Finding original concentration w%/w% A question from a grade 9 admission exam:
When 1 kg of salt is added to a solution of salt
and water, the solution becomes 33 1/3% salt by
mass. When 1 kg of water is added to the new solution, the resulting solution is 30% by mass. The percentage of salt in the original solution is...
I tried:
(Salt orig +1)/(Water orig + salt orig + 1) = 33 1/3. (S + 1)/(W+S+2) = 30. Which means 33 1/3*(W+S+1)=30*(W+S+2); 33 1/3 W+33 1/3 S +33 1/3 = 30W + 30S + 60; 3 1/3 W + 3 1/3 S = 26 1/3; S+W = 79. Putting that into the 1st eq gives S+1 / 80 = 33 1/3 meaning S= 26 2/3. Solving for W = 79-26 2/3 = 52 1/3. So the original concentration = 26 2/3 / (52 1/3 + 26 2/3) = 33 3/4%. But that's not one of the multiple choice answers (I did get further this time!).
$$\frac{S +1}{W + S + 1} = 1/3$$
$$\frac{ S + 1}{W+S+2}= 3/10$$



$$S +1=\frac {W + S + 1}3$$
$$S + 1=(W+S+2) \cdot 3/10$$



$$3S +3=W + S + 1$$
$$10S + 10=(W+S+2) \cdot 3$$



$$W=2S +2$$
$$10S + 10=(2S+2+S+2) \cdot 3$$
$$10S + 10=9S+12$$
$$S=2$$
$$W=6$$



The initial concentration is 25 % (w/w).