Question

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?

Accepted Answer

B. In $864$ units of M: - Quantity of X = $\frac{5}{9} 	imes 864 = 480$ units. - Quantity of Y = $\frac{4}{9} 	imes 864 = 384$ units. Content of raw material B in X = $\frac{3}{4} 	imes 480 = 360$ units. Content of raw material B in Y = $\frac{2}{3} 	imes 384 = 256$ units. Total B in Product M = $360 + 256 = 616$ units. Let $W$ be the units of water added. Since concentration of B in final mixture is $50\%$, total final mixture = $2 	imes 616 = 1232$ units. Water added $W = 1232 - 864 = 368$ units.

Chemical Mixture Ratio Calculation

Explanation

Practise this under exam conditions