.
A solution that is 80% acid and 20% water is mixed with a solution that is 50% acid and 50% water.
If twice as much 50% acid solution is used as 20% solution, what is the ratio of acid to water in the mixture of the solutions.
~~~~~~~~~~~~~~~
First what I want to say regarding this problem is that it is a PROVOCATION in its purest form.
Nobody, never and nowhere calls (20% water per 80% acid) as 20% solution.
It is called 80% solution --- THIS AND ONLY THIS way.
Therefore, the question formulation MUST BE
If twice as much 50% acid solution is used as 80% solution, what is the ratio of acid to water in the mixture of the solutions ?
For this formulation, my solution is as follows.
(I will produce a solution as simple as possible).
Let assume that we use 100 liters of the 80% acid solution.
Then the volume of the 50% acid solution is 200 liters, according to the problem's description.
100 liters of the 80% acid solution contain 0.8*100 = 80 liters of the pure acid.
200 liters of the 50% acid solution contain 0.5*200 = 100 liters of the pure acid.
Thus you have 100+200 = 300 liters of the resulting mixture, containing 80+100 = 180 liters of the pure acid.
Thus the concentration of the resulting mixture is = = 60%
and
the ratio under the question is = = . ANSWER
Solved.
----------------
I ask the author of the post do not submit such PROVOCATIVE formulations (!) (!) (!)