Question 1162778

A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 30% and the third contains 70%. He wants to use all three solutions to obtain a mixture of 81 liters containing 55% acid, using 3 times as much of the 70% solution as the 30% solution. How many liters of each solution should be used?
<pre>Quite a few tutors have been PREACHING on here, the need to solve problems like these using ONE VARIABLE. I guess some who continue to use multiple 
variables - the likes of the other person who responded - don't feel that they need to learn the most efficient and least complex way to solutions.
Again, I would suggest that you don't solve this problem his way!

Let the amount of 30% solution, be T
Then amount of 70% solution is: 3T
Therefore, the amount of 15% solution is: 81 - (T + 3T), or 81 - 4T
We then get the following equation: .3T + .7(3T) + .15(81 - 4T) = .55(81)
.3T + 2.1T + .15(81) - .6T = .55(81)
1.8T = .55(81) - .15(81)
1.8T = .4(81)
Amount of 30% solution to mix, or {{{highlight_green(matrix(1,6, T, "=", .4(81)/1.8, "=", 18, L))}}}
Do you think you can now find the 70% and 15% amounts?