SOLUTION: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 40% and the third contains 60%. He wants to use all three solutions to
Question 1157206: A chemist has three different acid solutions. The first acid solution contains 20% acid, the second contains 40% and the third contains 60%. He wants to use all three solutions to obtain a mixture of 84 liters containing 50% acid, using 3 times as much of the 60% solution as the 40% solution. How many liters of each solution should be used? Found 2 solutions by josgarithmetic, greenestamps:Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
VOLUME PURE ACID
20% x 0.2x
40% y 0.4y
60% 3y 0.6*3y
50% 84
Okay, I admit it, I'm a geek about mixture problems. The method I use is far easier for me than the traditional algebraic method, because mental arithmetic is easy for me.
I use this method for mixture problem using two ingredients all the time; it takes only a little more effort to use it on a problem like this involving three ingredients.
Take a look at my method and see if you think it might work for you.
The method only works on two ingredients at a time; so I work the problem in two steps.
(1) The mixture contains 3 times as much 60% acid as 40% acid. That means, of those two ingredients, 3/4 of the mixture is 60% acid. Using my general method, that means the percentage of the acid mixture is 3/4 of the way from 40% to 60%. Mental arithmetic tells me that percentage is 55%.
(2) So now my problem is mixing 55% acid and 20% acid to get 50% acid.
Using the same reasoning, 50% is 5/35 = 1/7 of the way from 55% to 20%; that means 1/7 of the total 84L is the 20% acid.
Now I'm ready to find the numbers.
20% acid: 1/7 of 84L, which is 12L
The other 72L is my 55% acid mixture -- of which 3/4 is 60% acid and 1/4 is 40% acid. So
60% acid: 3/4 of 72L = 54L
40% acid: 1/4 of 72L = 18L