Understanding the Problem

We are given the cost price (CP) of pure milk, the selling price (SP) of the milk-water mixture, and the profit percentage made on the sale. We need to find the ratio of water to milk in the mixture.

• CP of Milk = ₹6.40 per litre
• SP of Mixture = ₹8 per litre
• Profit % = 37.5%

Calculating Mixture Cost Price (CP_mix)

The formula relating SP, CP, and Profit % is:

$SP = CP \times (1 + \frac{Profit \%}{100})$

Substituting the given values:

$8 = CP_{mix} \times (1 + \frac{37.5}{100})$

$8 = CP_{mix} \times (1 + 0.375)$

$8 = CP_{mix} \times 1.375$

Solving for $CP_{mix}$:

$CP_{mix} = \frac{8}{1.375} = \frac{8}{11/8} = \frac{8 \times 8}{11} = \frac{64}{11}$

$CP_{mix} \approx ₹5.818$ per litre

Determining Milk and Water Quantities

The calculated $CP_{mix}$ represents the cost of the milk component within one litre of the mixture, as water is assumed to be free.

Let the quantity of milk in 1 litre of the mixture be $m$ litres and the quantity of water be $w$ litres. Therefore, $m + w = 1$ litre.

The cost of milk in the mixture is $m \times \text{CP of Milk}$.

Equating this cost to the $CP_{mix}$:

$m \times ₹6.40 = ₹\frac{64}{11}$

Solving for $m$:

$m = \frac{64}{11 \times 6.40} = \frac{64}{11 \times \frac{64}{10}} = \frac{64}{11} \times \frac{10}{64} = \frac{10}{11}$ litres

Now, find the quantity of water $w$:

$w = 1 - m = 1 - \frac{10}{11} = \frac{11 - 10}{11} = \frac{1}{11}$ litres

Finding the Water to Milk Ratio

The ratio of water to milk is $w : m$.

Ratio = $\frac{1}{11} : \frac{10}{11}$

Multiplying both sides by 11 to simplify:

Ratio = $1 : 10$

This corresponds to Option A.