using linear equations to solve mixture problems

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Solving Mixture Problems

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Mixture problems involve combining two or more things and determining some characteristic of either the ingredients or the resulting mixture. For example, we might want to know how much water to add to dilute a saline solution, or we might want to determine the percentage of concentrate in a jug of orange juice.

Introduction to Mixtures

Using a table to problem solve, practice problems.

We can use fractions, ratios, or percentages to describe quantities in mixtures.

If 200 grams of a saline solution has 40 grams of salt, what percentage of the solution is salt? Answer \(\frac{40}{200} = 0.20 = 20\%\) of the solution is salt.

If each can of orange juice concentrate contains the same amount of concentrate, which recipe will make the drink that is most orangey?

There is a general strategy for solving these mixture problems that uses simple algebra organized with a chart.

How much 40% rubbing alcohol do we need to add to 90% rubbing alcohol to make a 50% solution of rubbing alcohol? We could organize the data we are given in the following chart: Solution Type Concentration Amount of Solution Amount of Pure Alcohol 40% rubbing alcohol 0.4 ? liters ? liters 90% rubbing alcohol 0.9 ? liters ? liters 50% rubbing alcohol 0.5 10 liters \(0.5(10) = 5\) liters In general, the rows of the chart are the mixture types that you have. The columns describe the amount of each compound you have and the concentration of that component in each mixture (represented as a decimal). When you don't have some of the information needed to fill in the cart, use a variable instead. Solution Type Concentration Amount of Solution Amount of Pure Compound 40% rubbing alcohol 0.4 \(x\) \(0.4(x)\) 90% rubbing alcohol 0.9 \(10-x\) \(0.9(10-x)\) 50% rubbing alcohol 0.5 10 \(0.5(10) = 5\) The amount of 40% solution that we'll need is unknown (so make it \(x\)). The amount of 90% solution that we'll need is also unknown, but must be \(10-x\) liters so that we'll, in total, make 10 liters of the final solution. The amount of alcohol that each part of the mixture adds to the final result is equal to the amount of each solution mixed in, times the fraction of alcohol that that solution is made from. To use this chart to solve the problem, we will use the fourth column as an equation to solve for \(x.\) The 10 liters of our final mixture must have a total volume of 5 liters of alcohol in it in order to be 50% alcohol. Those 5 liters must come from a combination of the amount of 40% solution we mix in and the amount of 90% solution. If the volume (in liters) of 40% solution that we mix in is \(x,\) then \(0.4x\) will be the volume (in liters) of the amount of alcohol contributed. Similarly, \(0.9(10-x)\) will be the amount of alcohol contributed by the \(10-x\) liters of 90% alcohol solution that we add. Therefore, in total, \(0.4x + 0.9(10-x)\) must be equal to the 5 total liters we'll need in the final solution to make it 50% alcohol. Solving the equation, \[\begin{align} 0.4x + (9 - 0.9x) &= 5\\ -0.5x + 9 &= 5\\ 0.5x &= 4\\ \Rightarrow x &= \frac{4}{0.5}= 8. \end{align}\] Therefore, we need \(x = 8\) liters of the 40% solution.

How many grams of pure water must be added to 40 grams of a 10% saline solution to make a saline solution that is 5% salt? Answer Let's set up a table to solve this problem, using \(x\) to represent the number of grams of water that must be added. Solution Type Concentration Amount of Solution Total Amount of Salt water 0 \(x\) 0 10% solution 0.1 40 \((0.1)(40)\) 5% solution 0.05 40+\(x\) \((0.05)(40+x)\) Because the total amount of salt remains the same after we add the water, we can set up and solve the following equation: \[\begin{align} (0.1)(40)&=(0.05)(40+x) \\ 4 &= 2 + 0.05x \\ x&=40.\end{align}\] We need to add \(40 \) grams of pure water to create the solution that is \(5\%\) salt.

I have a 100 ml mixture that is 20% isopropyl alcohol (80 ml of water and 20 ml of isopropyl alcohol).

How much more alcohol do I need to add to make the mixture 25% alcohol?

There is a 40 litre solution of milk and water in which the concentration of milk is 72%. How much water must be added to this solution to make it a solution in which the concentration of milk is 60% ?

Let's practice using the strategies from above on a variety of problems.

Jill mixes 100 liters of A-beverage that contains 45% juice with 200 liters of B-beverage. The resulting C-beverage is 30% fruit juice. What is the percent of fruit juice in the 200 liters of the B-beverage? Answer Let's begin by making a table to show what we know. Beverage Liters of Beverage Concentration Liters of Juice Beverage A 100 0.45 \((0.45)(100) = 45\) Beverage B 200 \(x\) \(200x\) Beverage C 300 0.30 \((0.30)(300)=90\) The total number of liters of juice in Beverages A and B must equal 90, so \[\begin{align} 45 + 200x &= 90 \\ 200x &= 45 \\ 0.225 &= x.\end{align}\] The concentration of juice in Beverage B is 22.5%.

Unequal amounts of 40% and 10% acid solutions were mixed and the resulting mixture was 30% concentrated. However, the required concentration is 25%, so the Chemist added 300 cubic meters of 20% acid solution in order to get the required concentration. What was the original amount of 40% acid solution?

Write only the quantity without the units (cubic meters).

Legend: wt = weight

Strawberries contain about 15 wt% solids and 85 wt% water. To make strawberry jam, crushed strawberries and sugar are mixed in a 45:55 mass ratio, and the mixture is heated to evaporate water until the residue contains one-third water by mass.

Question: 1.) Calculate how many pounds of strawberries are needed to make a pound of jam.

In a mixture of 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:

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Here are some examples for solving mixture problems.

Coffee worth $1.05 per pound is mixed with coffee worth 85¢ per pound to obtain 20 pounds of a mixture worth 90¢ per pound. How many pounds of each type are used?

First, circle what you are trying to find— how many pounds of each type. Now, let the number of pounds of $1.05 coffee be denoted as x . Therefore, the number of pounds of 85¢‐per‐pound coffee must be the remainder of the twenty pounds, or 20 – x . Now, make a chart for the cost of each type and the total cost.

Now, set up the equation.

Therefore, five pounds of coffee worth $1.05 per pound are used. And 20 – x , or 20 – 5, or fifteen pounds of 85¢‐per‐pound coffee are used.

Solution A is 50% hydrochloric acid, while solution B is 75% hydrochloric acid. How many liters of each solution should be used to make 100 liters of a solution which is 60% hydrochloric acid?

First, circle what you're trying to find— liters of solutions A and B. Now, let x stand for the number of liters of solution A. Therefore, the number of liters of solution B must be the remainder of the 100 liters, or 100 – x . Next, make the following chart.

Therefore, using the chart, 60 liters of solution A and 40 liters of solution B are used.

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Chapter 6: Polynomials

6.8 Mixture and Solution Word Problems

Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below:

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Example 6.8.1

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

• The solution names are 50% (S 50 ), 60% (S 60 ), and 80% (S 80 ).
• The amounts are S 50 = 70 mL, S 80 , and S 60 = 70 mL + S 80 .
• The concentrations are S 50 = 0.50, S 60 = 0.60, and S 80 = 0.80.

The equation derived from this data is 0.50 (70 mL) + 0.80 (S 80 ) = 0.60 (70 mL + S 80 ).

Example 6.8.2

Sally and Terry blended a coffee mix that sells for [latex]\$2.50[/latex] by mixing two types of coffee. If they used 40 mL of a coffee that costs [latex]\$3.00,[/latex] how much of another coffee costing [latex]\$1.50[/latex] did they mix with the first?

The equation derived from this data is:

[latex]\begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \\ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \\ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \\ \hline &&-1.00(C_{1.50})&=&-20&& \\ \\ &&(C_{1.50})&=&\dfrac{-20}{-1}&& \\ \\ &&C_{1.50}&=&20&& \end{array}[/latex]

This means 20 mL of the coffee selling for [latex]\$1.50[/latex] is needed for the mix.

Example 6.8.3

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

The equation derived from this data is:

[latex]\begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \\ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \\ &-&7.56&&&&-7.56 \\ \hline &&&&0.06(B_{24})&=&0.84 \\ \\ &&&&B_{24}&=&\dfrac{0.84}{0.06} \\ \\ &&&&B_{24}&=&14 \end{array}[/latex]

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

Example 6.8.4

In Natasha’s candy shop, chocolate, which sells for [latex]\$4[/latex] a kilogram, is mixed with nuts, which are sold for [latex]\$2.50[/latex] a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for [latex]\$3.50[/latex] a kilogram. How much of each are used to make 30 kilograms of the mixture?

[latex]\begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \\ 4.00(C)&+&75&-&2.50(C)&=&105 \\ &-&75&&&&-75 \\ \hline &&&&1.50(C)&=&30 \\ \\ &&&&C&=&\dfrac{30}{1.50} \\ \\ &&&&C&=&20 \end{array}[/latex]

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Example 6.8.5

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

[latex]\begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \\ &&1.00A&=&45.5 \\ &&A&=&45.5 \\ \end{array}[/latex]

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

For questions 1 to 9, write the equations that define the relationship.

• A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
• How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
• You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
• How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
• How many litres of a blue dye that costs [latex]\$1.60[/latex] per litre must be mixed with 18 litres of magenta dye that costs [latex]\$2.50[/latex] per litre to make a mixture that costs [latex]\$1.90[/latex] per litre?
• How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
• A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
• A 20-gram alloy of platinum that costs [latex]\$220[/latex] per gram is mixed with an alloy that costs [latex]\$400[/latex] per gram. How many grams of the [latex]\$400[/latex] alloy should be used to make an alloy that costs [latex]\$300[/latex] per gram?
• How many kilograms of tea that cost [latex]\$4.20[/latex] per kilogram must be mixed with 12 kilograms of tea that cost [latex]\$2.25[/latex] per kilogram to make a mixture that costs [latex]\$3.40[/latex] per kilogram?

Solve questions 10 to 21.

• How many litres of a solvent that costs [latex]\$80[/latex] per litre must be mixed with 6 litres of a solvent that costs [latex]\$25[/latex] per litre to make a solvent that costs [latex]\$36[/latex] per litre?
• How many kilograms of hard candy that cost [latex]\$7.50[/latex] per kg must be mixed with 24 kg of jelly beans that cost [latex]\$3.25[/latex] per kg to make a mixture that sells for [latex]\$4.50[/latex] per kg?
• How many kilograms of soil supplement that costs [latex]\$7.00[/latex] per kg must be mixed with 20 kg of aluminum nitrate that costs [latex]\$3.50[/latex] per kg to make a fertilizer that costs [latex]\$4.50[/latex] per kg?
• A candy mix sells for [latex]\$2.20[/latex] per kg. It contains chocolates worth [latex]\$1.80[/latex] per kg and other candy worth [latex]\$3.00[/latex] per kg. How much of each are in 15 kg of the mixture?
• A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
• Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
• A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
• How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
• A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
• A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
• How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
• How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

Answer Key 6.8

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Applied Linear Equations: Mixture Problem - Concept

• Explanation

In order to understand graphing inverse functions, students should review the definition of inverse functions , how to find the inverse algebraically and how to prove inverse functions . The graphs of inverse functions and invertible functions have unique characteristics that involve domain and range. Techniques for graphing inverse functions can make it easier to graph certain functions by hand.

In this example we are going to be using linear equations to solve a mixture problem. And so for this particular word problem it's all about caffeine. We're in need of a little caffeine boost, we go to a local coffee shop and buy a drink that is 10 percent caffeine. You're still dragging a little bit after that so you go back buy another drink that is 10 percent, 10 ounces larger and 40 percent caffeine. So a lot more caffeine than that one, in total we just consumed 28 percent of caffeine. How large was our second drink? Okay, so there is our set up. And for these types of problems I find it most useful to first sort of make a diagram as to what is going on. Okay, so we have a first cup, have a second cup and we have a sort of ending result cup, your stomach if you will or some sort of collaborate collection of all that caffeine. So our first drink is 10 percent caffeine and it never tells us how large it is, so if we don't know how large something is, we can assign it a variable x. Okay, our second drink is 40 percent caffeine and we don't know how big that is either but we do know it is 10 ounces larger than the first. So if the first is 5 ounces, the second would be 15 we're just adding 10 onto that first number. So this just turns out to be x+10. And our sort of a collection is 20 percent caffeine and the total amount of liquid is these two previous cups added together. So we had our x, our x+10 throw those together, 2x+10. Okay. So we now have a diagram which we then need to turn into a equation for us to solve. Say this first cup was 10 percent, sorry, 10 ounces of liquid. It's 10 ounces of liquid and 10 percent caffeine, you actually ingested 10 percent of 10, 1 ounce. So we really take the percent of caffeine times the quantity figure out how much caffeine is actually in a glass. So this one turns out to be 0.1 times x. Remember when you're turning a percentage into a decimal, move your decimal place over two spots. Same idea for this one, 40 percent turns into 0.4 times a quantity, x+10. And the last one over here, 28 percent 0.28, times an amount of liquid, 2x+10. Okay, so, you could just solve this out right here through here, dealing with decimals. In general, I suggest getting rid of your decimals just because, decimals and fractions still sometimes have this sphere of and kind of freeze up. So multiply everything by a number to get rid of all our decimals. Here we have a tenth, here we have a tenth and here we have hundredths. So if you multiply everything by a hundred, all the decimals go away. By, by 100, 100 times a tenth is 10. 100 times fourth, 40 and 100 times 0.28 is 28, 2x+10. Okay, we now have a nice linear equation. No decimals, solve it out as we would any other linear equation. So distribute our terms to, so distribute that 40 and that 28. 10x stay the same, that gives us 40x, 40 times 10 is 400. 28 times 2, 56 and 28 times 10 is 280. Okay, combine like terms on each side. So 10x, 40x gives us 50x+400=56x+280. Okay, so now we have x's and other terms, combining all our x's. I try to keep my x's positive so I will subtract this 50x over to the other side. Get our 280 over to the other side as well, so subtract 280 and then simplify. So we've got rid of this 50x, 400-280=120, 56x-50x=6x and that 280 disappears. So we're left with 120=6x, to solve this out divide by six, x=20. Okay, now we need to always make sure we're actually solving for what the question asked for. The question asked for how large was our second drink. We found x here and that is actually the amount of our first drink. So if our first drink is 20 ounces that makes our second, twenty plus 10, 30 ounces. So our answer then is 30 ounces for our second drink. So given a word problem, sort of made a diagram, that get out sort of how to organize information. Change it into a linear equation which we then solved out and made sure we gave the proper answer.

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Module 16: Linear Systems

Mixture problems, learning outcomes.

• Solve mixture problems

Write a system of linear equations representing a mixture problem, solve the system and interpret the results

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution.  A solution is a mixture of two or more different substances like water and salt or vinegar and oil.  Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.  There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described  as a percentage.  This number comes from the ratio of how much mass is in a specific volume of liquid.  For example if you have [latex]50[/latex] grams of salt in a [latex]100mL[/latex] of water you have a [latex]50\%[/latex] salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths.  In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[/latex] of a [latex]9\%[/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Next we add the new volumes and new masses.

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution.  In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

A chemist has [latex]70 mL[/latex] of a [latex]50\%[/latex] methane solution. How much of an [latex]80\%[/latex] solution must she add so the final solution is [latex]60\%[/latex] methane?

Let’s use the problem solving process outlined previously to help us work through a solution to the problem.

Read and Understand:  We are looking for a new amount – in this case a volume –  based on the words “how much”.  We know two starting  concentrations and the final concentration, as well as one volume.

Define and Translate:  Solution 1 is the [latex]70 mL[/latex] of [latex]50\%[/latex] methane and solution 2 is the unknown amount with [latex]80\%[/latex] methane.  We can call our unknown amount [latex]x[/latex].

Write and Solve:   Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is [latex]0.6[/latex] because we want the final solution to be [latex]60\%[/latex] methane.

Add the total mass for solution 1 and solution 2 to get the total mass for the [latex]60\%[/latex] solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract [latex]35[/latex] from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by [latex]0.2[/latex]

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex] [latex]x=35[/latex]

[latex]35mL[/latex] of the [latex]80\%[/latex] solution must be added to the original [latex]70mL[/latex] of the [latex]50\%[/latex] solution to gain a solution with a concentration of [latex]60\%[/latex].

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

A farmer has two types of milk, one that is [latex]24\%[/latex] butterfat and another which is [latex]18\%[/latex] butterfat. How much of each should he use to end up with [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat?

Read and Understand:  We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is [latex]42[/latex] gallons. There are two unknowns in this problem.

Define and Translate:  We will call the unknown volume of the  [latex]24\%[/latex] solution [latex]x[/latex], and the unknown volume of the [latex]18\%[/latex] solution [latex]y[/latex].

Write and Solve:  Fill in the table with the information we know.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex] When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex] and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]-0.18(x+y) = (42)(-0.18)[/latex]

[latex]-0.18x - 0.18y = -7.56[/latex]

Now our system of equations looks like this:

[latex]0.24x + 0.18y = 8.4[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by [latex]0.06[/latex] on each side:

[latex]x = 14[/latex]

Now substitute the value for [latex]x[/latex] into one of the equations in order to solve for [latex]y[/latex].

[latex](14) + y = 42[/latex]

[latex]y = 28[/latex]

This can be interpreted as [latex]14[/latex] gallons of [latex]24\%[/latex] butterfat milk added to [latex]28[/latex] gallons of [latex]18\%[/latex] butterfat milk will give [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

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• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Ex: System of Equations Application - Mixture Problem. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/4s5MCqphpKo . License : CC BY: Attribution
• Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : . License : CC BY: Attribution
• Beginning and Intermediate Algebra Textbook. Authored by : Tyler Wallace. Located at : http://www.wallace.ccfaculty.org/book/book.html . License : CC BY: Attribution
• College Algebra. Authored by : Abramson, Jay. Provided by : OpenStax. Located at : https://openstaxcollege.org/textbooks/college-algebra . License : CC BY: Attribution