To solve the problem, we need to write equations that represent the amount of water remaining in each container over time.
Step 1: Define the variables.
- Let \( m \) represent the number of minutes that have passed.
Step 2: Write the equation for each container.
-
For Container A:
- Initial amount of water = 1200 oz
- Leaking rate = 6 oz/min
- Amount of water remaining after \( m \) minutes: \[ A = 1200 - 6m \]
-
For Container B:
- Initial amount of water = 1800 oz
- Leaking rate = 10 oz/min
- Amount of water remaining after \( m \) minutes: \[ B = 1800 - 10m \]
Step 3: Set the equations equal to each other. To find the time \( m \) when both containers have the same amount of water: \[ 1200 - 6m = 1800 - 10m \]
Step 4: Solve the equation.
-
First, simplify the equation: \[ 1200 - 6m = 1800 - 10m \]
-
Rearranging gives: \[ 10m - 6m = 1800 - 1200 \]
-
Simplifying further: \[ 4m = 600 \]
-
Dividing both sides by 4: \[ m = 150 \]
Step 5: Find the amount of water in each container at that time. Now that we know it takes \( m = 150 \) minutes for both containers to have the same amount of water, we can find out how much water is in each container at that time.
-
For Container A: \[ A = 1200 - 6(150) = 1200 - 900 = 300 \text{ oz} \]
-
For Container B: \[ B = 1800 - 10(150) = 1800 - 1500 = 300 \text{ oz} \]
Final Answer: It will take 150 minutes for the two containers to have the same amount of water. At that time, each container will contain 300 ounces of water.