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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!).

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  • $\begingroup$ We have a policy which states that ‎you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that ‎we aren't doing your homework for you, and that the Q/A is beneficial for broad audience. As "homework-like questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc. Please edit in your full reasoning or thoughts on this. See Homework. Otherwise, the question may get closed as "low effort homework-like question class." $\endgroup$
    – Poutnik
    Sep 29, 2019 at 16:46
  • $\begingroup$ I tried using algebra to solve for original concentration - using the formula salt orig +1/ salt orig + water orig + 1 = 33 1/3. Then S(o) + 1/ s(o) +1+w(o)+1 = 30. I went down a rabbit hole that told me w(o) = .77w(o) -1 which is clearly wrong! We haven't done anything like this in school and my science teacher says he has no idea. This question is from a practice exam for a science/math high school admission. Help! $\endgroup$
    – my_honey_7
    Sep 29, 2019 at 18:34
  • $\begingroup$ I should also say I'm in grade 8. Thanks. $\endgroup$
    – my_honey_7
    Sep 29, 2019 at 18:40
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    $\begingroup$ Put your failed solution attempts to the question, and you can ask for the reopening the question. Note that the needed algebra is very basic. $\endgroup$
    – Poutnik
    Sep 29, 2019 at 18:54
  • $\begingroup$ You should either calculate with mass fractions instead of percentages, either multiply the mass fractions by 100. $\endgroup$
    – Poutnik
    Sep 30, 2019 at 19:50

1 Answer 1

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


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


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


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


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

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  • $\begingroup$ Thanks!!! Like Sherlock Holmes once you explain it, it's so easy! Using 1/3 instead of 33.3% is brilliant. And the question isnt really chem, it's math! The original equation for w/w percent is chemistry though. $\endgroup$
    – my_honey_7
    Oct 2, 2019 at 10:26
  • $\begingroup$ You then can mark the answer as accepted. $\endgroup$
    – Poutnik
    Oct 2, 2019 at 14:33

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