MATH 90 CHAPTER 4PART I

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. continued.

• 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?

When given a total amount, we can use a single variable to represent two unknowns.

For example, I have $20 and a buy something. If you do not know how much I spent, there would be two unknowns, how much I spent and how much I have left.

• What if I spent $6, what would I have left? $14• What if I spent $12, what would I have left? $8

• You have a total of $20. • If s represents the amount you spent, • then 20 – s represents the amount you have.

Coin ProblemEx: A coin bank contains 27 coins in

dimes and quarters. The coins have a total value of $4.95. Find the number

of dimes and quarters in the bank.

1. What are we being asked to find?# of dimes and # of quarters in the bank

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

Coin ProblemEx: A coin bank contains 27 coins in

dimes and quarters. The coins have a total value of $4.95. Find the number

of dimes and quarters in the bank.

1. What are we being asked to find?# of dimes and # of quarters in the bank

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

A coin bank contains 27 coins in dimes and quarters. The coins have a total value of $4.95.

Find the number of dimes and quarters in the bank.

N# of coins coin value total value

V T

dimes

quarters

total

* =

Let d = # of dimes

d

27

27 – d = # of quarters

27 – d

0.10

0.25

0.10d

0.25(27 – d)

4.95

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

Total value in dimes + total value in quarters = total value

0.10d + 0.25(27 – d) = 4.95

       

       

       

       

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

0.10d + 0.25(27 – d) = 4.95

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

Yes! Yes!

0.10d + 6.75 – 0.25d = 4.95

6.75 – 0.15d = 4.95

– 0.15 d = – 1.8

d = 12

d = 12

Quarters: 27 - d

27 – d = 27 – 12 = 15

Solution: 12 dimes and 15 quarters

Check: 12 dimes + 15 quarters = 27 coins

12(.10) + 15(.25) = 1.20 + 3.75 = 4.95

Ex: A total of $7000 is deposited into two simple interest accounts. One account pays a simple interest of 10% annually. The other pays an

annual interest rate of 15% simple interest. How much should be invested in each account so that

the total interest earned is $800 ?

1. What are we being asked to find?How much should be invested in each account.

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 annual simple interest earned formula is:

I = Pr

I = simple interest, P = principal (initial investment), and r = interest rate (%)

We want to know how much to invest in each account.

Let p = principal for account #1.

We have two accounts:

Account #1 and Account #2.

A total of $7000 is deposited into two simple interest accounts. One account pays a simple interest of 10% annually. The other pays an annual interest rate of 15% simple interest. How much should be

invested in each account so that the total interest earned is $800.

Pprincipal interest rate Simple interest

r I

Acct #1

Acct #2

total

* =

Since p = Principal Acct #1

p

7000

7000 – p = Principal Acct #2

7000 – p

0.10

0.15

0.10p

0.15(7000 – p)

800

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

Interest Acct #1 + Interest Acct #2 = total Interest

0.10p + 0.15(7000 – p) = 800

       

       

       

       

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

0.10p + 0.15(7000 – p) = 800

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

Yes! Yes!

0.10p + 1050 – 0.15p = 800

1050 – 0.05p = 800

– 0.05 p = – 250

p = 5000

p = $5000

Principal Acct #2

7000–p = 7000 – 5000 = 2000

Solution: $5000 in Acct #1 and $2000 in Acct #2

Check: $5000 + $2000 = $7000 principal

.10(5000) = 500 and .15(2000) = 300

$500 + $300 = $800 interest

Investment word problem, set up a system of equations &

solve it.• 1st equation – amounts: what was

invested.

• 2nd equation – value: sum of percentages time amount equals the interest.

• Solve by one of the above methods.

Investment word problem, set up a system of equations & solve it.

• Bret has $50,000 to invest, some at 12% and the rest at 8%. If his annual interest earned is $4,740, how much did he invest at each rate?

8 50000

100 .12 .08 4740

8 8 400000

12 8 474000

4 74000

$18500 12%

$31500 8%

multiply by x y

multiply by x y

x y

x y

x

x invested at

y invested at