Question 1092352: How many gallons of distilled water must be mixed with 50 gallons of 30% alcohol solution to obtain a 25% solution?
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
There are many different ways to solve this kind of problem....
When one of the ingredients you are mixing is distilled water, here is one easy way to solve the problem.
The amount of alcohol in 50 gallons of 30% alcohol is
Since you are adding distilled water, that amount of alcohol will stay the same. If that same 15 gallons of alcohol is to be 25% of the final mixture, then
where x is the number of gallons in the final mixture.
Since we start with 50 gallons, and since we have to have 60 gallons in the final mixture, the number of gallons of distilled water we have to add is 10.
For a standard algebraic solution, you are mixing 50 gallons of 30% alcohol with x gallons of 0% alcohol; the mixture is to contain (50+x) gallons of 25% alcohol. So
And here is how I solve mixture problems. See if this makes sense to you....
You are starting with 50 gallons of 30% alcohol, and you are adding distilled water (0% alcohol) to end up with a mixture that is 25% alcohol. If you compare the percentage of the final mixture with the percentages of the two ingredients, you see that the percentage of the final mixture is "5 times as close to 30% as it is to 0%". That is, 30-25 = 5 and 25-0 = 25; 25/5 = 5.
But that "5 times as close to 30% as it is to 0%" means you need 5 times as much of the 30% ingredient as the 0% ingredient. If the number of gallons of the 30% ingredient is 50, the number of gallons of distilled water must be 1/5 of that, which is 10 gallons.
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