SOLUTION: How many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution?

Question 52078This question is from textbook intermediate algebra
: How many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution? This question is from textbook intermediate algebra

Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let x = the number of liters of the 10% alcohol solution.
From the problem description, you can write the equation: (Note: Change the percents to their decimal equivalents) The final solution will be (40+x) liters.
Simplify and solve for x.
Subtract 0.1x from both sides of the equation.
Subtract 16 from both sides.
Finally, divide both sides by 0.3

You will need to mix 13.33...(or 13 1/3) liters of 10% alcohol solution with 40 liters of 50% alcohol solution to obtain 53.33... (or 53 1/3) liters of 40% alcohol solution.
Check:
13.33(0.1) + 40(0.5) = 53.33(0.4)
1.333 + 20 = 21.333
21.333 = 21.333

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