There were three identical containers each containing the same amount of paint. Mr. Daniel poured  
216
216  ml of paint from container A into container C, and  
171
171  ml of paint from container B into container C. In the end, the amount of paint in container A was  
3
6
6
3
​
   of the amount of paint in container B.

(a)
How much paint was there in container A at the end?

(b)
How much paint was there in each container at first?

Let's solve this problem step by step.

### Step 1: Define Variables
Let the initial amount of paint in each container (A, B, and C) be \( x \) ml. Since all containers started with the same amount of paint, the amount of paint in each container at first is \( x \).

### Step 2: Track Changes in Paint
After Mr. Daniel poured paint from container A and B into container C:
- Container A lost 216 ml.
- Container B lost 171 ml.
- Container C gained a total of \( 216 + 171 = 387 \) ml.

Thus, after the transfers:
- The amount of paint in container A is \( x - 216 \) ml.
- The amount of paint in container B is \( x - 171 \) ml.
- The amount of paint in container C is \( x + 387 \) ml.

### Step 3: Set Up the Relationship Between A and B
According to the problem, the amount of paint in container A at the end was \( \frac{3}{6} \) (or half) of the amount of paint in container B. This gives us the equation:
\[
x - 216 = \frac{1}{2}(x - 171)
\]

### Step 4: Solve the Equation
Now let's solve this equation for \( x \).

1. Start by multiplying both sides by 2 to eliminate the fraction:
   \[
   2(x - 216) = x - 171
   \]
   
2. Expand both sides:
   \[
   2x - 432 = x - 171
   \]
   
3. Subtract \( x \) from both sides:
   \[
   x - 432 = -171
   \]
   
4. Add 432 to both sides:
   \[
   x = 432 - 171 = 261
   \]

### Step 5: Find the Amount of Paint in Each Container Initially
So, the initial amount of paint in each container was \( 261 \) ml.

### Step 6: Find the Amount of Paint in Container A at the End
From Step 2, the amount of paint in container A at the end is \( x - 216 \). Now substitute \( x = 261 \):
\[
261 - 216 = 45 \text{ ml}.
\]

Thus, at the end, container A had **45 ml** of paint.

### Final Answers:
(a) The amount of paint in container A at the end was **45 ml**.

(b) The amount of paint in each container at first was **261 ml**.

---

Would you like further details on any of the steps? Here are some related questions to help deepen understanding:

1. How did the transfers affect the total amount of paint in container C?
2. What would happen if Mr. Daniel poured a different amount from container A?
3. How could you check the final paint amounts in all three containers for consistency?
4. Could this problem be generalized for any starting amount?
5. What if the ratio between containers A and B had been different?

**Tip:** Always check relationships between variables by substituting back into the original conditions to confirm your solution!

There were three identical containers each containing the same amount of paint. Mr. Daniel poured 216 ml of paint from container A into container C, and 171 ml of paint from container B into container C. In the end, the amount of paint in container A was 3/6 of the amount of paint in container B. (a) How much paint was there in container A at the end? (b) How much paint was there in each container at first?

Learn how to solve a paint transfer problem involving three identical containers. This step-by-step solution covers key algebra concepts, including ratios and linear equations, to determine the final and initial paint quantities. Suitable for students in Grades 6-8.

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