An 80-litre mixture contains milk and water in the ratio 7:1. Some quantity of this mixture is removed and replaced with pure milk. If the final ratio of milk to water becomes 15:1, how many litres of the mixture were replaced?

Correct Answer: 40 litres

Step-by-Step Explanation:

Step 1: Determine initial quantities of milk and water

The total volume of the mixture is 80 litres, and the ratio of milk to water is 7 : 1.

Total ratio parts = 7 + 1 = 8 parts.

Initial quantity of milk = $\frac{7}{8}\times 80$ = 70 litres.

Initial quantity of water = $\frac{1}{8}\times 80$ = 10 litres.

Step 2: Express removed quantities in terms of removed mixture volume

Let x litres of the mixture be removed.

Since the mixture is homogeneous, the removed x litres contains milk and water in the same ratio of 7 : 1.

Quantity of milk removed = $\frac{7}{8}x$ litres.

Quantity of water removed = $\frac{1}{8}x$ litres.

Step 3: Account for replacing the removed mixture with pure milk

When x litres of pure milk is added, the new quantities of milk and water become:

New quantity of milk = 70 - $\frac{7}{8}x$ + x = 70 + $\frac{1}{8}x$ litres.

New quantity of water = 10 - $\frac{1}{8}x$ litres.

Step 4: Set up the equation using the final ratio

The final ratio of milk to water is given as 15 : 1.

$\frac{70+\frac{1}{8}x}{10-\frac{1}{8}x}=\frac{15}{1}$

Cross-multiplying both sides:

70 + $\frac{1}{8}x$ = 15 $\times$ (10 - $\frac{1}{8}x$)

70 + $\frac{1}{8}x$ = 150 - $\frac{15}{8}x$

Rearranging terms to solve for x:

$\frac{1}{8}x+\frac{15}{8}x$ = 150 - 70

$\frac{16}{8}x$ = 80

2x = 80

x = 40

Therefore, 40 litres of the mixture were removed and replaced with pure milk.