XAT 2015 Quant & DI Q7: Mixtures

Decode the given data

M is produced by mixing chemical X and chemical Y in the ratio 5 : 4

Hence, \\frac{5}{9}\\)^th of product M is chemical X and \\frac{4}{9}\\)^th of product M is chemical Y.

Chemical X has A and B in the ratio 1 : 3.
So, \\frac{3}{4}\\)^th of X is B.
Therefore, fraction of B in product M from chemical X = \\frac{5}{9}\\)×\\frac{3}{4}\\)

Chemical Y has B and C in the ratio 2 : 1.
So, \\frac{2}{3}\\)^rd of Y is B.
Therefore, fraction of B in product M from chemical Y = \\frac{4}{9}\\)×\\frac{1}{3}\\)

Adding the two, the fraction of B in Product M
= \\frac{5}{9}\\)×\\frac{3}{4}\\) + \\frac{4}{9}\\)×\\frac{2}{3}\\)
= \\frac{77}{108}\\)

Compute the amount of water added

The final product is obtained by mixing 864 units of product M with water.
In 864 units of Product M, amount of B = 864×\\frac{77}{108}\\) = 616

In the final mixture, concentration of B is 50%.

Therefore, the total quantity of final mixture = 616 × 2 = 1232
Water added = 1232 – 864 = 368

  The correct answer is Choice B.