Question 1184883

A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?
<pre>Let amount of 40% solution to mix, be F
Then amount of 85% solution to mix = 2F
Also, amount of 25% to mix = 135 - F - 2F = 135 - 3F
We then get: .4F + .85(2F) + .25(135 - 3F) = .35(135)
.4F + 1.7F + .25(135) - .75F = .35(135)
.4F + 1.7F - .75F = .35(135) - .25(135)
1.35F = .1(135)
Amount of 40% to be mixed, or {{{highlight_green(matrix(1,10, F, "=", .1(135)/1.35, "=", 13.5/1.35, "=", "1,350"/135, "=", 10, L))}}}
You should easily be able to calculate the amount of 25% and 85% solutions, to mix.