A mixture (milk + water + honey) contains ‘12m’ litres milk, ‘18n’ litres water and 42 litres more honey than water. If 25% of the mixture was replaced with 6 litres of milk, then the quantity of water in the resultant mixture becomes 10% less than that of milk. Instead, if 25% of the mixture is replaced with 72 litres of honey, then the ratio of the quantity of milk to that of honey will become 1:2. Find the initial quantity of water in the mixture.

According to the question, If 25% of the mixture was replaced with 6 litres of milk, then the quantity of milk in the resultant mixture = 12m × 0.75 + 6 = (9m + 6) litres. If 25% of the mixture was replaced with 6 litres of milk, then the quantity of water in the resultant mixture = 18n × 0.75 = ‘13.5n’ litres. ATQ; (9m + 6) × 0.90 = 13.5n Or, 9m + 6 = 13.5n ÷ 0.9 = 15n So, 9m = (15n – 6) – [equation I] If 25% of the mixture was replaced with 72 litres of honey, then the quantity of milk in the resultant mixture = 12m × 0.75 = ‘9m’ litres. If 25% of the mixture was replaced with 72 litres of honey, then the quantity of honey in the resultant mixture = (18n + 42) × 0.75 + 72 = 13.5n + 31.5 + 72 = (13.5n + 103.5) litres. So, 9m:(13.5n + 103.5) = 1:2 Or, 18m = 13.5n + 103.5 – [equation II] Multiplying equation (I) by ‘2’, we have: 18m = (30n – 12) So, (30n – 12) = (13.5n + 103.5) [from equation 2] Or, 16.5n = 115.5 So, n = 115.5 ÷ 16.5 = 7 Therefore, the initial quantity of water in the mixture = 18 × 7 = 126 litres.