There are three containers of equal capacity and all are completely filled with mixtures of acid and water in different ratios. The ratio of acid to water in the first container is 2 : 3, in the second container the ratio is 3:7 and in the third container it is 4:11. If the mixtures of all the three containers are mixed together, what will be the ratio of acid to water in it?

Answer

To solve the problem step by step, we will first define the volumes of acid and water in each container based on the given ratios. Then, we will find the total volumes of acid and water when all three containers are mixed together. Finally, we will calculate the ratio of acid to water in the resulting mixture.

### Step-by-Step Solution:

1. **Define the volumes based on the ratios:**
- For the **first container**, the ratio of acid to water is 2:3.
- Let the volume of acid be \(2x\) and the volume of water be \(3x\).
- Total volume = \(2x + 3x = 5x\).

- For the **second container**, the ratio of acid to water is 3:7.
- Let the volume of acid be \(3y\) and the volume of water be \(7y\).
- Total volume = \(3y + 7y = 10y\).

- For the **third container**, the ratio of acid to water is 4:11.
- Let the volume of acid be \(4z\) and the volume of water be \(11z\).
- Total volume = \(4z + 11z = 15z\).

2. **Set the total volumes equal:**
- Since all containers have equal capacity, we can set up the following equations:
\[ 5x = 10y = 15z \]

3. **Express \(x\), \(y\), and \(z\) in terms of a common variable:**
- From \(5x = 10y\):
\[ x = 2y \]
- From \(5x = 15z\):
\[ z = \frac{2y}{3} \]

4. **Calculate the total volumes of acid and water:**
- **Total acid** from all containers:
\[ \text{Total Acid} = 2x + 3y + 4z \]
Substitute \(x\) and \(z\):
\[ = 2(2y) + 3y + 4\left(\frac{2y}{3}\right) = 4y + 3y + \frac{8y}{3} \]
To combine, convert \(4y\) and \(3y\) into fractions:
\[ = \frac{12y}{3} + \frac{9y}{3} + \frac{8y}{3} = \frac{29y}{3} \]

- **Total water** from all containers:
\[ \text{Total Water} = 3x + 7y + 11z \]
Substitute \(x\) and \(z\):
\[ = 3(2y) + 7y + 11\left(\frac{2y}{3}\right) = 6y + 7y + \frac{22y}{3} \]
To combine, convert \(6y\) and \(7y\) into fractions:
\[ = \frac{18y}{3} + \frac{21y}{3} + \frac{22y}{3} = \frac{61y}{3} \]

5. **Find the ratio of acid to water:**
- The ratio of acid to water is:
\[ \text{Ratio} = \frac{\text{Total Acid}}{\text{Total Water}} = \frac{\frac{29y}{3}}{\frac{61y}{3}} = \frac{29y}{61y} = \frac{29}{61} \]
- Therefore, the ratio of acid to water in the mixture is:
\[ \text{Final Ratio} = 29 : 61 \]