Question:
A beaker contains a solution in which the concentration of milk is 80%. On the addition of some water, this ratio of milk to water becomes 8:5. On replacing 39 L of this solution with pure milk, the ratio of milk to water changes to 9:4. What is the volume of water added initially?
A beaker contains a solution in which the concentration of milk is 80%. On the addition of some water, this ratio of milk to water becomes 8:5. On replacing 39 L of this solution with pure milk, the ratio of milk to water changes to 9:4. What is the volume of water added initially?
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In replacement problems, it's easier to calculate the amount of each component removed from the mixture rather than working with the remaining volumes directly. Remember that the concentration of the removed part is the same as the concentration of the mixture it's taken from.
Updated On: Oct 2, 2025
- 30 litres
- 45 litres
- 50 litres
- 35 litres
- 60 litres
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This is a multi-step mixture problem. We need to track the quantities of milk and water through two distinct operations: addition of water and then replacement of the mixture with pure milk. Setting up equations based on the component quantities at each stage is the key.
Step 2: Detailed Explanation:
Let the initial volume of the solution be \(V\) litres.
Initial State:
Milk concentration = 80%. Water concentration = 20%.
Initial Milk = \(0.8V\).
Initial Water = \(0.2V\).
Operation 1: Addition of water
Let \(x\) litres of water be added.
New Milk = \(0.8V\) (Milk quantity doesn't change).
New Water = \(0.2V + x\).
The new ratio of milk to water is 8:5.
\[ \frac{\text{New Milk}}{\text{New Water}} = \frac{0.8V}{0.2V + x} = \frac{8}{5} \] Cross-multiply: \[ 5(0.8V) = 8(0.2V + x) \] \[ 4V = 1.6V + 8x \] \[ 2.4V = 8x \implies x = \frac{2.4V}{8} = 0.3V \quad \text{(Equation 1)} \] The total volume of the solution after this step is \(V_{new} = V_{milk} + V_{water} = 0.8V + (0.2V + 0.3V) = 1.3V\).
The concentration of milk in this new mixture is \(\frac{8}{8+5} = \frac{8}{13}\), and water is \(\frac{5}{13}\).
Operation 2: Replacement of solution
39 L of this solution is removed and replaced with 39 L of pure milk.
Amount of milk removed = \(39 \times (\text{milk concentration}) = 39 \times \frac{8}{13} = 3 \times 8 = 24\) litres.
Amount of water removed = \(39 \times (\text{water concentration}) = 39 \times \frac{5}{13} = 3 \times 5 = 15\) litres.
Quantities after removing 39 L:
Milk remaining = \(0.8V - 24\).
Water remaining = \((0.2V + x) - 15 = (0.2V + 0.3V) - 15 = 0.5V - 15\).
Now, 39 L of pure milk is added.
Final Milk = (Milk remaining) + 39 = \((0.8V - 24) + 39 = 0.8V + 15\).
Final Water = (Water remaining) = \(0.5V - 15\).
The final ratio of milk to water is 9:4.
\[ \frac{\text{Final Milk}}{\text{Final Water}} = \frac{0.8V + 15}{0.5V - 15} = \frac{9}{4} \] Cross-multiply: \[ 4(0.8V + 15) = 9(0.5V - 15) \] \[ 3.2V + 60 = 4.5V - 135 \] \[ 4.5V - 3.2V = 60 + 135 \] \[ 1.3V = 195 \] \[ V = \frac{195}{1.3} = \frac{1950}{13} = 150 \text{ litres} \] Step 3: Final Answer:
The question asks for the volume of water added initially, which is \(x\).
From Equation 1, we have \(x = 0.3V\).
\[ x = 0.3 \times 150 = 45 \text{ litres} \] The volume of water added initially was 45 litres.
This is a multi-step mixture problem. We need to track the quantities of milk and water through two distinct operations: addition of water and then replacement of the mixture with pure milk. Setting up equations based on the component quantities at each stage is the key.
Step 2: Detailed Explanation:
Let the initial volume of the solution be \(V\) litres.
Initial State:
Milk concentration = 80%. Water concentration = 20%.
Initial Milk = \(0.8V\).
Initial Water = \(0.2V\).
Operation 1: Addition of water
Let \(x\) litres of water be added.
New Milk = \(0.8V\) (Milk quantity doesn't change).
New Water = \(0.2V + x\).
The new ratio of milk to water is 8:5.
\[ \frac{\text{New Milk}}{\text{New Water}} = \frac{0.8V}{0.2V + x} = \frac{8}{5} \] Cross-multiply: \[ 5(0.8V) = 8(0.2V + x) \] \[ 4V = 1.6V + 8x \] \[ 2.4V = 8x \implies x = \frac{2.4V}{8} = 0.3V \quad \text{(Equation 1)} \] The total volume of the solution after this step is \(V_{new} = V_{milk} + V_{water} = 0.8V + (0.2V + 0.3V) = 1.3V\).
The concentration of milk in this new mixture is \(\frac{8}{8+5} = \frac{8}{13}\), and water is \(\frac{5}{13}\).
Operation 2: Replacement of solution
39 L of this solution is removed and replaced with 39 L of pure milk.
Amount of milk removed = \(39 \times (\text{milk concentration}) = 39 \times \frac{8}{13} = 3 \times 8 = 24\) litres.
Amount of water removed = \(39 \times (\text{water concentration}) = 39 \times \frac{5}{13} = 3 \times 5 = 15\) litres.
Quantities after removing 39 L:
Milk remaining = \(0.8V - 24\).
Water remaining = \((0.2V + x) - 15 = (0.2V + 0.3V) - 15 = 0.5V - 15\).
Now, 39 L of pure milk is added.
Final Milk = (Milk remaining) + 39 = \((0.8V - 24) + 39 = 0.8V + 15\).
Final Water = (Water remaining) = \(0.5V - 15\).
The final ratio of milk to water is 9:4.
\[ \frac{\text{Final Milk}}{\text{Final Water}} = \frac{0.8V + 15}{0.5V - 15} = \frac{9}{4} \] Cross-multiply: \[ 4(0.8V + 15) = 9(0.5V - 15) \] \[ 3.2V + 60 = 4.5V - 135 \] \[ 4.5V - 3.2V = 60 + 135 \] \[ 1.3V = 195 \] \[ V = \frac{195}{1.3} = \frac{1950}{13} = 150 \text{ litres} \] Step 3: Final Answer:
The question asks for the volume of water added initially, which is \(x\).
From Equation 1, we have \(x = 0.3V\).
\[ x = 0.3 \times 150 = 45 \text{ litres} \] The volume of water added initially was 45 litres.
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