A vessel contains liquids A and B in ratio 5 : 3. If 16 litres of the mixture are removed and the same quantity of liquid B is added the ratio becomes 3 : 5 What quantity does the vessel hold?

40 ltrs

Let the vessel contains 5x litres and 3x litres of liquids A and B respectively.

The removed quantity contains = \(\frac{16}{5 + 3}\) × 5 = 10 litres of A and 16 - 10 = 6 litres of B

Now, (5x - 10) : (3x - 6 + 16) = 3 : 5

⇒ \(\frac{(5x - 10) }{3x + 10}\) = \(\frac{3}{5}\)

⇒ 25x - 50 = 9x + 30

⇒ 16x = 80

⇒ x = 5

∴ Capacity of vessel = 8x = 8 × 5 = 40