The liquids, X and Y are mixed in the ratio of 3: 2 and the mixture is sold at ₹11 per litre at a profit of 10%. If the liquid X costs ₹2 more per litre then Y, the cost of X per litre is (in ₹):

Selling price of mixture = Rs. 11 at 10% profit

=> Cost price of 1 litre of mixture = $$\frac{11}{100+10}\times100=Rs.$$ $$10$$

Let cost of liquid X = Rs. $$x$$, => Cost of liquid Y = Rs. $$(x-2)$$

The liquids are mixed in the ratio of 3 : 2. In 5 liters of the mixture, 3 liters will be first liquid and 2 liters will be the second liquid.

Thus, cost of 5 litres of mixture = $$3x+2(x-2)=5\times10$$

=> $$3x+2x-4=50$$

=> $$5x=50+4=54$$

=> $$x=\frac{54}{5}=Rs.$$ $$10.80$$

=> Ans - (B)