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?
Answer: B
In 864 units of M,
X = 5/9 × 864 = 480 units
Y = 864 – 480 = 384 units
B in 480 units of X = 3/4 × 480 = 360 units
B in 384 units of Y = 2/3 × 384 = 256 units
Total units of B = 360 + 256 = 616
Concentration of B in the final mixture is 50%
Thus, water in the final mixture = (2 × 616) – 864 = 368 units.
Hence, option (b).