Question 405382: How many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution? Can you please show me how to work out this problem also:) Thanks.
Found 2 solutions by josmiceli, IWork4Dessert:
Answer by josmiceli(19441)   (Show Source):
Answer by IWork4Dessert(60)  (Show Source):
You can put this solution on YOUR website! The best way to work out this problem is by using a chart. Since I can't really illustrate one for you, I'll try to explain it.
There are three rows and three columns. One column is labeled "percent", one "amount", one "total".
On the side of the first row, write "acid". On the side of the second row write "solution 1". On the side of the third row write "solution 2".
In the first column, the percent column, you'll want to put the percentage of acidity. So "acid" should be 40% acidic, the "solution 1" 20% acidic, and the "solution 2" 25% acidic. Under amount, you'll want to put x for "acid", 12 for "solution 1", and x+12 for "solution 2", as they're being combined.
In the last column, multiply the first by the second rows. "Acid" will be 40x, "solution 1" will be 240, and "solution 2" will be 300+25x.
Now, you know that your acid and your solution 1 combined will equal solution 2, so all you've got to do is add them together. Your equation will look like this:
40x+240=300+25x
Solve from there.
15x=60
x=4 L
You can now plug it back in to see if it works properly by sticking it into your equation.
40(4)+240=300+25(4)
160+240=300+100
400=400
Voila.
Acid is 4 liters, and there's your answer.
Questions, let me know.
Hope this helps!
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