1. Product M is produced by mixing chemical X and chemical Y in the ratio of 5 : 4. Chemical X is prepared by mixing two raw materials, A and B, in the ratio of 1 : 3. Chemical Y is prepared by mixing raw materials, B and C, in the ratio of 2 : 1. Then the final mixture is prepared by mixing 864 units of product M with water. If the concentration of the raw material B in the final mixture is 50%, how much water had been added to product M?
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By: anil on 05 May 2019 02.37 am
Let the quantities of the chemicals X and Y, mixed to produce product M be $$5c$$ and $$4c$$ respectively. X is prepared by mixing A and B in the ratio = 1 : 3 => Quantity of B in X = $$frac{3}{4} imes 5c = frac{15 c}{4}$$ Y is prepared by mixing B and C in the ratio = 2 : 1 Quantity of B in Y = $$frac{2}{3} imes 4c = frac{8 c}{3}$$ Quantity of B in M = $$frac{15 c}{4} + frac{8 c}{3} = frac{77 c}{12}$$ Now, 864 units of M was mixed with water to prepare the final mixture. => Total quantity of M = $$9c = 864$$ => $$c = frac{864}{9} = 96$$ Concentration of raw material B in the final mixture is 50 % => Quantity of final mixture = $$frac{100}{50} imes frac{77}{12} imes 96 = 1232$$ $$ herefore$$ Quantity of water added to M = $$1232 - 864 = 368$$ units
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