Yellow, blue, and white colours are mixed in the ratio 7:6:3. How much of each colour is required in a can of 960 L of colour?

Answer

To solve the problem of how much of each color is required in a can of 960 L of color mixed in the ratio of 7:6:3, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Ratio**: The colors are mixed in the ratio of Yellow: Blue: White = 7:6:3.

2. **Assign Variables**: Let the amount of Yellow be \( 7x \), Blue be \( 6x \), and White be \( 3x \), where \( x \) is a common multiplier.

3. **Set Up the Equation**: The total volume of the mixture is given as 960 L. Therefore, we can write the equation:
\[ 7x + 6x + 3x = 960 \]

4. **Combine Like Terms**: Combine the terms on the left side:
\[ 16x = 960 \]

5. **Solve for \( x \)**: Divide both sides of the equation by 16 to find \( x \):
\[ x = \frac{960}{16} = 60 \]

6. **Calculate the Amount of Each Color**:
- **Yellow**: \( 7x = 7 \times 60 = 420 \) L
- **Blue**: \( 6x = 6 \times 60 = 360 \) L
- **White**: \( 3x = 3 \times 60 = 180 \) L

7. **Final Answer**: The amounts of each color required are:
- Yellow: 420 L
- Blue: 360 L
- White: 180 L