- The Five-Step Method
- Number Problems
- Value and Rate Problems: Coin Problems Problems and Mixture Problems

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Beginning with the graphical solution of systems, this chapter includes an interpretation of independent, inconsistent, and dependent systems and examples to illustrate the applications for these systems. The substitution method and the addition method of solving a system by elimination are explained, noting when to use each method. The five-step method is again used to illustrate the solutions of value and rate problems (coin and mixture problems), using drawings that correspond to the actual situation. Objectives of this module: become more familiar with the five-step method for solving applied problems, be able to solve number problems, be able to solve value and rate problems.

- The Five-Step Method
- Number Problems
- Value and Rate Problems: Coin Problems Problems and Mixture Problems

When solving practical problems, it is often more convenient to introduce two variables rather than only one. Two variables should be introduced only when two relationships can be found within the problem. Each relationship will produce an equation, and a system of two equations in two variables will result.

We will use the five-step method to solve these problems.

- Introduce two variables, one for each unknown quantity.
- Look for two relationships within the problem. Translate the verbal phrases into mathematical expressions to form two equations.
- Solve the resulting system of equations.
- Check the solution.
- Write a conclusion.

The sum of two numbers is 37. One number is 5 larger than the other. What are the numbers?

Step 2: There are two relationships.

(a) The Sum is 37.

(b) One is 5 larger than the other.

Step 3:

We can easily solve this system by substitution. Substitute

Step 4: The Sum is 37.

One is 5 larger than the other.

Step 5: The two numbers are 16 and 21.

The difference of two numbers is 9, and the sum of the same two numbers is 19. What are the two numbers?

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

The two numbers are 14 and 5.

The problems in Sample Sets B and C are *value problems*. They are referred to as value problems because one of the equations of the system used in solving them is generated by considering a value, or rate, or amount times a quantity.

A parking meter contains 27 coins consisting only of dimes and quarters. If the meter contains $4.35, how many of each type of coin is there?

Step 2: There are two relationships.

(a) There are 27 coins.

(b) Contribution due to dimes =

Contribution due to quarters =

Step 3:

We can solve this system using elimination by addition. Multiply both sides of equation

Step 4: 16 dimes and 11 quarters is 27 coins.

The solution checks.

Step 5: There are 11 quarters and 16 dimes.

A bag contains only nickels and dimes. The value of the collection is $2. If there are 26 coins in all, how many of each coin are there?

There are 14 dimes and 12 nickels.

A chemistry student needs 40 milliliters (ml) of a 14% acid solution. She had two acid solutions, A and B, to mix together to form the 40 ml acid solution. Acid solution A is 10% acid and acid solution B is 20% acid. How much of each solution should be used?

Step 2: There are two relationships.

(a) The sum of the number of ml of the two solutions is 40.

(b) To determine the second equation, draw a picture of the situation.

The equation follows directly from the drawing if we use the idea of amount times quantity.

Solve this system by addition. First, eliminate decimals in equation 2 by multiplying both sides by 100.

Eliminate

Step 4: 24 ml and 16 ml to add to 40 ml.

The solution checks.

Step 5: The student should use 24 ml of acid solution A and 16 ml of acid solution B.

A chemistry student needs 60 ml of a 26% salt solution. He has two salt solutions, A and B, to mix together to form the 60 ml solution. Salt solution A is 30% salt and salt solution B is 20% salt. How much of each solution should be used?

The student should use 36 ml of salt solution A and 24 ml of salt solution B.

The sum of two numbers is 22. One number is 6 more than the other. What are the numbers?

The two numbers are 14 and 8.

The sum of two numbers is 32. One number is 8 more than the other. What are the numbers?

The difference of two numbers is 12 and one number is three times as large as the other. Whatare the numbers?

The two numbers are 18 and 6.

The difference of two numbers is 9 and one number is 10 times larger than the other. Whatare the numbers?

Half the sum of two numbers is 14 and half the difference is 2. What are the numbers?

The numbers are 16 and 12.

One third of the sum of two numbers is 6 and one fifth of the difference is 2. What are the numbers?

A 14 pound mixture of grapes sells for

6 pounds of Type 1 and 8 pounds of Type 2.

The cost of 80 liters of a blended cleaning solution is $28. Type 1 solution costs 20¢ a liter and type 2 solution costs 40¢ a liter. How many liters of each solution were used to form the blended solution?

The cost of 42 grams of a certain chemical compound is

12 grams of Type 1 and 30 grams of Type 2.

A play was attended by 342 people, some adults and some children. Admission for adults was

200 tickets were sold to a college’s annual musical performance. Tickets for students were

37 non-student tickets.

A chemistry student needs 22 ml of a 38% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 40% acid and acid solution B is 30% acid. How much of each solution should be used?

A chemistry student needs 50 ml of a 72% salt solution. He has two salt solutions, A and B, to mix together to form the solution. Salt solution A is 60% salt and salt solution B is 80% salt. How much of each solution should be used?

30 ml of 80% solution; 20 ml of 60% solution.

A chemist needs 2 liters of an 18% acid solution. He has two solutions, A and B, to mix together to form the solution. Acid solution A is 10% acid and acid solution B is 15% acid. Can the chemist form the needed 18% acid solution? (Verify by calculation.) If the chemist locates a 20% acid solution, how much would have to be mixed with the 10% solution to obtain the needed 2-liter 18% solution?

A chemist needs 3 liters of a 12% acid solution. She has two acid solutions, A and B, to mix together to form the solution. Acid solution A is 14% acid and acid solution B is 20% acid. Can the chemist form the needed 12% solution? (Verify by calculation.) If the chemist locates a 4% acid solution, how much would have to be mixed with the 14% acid solution to obtain the needed 3-liter 12% solution?

A chemistry student needs 100 ml of a 16% acid solution. He has a bottle of 20% acid solution. How much pure water and how much of the 20% solution should be mixed to dilute the 20% acid solution to a 16% acid solution?

A chemistry student needs 1 liter of a 78% salt solution. She has a bottle of 80% salt solution. How much pure water and how much of the 80% salt solution should be mixed to dilute the 80% salt solution to a 78% salt solution?

25 ml of pure water; 975 ml of 80% salt solution.

A parking meter contains 42 coins. The total value of the coins is

A child’s bank contains 78 coins. The coins are only pennies and nickels. If the value of the coins is

18 nickels; 60 pennies.

*((Reference))* Simplify

*((Reference))* Find the product:

*((Reference))* Find the difference:

*((Reference))* Use the substitution method to solve

*((Reference))* Use the addition method to solve

Comments:"Reviewer's Comments: 'I recommend this book for courses in elementary algebra. The chapters are fairly clear and comprehensible, making them quite readable. The authors do a particularly nice job […]"