A weighted average approach to solving mixture problems Tyler Olsen
Abstract In high school and college algebra classes, it is typical for mixture problems to be solved by setting up a system of linear equations in 2 variables. The most difficult part of solving a mixture problem for students is setting up the correct equations. This paper explores the use of a weighted average to solve such problems, giving students an alternative formulaic approach to solving mixture problems.
Mixture Problems in Algebra A typical mixture problem looks something like this: A chemist needs 100L of a 15% nitric acid solution, but in his lab, only a 10% and a 35% solution are available. How much of each available solution will be required to obtain 100L of 15% nitric acid solution?
To solve a problem like this a student must set up a system of linear equations in two variables. This is the point where most students run into difficulties, and some ultimately give up. Most textbooks authors use a table approach that they believe promotes understanding:
Amount of solution
% nitrtic acid
Amount pure nitric acid in the solution
Solution 1
x
10%
0.10x
Solution 2
y
35%
0.35y
Mixture
100
15%
0.15(100) =15
The first equation is formed from the amount of each solution being mixed: x + y = 100 The second equation comes from the amount of pure nitric acid being combined: 0.10x + 0.35y = 15 These equations are then solved simultaneously using the substitution method as follows:
So, 80L of 10% solution must be mixed with 20L of 35% solution to obtain 100L of a 15% solution. Solving the equations is routine for most students, but forming the equations that must be solved often causes confusion, frustration, and ultimately hopelessness. Many students convince themselves that they cannot do mixture problems and they simply skip over them on exams. There is always more than one way to climb a mountain, and solving a mixture problem is no exception.
What is an average? The word “average” has been in use since the late 15th century. From the Arabic word awar meaning ‘damage to goods’ it became avarie in French and avaria in Italian. Originally denoting a charge or customs duty payable by the owner of goods to be shipped, the term later denoted the financial liability from goods lost or damaged at sea, and specifically the equitable apportionment of this between the owners of the vessel and the cargo (Oxford). Mathematically, average is a number expressing the central or typical value in a set of data, in particular the mode (most common value in the date set), median (middle value when the data set is arranged from smallest to largest), and, most commonly, the mean. In elementary school, students are taught that the average or mean of a given set of numbers can be calculated by adding up all the numbers in the set then dividing by the count of numbers that were added together: For example, the average of the numbers 1; 2; 9 is found by adding 1+2+9=12, then dividing this sum by the count of numbers, 3, to get an average; 12/3=4. Weighted average is not typically introduced until a student reaches college.
.
When there are repeated numbers in the set of numbers being averaged,represents the “number of times” the same elementappears in the set of numbers. The greater the absolute value of the multiplier,, the more affect it has on the average. Hence the “weight” of the numbers being added changes the overall average. This equation is modeled in calculus as:
Weighted average is something most people have experience with without realizing it. It is used to calculate the average test score of a class; grade point average; number of gallons of gas that are needed to maintain the proper octane in a vehicle; average rainfall over a given time; and yet it is something that few “average” people actually know about or understand. Average is a designation that represents two or more non-like numbers being manipulated to represent a “center” for those numbers. This center is the base of all studied information; an “average “ person has an IQ of about 100; meteorologists concern themselves with average snow or rainfalls and the consequences of how far above or below average a certain region is; food packers claim a weight on their packaging when that weight could be within a certain tolerance above or below the stated average. When describing the term average, words such as “quantities”, “central”, and “set of data” are used, but what does all that mean to the “average” person? Looking at a number line that begins at the number 1 and ends at the number 9, the average becomes visible by applying simple mental calculations. The center (or median) is 5 because there are an equal number of values on either side of 5. Because of the symmetry of the data, 5 is also the mean. This can be determined by adding up all 9 whole numbers to get a sum of 45 then diving by 9, or by averaging the end points 1 and 9. However, it is not only this “true” mean that has a relationship with these end values, but every single number on the number line has a relationship with these end numbers, which we will refer to as the origins. Every number between the origins (not limited to whole numbers, but the infinite number of decimals as well) has a relationship to these origins. The location or relationship of these numbers can be expressed as a weighted average of the origins and can thus be considered “off center means”. According to Chinn (2011), “Everyone should have some idea about statistics.” The statistics he is referring to are averages. If averages are learned at a young age, the concepts have years to grow and develop. The step from basic average in the 6th grade to weighted average is a small step, but an important concept that could become just another everyday tool. The relationship between mixture problems and weighted average The word “mixture” by definition is “a combination of different qualities, things, or emotions in which the component elements are individually distinct” (OED). When mixing things together, the same thing is being added to the other thing over and over, even though they may appear distinct. This means that mixture is a weighted average. It is an average of two numbers combined in such a way that their output is a number that lies between two end points, and has a relationship with respect to these points. Consider this problem thought out intuitively by a student: A vehicle can hold 10 gallons of fuel, and this vehicle requires 88-octanegas in order to function properly. However, the local service station is out of the 88, but has 87- and 91-octanes still available. How many gallons of each fuel type would be needed to obtain and maintain an 88-octane level?
These mixtures are the same thing, fuel, just with different octane ratings (distinct). If one gallon of 87-octane gas has a rating of 87, then two gallons still have a rating of 87, because the average of 87 and 87 is 87 (87+87 / 2 = 87). No matter how many gallons of fuel are mixed, the rating never changes if the ratings are the same. If one gallon of 87 and one gallon of 91 are mixed then the two gallons combined have an average rating of (87+91) / 2 = 89. When one of each value, or one of each “origin” is used, the “true” average of those origins is returned. The true average here is 89-octane, which is higher than required, so an adjustment to the mixture is needed to obtain the desired OCM of 88. Mixing two gallons of 87-octane with one gallon of 91-octane results in three gallons of (87+87+91) / 3 = 88.333-octane. This is still too high, so we need to mix in more of the lower octane fuel. Mixing three gallons of 87-octane with one gallon of 91-octane gives four gallons of (87+87+87+91) / 4 = 88-octane, exactly what is needed. So, for every three gallons of 87-octane, one gallon of 91-octane is needed to maintain an 88-octane mixture. This translates to 3/4 of the tank (7.5 gallons) filled with 87-octane fuel, and ¼ of the tank (2.5 gallons) with 91-octane fuel. Seeing this intuitive approach creates the relationship between mixture and weighted average and allows students to look at an alternative, formulaic way to solve mixture problems in algebra. The formulaic approach Let’s reconsider the mixture problem already solved at the start of this paper, this time using unknown values: A chemist needs T Liters of a D% nitric acid solution, but in his lab, only an m% and an M% solution are available. How much of each available solution will be required to obtain T Liters of D% nitric acid solution?
The equations are set up as before with Q(m) = Quantity of m% solution and therefore T – Q(m) = quantity of M% solution. Q(m) · m% + (T – Q(m)) · M% = T · D% Multiplying this equation by 100, removes the % signs: Q(m) · m + (T – Q(m)) · M = T · D Simplifying and solving the equation for Q(m) is now a simple process: Q(m) · m + T · M – Q(m) · M = T · D Q(m)(m – M) = TD – TM
In this formula: m = minimum % solution available M = maximum % solution available Q(m) = Quantity of m% solution required T = Total quantity of mixture required D = Desired % of mixture required With this simple proof, the relationship between weighted average and mixture becomes clear. Once this relationship is understood, students will develop a deeper understanding of averages, and what they truly represent. For the original example solved earlier: A chemist needs 100L of a 15% nitric acid solution, but in his lab, only a 10% and a 35% solution are available. How much of each available solution will be required to obtain 100L of 15% nitric acid solution?
The formula delivers:
Thus, we achieve the same answer as before that 80 liters of 10% solution must be combined with 100 – 80 = 20 liters of 35% solution to achieve 100 liters of 15% solution. For the gasoline problem: A vehicle can hold 10 gallons of fuel, and this vehicle requires 88-octanegas in order to function properly. However, the local service station is out of the 88, but has 87- and 91-octanes still available. How many gallons of each fuel type would be needed to obtain and maintain an 88-octane level?
The formula gives us:
Once again, we get the same answer as before. 7.5 gallons of 87-octane gas needs to be mixed with 2.5 gallons of 91-octane gas to achieve 10 gallons of 88-octane. Variations The formula has two important variations. The first variation solves the mixture problem for the amount of solution required at the higher concentration by simply replacing M with m and vice-versa: The second variation is used in mixture problems in which the desired value, D, is unknown. Instead, the variable V is defined to be the product of TD: , where V = TD. This variation comes into play when mixture problems using dollar amounts rather than concentrations are being considered. A typical example would be: INSERT EXAMPLE HERE Conclusion Tools are created almost everyday to help the human race overcome obstacles and challenges; the tools provided for use in the math world are no different. From the creation of hammers, to the use of calculators, the every day world as well as the world of mathematics has changed and adapted. Giving students the opportunity to develop a deeper knowledge of average and weighted average is important. Opening the door from average to weighted average with the use of this formula will allow students to create their own relationship between weighted average and mixture. For students that simply cannot see how to set up a system of linear equations that can be used to solve a mixture problem, this formula could save them many hours of frustration. It is a tool that can be used to obtain a solution to any mixture problem. Any time a student’s mental block can be removed, by offering an alternative way of thinking about a problem, is a good thing. Using a weighted average approach allows them to look at the problem from a different angle and may result in a deeper understanding of what is actually being asked, as well as giving these students a better outlook on math. The formula allows students to understand these problems by setting up the problem in a simplified form, and opens their minds to the definition of weighted average. Now, instead of skipping mixture problems, they have the tool to solve the problem and move on.
Tyler Olsen
Abstract
In high school and college algebra classes, it is typical for mixture problems to be solved by setting up a system of linear equations in 2 variables. The most difficult part of solving a mixture problem for students is setting up the correct equations. This paper explores the use of a weighted average to solve such problems, giving students an alternative formulaic approach to solving mixture problems.
Mixture Problems in Algebra
A typical mixture problem looks something like this:
A chemist needs 100L of a 15% nitric acid solution, but in his lab, only a 10% and a 35% solution are available. How much of each available solution will be required to obtain 100L of 15% nitric acid solution?
To solve a problem like this a student must set up a system of linear equations in two variables. This is the point where most students run into difficulties, and some ultimately give up. Most textbooks authors use a table approach that they believe promotes understanding:
The first equation is formed from the amount of each solution being mixed: x + y = 100
The second equation comes from the amount of pure nitric acid being combined:
0.10x + 0.35y = 15
These equations are then solved simultaneously using the substitution method as follows:
So, 80L of 10% solution must be mixed with 20L of 35% solution to obtain 100L of a 15% solution.
Solving the equations is routine for most students, but forming the equations that must be solved often causes confusion, frustration, and ultimately hopelessness. Many students convince themselves that they cannot do mixture problems and they simply skip over them on exams. There is always more than one way to climb a mountain, and solving a mixture problem is no exception.
What is an average?
The word “average” has been in use since the late 15th century. From the Arabic word awar meaning ‘damage to goods’ it became avarie in French and avaria in Italian. Originally denoting a charge or customs duty payable by the owner of goods to be shipped, the term later denoted the financial liability from goods lost or damaged at sea, and specifically the equitable apportionment of this between the owners of the vessel and the cargo (Oxford). Mathematically, average is a number expressing the central or typical value in a set of data, in particular the mode (most common value in the date set), median (middle value when the data set is arranged from smallest to largest), and, most commonly, the mean.
In elementary school, students are taught that the average or mean of a given set of numbers can be calculated by adding up all the numbers in the set then dividing by the count of numbers that were added together:
For example, the average of the numbers 1; 2; 9 is found by adding 1+2+9=12, then dividing this sum by the count of numbers, 3, to get an average; 12/3=4. Weighted average is not typically introduced until a student reaches college.
When there are repeated numbers in the set of numbers being averaged,
Weighted average is something most people have experience with without realizing it. It is used to calculate the average test score of a class; grade point average; number of gallons of gas that are needed to maintain the proper octane in a vehicle; average rainfall over a given time; and yet it is something that few “average” people actually know about or understand.
Average is a designation that represents two or more non-like numbers being manipulated to represent a “center” for those numbers. This center is the base of all studied information; an “average “ person has an IQ of about 100; meteorologists concern themselves with average snow or rainfalls and the consequences of how far above or below average a certain region is; food packers claim a weight on their packaging when that weight could be within a certain tolerance above or below the stated average. When describing the term average, words such as “quantities”, “central”, and “set of data” are used, but what does all that mean to the “average” person?
Looking at a number line that begins at the number 1 and ends at the number 9, the average becomes visible by applying simple mental calculations. The center (or median) is 5 because there are an equal number of values on either side of 5. Because of the symmetry of the data, 5 is also the mean. This can be determined by adding up all 9 whole numbers to get a sum of 45 then diving by 9, or by averaging the end points 1 and 9. However, it is not only this “true” mean that has a relationship with these end values, but every single number on the number line has a relationship with these end numbers, which we will refer to as the origins.
Every number between the origins (not limited to whole numbers, but the infinite number of decimals as well) has a relationship to these origins. The location or relationship of these numbers can be expressed as a weighted average of the origins and can thus be considered “off center means”.
According to Chinn (2011), “Everyone should have some idea about statistics.” The statistics he is referring to are averages. If averages are learned at a young age, the concepts have years to grow and develop. The step from basic average in the 6th grade to weighted average is a small step, but an important concept that could become just another everyday tool.
The relationship between mixture problems and weighted average
The word “mixture” by definition is “a combination of different qualities, things, or emotions in which the component elements are individually distinct” (OED). When mixing things together, the same thing is being added to the other thing over and over, even though they may appear distinct. This means that mixture is a weighted average. It is an average of two numbers combined in such a way that their output is a number that lies between two end points, and has a relationship with respect to these points. Consider this problem thought out intuitively by a student:
A vehicle can hold 10 gallons of fuel, and this vehicle requires 88-octane gas in order to function properly. However, the local service station is out of the 88, but has 87- and 91-octanes still available. How many gallons of each fuel type would be needed to obtain and maintain an 88-octane level?
These mixtures are the same thing, fuel, just with different octane ratings (distinct). If one gallon of 87-octane gas has a rating of 87, then two gallons still have a rating of 87, because the average of 87 and 87 is 87 (87+87 / 2 = 87). No matter how many gallons of fuel are mixed, the rating never changes if the ratings are the same. If one gallon of 87 and one gallon of 91 are mixed then the two gallons combined have an average rating of (87+91) / 2 = 89. When one of each value, or one of each “origin” is used, the “true” average of those origins is returned. The true average here is 89-octane, which is higher than required, so an adjustment to the mixture is needed to obtain the desired OCM of 88. Mixing two gallons of 87-octane with one gallon of 91-octane results in three gallons of (87+87+91) / 3 = 88.333-octane. This is still too high, so we need to mix in more of the lower octane fuel. Mixing three gallons of 87-octane with one gallon of 91-octane gives four gallons of (87+87+87+91) / 4 = 88-octane, exactly what is needed. So, for every three gallons of 87-octane, one gallon of 91-octane is needed to maintain an 88-octane mixture. This translates to 3/4 of the tank (7.5 gallons) filled with 87-octane fuel, and ¼ of the tank (2.5 gallons) with 91-octane fuel.
Seeing this intuitive approach creates the relationship between mixture and weighted average and allows students to look at an alternative, formulaic way to solve mixture problems in algebra.
The formulaic approach
Let’s reconsider the mixture problem already solved at the start of this paper, this time using unknown values:
A chemist needs T Liters of a D% nitric acid solution, but in his lab, only an m% and an M% solution are available. How much of each available solution will be required to obtain T Liters of D% nitric acid solution?
The equations are set up as before with Q(m) = Quantity of m% solution and therefore T – Q(m) = quantity of M% solution.
Q(m) · m% + (T – Q(m)) · M% = T · D%
Multiplying this equation by 100, removes the % signs:
Q(m) · m + (T – Q(m)) · M = T · D
Simplifying and solving the equation for Q(m) is now a simple process:
Q(m) · m + T · M – Q(m) · M = T · D
Q(m)(m – M) = TD – TM
In this formula:
m = minimum % solution available
M = maximum % solution available
Q(m) = Quantity of m% solution required
T = Total quantity of mixture required
D = Desired % of mixture required
With this simple proof, the relationship between weighted average and mixture becomes clear. Once this relationship is understood, students will develop a deeper understanding of averages, and what they truly represent.
For the original example solved earlier:
A chemist needs 100L of a 15% nitric acid solution, but in his lab, only a 10% and a 35% solution are available. How much of each available solution will be required to obtain 100L of 15% nitric acid solution?
The formula delivers:
Thus, we achieve the same answer as before that 80 liters of 10% solution must be combined with 100 – 80 = 20 liters of 35% solution to achieve 100 liters of 15% solution.
For the gasoline problem:
A vehicle can hold 10 gallons of fuel, and this vehicle requires 88-octane gas in order to function properly. However, the local service station is out of the 88, but has 87- and 91-octanes still available. How many gallons of each fuel type would be needed to obtain and maintain an 88-octane level?
The formula gives us:
Once again, we get the same answer as before. 7.5 gallons of 87-octane gas needs to be mixed with 2.5 gallons of 91-octane gas to achieve 10 gallons of 88-octane.
Variations
The formula
The second variation is used in mixture problems in which the desired value, D, is unknown. Instead, the variable V is defined to be the product of TD:
INSERT EXAMPLE HERE
Conclusion
Tools are created almost everyday to help the human race overcome obstacles and challenges; the tools provided for use in the math world are no different. From the creation of hammers, to the use of calculators, the every day world as well as the world of mathematics has changed and adapted. Giving students the opportunity to develop a deeper knowledge of average and weighted average is important. Opening the door from average to weighted average with the use of this formula will allow students to create their own relationship between weighted average and mixture.
For students that simply cannot see how to set up a system of linear equations that can be used to solve a mixture problem, this formula could save them many hours of frustration. It is a tool that can be used to obtain a solution to any mixture problem.
Any time a student’s mental block can be removed, by offering an alternative way of thinking about a problem, is a good thing. Using a weighted average approach allows them to look at the problem from a different angle and may result in a deeper understanding of what is actually being asked, as well as giving these students a better outlook on math. The formula allows students to understand these problems by setting up the problem in a simplified form, and opens their minds to the definition of weighted average. Now, instead of skipping mixture problems, they have the tool to solve the problem and move on.