pat thompson misture problem

7/24/2019 Pat Thompson Misture Problem

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How to think about "Mixture" Problems

Math Student Teaching Spring 2000

Pat Thompson,Vanderbilt University

We want to make 300 g of a salt solution containing 6% salt. We will

make it by pouring together two other salt solutions, one containing8% salt and the other containing 5% salt. How many grams of the 5%

solution and how many grams of the 8% solutions should we mix?

Things to notice about the problem

There are only two substances involved -- salt and water. We mix salt and water in various proportions to make solutions of different

strengths. Any particular solution is named by the percent of its weight (or

mass) that is salt. A solution is called "5% solution" is called that because 5%

of it is salt. When we pour together two different solutions, we get a new solution. The

new solution is made of water and salt only. All the water in the newsolution came from the water in the two solutions we mixed. All the salt in

the new solution came from the two solutions we mixed.

The weight of the salt in the new solution makes some percent of the newsolution's weight.

We can organize our understanding of this problem as

such:

Start off with two mixtures:
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Combine them to make a new mixture

Things we can deduce from the information in this problem

We want to make 300 g of a salt solution containing 6% salt.

So, the amount of salt in the new solution is 6% of 300 g, or 18 grams.


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We will make it by pouring together two other salt solutions, one containing8% salt and the other containing 5% salt.

However much 8% solution we use, 8 percent of it will be salt. If weusexgrams of 8% solution, it will be made of .08xgrams of salt. If weuse .08xgrams of salt from the 8% solution, we must get the rest of the salt

from the 5% solution. So the amount of 5% solution we use will contain 18-

.08xgrams of salt.

How many grams of the 5% solution and how many grams of the 8%

solutions should we mix?

Any amount of 5% solution is 20 times as large as the amount of salt in it.So, if we get 18-.08xgrams of salt from the 5% solution, we must use

20(18-.08x) grams of 5% solution.

We are making 300 grams of 6% solution, and that comes from combining

the two amounts we start with. So, 300 = (number of grams of 8% solution)+ (number of grams of 5% solution).

Therefore:300 =x+ 20(18-.08x)

300 =x+ 360 - 1.6x-60 = -.6x

100 =x. Use 100 grams of 8% solution and 20(18-8), or 200 grams of 5%

solution.

We can use a table to organize our thinking

1.

We know we will have 300 grams of solution that is 6% salt.2. 300 grams of solution that is 6% salt contains 18 grams of salt.3. Assume we havexgrams of 8% solution.

4.xgrams of solution that is 8% salt will contain .08xgrams of salt.

5. Since there are 18 grams of salt altogether and .08xgrams of it comes fromthe 8% solution, the rest of it, 18-.08xgrams, must come from the 5%

solution.6. For the 5% solution to be 5% salt, the entire amount of solution must weigh

20 times as much as the salt in it. Therefore, there is 20(18-.08x) grams of 5%

solution.


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7. The two amounts of solution must total 300 grams when combined.Therefore,

300 =x+ 20(18 - .08x).

We can generalize this understanding(after having solved several problems like this)

Start off with two amounts of two mixtures

Combine them to make a new mixture


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Use a table to organize your thinking
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pat thompson misture problem

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