Solving the Milk Solution Percentage Problem

This problem involves calculating the amount of pure milk to add to a solution to change its concentration percentage. We start with a known volume and milk percentage and aim for a higher milk percentage by adding only milk.

Understanding the Initial Milk Solution

We are given an initial milk solution of 80 litres. This solution contains a certain percentage of milk and the rest is assumed to be water (or another solvent, but milk solution problems usually imply water). The initial percentage of milk is 60%.

• Total initial volume = 80 litres
• Initial milk percentage = 60%
• Initial water percentage = 100% - 60% = 40%

Let's calculate the initial amounts of milk and water in the solution:

• Amount of milk initially = 60% of 80 litres
• $\text{Initial Milk} = \frac{60}{100} \times 80$ litres
• $\text{Initial Milk} = 0.60 \times 80$ litres
• $\text{Initial Milk} = 48$ litres

• Amount of water initially = 40% of 80 litres
• $\text{Initial Water} = \frac{40}{100} \times 80$ litres
• $\text{Initial Water} = 0.40 \times 80$ litres
• $\text{Initial Water} = 32$ litres

We can verify this: $48$ litres (milk) + $32$ litres (water) = $80$ litres (total solution).

Adding Milk to the Solution

Pure milk is added to the solution. Let's say 'x' litres of milk are added. When pure milk is added, the amount of milk in the solution increases, and the total volume of the solution also increases. However, the amount of water in the solution remains unchanged.

• Amount of milk after adding 'x' litres = Initial Milk + x = $48 + x$ litres
• Amount of water after adding 'x' litres = Initial Water = $32$ litres (This is the key quantity that remains constant)
• New total volume of the solution = Initial Total Volume + x = $80 + x$ litres

Setting Up the Equation for the New Percentage

After adding 'x' litres of milk, the new solution should have 80% milk. This means the remaining percentage is water, which is 100% - 80% = 20%.

The amount of water (which is 32 litres) now represents 20% of the new total volume ($80 + x$ litres).

We can set up the equation based on the constant amount of water:

$\text{Amount of Water} = \text{New Water Percentage} \times \text{New Total Volume}$

$32 = 20\% \times (80 + x)$

Convert the percentage to a decimal:

$32 = 0.20 \times (80 + x)$

Solving for the Unknown Quantity (x)

Now, we need to solve the equation $32 = 0.20 \times (80 + x)$ for 'x'.

Divide both sides by 0.20:

$\frac{32}{0.20} = 80 + x$

Calculating the left side:

$\frac{32}{0.20} = \frac{32}{\frac{20}{100}} = 32 \times \frac{100}{20} = 32 \times 5 = 160$

So the equation becomes:

$160 = 80 + x$

To find x, subtract 80 from both sides:

$x = 160 - 80$

$x = 80$ litres

This means that 80 litres of pure milk should be added to the solution.

Verifying the Solution

Let's check if adding 80 litres of milk results in an 80% milk solution:

• Initial milk = 48 litres
• Milk added = 80 litres
• New total milk = $48 + 80 = 128$ litres

• Initial water = 32 litres
• Water added = 0 litres
• New total water = 32 litres

• New total volume = $80 + 80 = 160$ litres

New milk percentage = $\frac{\text{New Milk}}{\text{New Total Volume}} \times 100\%$

New milk percentage = $\frac{128}{160} \times 100\%$

$\frac{128}{160} = \frac{128 \div 32}{160 \div 32} = \frac{4}{5}$

New milk percentage = $\frac{4}{5} \times 100\% = 0.8 \times 100\% = 80\%$

The calculation is correct. Adding 80 litres of milk makes the solution 80% milk.

Revision Table: Milk Solution Problem

Additional Information: Mixture Problems

Mixture problems often involve changing the concentration of a component in a solution by adding more of the component or adding more of the other component (like water) or draining some mixture and replacing it. A common strategy is to focus on the quantity that remains constant throughout the process. In this problem, the amount of water stayed the same, making it a useful basis for setting up the equation.

Key concepts:

• Concentration: The amount of a substance divided by the total amount of the mixture, often expressed as a percentage.
• Constant Quantity: Identify which component's amount doesn't change during the mixing process.
• Setting up Equations: Use the constant quantity and the new percentage/ratio to form an equation involving the unknown amount added or removed.

Alternative approach: Set up the equation based on the milk percentage:

New Milk % = $\frac{\text{New Milk Volume}}{\text{New Total Volume}} \times 100\%$

$80\% = \frac{48 + x}{80 + x} \times 100\%$

$0.80 = \frac{48 + x}{80 + x}$

$0.80 \times (80 + x) = 48 + x$

$64 + 0.8x = 48 + x$

$64 - 48 = x - 0.8x$

$16 = 0.2x$

$x = \frac{16}{0.2} = \frac{160}{2} = 80$

This confirms the previous result. Both methods yield the same answer, focusing on the constant quantity (water) or the changing quantity (milk) both work, but the constant quantity method can sometimes be simpler.