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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:

Example Mixture Problem Solution Table
Name Amount Value Equation

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.
Name Amount Value Equation
S 70 mL 0.50 0.50 (70 mL)
S S 0.80 0.80 (S )
S 70 mL + S 0.60 0.60 (70 mL + S )

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?

Name Amount Value Equation
C C [latex]\$1.50[/latex] [latex]\$1.50[/latex] (C )
C 40 mL [latex]\$3.00[/latex] [latex]\$3.00[/latex] (40 mL)
C 40 mL + C [latex]\$2.50[/latex] [latex]\$2.50[/latex] (40 mL + C )

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?

Name Amount Value Equation
B B 0.24 0.24 (B )
B 42 L − B 0.18 0.18 (42 L − B )
B 42 L 0.20 0.20 (42 L)

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?

Name Amount Value Equation
Chocolate C [latex]\$4.00[/latex] [latex]\$4.00[/latex] (C)
Nuts 30 kg − C [latex]\$2.50[/latex] [latex]\$2.50[/latex] (30 kg − C)
Mix 30 kg [latex]\$3.50[/latex] [latex]\$3.50[/latex] (30 kg)

[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?

Name Amount Value Equation
Antifreeze (A) A 1.00 1.00 (A)
Water (W) 70 L − A 0.00 0.00 (70 L − A)
65% Solution 70 L 0.65 0.65 (70 L)

[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 Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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how to solve a value mixture problem using a linear equation

5.5 Solve Mixture Applications with Systems of Equations

Learning objectives.

By the end of this section, you will be able to:

  • Solve mixture applications
  • Solve interest applications

Be Prepared 5.5

Before you get started, take this readiness quiz.

  • Multiply 4.025 (1,562). 4.025 (1,562). If you missed this problem, review Example 1.98 .
  • Write 8.2% as a decimal. If you missed this problem, review Example 1.106 .
  • Earl’s dinner bill came to $32.50 and he wanted to leave an 18% tip. How much should the tip be? If you missed this problem, review Example 3.15 .

Solve Mixture Applications

When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this:

Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let n be the number of nickels and then write the number of dimes in terms of n , or if we would let d be the number of dimes and write the number of nickels in terms of d .

Now that we know how to solve systems of equations with two variables, we’ll just let n be the number of nickels and d be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column.

For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.

Example 5.45

Translate to a system of equations and solve:

The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled $1,302. How many $11 adult and how many $8 child tickets were sold?

the problem. We will create a table to organize the information.
what we are looking for. We are looking for the number of adult tickets
and the number of child tickets sold.
what we are looking for. Let the number of adult tickets.
the number of child tickets
A table will help us organize the data.
We have two types of tickets: adult and child.
Write and for the number of tickets.
Write the total number of tickets sold at the
bottom of the Number column.
Altogether 147 were sold.
Write the value of each type of ticket in the
Value column.
The value of each adult ticket is $11.
The value of each child tickets is $8.
The number times the value gives the total
value, so the total value of adult tickets is
, and the total value of child
tickets is
Altogether the total value of the tickets was
$1,302.
Fill in the Total Value column.
into a system of equations.
The Number column and the Total Value
column give us the system of equations.
We will use the elimination method to solve
this system.
Multiply the first equation by −8.
Simplify and add, then solve for .
Substitute = 42 into the first equation,
then solve for .
the answer in the problem.

  42 adult tickets at $11 per ticket makes $462
  105 child tickets at $8 per ticket makes $840.
  The total receipts are $1,302.✓
the question. The movie theater sold 42 adult tickets and 105 child tickets.

Try It 5.89

The ticket office at the zoo sold 553 tickets one day. The receipts totaled $3,936. How many $9 adult tickets and how many $6 child tickets were sold?

Try It 5.90

A science center sold 1,363 tickets on a busy weekend. The receipts totaled $12,146. How many $12 adult tickets and how many $7 child tickets were sold?

In Example 5.46 we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.

Example 5.46

Priam has a collection of nickels and quarters, with a total value of $7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?

the problem. We will create a table to organize the information.
what we are looking for. We are looking for the number of nickels
and the number of quarters.
what we are looking for. Let the number of nickels.
the number of quarters
A table will help us organize the data.
We have two types of coins, nickels
and quarters.
Write and for the number of each type of coin.
Fill in the Value column with the value of each
type of coin.
The value of each nickel is $0.05.
The value of each quarter is $0.25.
The number times the value gives the total
value, so, the total value of the nickels is
(0.05) = 0.05 and the total value of
quarters is (0.25) = 0.25 .
Altogether the total value of the coins
is $7.30.
into a system of equations.
The Total value column gives one equation.
We also know the number of nickels is six less
than three times the number of quarters.
Translate to get the second equation.
Now we have the system to solve.
the system of equations
We will use the substitution method.
Substitute = 3 − 6 into the first equation.
Simplify and solve for .
To find the number of nickels, substitute
= 19 into the second equation.
the answer in the problem.

the question. Priam has 19 quarters and 51 nickels.

Try It 5.91

Matilda has a handful of quarters and dimes, with a total value of $8.55. The number of quarters is 3 more than twice the number of dimes. How many dimes and how many quarters does she have?

Try It 5.92

Juan has a pocketful of nickels and dimes. The total value of the coins is $8.10. The number of dimes is 9 less than twice the number of nickels. How many nickels and how many dimes does Juan have?

Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.

Example 5.47

Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $7.60 per pound. Nuts cost $9.00 per pound and chocolate chips cost $2.00 per pound. How many pounds of nuts and how many pounds of chocolate chips should he use?

the problem. We will create a table to organize the information.
what we are looking for. We are looking for the number of pounds of nuts
and the number of pounds of chocolate chips.
what we are looking for. Let the number of pound of nuts.
the number of pounds of chips
Carson will mix nuts and chocolate chips
to get trail mix.
Write in and for the number of pounds
of nuts and chocolate chips.

There will be 20 pounds of trail mix.
Put the price per pound of each item in
the Value column.
Fill in the last column using
Number · Value = Total Value
into a system of equations.
We get the equations from the Number
and Total Value columns.
Solve the system of equations
We will use elimination to solve the system.
Multiply the first equation by −2 to eliminate .
Simplify and add. Solve for .
To find the number of pounds of
chocolate chips, substitute = 16 into
the first equation, then solve for .

the answer in the problem.

the question. Carson should mix 16 pounds of nuts with
4 pounds of chocolate chips to create the trail mix.

Try It 5.93

Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her $6 per pound. Peanuts are $4 per pound and cashews are $9 per pound. How many pounds of peanuts and how many pounds of cashews should she use?

Try It 5.94

Sammy has most of the ingredients he needs to make a large batch of chili. The only items he lacks are beans and ground beef. He needs a total of 20 pounds combined of beans and ground beef and has a budget of $3 per pound. The price of beans is $1 per pound and the price of ground beef is $5 per pound. How many pounds of beans and how many pounds of ground beef should he purchase?

Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water.

For these kinds of mixture problems, we’ll use percent instead of value for one of the columns in our table.

Example 5.48

Sasheena is a lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution?

the problem. A figure may help us visualize the situation, then we
will create a table to organize the information.
Sasheena must mix some of the 25%
solution and some of the 50% solution
together to get 200 ml of the 40% solution.
what we are looking for. We are looking for how much of each solution
she needs.
what we are looking for. Let number of ml of 25% solution.
number of ml of 50% solution
A table will help us organize the data.

She will mix ml of 25% with ml of
50% to get 200 ml of 40% solution.

We write the percents as decimals in
the chart.

We multiply the number of units times
the concentration to get the total
amount of sulfuric acid in each solution.
into a system of
equations. We get the equations from
the Number column and the Amount
column.
Now we have the system.
the system of equations.
We will solve the system by elimination.
Multiply the first equation by −0.5 to
eliminate .
Simplify and add to solve for .
To solve for , substitute = 80 into the
first equation.
the answer in the problem.

the question. Sasheena should mix 80 ml of the 25% solution
with 120 ml of the 50% solution to get the 200 ml
of the 40% solution.

Try It 5.95

LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution?

Try It 5.96

Anatole needs to make 250 milliliters of a 25% solution of hydrochloric acid for a lab experiment. The lab only has a 10% solution and a 40% solution in the storeroom. How much of the 10% and how much of the 40% solutions should he mix to make the 25% solution?

Solve Interest Applications

The formula to model interest applications is I = Prt . Interest, I , is the product of the principal, P , the rate, r , and the time, t . In our work here, we will calculate the interest earned in one year, so t will be 1.

We modify the column titles in the mixture table to show the formula for interest, as you’ll see in Example 5.49 .

Example 5.49

Adnan has $40,000 to invest and hopes to earn 7.1% interest per year. He will put some of the money into a stock fund that earns 8% per year and the rest into bonds that earns 3% per year. How much money should he put into each fund?

the problem. A chart will help us organize the information.
what we are looking for. We are looking for the amount to invest in each fund.
what we are looking for. Let the amount invested in stocks.
the amount invested in bonds.
Write the interest rate as a decimal for
each fund.
Multiply:
Principal · Rate · Time
to get the Interest.
into a system of
equations.
We get our system of equations from
the Principal column and the
Interest column.
the system of equations
Solve by elimination.
Multiply the top equation by −0.03.
Simplify and add to solve for .
To find , substitute = 32,800 into the first equation.
the answer in the problem. We leave the check to you.
the question. Adnan should invest $32,800 in stock and
$7,200 in bonds.

Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, s + b = 40,000, represents the total amount of money invested and the second equation, 0.08 s + 0.03 b = 0.071(40,000), represents the interest earned.

Try It 5.97

Leon had $50,000 to invest and hopes to earn 6.2 % interest per year. He will put some of the money into a stock fund that earns 7% per year and the rest in to a savings account that earns 2% per year. How much money should he put into each fund?

Try It 5.98

Julius invested $7,000 into two stock investments. One stock paid 11% interest and the other stock paid 13% interest. He earned 12.5% interest on the total investment. How much money did he put in each stock?

Example 5.50

Rosie owes $21,540 on her two student loans. The interest rate on her bank loan is 10.5% and the interest rate on the federal loan is 5.9%. The total amount of interest she paid last year was $1,669.68. What was the principal for each loan?

the problem. A chart will help us organize the information.
what we are looking for. We are looking for the principal of each loan.
what we are looking for. Let the principal for the bank loan.
the principal on the federal loan
The total loans are $21,540.
Record the interest rates as decimals
in the chart.
Multiply using the formula = Pr to
get the Interest.
into a system of
equations.
The system of equations comes from
the Principal column and the Interest
column.
the system of equations
We will use substitution to solve.
Solve the first equation for .
Substitute = − + 21,540 into the
second equation.
Simplify and solve for .
To find , substitute = 12,870 into
the first equation.

the answer in the
problem.
We leave the check to you.
the question. The principal of the bank loan is $12,870 and
the principal for the federal loan is $8,670.

Try It 5.99

Laura owes $18,000 on her student loans. The interest rate on the bank loan is 2.5% and the interest rate on the federal loan is 6.9 %. The total amount of interest she paid last year was $1,066. What was the principal for each loan?

Try It 5.100

Jill’s Sandwich Shoppe owes $65,200 on two business loans, one at 4.5% interest and the other at 7.2% interest. The total amount of interest owed last year was $3,582. What was the principal for each loan?

Access these online resources for additional instruction and practice with solving application problems with systems of linear equations.

  • Cost and Mixture Word Problems
  • Mixture Problems

Section 5.5 Exercises

Practice makes perfect.

In the following exercises, translate to a system of equations and solve.

Tickets to a Broadway show cost $35 for adults and $15 for children. The total receipts for 1650 tickets at one performance were $47,150. How many adult and how many child tickets were sold?

Tickets for a show are $70 for adults and $50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled $17,200. How many adult and how many child tickets were sold?

Tickets for a train cost $10 for children and $22 for adults. Josie paid $1,200 for a total of 72 tickets. How many children’s tickets and how many adult tickets did Josie buy?

Tickets for a baseball game are $69 for Main Level seats and $39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of $804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy?

Tickets for a dance recital cost $15 for adults and $7 for children. The dance company sold 253 tickets and the total receipts were $2,771. How many adult tickets and how many child tickets were sold?

Tickets for the community fair cost $12 for adults and $5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of $2,204. How many adult tickets and how many child tickets were sold?

Brandon has a cup of quarters and dimes with a total value of $3.80. The number of quarters is four less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is $0.95. The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse?

Peter has been saving his loose change for several days. When he counted his quarters and dimes, he found they had a total value $13.10. The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have?

Lucinda had a pocketful of dimes and quarters with a value of $ $6.20. The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have?

A cashier has 30 bills, all of which are $10 or $20 bills. The total value of the money is $460. How many of each type of bill does the cashier have?

A cashier has 54 bills, all of which are $10 or $20 bills. The total value of the money is $910. How many of each type of bill does the cashier have?

Marissa wants to blend candy selling for $1.80 per pound with candy costing $1.20 per pound to get a mixture that costs her $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

How many pounds of nuts selling for $6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him $5 per pound?

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $1.79 per gallon and the cost of the fruit drink is $2.49 per gallon. Hannah’s budget requires that the punch cost $2.21 per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

Joseph would like to make 12 pounds of a coffee blend at a cost of $6.25 per pound. He blends Ground Chicory at $4.40 a pound with Jamaican Blue Mountain at $8.84 per pound. How much of each type of coffee should he use?

Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost $7.80 per pound with French Roast Columbian coffee that cost $8.10 per pound to make a 20 pound blend. Their blend should cost them $7.92 per pound. How much of each type of coffee should they buy?

Melody wants to sell bags of mixed candy at her lemonade stand. She will mix chocolate pieces that cost $4.89 per bag with peanut butter pieces that cost $3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her $4.23 a bag to make. How many bags of chocolate pieces and how many bags of peanut butter pieces should she use?

Jotham needs 70 liters of a 50% alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution?

Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution?

A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution?

A scientist needs 120 liters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution?

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used?

A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of a 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used?

Hattie had $3,000 to invest and wants to earn 10.6% interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

Carol invested $2,560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25% interest on the total investment. How much money did she put in each account?

Sam invested $48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received $4,000 in interest in one year?

Arnold invested $64,000, some at 5.5% interest and the rest at 9%. How much did he invest at each rate if he received $4,500 in interest in one year?

After four years in college, Josie owes $65,800 in student loans. The interest rate on the federal loans is 4.5% and the rate on the private bank loans is 2%. The total interest she owed for one year was $2,878.50. What is the amount of each loan?

Mark wants to invest $10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn $1,095 in interest in one year?

A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25% interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be $1150 in one year. How much money should be invested at each rate?

A business has two loans totaling $85,000. One loan has a rate of 6% and the other has a rate of 4.5%. This year, the business expects to pay $4650 in interest on the two loans. How much is each loan?

Everyday Math

Laurie was completing the treasurer’s report for her son’s Boy Scout troop at the end of the school year. She didn’t remember how many boys had paid the $15 full-year registration fee and how many had paid the $10 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If $250 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

As the treasurer of her daughter’s Girl Scout troop, Laney collected money for some girls and adults to go to a three-day camp. Each girl paid $75 and each adult paid $30. The total amount of money collected for camp was $765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?

Writing Exercises

Take a handful of two types of coins, and write a problem similar to Example 5.46 relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

In Example 5.50 we solved the system of equations { b + f = 21,540 0.105 b + 0.059 f = 1669.68 { b + f = 21,540 0.105 b + 0.059 f = 1669.68 by substitution. Would you have used substitution or elimination to solve this system? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Access for free at https://openstax.org/books/elementary-algebra/pages/1-introduction
  • Authors: Lynn Marecek, MaryAnne Anthony-Smith
  • Publisher/website: OpenStax
  • Book title: Elementary Algebra
  • Publication date: Feb 22, 2017
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/elementary-algebra/pages/1-introduction
  • Section URL: https://openstax.org/books/elementary-algebra/pages/5-5-solve-mixture-applications-with-systems-of-equations

© Feb 9, 2022 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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    how to solve a value mixture problem using a linear equation

  2. how to solve a value mixture problem using a linear equation

    how to solve a value mixture problem using a linear equation

  3. how to solve a value mixture problem using a linear equation

    how to solve a value mixture problem using a linear equation

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    how to solve a value mixture problem using a linear equation

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    how to solve a value mixture problem using a linear equation

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    how to solve a value mixture problem using a linear equation

VIDEO

  1. Solving a value mixture problem using a system of linear equations

  2. How to Solve Mixture Problems

  3. Solving a Value Mixture Problem using a Linear Equation ALEKS Math 95

  4. Solving a value mixture problem using a linear equation

  5. Solving a value mixture problem using a system of linear equations 

  6. Value Mixture Problem 2

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  1. ALEKS: Solving a value mixture problem using a linear ...

    Learn how to solve a value mixture problem using a linear equation with ALEKS, a personalized learning platform for math and chemistry.

  2. Solving a value mixture problem using a linear equation

    Timely Math Tutor. 8.17K subscribers. Subscribed. 21. 2.8K views 2 years ago Linear Equations - Applications. For a complete list of Timely Math Tutor videos by course: www.timelymathtutor.com...

  3. Solving a Value Mixture Problem Using a Linear Equation

    Learn how to solve a value mixture problem using a linear equation, and see examples that walk through sample problems step-by-step for you to improve your math knowledge and...

  4. 4.4: Solve Mixture Applications with Systems of Equations

    We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column. For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.

  5. 5.5: Solve Mixture Applications with Systems of Equations

    Access these online resources for additional instruction and practice with solving application problems with systems of linear equations. Cost and Mixture Word Problems; Mixture Problems

  6. Solving a Value Mixture Problem Using a System of Linear ...

    Learn how to use a system of linear equations to solve value mixture problems, and see examples that walk through sample problems step-by-step for you to improve your math knowledge and...

  7. 2.3: Solve Mixture Applications with Systems of Equations

    Fill in the last column using \(\text{Number}\cdot\text{Value}=\text{Total Value}\) Translate into a system of equations. We get the equations from the Number and Total Value columns. \(\left\{\begin{aligned}n+c&=20\\ 9n+2c&=152\end{aligned}\right.\) Solve the system of equations. We will use elimination to solve the system.

  8. 4.3 Solve Mixture Applications with Systems of Equations

    2.1 Use a General Strategy to Solve Linear Equations; 2.2 Use a Problem Solving Strategy; 2.3 Solve a Formula for a Specific Variable; 2.4 Solve Mixture and Uniform Motion Applications; 2.5 Solve Linear Inequalities; 2.6 Solve Compound Inequalities; 2.7 Solve Absolute Value Inequalities

  9. 6.8 Mixture and Solution Word Problems – Intermediate Algebra

    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:

  10. 5.5 Solve Mixture Applications with Systems of Equations

    2.1 Solve Equations Using the Subtraction and Addition Properties of Equality; 2.2 Solve Equations using the Division and Multiplication Properties of Equality; 2.3 Solve Equations with Variables and Constants on Both Sides; 2.4 Use a General Strategy to Solve Linear Equations; 2.5 Solve Equations with Fractions or Decimals