Algebra Mixture Problems

Question

How many liters of a 5% acid solution should be mixed with 40 liters of 10% acid solution to obtain a mixture that is 8% of acid?

Answer 1

Zyan, the best way to find the solution to mixture problems is to create a table. For this problem, column 1 contains the percent of acid solution, column 2 contains the liter value, column 3 contains the value from multiplying column 1 and 2.

 

% Acid                Liters        %Acid * Liters

--------------------------------------------

5% or .05               x              .05 * x

 

10% or .10            40             .10 * 40

 

8% or .08           40 + x         .08(40 + x)

 

Note: we use "x" to denote the number of liters of 5% acid we need. The total number of liters is then 40 + x.

 

We create an equation that shows the relationship of the 5% acid and 10% acid to the 8% acid. Here is the equation:

 

.05x + .10*40 = .08(40 + x), note these are the values from column 3 of our table.

 

I do not like to work with fractions or decimals, so I will multiply the entire equation by 100 to remove the decimals:

 

5x + 400 = 8(40 + x)

 

5x + 400 = 320 + 8x

 

Combine like terms by subtracting 5x from both sides.

 

5x - 5x + 400 = 320 + 8x - 5x

 

400 = 320 + 3x

 

Next, subtract 320 from both sides of the equation:

 

400 - 320 = 320 - 320 + 3x

 

80 = 3x

 

Divide both sides of the equation by 3 giving:

 

x = 26.67

 

The answer to the question is you will need to add 26.67 liters of 5% acid solution to the 40 liters of 10% acid solution to get an 8% acid solution.

 

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