SOLUTION: A mixture of 8% disinfectant solution is to be made from 12% and 5% disinfectant solutions. How much of each solution should be used if 63 gallons of 8% solution are needed?

Question 1103821: A mixture of 8% disinfectant solution is to be made from 12% and 5% disinfectant solutions. How much of each solution should be used if 63 gallons of 8% solution are needed?

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

x + y = 63, (1) (total volume) 0.12*x + 0.05*y = 0.08*63. (2) (pure disinfectant) Multiply eq(1) by 0.05. The modified system is 0.05*x + 0.05y = 0.05*63, (3) 0.12*x + 0.05*y = 0.08*63. (4) From eq(4) subtract eq(3). The terms "0.05*y" will cancel each other, and you will get a single equation for only one unknown x. (It is how the Elimination method works). 0.012x - 0.05x = 0.08*63 - 0.05*63 0.07x = 0.03*63 ====> x = = 3*9 = 27. Answer. 27 gallons of the 12% disinfectant and (63-27) = 36 gallons of the 5% disinfectant.

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There is entire bunch of introductory lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Here is a method for solving mixture problems like this that is much faster and easier then the standard algebraic method usually shown.

The method uses a diagram like this:

The 12 and 5 in the first column are the percentages of the two ingredients.
The 8 in the middle column is the percentage of the mixture.
The 3 and 4 in the last column are the differences, computed diagonally, between the numbers in the first and second columns: 12-8=4; 8-5=3.

When the numbers given in the problem are used in this way, the numbers in the last column give the ratio in which the two ingredients need to be mixed.

In this problem, those numbers tell us that the 12% ingredient and the 5% ingredient need to be mixed in the ratio 3:4 to make an 8% mixture.

A ratio of 3:4 means 3/7 of the mixture should be the 12% solution and 4/7 should be the 5% solution. So
12%: 4/7 of 63 gallons = 36 gallons
5%: 3/7 of 63 gallons = 27 gallons
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