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?

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 = $$\frac{3}{4} \times 5c = \frac{15 c}{4}$$

Y is prepared by mixing B and C in the ratio = 2 : 1

Quantity of B in Y = $$\frac{2}{3} \times 4c = \frac{8 c}{3}$$

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 = $$\frac{100}{50} \times \frac{77}{12} \times 96 = 1232$$

$$\therefore$$ Quantity of water added to M = $$1232 - 864 = 368$$ units