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. 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?)

Answer 1

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.