In a mixture of 60 litres the ratio of milk and water is 2 : 1. The amount of…

Concept

In a mixture-adjustment problem where only one component is added (here, water), the quantity of the other component (milk) stays fixed throughout. So the required addition can be found by fixing the constant quantity, writing it against the new total for the desired ratio, and solving for the unknown amount added.

Application

1. The 60 litre mixture has milk and water in the ratio 2 : 1, so milk = (2/3) × 60 = 40 litres and water = (1/3) × 60 = 20 litres.

2. Only water is added, so the milk quantity stays fixed at 40 litres; let x litres of water be added.

3. The new ratio of milk to water must be 1 : 2, so 40 / (20 + x) = 1 / 2.

4. Cross-multiplying: 20 + x = 80, so x = 60.

Cross-check

Adding 60 litres of water gives a new water quantity of 20 + 60 = 80 litres, with milk still 40 litres. The ratio 40 : 80 simplifies to 1 : 2, which matches the required ratio, confirming the value.