One of the many applications of percent is toward mixture problems. Mixture problems involve a situation where chemicals or solutions with different concentration are being combined or mixed then we determine the unknown concentration or volume if the other two components are known. The diagram below illustrates the process on how to solve mixture word problems. As figured, mixture problem always contains two sets of quantities or equations: the volume and the concentration content.
1. A pharmacist has 30L of a 10% drug solution. How many liters of 5% solution must be added to get a mixture that is 8%?
To solve mixture problems, we establish two equations. The first equation pertains to the volume of liquids to be mixed or added. The second equation pertains to the concentration of acid, alcohol, solution or any chemicals found in the liquids to be mixed. Total concentration of such chemicals is obtained by multiplying the volume and the concentration level or content. Then solve these equations simultaneously by using any algebraic method.
Let \(x\) be the unknown volume (in liters) of the 5% solution so \(30+x\) is the volume (in liters) of the total mixture which is 8% solution. The equations are
2. A manufacturer of a certain brand of soft drinks claimed that their orange soda is “naturally flavored”, though it has only 5% orange juice. A new regulation specifies that to be “natural” any drink should have at least 10% fruit juice. How much pure orange juice must this manufacturer add to a 1000 gal orange soda to conform the said regulation?
Volume: 1000 gallons + \(x\) gallons = 1000 + \(x\) gallons
Amount of orange juice: 5% of 1000 gallons + 100% of \(x\) gallons = 10% of 1000 + \(x\) gallons
From these data, we can now formulate the equations. These are,
Solve for \(x\) in the second equation.