SOLUTION: 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

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Question 1149745: 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.
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39613)   (Show Source):
You can put this solution on YOUR website!
x, amount of the 80% acid
2x, amount of the 50% acid

Account for the acid of the mix

---------amount of the acid

--------------amount of the mixture

Quantity of just the water

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what is the ratio of acid to water in the mixture of the solutions.?
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You have the numbers calculated to answer this question.

Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website!
.
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.
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            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.

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I ask the author of the post do not submit such PROVOCATIVE formulations (!) (!) (!)

Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!

You are mixing an 80% acid solution with a 50% acid solution, using twice as much of the 50% acid solution.

The easiest and fastest way to a solution to the problem is using the fact that, with twice as much of the 50% acid solution as the 80% acid solution, the percentage of the mixture will be twice as close to 50% as it is to 80%.

Use any of a number of different ways to determine that the mixture is 60% acid.

Then, since the question is the ratio of acid to water in the final mixture, the answer is 60:40, or 3:2.