The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:

Let total quantity of the mixture = x liters

• Initial ratio of petrol to kerosene = 3:2
• So petrol = (3/5)x, kerosene = (2/5)x

Now 10 liters is removed

Amount of petrol removed = (3/5) × 10 = 6 liters

Amount of kerosene removed = (2/5) × 10 = 4 liters

Remaining:

• Petrol left = (3x/5 - 6)
• Kerosene left = (2x/5 - 4)

10 liters of kerosene added, so:

Final kerosene = (2x/5 - 4) + 10 = (2x/5 + 6)

New ratio = 2:3

(3x/5 - 6) / (2x/5 + 6) = 2 / 3

Cross multiplying:

3(3x/5 - 6) = 2(2x/5 + 6)
=> (9x - 90) = (4x + 60)
=> 9x - 90 = 4x + 60
=> 5x = 150
=> x = 30

Answer: 30 liters