A mixture (milk + water + honey) contains ‘10m’ litres milk, ‘15n’ litres water and 35 litres more honey than water. If 25% of the mixture was replaced with 5 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 60 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 5 litres of milk, then the quantity of milk in the resultant mixture = 10m × 0.75 + 5 = (7.5m + 5) litres. If 25% of the mixture was replaced with 5 litres of milk, then the quantity of water in the resultant mixture = 15n × 0.75 = ‘11.25n’ litres. ATQ; (7.5m + 5) × 0.90 = 11.25n Or, 7.5m + 5 = 11.25n ÷ 0.9 = 12.5n So, 7.5m = (12.5n – 5) – [equation I] If 25% of the mixture was replaced with 60 litres of honey, then the quantity of milk in the resultant mixture = 10m × 0.75 = ‘7.5m’ litres. If 25% of the mixture was replaced with 60 litres of honey, then the quantity of honey in the resultant mixture = (15n + 35) × 0.75 + 60 = 11.25n + 26.25 + 60 = (11.25n + 86.25) litres. So, 7.5m:(11.25n + 86.25) = 1:2 Or, 15m = 11.25n + 86.25 – [equation II] Multiplying equation (I) by ‘2’, we have: 15m = (25n – 10) So, (25n – 10) = (11.25n + 86.25) [from equation 2] Or, 13.75n = 96.25 So, n = 96.25 ÷ 13.75 ≈ 7 Therefore, the initial quantity of water in the mixture = 15 × 7 = 105 litres.