounces will be in each container at that time?) .

Answer 1

Let's start by defining the amounts of water in each container over time.

Container A:

• Initial amount of water: 1200 oz
• Water lost per minute: 6 oz
• Amount of water after \( m \) minutes: \[ A(m) = 1200 - 6m \]

Container B:

• Initial amount of water: 1800 oz
• Water lost per minute: 10 oz
• Amount of water after \( m \) minutes: \[ B(m) = 1800 - 10m \]

We want to find the time \( m \) when the amounts of water in both containers are equal: \[ A(m) = B(m) \]

Now we can set up the equation: \[ 1200 - 6m = 1800 - 10m \]

Next, we will solve for \( m \):

1. Isolate the variable \( m \): \[ 1200 - 6m + 10m = 1800 \] \[ 1200 + 4m = 1800 \]

2. Subtract 1200 from both sides: \[ 4m = 1800 - 1200 \] \[ 4m = 600 \]

3. Divide both sides by 4: \[ m = \frac{600}{4} \] \[ m = 150 \]

Now we have determined that it will take 150 minutes for the two containers to have the same amount of water.

Next, we'll find out how much water will be in each container at that time.

Calculate the amount of water in each container after 150 minutes:

• For Container A: \[ A(150) = 1200 - 6 \times 150 \] \[ A(150) = 1200 - 900 \] \[ A(150) = 300 \]

• For Container B: \[ B(150) = 1800 - 10 \times 150 \] \[ B(150) = 1800 - 1500 \] \[ B(150) = 300 \]

Conclusion: It will take \( m = 150 \) minutes for both containers to have the same amount of water, which will be 300 ounces each.

Final answers:

• Time: 150 minutes
• Amount of water in each container at that time: 300 ounces