# MATH 90 CHAPTER 4 PART I MSJC ~ San Jacinto Campus

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MATH 90 CHAPTER 4 PART I MSJC ~ San Jacinto Campus
Math Center Workshop Series Janice Levasseur

Polya’s 4 Steps to Problem Solving
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. 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.

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?

What if I spent \$6, what would I have left? \$14
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 Problem Ex: A coin bank contains 27 coins in dimes and quarters
Coin Problem Ex: A coin bank contains 27 coins in dimes and quarters. The coins have a total value of \$ 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 Problem Ex: A coin bank contains 27 coins in dimes and quarters
Coin Problem Ex: A coin bank contains 27 coins in dimes and quarters. The coins have a total value of \$ 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
A coin bank contains 27 coins in dimes and quarters. The coins have a total value of \$ Find the number of dimes and quarters in the bank. T N V = * # of coins coin value total value d 0.10 0.10d dimes quarters 27 – d 0.25 0.25(27 – d) total 27 4.95 Let d = # of dimes  27 – d = # of quarters Use the information in the table to write an algebraic equation: Total value in dimes + total value in quarters = total value 0.10d (27 – d) = 4.95

3. Using the plan devised in Step 2, solve the (algebraic) problem
0.10d (27 – d) = 4.95 d = 12 0.10d – 0.25d = 4.95 Quarters: 27 - d 6.75 – 0.15d = 4.95 27 – d = 27 – 12 = 15 – 0.15 d = – 1.8 Solution: 12 dimes and 15 quarters  d = 12 4. Did we answer the question being asked? Is our answer complete? Check the solutions. Yes! Yes! Check: 12 dimes + 15 quarters = 27 coins 12(.10) + 15(.25) = = 4.95

Ex: A total of \$7000 is deposited into two simple interest accounts
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 have two accounts: Account #1 and Account #2. We want to know how much to invest in each account. Let p = principal for account #1.

I P r = * p 0.10 0.10p Acct #1 Acct #2 0.15 total 7000 800 7000 – p
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. I P r = * principal interest rate Simple interest p 0.10 0.10p Acct #1 Acct #2 7000 – p 0.15 0.15(7000 – p) total 7000 800 Since p = Principal Acct #1  7000 – p = Principal Acct #2 Use the information in the table to write an algebraic equation: Interest Acct #1 + Interest Acct #2 = total Interest 0.10p (7000 – p) = 800

3. Using the plan devised in Step 2, solve the (algebraic) problem
0.10p (7000 – p) = 800 p = \$5000 0.10p – 0.15p = 800 Principal Acct #2 1050 – 0.05p = 800 7000–p = 7000 – 5000 = 2000 – 0.05 p = – 250 Solution: \$5000 in Acct #1 and \$2000 in Acct #2  p = 5000 4. Did we answer the question being asked? Is our answer complete? Check the solutions. Check: \$ \$2000 = \$7000 principal .10(5000) = 500 and .15(2000) = 300 \$500 + \$300 = \$800 interest Yes! Yes!

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?

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