Answer:
- 360 mL of pure alcohol
- 600 mL of 70% solution
Explanation:
Every mixture problem is set up the same way. It doesn't matter whether it is a mixture of solutions, types of coffee beans, candy, nuts, coins, or anything else.
One measure involved is the quantity of the mix.
The other measure involved is some characteristic of the components of the mix.
An equation is written for each of these measures.
_____
Here, the first measure, quantity of the mix, is the total amount of solution. Apparently, we are to assume that this includes all of the 240 mL of the 25% mixture. It will also include the amount of pure alcohol that we add.
The other measure of interest in this problem is the amount of alcohol, expressed as a percentage of the amount of solution. (In many mixture problems, it is the total monetary value.)
We are asked to find the amount of 100% alcohol added, and the total amount of solution. We can assign variables to these quantities:
- let "a" represent the amount of 100% alcohol added (in mL)
- let "q" represent the total quantity of solution (in mL)
Then the two equations we can write are ...
240 + a = q . . . . . . . . . . . . . . . . . . . . . . the total quantity of the mix
25% × 240 + 100% × a = 70% × q . . . . the amount of alcohol present
Substituting the first expression for q into the second equation and writing percentages as decimals, we have ...
0.25×240 +a = 0.70×(240 +a)
60 +a = 168 + .70a . . . . eliminate parentheses
0.30a = 108 . . . . . . . . . .add -0.70a-60 to both sides
108/0.30 = a = 360 . . . . divide by the coefficient of a
Then the total quantity of mix is ...
240 +360 = q = 600
360 mL of pure alcohol must be added to the 240 mL of 25% stock to obtain 600 mL of 70% solution.