Let the amount of water in the mixture be $x$ liters. According to the problem, the amount of milk is $x + 20$ liters, since it is 20 liters more than the amount of water.

The cost of pure milk is Rs. 44 per liter, and the cost of the mixture is Rs. 33 per liter.

We can set up the equation for the total cost of the mixture based on the quantities of milk and water and their costs.

• Cost of milk in the mixture: $44 \times (x + 20)$
• Cost of water in the mixture: $0 \times x = 0$ (since water is free)

Now, the total cost of the mixture is equal to $33$ per liter times the total quantity of the mixture, which is $x + (x + 20) = 2x + 20$.

We can set up the following equation:

$\frac{44(x + 20)}{2x + 20} = 33$

Now, let's solve for $x$.

Step 1: Cross-multiply

$44(x + 20) = 33(2x + 20)$

Step 2: Expand both sides

$44x + 880 = 66x + 660$

Step 3: Bring all terms involving $x$ to one side

$880 - 660 = 66x - 44x$

$220 = 22x$

Step 4: Solve for $x$

$x = \frac{220}{22} = 10$

Step 5: Calculate the total quantity of the mixture

The total quantity of the mixture is:

$2x + 20 = 2(10) + 20 = 20 + 20 = 40 \, \text{liters}$

Thus, the total quantity of the mixture is $40$ liters.

Would you like more details on any step, or have any other questions?

Here are 5 related questions:

1. How would the equation change if water had a cost per liter?
2. What happens to the mixture cost if the price of milk changes?
3. How can you verify that the given price of the mixture is correct?
4. How would you calculate the ratio of milk to water in the mixture?
5. What if the total mixture cost was given instead of per liter? How would that affect the problem?

Tip: In problems involving mixtures, setting up equations based on cost or concentration is key to finding the solution.