Question 1157206
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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.<br>
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.<br>
Take a look at my method and see if you think it might work for you.<br>
The method only works on two ingredients at a time; so I work the problem in two steps.<br>
(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%.<br>
(2) So now my problem is mixing 55% acid and 20% acid to get 50% acid.<br>
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.<br>
Now I'm ready to find the numbers.<br>
20% acid: 1/7 of 84L, which is 12L<br>
The other 72L is my 55% acid mixture -- of which 3/4 is 60% acid and 1/4 is 40% acid.  So<br>
60% acid: 3/4 of 72L = 54L
40% acid: 1/4 of 72L = 18L<br>
ANSWERS:
12L of 20%
18L of 40%
54L of 60%<br>
CHECK:
.2(12)+.4(18)+.6(54) = 2.4+7.22+32.44 = 42
.50(84) = 42<br>