This question is from Mixtures and alligation. Two alcohol solutions are mixed in certain proportions. After that, some alteration has been done. We need to find the percentage of Alcohol in one of these two solutions. Mixtures and alligation is fairly simple idea. One can expect 2-3 questions from this topic in **CAT Exam**. Make sure you do not miss out on this easy topic your **CAT Preparation**. Practice **CAT Question Paper** with detailed video and text solutions, and take your Quant score a notch above.

Question 18 : Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

- 94%
- 92%
- 90%
- 89%

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We are told that the A and B are mixed in the ratio 1:3

That means, one unit of A is mixed with 3 units of B

Observe that the total Volume is 1 + 3 = 4 units.

The volume is doubled by adding A, this implies that a quantity of 4 units of A is added to the existing 4units of the mixture.

The total volume will now be 4 + 4 = 8 units.

Of which three units are B and the rest 5 units are A. Hence the ratio of A to B in the final mixture is 5 : 3

We are told that A has a concentration of 60% alcohol and the final mixture has a concentration of 72% alcohol

Therefore, \\frac{5 × Concentration of A + 3 × Concentration of B }{5 + 3}) = 72

\\frac{5 × 60 + 3 × Concentration of B }{5 + 3}) = 72

\\frac{300 + 3 × Concentration of B }{8}) = 72

300 + 3 × Concentration of B = 576

3 × Concentration of B = 576 - 300 = 276

3 × Concentration of B = 276 = 270 + 6

3 × Concentration of B = 270 + 6

Concentration of B = \\frac{270}{3}) + \\frac{6}{3})

Concentration of B = 90 + 2 = 92.

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