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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

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## "Mixture" Word Problems

Explanation Examples

## What are "mixture" word problems?

Mixture word problems are exercises which involve creating a mixture from two or more different things, and then determining some quantity (such as percentage, price, number of liters, etc) of the resulting mixture. There will always be a "rate" of some sort, such as miles per hour or cost per pound.

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## MathHelp.com

Mixture Word Problems

Here is an example:

Your school is holding a family-friendly event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for kids six years old and under) are $2.50 . From past experience, you expect about 13,000 people to attend the event.

You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket.

But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460 . How much revenue from each of child and adult tickets can you expect?

To solve this, we need to figure out the ratio of tickets that have already been sold. If we work methodically, we can find the answer.

Let A stand for the number of adult tickets pre-sold. Since a total of 548 tickets have been sold, then the number of children tickets pre-sold thus far must be 548 − A .

this construction is important! When you've got a variable that stands for part of whatever it is that you're working with, then the amount that is left for the other part of whatever you're working with is found by subtracting the variable from the total. That is, (the total amount) less (the amount that is being represented by the variable) is (the amount that is left for ther other amount). This "how much is left" construction is something you will need to understand and use.

Since each adult ticket cost $5.00 , then 5A stands for the revenue brought in from the adult tickets pre-sold; likewise, 2.5(548 − A) stands for the revenue brought in from the child tickets. (Note: The per-ticket cost is the "rate" for this exercise.)

Organizing this information in a grid, we get:

From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:

5A + 2.5((548 − A) = 2,460

5A + 1,370 − 2.5A = 2,460

1,370 + 2.5A = 2,460

2.5A = 1,090

A = (1,090)/(2.5) = 436

So 436 adult tickets were pre-sold, so the number of child tickets pre-sold is:

C = 548 − 436 = 112

So 112 child tickets were pre-sold.

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We expect about 13,000 people in total to attend this event. We have a ratio of adult pre-sold tickets to total pre-sold tickets. Assuming that the pre-sold rate (or ratio) is representative of the total number of adult tickets, we can set up a proportion (of pre-sold adult tickets to total pre-sold tickets), using a variable for the unknown total number of adult tickets expected to be sold:

436/548 = x /13,000

[(436)(13,000)]/548 = x

10,343.0656934... = x

This works out to about 10,343 adult tickets. The remaining 2,657 tickets, of the expected total of 13,000 , will be child tickets. Then the expected total ticket revenue is given by:

5(10,343) + 2.5(2,657) = 58,357.5

So the expected total revenues from ticket sales is $58,357.50 .

Let's try another one. This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Let w stand for the number of liters of the weaker 10% solution. Since the total number of liters is going to be 10 , then the number of liters left, after the first w liters have been poured, will be 10 − w liters needed of the 30% solution.

(The clear labeling of the variable is important. While I picked w to stand for the w eaker acid, I might not remember this by the end of the exercise. If I don't label, I might not be able correctly to interpret my answer in the end.)

For mixture problems, it is often very helpful to do a grid, so let's do that here:

When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution. So, adding down the right-hand column, setting the inputs equal to the mixed output, we get:

0.10 w + 0.30(10 − w ) = 1.5

0.10 w + 3 − 0.30 w = 1.5

3 − 0.2 w = 1.5

1.5 = 0.2 w

(1.5)/(0.2) = w = 7.5

Looking back and confirming that the variable stands for the number of liters of the weaker acid, we see that we would need 7.5 liters of 10% acid . This means that we would need another 10 − 7.5 = 2.5 liters of the stronger 30% acid .

(If you think about it, this makes sense. Fifteen percent is closer to 10% than it is to 30% , so we ought to need more 10% solution in our mix.)

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## Algebra: Mixture Word Problems

In these lessons, we will learn how to solve mixture word problems using algebra.

Related Pages Mixture Problems Solving Mixture Word Problems Using Algebra More Algebra Lessons

## What Are Mixture Problems?

Mixture problems are word problems where items or quantities of different values are mixed together.

Sometimes different liquids are mixed together changing the concentration of the mixture as shown in example 1 , example 2 and example 3 .

At other times, quantities of different costs are mixed together as shown in [example 4](#mix

We recommend using a table to organize your information for mixture problems. Using a table allows you to think of one number at a time instead of trying to handle the whole mixture problem at once.

We will show you how it is done by the following examples of mixture problems:

- Adding to the Solution
- Removing from the Solution
- Replacing the Solution
- Mixing Quantities of Different Costs

## How To Solve Mixture Problems When We Are Adding To The Solution?

Mixture Problems: Example 1: John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?

Solution: Set up a table for salt using the information from the question.

## How To Solve Mixture Problems When We Are Removing From The Solution?

Mixture Problems: Example 2: John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 3 0% solution?

Solution: Set up a table for water . The water is removed from the original solution.

## How To Solve Mixture Problems When We Are Replacing The Solution?

Mixture Problems: Example 3: A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution?

Solution: Set up a table for alcohol . The alcohol is replaced: an amount of 15% alcohol is removed and the same amount of 80% alcohol is added.

## Mixture Word Problems: Mixing Quantities Of Different Costs

Mixture Problems: Example 4: How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?

Solution: Set up a table for the chocolates with different costs.

## How To Set Up And Solve Mixture Word Problems?

Some word problems using systems of equations involve mixing two quantities with different prices. To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables. Other types of word problems using systems of equations include rate word problems and work word problems.

## How To Solve Acid Solution Problems?

Example: The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid?

Example: How much pure acid must be mixed with 200 mL of 5% acid to get a 25% acid?

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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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## How to Solve Mixture Word Problems

Last Updated: March 29, 2019

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been viewed 102,402 times.

Mixture word problems involve creating a mixture from two ingredients. A common type of problem is creating a solution of a certain strength, such as a 20% saline solution, from two solutions of varying strengths. Since these are multi-step problems involving a bit of logic, they can sometimes be confusing to solve. It is helpful to begin these types of problems by setting up a table that can help you keep track of variables. From there you can use algebra to find the missing information.

## Setting Up a Table

- For example, you might have a 20% saline solution, and a 15% saline solution. If you need to make 5 liters of an 18% saline solution, how many liters of each solution do you need to combine?
- For this problem you would label the three rows “20% solution,” “15% solution,” and “18% Mixture.”

- For example, you would label the second column “Percent Saline.” Since the first ingredient is 20% saline, in the first row you will write .20. Since the second solution is 15% saline, in the second row you will write .15. Since the final mixture needs to be 18% saline, in the third row you will write .18.

## Setting Up an Equation

## Solving the Problem

## Applying the Concept to Price Problems

- For example, you might be trying to solve the following problem: The student council is selling 100 cups of punch at a school dance. The punch is made from a combination of fruit juice and lemon-lime soda. They want to sell each cup of punch for $1.00. Normally they would sell a cup of fruit juice for $1.15 and a cup of lemon lime soda for $0.75. How many cups of each ingredient should the student council use to make the punch?
- In this problem, fruit juice and lemon-lime soda are the two ingredients.

- For example, since you know the student council plans on making 100 cups of punch, you would write 100 in the third row of the first column.

- For example, you know that the punch will be sold for $1.00 per cup, so write a 1 in the second column for the mixture. The fruit juice sells for $1.15 per cup, so write 1.15 in the second column for this ingredient. The soda sells for $0.75 per cup, so write 0.75 in the second column for lemon-lime soda.

## Expert Q&A

## You Might Also Like

- ↑ http://www.algebralab.org/Word/Word.aspx?file=Algebra_Mixtures.xml
- ↑ https://www.khanacademy.org/math/algebra/one-variable-linear-equations/alg1-linear-equations-word-problems/v/mixture-problems-2
- ↑ http://purplemath.com/modules/mixture.htm

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## Mixture Word Problems Lesson

- Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
- Demonstrate an understanding of the six-step method used for solving applications of linear equations
- Learn how to solve mixture word problems

## How to Solve Mixture Word Problems

Six-step method for solving word problems with linear equations in one variable.

- Read the problem and determine what you are asked to find
- If more than one unknown exists, we express the other unknowns in terms of this variable
- Write out an equation that describes the given situation
- Solve the equation
- State the answer using a nice clear sentence
- We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.

## Mixture Word Problems

Skills check:.

Solve each word problem.

A metallurgist needs to make 12 oz. of an alloy containing 71% copper. She is going to melt and combine one metal that is 78% copper with another metal that is 66% copper. How much of each should she use?

Please choose the best answer.

3 ounces of Ryan's Premium Molasses was made by combining 2 ounces of cane molasses which costs $1 per ounce with 1 ounce of beet molasses which costs $4 per ounce. Find the cost per ounce of the mixture.

How many gallons of an 80% sugar solution must be mixed with 7 gallons of a 35% sugar solution to make a 59% sugar solution?

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