Let the initial quantity of milk be $8x$ litres and the initial quantity of water be $x$ litres.
Thus, the total initial quantity of the mixture is:

$8x + x = 9x \text{ litres}$

Step 1: Removing 18 litres of mixture

When 18 litres of the mixture is removed, the quantity of milk and water removed will be in the same ratio of 8:1.

Milk removed:

$\frac{8}{9} \times 18 = 16 \text{ litres}$

Water removed:

$\frac{1}{9} \times 18 = 2 \text{ litres}$

Step 2: Updating the remaining quantities

After removing 18 litres, the remaining milk and water are:

$\text{Remaining milk} = 8x - 16$

$\text{Remaining water} = x - 2$

Step 3: Adding 4 litres of water

After adding 4 litres of water, the new quantity of water is:

$x - 2 + 4 = x + 2$

Step 4: Setting up the given condition

It is given that the quantity of milk is 24 litres more than the quantity of water:

$(8x - 16) = (x + 2) + 24$

Step 5: Solving for $x$

$8x - 16 = x + 26$

$8x - x = 26 + 16$

$7x = 42$

$x = 6$

Step 6: Finding the initial quantity of milk

$\text{Initial milk} = 8x = 8 \times 6 = 48 \text{ litres}$

Thus, the initial quantity of milk in the mixture is 48 litres.

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Would you like a step-by-step breakdown with a diagram?

Here are some related problems to consider:

1. If 10 litres of mixture were removed instead of 18, what would be the new ratio of milk to water?
2. How much water should be added to make the milk-to-water ratio 1:1 after removal?
3. If the total initial quantity was 90 litres, what was the water content?
4. What would happen to the ratio if an additional 5 litres of milk were added after removal?
5. How would the final composition change if only 12 litres were removed instead of 18?

Tip: When removing a portion of a mixture, always use the initial ratio to determine how much of each component is taken out!