Q. 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
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
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 = 270/3 + 6/3
Concentration of B = 90 + 2 = 92.