Mixture problems are common in high school math.<\/p>\n
Example:<\/strong><\/p>\n A grocer has two nut mixes. \u00a0One is 25% cashews; the other, 50% cashews. \u00a0He wants to concoct 10 kg of a mixture that is 35% cashews. \u00a0How many kg of each mixture should s\/he use?<\/p>\n Solution:<\/strong><\/p>\n Let x = kg of the 25% mixture.<\/p>\n Then, 10 – x = kg of the 50% mixture.<\/p>\n Now we track the cashews in each by percentage, leading to the total:<\/p>\n 0.25x + 0.50(10-x) = 0.35(10)<\/p>\n Notice that in math, we don’t use percents; rather, we use decimals. \u00a0To get the decimal, divide the percent by 100. \u00a0Therefore, 25%=0.25, and so on.<\/p>\n Next, we simplify the equation:<\/p>\n 0.25x + 5 – 0.5x = 3.5<\/p>\n -0.25x + 5 = 3.5<\/p>\n Subtract 5 from both sides:<\/p>\n -0.25x = 3.5 – 5<\/p>\n -0.25x = -1.5<\/p>\n Divide both sides by -0.25:<\/p>\n x=6<\/p>\n 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.<\/p>\n The one variable method is elegant. I’ll do the two variable method in a future post.<\/p>\n