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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

  1. 94%
  2. 92%
  3. 90%
  4. 89%

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Explanatory Answer

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.


The answer is, "92%"

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