MATH 90 CHAPTER 4PART II

MSJC ~ San Jacinto CampusMath Center Workshop Series

Janice Levasseur

Polya’s 4 Steps to Problem Solving

1. Understand the problem

2. Devise a plan to solve the problem

3. Carry out and monitor your plan

4. Look back at your work and check your results

• We will keep these steps in mind as we tackle the application problems from the infamous Chapter 4!

1. Understand the problem

• Read the problem carefully at least twice. – In the first reading, get a general overview of the

problem. – In the second reading, determine (a) exactly what you

are being asked to find and (b) what information the problem provides.

• Try to make a sketch to illustrate the problem. Label the information given.

• Make a list of the given facts. Are they all pertinent to the problem?

• Determine if the information you are given is sufficient to solve the problem.

2. Devise a Plan to Solve the Problem

• Have you seen the problem or a similar problem before? Are the procedures you used to solve the similar problem applicable to the new problem?

• Can you express the problem in terms of an algebraic equation?

• Look for patterns or relationships in the problem that may help in solving it.

• Can you express the problem more simply?• Will listing the information in a table help?

2. cont.

• Can you substitute smaller or simpler numbers to make the problem more understandable?

• Can you make an educated guess at the solution? Sometimes if you know an approximate solution, you can work backwards and eventually determine the correct procedure to solve the problem.

3. Carry Out and Monitor Your Plan

• Use the plan you devised in step 2 to solve the problem.

• Check frequently to see whether it is productive or is going down a dead-end street. If unproductive, revisit Step 2.

4. Look Back at Your Work and Check Your Results

• Ask yourself, “Does the answer make sense?” and “Is the answer reasonable?” If the answer is not reasonable, recheck your method for solving the problem and your calculations.

• Can you check the solution using the original statement?

• Is there an alternative method to arrive at the same conclusion?

• Can the results of this problem be used to solve other problems?

Mixture ProblemEx: A caterer made an ice cream punch by

combining fruit juice that costs $4.50/gallon with ice cream that costs $6.50/gallon. How many gallons of each were used to make 100 gallons of punch

that cost $5.00/gallon?

1. What are we being asked to find?How many gallons of ice cream and juice were used.

2. Can you express the problem in terms of an algebraic equation? Is there an applicable formula? Can you list the information given in a table to help solve the problem?

The value mixture formula is:

V = AC

V = value of the ingredient, A = amount of the ingredient, and C = cost per unit of the ingredient (aka “unit cost”)

We want to know how much of each ingredient was used.

Let j = gallons of juice.

We have two ingredients:

Juice and Ice Cream.

A caterer made an ice cream punch by combining fruit juice that costs $4.50/gallon with ice cream that costs

$6.50/gallon. How many gallons of each were used to make 100 gallons of punch that cost $5.00/gallon?

Aamount Unit cost value

C V

juice

Ice cream

total

* =

Since j = gal of juice

j

100

100 – j = gal of ice cream

100 – j

4.50

6.50

4.50j

6.50(100 – j)

5.00(100)

Use the information in the table to write an algebraic equation:

Juice value + ice cream value = punch (mixture) value

4.50j + 6.50(100 – j) = 5.00(100)

       

       

       

       5.00

3. Using the plan devised in Step 2, solve the (algebraic) problem

4.50j + 6.50(100 – j) = 5.00(100)

4. Did we answer the question being asked? Is our answer complete? Check the solutions.

Yes! Yes!

4.50j + 650 – 6.50j = 500

650 – 2.0j = 500

– 2.0j = – 150

j = 75

j = 75 gal

Gal of ice cream

100 – j = 100 – 75 = 25

Solution: 75 gallons of juice and 25 gallons of ice cream

Check: 75 + 25 = 100 gallons

4.50(75) = 337.50 and 6.50(25) = 162.50

$337.50 + $162.50 = $500 for 100 gallons $5/gallon

Mixture word problem, set up a system of equations and solve it.

• 1st equation – amounts: ounces, pounds, liter, etc.

• 2nd equation – value: $ or % multiplied to the amount added together and equal to the $ or % multiplied to the total amount.

• Solve by one of the above methods.

Set up a system of equations & solve it. Victor mixes a 20% acid solution with a 45% acid

solution to create 100 ml of a 40% acid solution. How many ml of each solution did he use?

100 20

.20 .45 40 100

20 20 2000

20 45 4000

25 2000 80 45%

20 20%

x y multiply by

x y multiply by

x y

x y

y y ml of acid

x ml of acid

To be a successful “word” problem solver:

1. Don’t say, “I hate word problems!”

2. Take a deep breath and tackle the word problem using Poyla’s 4 steps

3. PRACTICE, PRACTICE, PRACTICE

4. Get help (Instructor, Learning/Math Center, study buddy/group, SI)

5. PRACTICE, PRACTICE, PRACTICE Good Luck . . . You can do it!