A beaker contains a mixture of two liquids X and Y in the ratio 5 : 3. When 6 litres of the mixture is drawn off and then replaced with Y, the ratio of X and Y becomes 5 : 7. How many litres of liquid X was contained in the beaker initially?

11.25

Let the beaker initially 5z and 3z of mixtures of X and Y respectively,

⇒ Quantity of X in mixture left = (5z - \(\frac{5}{8}\)) x 6) = (5z - \(\frac{15}{4}\)) L

⇒ Quantity of X in mixture left = (3z - \(\frac{3}{8}\)) x 6) = (3z - \(\frac{9}{4}\)) L

⇒ \(\frac{20z - 15}{12z + 15}\) = \(\frac{5}{7}\)

⇒ (140z - 6z) = (105 + 75)

⇒ 80z = 180,

⇒ z = \(\frac{180}{80}\)

⇒ Liters of liquidity initially in X = 5z = 5 x \(\frac{180}{80}\) = 11.25.

Hence, initial liquid X was 11.25