Milk and water in a solution of 56 ml is such that after replacing 16 ml of the solution and adding 5 ml of water, the milk and water ratio is 5 ∶ 4. What was the concentration of milk in the mixture initially?

Question

Question

Answer 1

Let the quantity of milk and water in the mixture be x and 56 – x

Now, 16ml of the solution is is removed

Quantity of solution replaced will be removed in the same ratio as the contents in the solution

∴ Quantity of milk left after removal = x – (16 × x/56) = 5x/7

Quantity of water left after removal = 56 – x – {16 × (56 – x)/56} = 40 – 5x/7

Ratio of milk and water after adding 5 ml of water is 5 ∶ 4

⇒ (5x/7)/(40 + 5 – 5x/7) = 5/4

⇒ 20x = 45 × 7 × 5 – 25x

⇒ x = 7 × 5 = 35

∴ Concentration of milk in the mixture initially was 35 ml

Alternate Method

Quantity of Solution = 56 ml (Given)

16 ml of the solution replaced with 5 ml of water

then the solution quantity will be = (56 – 16) + 5 = 45 ml

Milk : Water = 5 : 4

M + W = 45 (will have to be distributed in the ratio of 5 : 4)

In the Final Mixture,

Milk = 25 ml, Water = 20 ml

In 20 ml Water, 5 litre was added after replacement.

∴ Before Replacement, Water = 20 - 5 = 15 ml

Ratio of Milk and Water (Initially) = 25 : 15 = 5 : 3

16 ml of solution was replaced (will be calculated in the ratio of 5 : 3)

Replaced Quantity of Milk and Water = 10 ml and 6 ml

Hence, Initial Quantity of Milk = 25 ml + 10 ml = 35 ml