A mixture (milk + water + honey) contains ‘16m’

Solution

According to the question, If 25% of the mixture was replaced with 8 litres of milk, then the quantity of milk in the resultant mixture = 16m × 0.75 + 8 = (12m + 8) litres. If 25% of the mixture was replaced with 8 litres of milk, then the quantity of water in the resultant mixture = 24n × 0.75 = ‘18n’ litres. ATQ; (12m + 8) × 0.90 = 18n Or, 12m + 8 = 18n ÷ 0.9 = 20n So, 12m = (20n – 8) – [equation I] If 25% of the mixture was replaced with 96 litres of honey, then the quantity of milk in the resultant mixture = 16m × 0.75 = ‘12m’ litres. If 25% of the mixture was replaced with 96 litres of honey, then the quantity of honey in the resultant mixture = (24n + 56) × 0.75 + 96 = 18n + 42 + 96 = (18n + 138) litres. So, 12m:(18n + 138) = 1:2 Or, 24m = 18n + 138 – [equation II] Multiplying equation (I) by ‘2’, we have: 24m = (40n – 16) So, (40n – 16) = (18n + 138) [from equation 2] Or, 22n = 154 So, n = 154 ÷ 22 ≈ 7 Therefore, the initial quantity of water in the mixture = 24 × 7 = 168 litres.