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

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (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.
x%280.1%29+%2B+40%280.5%29+=+%2840%2Bx%29%280.4%29 Simplify and solve for x.
0.1x+%2B+20+=+16+%2B+0.4x Subtract 0.1x from both sides of the equation.
20+=+16+%2B+0.3x Subtract 16 from both sides.
4+=+0.3x Finally, divide both sides by 0.3
13.33+=+x
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