A mixture (milk + water + honey) contains ‘14m’

Solution

According to the question, If 25% of the mixture was replaced with 7 litres of milk, then the quantity of milk in the resultant mixture = 14m × 0.75 + 7 = (10.5m + 7) litres. If 25% of the mixture was replaced with 7 litres of milk, then the quantity of water in the resultant mixture = 21n × 0.75 = ‘15.75n’ litres. ATQ; (10.5m + 7) × 0.90 = 15.75n Or, 10.5m + 7 = 15.75n ÷ 0.9 = 17.5n So, 10.5m = (17.5n – 7) – [equation I] If 25% of the mixture was replaced with 84 litres of honey, then the quantity of milk in the resultant mixture = 14m × 0.75 = ‘10.5m’ litres. If 25% of the mixture was replaced with 84 litres of honey, then the quantity of honey in the resultant mixture = (21n + 49) × 0.75 + 84 = 15.75n + 36.75 + 84 = (15.75n + 120.75) litres. So, 10.5m:(15.75n + 120.75) = 1:2 Or, 21m = 15.75n + 120.75 – [equation II] Multiplying equation (I) by ‘2’, we have: 21m = (35n – 14) So, (35n – 14) = (15.75n + 120.75) [from equation 2] Or, 19.25n = 134.75 So, n = 134.75 ÷ 19.25 ≈ 7 Therefore, the initial quantity of water in the mixture = 21 × 7 = 147 litres