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8.8: Problem-Solving Strategies

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Learning Objectives

  • Read and understand given problem situations.
  • Make tables and identify patterns.
  • Solve real-world problems using selected strategies as part of a plan.

Introduction

Problem solving appears everywhere, in your regular life as well as in all jobs and careers. Of course, in this manual we concentrate on solving problems that involve algebra. From previous sections, remember our problem solving plan.

Step 1

Understand the problem.

Read the problem carefully. Once the problem is read, list all the components and data that are involved. This is where you will be assigning your variables.

Step 2

Devise a plan – Translate.

Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table as a start to solving your problem.

Step 3

Carry out the plan – Solve.

This is where you solve the equation you came up with in Step 2.

Step 4

Look – Check and Interpret.

Check to see if you used all your information and that the answer makes sense.

Examples of Exponential Word Problems

In this section, we will be applying this problem solving strategy to solving real-world problems where exponential functions appear. Compound interest, loudness of sound, population increase, population decrease or radioactive decay are all applications of exponential functions. In these problems, we will use the methods of constructing a table and identifying a pattern to help us devise a plan for solving the problems.

Example 1 Compound Interest

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

Solution

Step 1

Read the problem and summarize the information.

$4000 is invested at 6% interest compounded annually

We want to know how much money we have after five years.

Assign variables.

Let x= time in years

Let y= amount of money in investment account

Step 2

Look for a pattern.

We start with $4000 and each year we apply a 6% interest on the amount in the bank

\text{Start} && \$4000 \\1^{st}\ \text{year} && \text{Interest} & = 4000 \times (0.06) = \$240 \\&& \text{This is added to the previous amount} & = \$4000 + \$4000 \times (0.06) \\& && = \$4000(1 + 0.06) \\& && = \$4000 (1.06) \\& && = \$4240 \\2^{nd}\ \text{year} && \text{Previous amount + interest on the previous amount}. & = \$4240(1 + 0.06) \\& && =\$4240 (1.06) \\& && = \$4494.40

The pattern is that each year we multiply the previous amount by the factor of 1.06.

Let’s fill in a table of values.

& \text{Time (Years)} && 0 && 1 && 2 && 3 && 4 && 5 \\& \text{Investments Amount} (\$) && 4000 && 4240 && 4494.4 && 4764.06 && 5049.90 && 5352.9

Answer We see that at the end of five years we have $5352.90 in the investment account.

Step 3 In the case of 5 years, we don’t need an equation to solve the problem. However, if we want the amount at the end of 20 years, we get tired of multiplying by 1.06, and we want a formula.

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 5 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} = \$12828.54

Step 4 Looking back over the solution, we see that we obtained the answers to the questions we were asked and the answers make sense.

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, && y & =4000 \cdot (1.06)^0 = 4000 \\x=1, && y &= 4000 \cdot (1.06)^1 = 4240 \\x=2, && y &= 4000 \cdot (1.06)^2 = 4494.4

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.

Example 2 Population decrease

In 2002, the population of school children in a city was 90,000. This population decreases at a rate of 5% each year. What will be the population of school children in year 2010?

Solution

Step 1

Read the problem and summarize the information.

In 2002, population = >90,000.

Rate of decrease = 5\% each year.

What is the population in year 2010?

Assign variables.

Let x= time since 2002 (in years)

Let y= population of school children

Step 2

Look for a pattern.

Let’s start in 2002.

Population = 90,000

Rate of decrease is 5% each year, so we need to find the amount of increase by 90,000 \times 0.05 and subtract this increase from the original number 90,000 - 90,000 \times 0.05 = 90,000(1 - 0.05) = 90,000 \times 0.95.

\text{In}\ 2003 && \text{Population} & = 90,000 \times 0.95 \\\text{In}\ 2004 && \text{Population} & = 90,000 \times 0.95 \times 0.95

The pattern is that for each year we multiply by a factor of 0.95

Let’s fill in a table of values:

 & \text{Year} && 2002 && 2003 && 2004 && 2005 && 2006 && 2007 \\& \text{Population} && 90,000 && 85,500 && 81,225 && 77,164 && 73,306 && 69,640

Step 3

Let’s find a formula for this relationship.

Since we take the original population and keep multiplying by the same factor of 0.95, this pattern fits to an exponential formula.

y = 90000 \cdot (0.95)^x

To find the population in year 2010, plug in x=8 (number of years since 2002)

y= 90000 \cdot (0.95)^8 = 59,708 school children

Step 4

Looking back over the solution, we see that we answered the question we were asked and that it makes sense.

The answer makes sense because the numbers decrease each year as we expected. We can check that the formula is correct by plugging in the values of x from the table to see if the values match those given by the formula.

\text{Year}\ 2002, x=0 && \text{Population} & = y=90000 \cdot (0.95)^0 = 90,000 \\\text{Year}\ 2003, x=1 && \text{Population} & = y = 90000 \cdot (0.95)^1 = 85,500 \\\text{Year}\ 2004, x=2 && \text{Population} & = y = 90000 \cdot (0.95)^2 = 81,225

Example 3 Loudness of sound

Loudness is measured in decibels (dB). An increase in loudness of 10 decibels means the sound intensity increases by a factor of 10. Sound that is barely audible has a decibel level of 0 dB and an intensity level of 10^{-12} \ W/m^2. Sound painful to the ear has a decibel level of 130 dB and an intensity level of 10\ W/m^2.

(a) The decibel level of normal conversation is 60 dB. What is the intensity of the sound of normal conversation?

(b) The decibel level of a subway train entering a station is 100 dB. What is the intensity of the sound of the train?

Solution:

Step 1

Read the problem and summarize the information.

For 10 decibels, sound intensity increases by a factor of 10.

Barely audible sound =0 \ dB = 10^{-12} \ W/m^2

Ear-splitting sound = 130 \ dB =10 \ W/m^2

Find intensity at 60 dB and find intensity at 100 dB.

Assign variables.

Let x = sound level in decibels (dB)

Let y = intensity of sound (W/m^2)

Step 2

Look for a pattern.

Let’s start at 0 dB

\text{For}\ 0\ dB && \text{Intensity} = 10^{-12}\ W/m^2

For each decibel the intensity goes up by a factor of ten.

\text{For}\ 10\ dB && \text{Intensity} & = 10^{-12}\times 10\ W/m^2 \\\text{For}\ 20\ dB && \text{Intensity} & = 10^{-12} \times 10 \times 10\ W/m^2\\\text{For}\ 30\ dB && \text{Intensity} & = 10^{-12}\times 10 \times 10 \times 10\ W/m^2

The pattern is that for each 10 decibels we multiply by a factor of 10.

Let’s fill in a table of values.

& \text{Loudness\ (dB)} && 0 && 10 && 20 && 30 && 40 && 50 \\& \text{Intensity}\ (W/m^2) && 10^{-12} && 10^{-11} && 10^{-10} && 10^{-9} && 10^{-8} && 10^{-7}

Step 3

Let’s find a formula for this relationship.

Since we take the original sound intensity and keep multiplying by the same factor of 10, that means we can use exponential notation.

y=10^{-12} \cdot 10^{\frac{x}{10}}

The power is \frac{x}{10}, since we go up by 10 dB each time.

To find the intensity at 60 dB we use x=60 in the equation.

y=10^{-12} \cdot (10)^{\left(\frac{60}{10}\right )}=10^{-12}\cdot (10)^6=10^{-6}\ W/m^2

To find the intensity at 100 dB we use x=100 in the equation.

y=10^{-12}\cdot (10)^{\left (\frac{100}{10}\right )}=10^{-12}\cdot (10)^{10}=10^{-2}\ W/m^2

Step 4

Looking back over the solution, we see that we did not use all the information we were given. We still have the fact that a decibel level of 130 dB has an intensity level of 10 \ W/m^2.

We can use this information to see if our formula is correct. Use x=130 in our formula.

y=10^{-12}\cdot (10)^{ \left( \frac{130}{10} \right )}=10^{-12} \cdot (10)^{13}=10\ W/m^2

The formula confirms that a decibel level of 130 dB corresponds to an intensity level of 10 \ W/m^2.

Review Questions

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 6 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 (ten 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?

Review Answers

  1. 100 (.965)^x = 100 (.965)^6= 80.75\%
  2. 1200 (.985)^x = 1200 (.985)^3 = 1147
  3. 200 (1.08)^x = 200 (1.08)^8 = 370
  4. 360 (1.0725)^x = 360 (1.0725)^{12}= \$833.82

Texas Instruments Resources

In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9618.

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