Question 289107
If you think in terms of how much antifreeze you are dealing with, I think that the problem becomes a little easier. For example, 10 liters of a 90% antifreeze solution contains (10*(0.9) = 9) liters of antifreeze, so...let x equal the required number of liters of the 90% antifreeeze solution.
Since you want to end up with 30 liters, you will also need (30-x) liters of the 75% antifreeze solution. In algebra this can be expressed as:
0.9x+0.75(30-x) = 0.8(30) Simplify and solve for x.
0.9x+22.5-0.75x = 24 Combine like-terms.
0.15x+22.5 = 24 Subtract 22.5 from both sides.
0.15x = 1.5 Finally, divide both sides by 0.15
x = 10
You will need to mix 10 liters of 90% antifreeze solution with (30-10 = 20) liters of 75% antifreeze solution to obtain 30 liters of 80% antifreeze solution.