Question
A mixture (milk + water + honey) contains β14mβ
litres milk, β21nβ litres water and 49 litres more honey than water. If 25% of the mixture was replaced with 7 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 84 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.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 Γ 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 Γ 0.75 = β15.75nβ litres. ATQ; (10.5m + 7) Γ 0.90 = 15.75n Or, 10.5m + 7 = 15.75n Γ· 0.9 = 17.5n So, 10.5m = (17.5n β 7) β [equation I] If 25% of the mixture was replaced with 84 litres of honey, then the quantity of milk in the resultant mixture = 14m Γ 0.75 = β10.5mβ litres. If 25% of the mixture was replaced with 84 litres of honey, then the quantity of honey in the resultant mixture = (21n + 49) Γ 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 β [equation II] Multiplying equation (I) by β2β, we have: 21m = (35n β 14) So, (35n β 14) = (15.75n + 120.75) [from equation 2] Or, 19.25n = 134.75 So, n = 134.75 Γ· 19.25 β 7 Therefore, the initial quantity of water in the mixture = 21 Γ 7 = 147 litres
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