<meta http-equiv="refresh" content="1; url=/nojavascript/"> Problem-Solving Strategies | CK-12 Foundation
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version.

8.8: Problem-Solving Strategies

Created by: CK-12

We have to deal with problem solving in everyday life. Therefore, it is important to know the steps you must take when problem solving.

Example: Suppose $4000 is invested at a 6% interest rate compounded annually. How much money will there be in the bank at the end of five years? At the end of 20 years?

Solution: Read the problem and summarize the information.

$4000 is invested at a 6% interest rate compounded annually. We want to know how much money we have after five years.

  • Assign variables. Let x= time in years and y= amount of money in investment account.
  • We start with $4000 and each year we apply a 6% interest rate on the amount in the bank.
  • The pattern is that each year we multiply the previous amount by the factor of 100\%+6\%=106\%=1.06.
  • Complete a table of values.
Time (years) 0 1 2 3 4 5
Investment Amount ($) 4000 4240 4494.40 4764.06 5049.91 5352.90

Using the table, we see that at the end of five years we have $5352.90 in the investment account.

In the case of five years, we don’t need an equation to solve the problem. However, if we want the amount at the end of 20 years, it becomes too difficult to constantly multiply. We can use a formula instead.

Since we take the original investment and keep multiplying by the same factor of 1.06, that means we can use exponential notation.

y=4000 \cdot (1.06)^x

To find the amount after five years we use x=5 in the equation.

y=4000 \cdot (1.06)^5=\$ 5352.90

To find the amount after 20 years we use x=20 in the equation.

y=4000 \cdot (1.06)^{20}=\$ 12,828.54

To check our answers we can plug in some low values of x to see if they match the values in the table:

& x = 0 && 4000 \cdot (1.06)0 = 4000\\& x = 1	&& 4000 \cdot (1.06)1 = 4240\\& x = 2	&& 4000 \cdot (1.06)2 = 4494.40

The answers make sense because after the first year, the amount goes up by $240 (6% of $4000). The amount of increase gets larger each year and that makes sense because the interest is 6% of an amount that is larger and larger every year.

Multimedia Link: To learn more about how to use the correct exponential function, visit the http://regentsprep.org/REgents/math/ALGEBRA/AE7/ExpDecayL.htm - algebra lesson page by RegentsPrep.

Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

CK-12 Basic Algebra: Word Problem Solving (7:21)

Apply the problem-solving techniques described in this section to solve the following problems.

  1. Half-life Suppose a radioactive substance decays at a rate of 3.5% per hour. What percent of the substance is left after six hours?
  2. Population decrease In 1990, a rural area has 1200 bird species. If species of birds are becoming extinct at the rate of 1.5% per decade (10 years), how many bird species will there be left in year 2020?
  3. Growth Nadia owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007?
  4. Investment Peter invests $360 in an account that pays 7.25% compounded annually. What is the total amount in the account after 12 years?

Image Attributions

Files can only be attached to the latest version of None


Please wait...
You need to be signed in to perform this action. Please sign-in and try again.
Please wait...
Image Detail
Sizes: Medium | Original

Original text