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. Next, show each step necessary to solve the equation and answer the following: 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?)

1 answer

To determine how many minutes \( m \) it will take for Container A and Container B to have the same amount of water, we can set up equations based on the initial amounts of water and the rates at which they leak.

Step 1: Define the Initial Conditions

  • Container A:

    • Initial amount of water: 1200 oz
    • Rate of leakage: 6 oz/min
    • Amount of water after \( m \) minutes: \[ A(m) = 1200 - 6m \]
  • Container B:

    • Initial amount of water: 1800 oz
    • Rate of leakage: 10 oz/min
    • Amount of water after \( m \) minutes: \[ B(m) = 1800 - 10m \]

Step 2: Set the Equations Equal to Each Other

To find out when the amounts of water in both containers will be equal, set \( A(m) \) equal to \( B(m) \):

\[ 1200 - 6m = 1800 - 10m \]

Step 3: Solve the Equation

  • First, simplify by rearranging terms: \[ 1200 - 6m + 10m = 1800 \] Combine like terms: \[ 1200 + 4m = 1800 \] Subtract 1200 from both sides: \[ 4m = 600 \] Next, divide both sides by 4 to solve for \( m \): \[ m = 150 \]

Step 4: Calculate the Amount of Water in Each Container at \( m = 150 \)

Now, substitute \( m = 150 \) back into the equations for \( A(m) \) and \( B(m) \):

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

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

Final Answers:

  1. It will take \( m = 150 \) minutes for the two containers to have the same amount of water.
  2. At that time, each container will have 300 ounces of water.