Tutoring math, this method comes up. The two variable method is more popular, but I’ve always preferred this one. The tutor offers it “for your entertainment.”

Mixture problems are common in high school math.

Example:

A grocer has two nut mixes.  One is 25% cashews; the other, 50% cashews.  He wants to concoct 10 kg of a mixture that is 35% cashews.  How many kg of each mixture should s/he use?

Solution:

Let x = kg of the 25% mixture.

Then, 10 – x = kg of the 50% mixture.

Now we track the cashews in each by percentage, leading to the total:

0.25x + 0.50(10-x) = 0.35(10)

Notice that in math, we don’t use percents; rather, we use decimals.  To get the decimal, divide the percent by 100.  Therefore, 25%=0.25, and so on.

Next, we simplify the equation:

0.25x + 5 – 0.5x = 3.5

-0.25x + 5 = 3.5

Subtract 5 from both sides:

-0.25x = 3.5 – 5

-0.25x = -1.5

Divide both sides by -0.25:

x=6

Looking back at our “let” statements (which we always need to write), we recall that x stands for the kg of the 25% mixture. Therefore, we need 6kg of the 25% mixture, and 4kg of the 50% one. The yield will be 10kg that is 35% cashews.

The one variable method is elegant. I’ll do the two variable method in a future post.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.