MATH 90 CHAPTER 4
PART II
MSJC ~ San Jacinto Campus
Math Center Workshop Series
Janice Levasseur
Polya’s 4 Steps to Problem Solving
1.
2.
3.
4.
•
Understand the problem
Devise a plan to solve the problem
Carry out and monitor your plan
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 Problem
Ex: 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 have two ingredients:
Juice and Ice Cream.
We want to know how much of each ingredient
was used.
Let j = gallons of juice.
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?
A
amount
juice
j
Ice cream 100 – j
total
100
Since j = gal of juice
*
C
=
V
Unit cost
value
4.50
4.50j
6.50
6.50(100 – j)
5.00
5.00(100)
 100 – j = gal of ice cream
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)
3. Using the plan devised in Step 2, solve the
(algebraic) problem
4.50j + 6.50(100 – j) = 5.00(100)
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
4. Did we answer the question being asked? Is our
answer complete? Check the solutions.
Yes! Check: 75 + 25 = 100 gallons
Yes! 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?
x  y  100
multiply by  20 
.20 x  .45 y  40
multiply by 100 
20 x  20 y  2000
20 x  45 y  4000
25 y  2000
y  80ml of 45% acid
x  20ml of 20% 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!