A mixture (milk + water + honey) contains ‘18m’

Solution

According to the question, If 25% of the mixture was replaced with 9 litres of milk, then the quantity of milk in the resultant mixture = 18m × 0.75 + 9 = (13.5m + 9) litres. If 25% of the mixture was replaced with 9 litres of milk, then the quantity of water in the resultant mixture = 27n × 0.75 = ‘20.25n’ litres. ATQ; (13.5m + 9) × 0.90 = 20.25n Or, 13.5m + 9 = 20.25n ÷ 0.9 = 22.5n So, 13.5m = (22.5n – 9) – [equation I] If 25% of the mixture was replaced with 108 litres of honey, then the quantity of milk in the resultant mixture = 18m × 0.75 = ‘13.5m’ litres. If 25% of the mixture was replaced with 108 litres of honey, then the quantity of honey in the resultant mixture = (27n + 63) × 0.75 + 108 = 20.25n + 47.25 + 108 = (20.25n + 155.25) litres. So, 13.5m:(20.25n + 155.25) = 1:2 Or, 27m = 20.25n + 155.25 – [equation II] Multiplying equation (I) by ‘2’, we have: 27m = (45n – 18) So, (45n – 18) = (20.25n + 155.25) [from equation 2] Or, 24.75n = 173.25 So, n = 173.25 ÷ 24.75 ≈ 7 Therefore, the initial quantity of water in the mixture = 27 × 7 = 189 litres.