Two containers A and B contain solutions made up of only alcohol and water. The solution in container A has 30 percent alcohol by volume. They are both poured into an empty container which resulted in a solution having 25 percent water by volume. assume. Assume that the volume of the solution in container A is 3 times as large as that of container B. What is the percentage of water by volume of the solution in container B?
I have given this a try, but it doesn't seem to work.
$ a_C = \text{amount of alcohol in container} \, C$
$w_c = \text{amount of water in container} \, C$
$V_c = \text{volume of solution in container} \, C$
$a_A = 0.3 V_A$
$w_a = 0.7 V_A$ (shouldn't it be?)
$V_A = 3 V_B$
Then I tried this, but I am getting a negative value which doesn't seem to make sense unless you know what it means.
$w_A + w_b = 0.25 (V_a + V_b)$
$0.7 V_a + w_b = 0.25 ( 3 V_b + V_b ) $
$2.1 V_b + w_b = 0.25 (4 V_b) $
$2.1 V_b + w_b = V_b$
$w_b = -1.1 V_b$
I also tried assuming that the amount of water inside the new container is not $w_A + w_B$, say $W_{AB}$, but I don't think it is solvable then.
$W_{AB} = 0.25 (V_A + V_B) $
$W_{AB} = 0.25 (3 V_B + V_B) $
$W_{AB} = 0.25 (4 V_B)$
$W_{AB} = V_B$
This makes it seem that the amount of water inside the new container is just the amount of water inside container B. Thus, the percentage of water by volume of the solution in container B is 100%?
It is a multiple-choice question too.
a) 8
b) 9
c) 10
d) 12
e) none of these
Please help. Thank you.
They mean to say the final mix is $25\%$ alcohol by volume, not water. You should then get $10\%$ for an answer. The final mix is $4$ times the volume of $B$, so it needs the volume of $B$ of alcohol. $A$ supplies $3 \cdot 30\%$ of the volume of $B$ of alcohol, so you need $10\%$ of the volume of $B$ of alcohol to come from $B$.