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Sagot :
First, let's denote the volumes of the three containers. Given that their volumes are in the ratio 3:4:5, we can represent the containers' volumes as [tex]\( V_1 = 3x \)[/tex], [tex]\( V_2 = 4x \)[/tex], and [tex]\( V_3 = 5x \)[/tex], where [tex]\( x \)[/tex] is a common multiplying factor.
Next, let's consider the ratios of milk to water in each of the containers:
1. The first container has milk and water in the ratio 4:1.
2. The second container has milk and water in the ratio 3:1.
3. The third container has milk and water in the ratio 5:2.
We now calculate the individual volumes of milk and water in each container.
### First Container
- Total volume, [tex]\( V_1 = 3x \)[/tex].
- The ratio of milk to water is 4:1. Thus, the total parts = 4 + 1 = 5 parts.
- Milk volume in the first container = [tex]\( \frac{4}{5} \times 3x = \frac{12x}{5} \)[/tex].
- Water volume in the first container = [tex]\( \frac{1}{5} \times 3x = \frac{3x}{5} \)[/tex].
### Second Container
- Total volume, [tex]\( V_2 = 4x \)[/tex].
- The ratio of milk to water is 3:1. Thus, the total parts = 3 + 1 = 4 parts.
- Milk volume in the second container = [tex]\( \frac{3}{4} \times 4x = 3x \)[/tex].
- Water volume in the second container = [tex]\( \frac{1}{4} \times 4x = x \)[/tex].
### Third Container
- Total volume, [tex]\( V_3 = 5x \)[/tex].
- The ratio of milk to water is 5:2. Thus, the total parts = 5 + 2 = 7 parts.
- Milk volume in the third container = [tex]\( \frac{5}{7} \times 5x = \frac{25x}{7} \)[/tex].
- Water volume in the third container = [tex]\( \frac{2}{7} \times 5x = \frac{10x}{7} \)[/tex].
Now, let's combine the contents of all three containers into the fourth container.
### Total Milk in the Fourth Container
[tex]\[ \text{Total Milk} = \frac{12x}{5} + 3x + \frac{25x}{7} \][/tex]
Converting all terms to a common denominator ([tex]\(35\)[/tex]):
[tex]\[ \text{Total Milk} = \left( \frac{12x}{5} \times \frac{7}{7} \right) + \left( 3x \times \frac{35}{35} \right) + \left( \frac{25x}{7} \times \frac{5}{5} \right) \][/tex]
[tex]\[ \text{Total Milk} = \frac{84x}{35} + \frac{105x}{35} + \frac{125x}{35} \][/tex]
[tex]\[ \text{Total Milk} = \frac{314x}{35} \][/tex]
### Total Water in the Fourth Container
[tex]\[ \text{Total Water} = \frac{3x}{5} + x + \frac{10x}{7} \][/tex]
Converting all terms to a common denominator ([tex]\(35\)[/tex]):
[tex]\[ \text{Total Water} = \left( \frac{3x}{5} \times \frac{7}{7} \right) + \left( x \times \frac{35}{35} \right) + \left( \frac{10x}{7} \times \frac{5}{5} \right) \][/tex]
[tex]\[ \text{Total Water} = \frac{21x}{35} + \frac{35x}{35} + \frac{50x}{35} \][/tex]
[tex]\[ \text{Total Water} = \frac{106x}{35} \][/tex]
Therefore, the total volumes in the fourth container are:
[tex]\[ \text{Total Milk} = \frac{314x}{35} \][/tex]
[tex]\[ \text{Total Water} = \frac{106x}{35} \][/tex]
### Ratio of Milk to Water in the Fourth Container
[tex]\[ \text{Ratio of Milk to Water} = \frac{\frac{314x}{35}}{\frac{106x}{35}} = \frac{314}{106} \][/tex]
This simplifies as follows:
[tex]\[ \frac{314}{106} = 2.9622641509433962 \][/tex]
Thus, the ratio of milk to water in the fourth container is approximately [tex]\(2.96:1\)[/tex].
Next, let's consider the ratios of milk to water in each of the containers:
1. The first container has milk and water in the ratio 4:1.
2. The second container has milk and water in the ratio 3:1.
3. The third container has milk and water in the ratio 5:2.
We now calculate the individual volumes of milk and water in each container.
### First Container
- Total volume, [tex]\( V_1 = 3x \)[/tex].
- The ratio of milk to water is 4:1. Thus, the total parts = 4 + 1 = 5 parts.
- Milk volume in the first container = [tex]\( \frac{4}{5} \times 3x = \frac{12x}{5} \)[/tex].
- Water volume in the first container = [tex]\( \frac{1}{5} \times 3x = \frac{3x}{5} \)[/tex].
### Second Container
- Total volume, [tex]\( V_2 = 4x \)[/tex].
- The ratio of milk to water is 3:1. Thus, the total parts = 3 + 1 = 4 parts.
- Milk volume in the second container = [tex]\( \frac{3}{4} \times 4x = 3x \)[/tex].
- Water volume in the second container = [tex]\( \frac{1}{4} \times 4x = x \)[/tex].
### Third Container
- Total volume, [tex]\( V_3 = 5x \)[/tex].
- The ratio of milk to water is 5:2. Thus, the total parts = 5 + 2 = 7 parts.
- Milk volume in the third container = [tex]\( \frac{5}{7} \times 5x = \frac{25x}{7} \)[/tex].
- Water volume in the third container = [tex]\( \frac{2}{7} \times 5x = \frac{10x}{7} \)[/tex].
Now, let's combine the contents of all three containers into the fourth container.
### Total Milk in the Fourth Container
[tex]\[ \text{Total Milk} = \frac{12x}{5} + 3x + \frac{25x}{7} \][/tex]
Converting all terms to a common denominator ([tex]\(35\)[/tex]):
[tex]\[ \text{Total Milk} = \left( \frac{12x}{5} \times \frac{7}{7} \right) + \left( 3x \times \frac{35}{35} \right) + \left( \frac{25x}{7} \times \frac{5}{5} \right) \][/tex]
[tex]\[ \text{Total Milk} = \frac{84x}{35} + \frac{105x}{35} + \frac{125x}{35} \][/tex]
[tex]\[ \text{Total Milk} = \frac{314x}{35} \][/tex]
### Total Water in the Fourth Container
[tex]\[ \text{Total Water} = \frac{3x}{5} + x + \frac{10x}{7} \][/tex]
Converting all terms to a common denominator ([tex]\(35\)[/tex]):
[tex]\[ \text{Total Water} = \left( \frac{3x}{5} \times \frac{7}{7} \right) + \left( x \times \frac{35}{35} \right) + \left( \frac{10x}{7} \times \frac{5}{5} \right) \][/tex]
[tex]\[ \text{Total Water} = \frac{21x}{35} + \frac{35x}{35} + \frac{50x}{35} \][/tex]
[tex]\[ \text{Total Water} = \frac{106x}{35} \][/tex]
Therefore, the total volumes in the fourth container are:
[tex]\[ \text{Total Milk} = \frac{314x}{35} \][/tex]
[tex]\[ \text{Total Water} = \frac{106x}{35} \][/tex]
### Ratio of Milk to Water in the Fourth Container
[tex]\[ \text{Ratio of Milk to Water} = \frac{\frac{314x}{35}}{\frac{106x}{35}} = \frac{314}{106} \][/tex]
This simplifies as follows:
[tex]\[ \frac{314}{106} = 2.9622641509433962 \][/tex]
Thus, the ratio of milk to water in the fourth container is approximately [tex]\(2.96:1\)[/tex].
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