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?
Answer & explanation
Correct answer: option 3
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