Mixture Word Problem -- Algebra 1

Question

Professor Baird wants to mix a solution containing 10% acid with one containing 15% acid to obtain a 20-ounce solution containing 12% acid. How many ounces of the 10% solution should the professor use?

 

Please show the work in a clear and understandable way. I need to understand exactly HOW you got the answer. Thank you!

Answer 1

Let A be the solution containing 10% acid and B be the solution with 15 % acid. 

 

Now let's set up a system of equations using the above information. 

 

Equation 1 :

 

A + B = 20

 

I added the weights of each solution  to get the total weight of the combined solution. 

 

Equation 2 :

 

0.10A + 0.15B = 20 (0.12)

 

In this case, I derived the equation using the weight of each solution and the percentage of acid in each solution. 

 

Finally I'm going to use substitution to solve. 

 

A = 20 - B

 

0.10 (20-B)  + 0.15B = 20 (0.12)

 

2 + 0.05 B = 2.4

 

0.05B = 0.4

 

B = 8

 

A = 20 - 8 

 

= 12

 

So we needed 12 ounces of the 10 percent solution to form the mixture 

 

 

Answer 2

The answer is that you need to mix 8 ounces of a 15% acid solution with 12 ounces of a 10% acid solution to obtain a 20 ounce solution that will be 12% acid. Here is how I solved it.

 

Let x represent the number of ounces of the 15% acid solution to be used.

Let y represent the number of ounces of the 10% acid solution to be used.

We have made the assumption that the required quantities of each are within our grasp.

 

x + y represents the total number of ounces in our solution. This is stated as 20.

Thus, x + y = 20  is one equation that describes the situation. (Eq. 1)

 

In x ounces of a 15% acid solution, the amount of acid is 15% of x or 0.15x

In y ounces of a 10% acid solution, the amount of acid is 10% of y or 0.10y

In 20 ounces of a 12% acid solution, the amount of acid is 12% of 20 or 2.4

 

Thus, 0.15x + 0.10y = 2.4 is another equation that describes the situation. (Eq. 2)

 

So, we have two equations with two variables, so we can solve this a variety of ways. I prefer the elimination method as follows:

 

Multiply Eq. 1 above by -0.10 to obtain -0.10x - 0.10y = -0.10(20) = -2.0

Now, add this equation to Eq. 2 above to obtain 0.05x = 0.4

To solve for x divide both sides by 0.05 to obtain x = 8. This means that you need 8 ounces of a 15% acid solution. By replacing x with 8 in Eq. 1 yields 8 + y = 20. One may conclude that y = 12. That means you need 12 ounces of a 10% acid solution.

 

To check our "solution" (pun intended) : Please note that equal amounts of acid (1.2 ounces) are obtained from each source. (i.e. 0.15(8) = 0.10(12) = 1.2) for a total amount of 2.4 ounces of acid. Also, recall that 12% = 0.12 and that 0.12(20) = 2.4

 

Answer: One may mix 8 ounces of a 15% acid solution with 12 ounces of a 10% acid solution to obtain 20 ounces of a 12% acid solution.

 

I hope my explanation was clear.