application homework problem

2.3 Models and Applications

Learning objectives.

In this section, you will:

• Set up a linear equation to solve a real-world application.
• Use a formula to solve a real-world application.

Neka is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum number of points that can be earned is 100. Is it possible for Neka to end the course with an A? A simple linear equation will give Neka his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

Setting up a Linear Equation to Solve a Real-World Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10 x . 0.10 x . This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C . C .

When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions.

Given a real-world problem, model a linear equation to fit it.

• Identify known quantities.
• Assign a variable to represent the unknown quantity.
• If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
• Write an equation interpreting the words as mathematical operations.
• Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 17 and their sum is 31. 31. Find the two numbers.

Let x x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign.

The two numbers are 7 7 and 24. 24.

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36 , 36 , find the numbers.

Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

• ⓐ Write a linear equation that models the packages offered by both companies.
• ⓑ If the average number of minutes used each month is 1,160, which company offers the better plan?
• ⓒ If the average number of minutes used each month is 420, which company offers the better plan?
• ⓓ How many minutes of talk-time would yield equal monthly statements from both companies?
• ⓐ The model for Company A can be written as A = 0.05 x + 34. A = 0.05 x + 34. This includes the variable cost of 0.05 x 0.05 x plus the monthly service charge of $34. Company B ’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04 x . 0.04 x . Company B ’s model can be written as B = 0.04 x + $ 40. B = 0.04 x + $ 40.

If the average number of minutes used each month is 1,160, we have the following:

So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.

If the average number of minutes used each month is 420, we have the following:

If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B ’s monthly cost of $56.80.

To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of ( x , y ) ( x , y ) coordinates: At what point are both the x- value and the y- value equal? We can find this point by setting the equations equal to each other and solving for x.

Check the x- value in each equation.

Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

Using a Formula to Solve a Real-World Application

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = L W ; A = L W ; the perimeter of a rectangle, P = 2 L + 2 W ; P = 2 L + 2 W ; and the volume of a rectangular solid, V = L W H . V = L W H . When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Solving an Application Using a Formula

It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

This is a distance problem, so we can use the formula d = r t , d = r t , where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or 1 2 1 2 h at rate r . r . His drive home takes 40 min, or 2 3 2 3 h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance d . d . A table, such as Table 2 , is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

As both equations equal the same distance, we set them equal to each other and solve for r .

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.

The distance between home and work is 20 mi.

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r . r .

On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Solving a Perimeter Problem

The perimeter of a rectangular outdoor patio is 54 54 ft. The length is 3 3 ft greater than the width. What are the dimensions of the patio?

The perimeter formula is standard: P = 2 L + 2 W . P = 2 L + 2 W . We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3 .

Now we can solve for the width and then calculate the length.

The dimensions are L = 15 L = 15 ft and W = 12 W = 12 ft.

Find the dimensions of a rectangle given that the perimeter is 110 110 cm and the length is 1 cm more than twice the width.

Solving an Area Problem

The perimeter of a tablet of graph paper is 48 in. The length is 6 6 in. more than the width. Find the area of the graph paper.

The standard formula for area is A = L W ; A = L W ; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as L = W + 6. L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

Now, we find the area given the dimensions of L = 15 L = 15 in. and W = 9 W = 9 in.

The area is 135 135 in. 2 .

A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft 2 of new carpeting should be ordered?

Solving a Volume Problem

Find the dimensions of a shipping box given that the length is twice the width, the height is 8 8 inches, and the volume is 1,600 in. 3 .

The formula for the volume of a box is given as V = L W H , V = L W H , the product of length, width, and height. We are given that L = 2 W , L = 2 W , and H = 8. H = 8. The volume is 1,600 1,600 cubic inches.

The dimensions are L = 20 L = 20 in., W = 10 W = 10 in., and H = 8 H = 8 in.

Note that the square root of W 2 W 2 would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

Access these online resources for additional instruction and practice with models and applications of linear equations.

• Problem solving using linear equations
• Problem solving using equations
• Finding the dimensions of area given the perimeter
• Find the distance between the cities using the distance = rate * time formula
• Linear equation application (Write a cost equation)

2.3 Section Exercises

To set up a model linear equation to fit real-world applications, what should always be the first step?

Use your own words to describe this equation where n is a number: 5 ( n + 3 ) = 2 n 5 ( n + 3 ) = 2 n

If the total amount of money you had to invest was $2,000 and you deposit x x amount in one investment, how can you represent the remaining amount?

If a carpenter sawed a 10-ft board into two sections and one section was n n ft long, how long would the other section be in terms of n n ?

If Bill was traveling v v mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?

Real-World Applications

For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked.

Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?

Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?

Ruden originally filled out 8 more applications than Hanh. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?

For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls.

Find the model of the total cost of Company A’s plan, using m m for the minutes.

Find the model of the total cost of Company B’s plan, using m m for the minutes.

Find out how many minutes of calling would make the two plans equal.

If the person makes a monthly average of 200 min of calls, which plan should for the person choose?

For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 8 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P P for the number of devices that need data plans as part of their cost.

Find the model of the total cost of the Family Plan.

Find the model of the total cost of the Mobile Share Plan.

Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost.

If a family has 3 smart phones, which plan should they choose?

For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%.

If we let x x be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD?

Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return.

Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart?

Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben?

Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h?

A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?

Raúl has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Raúl need to invest in each option to make get a total 11% return on his $20,000?

For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven.

Write the model equation for the cost of renting a truck with plan A.

Write the model equation for the cost of renting a truck with plan B.

Find the number of miles that would generate the same cost for both plans.

If Tim knows he has to travel 300 mi, which plan should he choose?

For the following exercises, use the formula given to solve for the required value.

A = P ( 1 + r t ) A = P ( 1 + r t ) is used to find the principal amount P deposited, earning r % interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $ 8,000. A = $ 8,000.

The formula F = m v 2 R F = m v 2 R relates force ( F ) ( F ) , velocity ( v ) ( v ) , mass , and resistance ( m ) ( m ) . Find R R when m = 45 , m = 45 , v = 7 , v = 7 , and F = 245. F = 245.

F = m a F = m a indicates that force ( F ) equals mass ( m ) times acceleration ( a ). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it.

S u m = 1 1 − r S u m = 1 1 − r is the formula for an infinite series sum. If the sum is 5, find r . r .

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question.

Solve for W : P = 2 L + 2 W P = 2 L + 2 W

Use the formula from the previous question to find the width, W , W , of a rectangle whose length is 15 and whose perimeter is 58.

Solve for f : 1 p + 1 q = 1 f f : 1 p + 1 q = 1 f

Use the formula from the previous question to find f f when p = 8 and  q = 13. p = 8 and  q = 13.

Solve for m m in the slope-intercept formula: y = m x + b y = m x + b

Use the formula from the previous question to find m m when the coordinates of the point are ( 4 , 7 ) ( 4 , 7 ) and b = 12. b = 12.

The area of a trapezoid is given by A = 1 2 h ( b 1 + b 2 ) . A = 1 2 h ( b 1 + b 2 ) . Use the formula to find the area of a trapezoid with h = 6 , b 1 = 14 , and  b 2 = 8. h = 6 , b 1 = 14 , and  b 2 = 8.

Solve for h: A = 1 2 h ( b 1 + b 2 ) A = 1 2 h ( b 1 + b 2 )

Use the formula from the previous question to find the height of a trapezoid with A = 150 , b 1 = 19 A = 150 , b 1 = 19 , and  b 2 = 11. b 2 = 11.

Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2 L + 2 W . P = 2 L + 2 W .

Distance equals rate times time, d = r t . d = r t . Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h.

Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h.

What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h?

If the area model for a triangle is A = 1 2 b h , A = 1 2 b h , find the area of a triangle with a height of 16 in. and a base of 11 in.

Solve for h: A = 1 2 b h A = 1 2 b h

Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215.

The volume formula for a cylinder is V = π r 2 h . V = π r 2 h . Using the symbol π π in your answer, find the volume of a cylinder with a radius, r , r , of 4 cm and a height of 14 cm.

Solve for h: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16 π 16 π

Solve for r: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of 324 π . 324 π .

The formula for the circumference of a circle is C = 2 π r . C = 2 π r . Find the circumference of a circle with a diameter of 12 in. (diameter = 2 r ). Use the symbol π π in your final answer.

Solve the formula from the previous question for π . π . Notice why π π is sometimes defined as the ratio of the circumference to its diameter.

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Math Review of Solving Application Problems

• July 24, 2014
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Solving application problems is a process that includes understanding the problem, translating it into an equation, solving the equation, checking the answer, and answering the question. This process can be used to solve many different types of problems.

Understanding and Setting up The Application Problem

The first part of the process involves understanding what is being asked. This includes noticing any key words that refer to operations and any quantities that are in relationship to one another. It is important to have an idea of what sort of quantity will represent a solution. Suppose a problem asks how many points the first-place winner had. That would require just one answer. If the problem described that the first-place winner had 10 points more than the second-place winner, and 17 points more than the third-place winner, all three point values would be necessary to completely answer the question.

Figure 1: Setting up the application problem involves finding quantities in relationship to one another.

Translating the Problem into an Equation

The second part of the problem involves putting the problem in symbol form. This includes choosing a letter to represent a variable, and writing down exactly what the variable stands for. Suppose that the point values of the three winners totaled 114. Let the point value of the first-place winner be x, the second-place winner x – 10, and the third-place winner x – 17. Then, x + x – 10 + x – 17 = 114.

Figure 2: Translating the problem into an equation involves putting the problem into symbol form.

Solving and Checking the Equation

If x + x – 10 + x – 17 = 114, then 3x = 114 + 27, or 3x = 141. So, x = 47, x – 10 = 37, and x -17 = 30. To check, 47 + 37 + 30 = 114. The answer makes sense, and fits the parameters of the problem. In this case, the original equation was x + x -10 + x – 17 = 114.

Answer the Question Asked

If the work is done to understand the problem before it is set up and solved, it is easier to answer the appropriate question. In this case, all three point values were necessary to completely answer the question. Suppose that more information were added to the problem. During the same contest last year, the point values of the first-place, second-place and third-place winners totaled 100, but the second place winner had 14 points less than the first-place winner, and the third-place winner had 18 points less than the first place winner. Which year did the third-place winner earn more points, and what was the difference? Last year, the equation was x + x – 14 + x – 18 = 100, so 3x = 132. The first-place winner earned 44 points, the second-place winner earned 44-14 or 30 points, and the third-place winner earned 28 points. However, the question asked has 2 parts. Last year, the third-place winner earned 28 points, and this year, the third-place winner earned 30 points. This year, the third-place winner earned more points, and the difference was a gain of 2 points.

Figure 3: Answering the question asked brings all the parts of the problem into balance.

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Math Review of Changing Application Problems into Equations

Math review of factoring a monomial from a polynomial.

Module 4: Equations and Inequalities

Models and applications, learning outcomes.

• Write a linear equation to express the relationship between unknown quantities.
• Write a linear equation that models two different cell phone packages.
• Use a linear model to answer questions.
• Set up a linear equation involving distance, rate, and time.
• Find the dimensions of a rectangle given the area.
• Find the dimensions of a box given information about its side lengths.

Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.

College students taking an exam. Credit: Kevin Dooley

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

Writing a Linear Equation to Solve an Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[/latex]. This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[/latex].

When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table lists some common verbal expressions and their equivalent mathematical expressions.

How To: Given a real-world problem, model a linear equation to fit it

• Identify known quantities.
• Assign a variable to represent the unknown quantity.
• If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
• Write an equation interpreting the words as mathematical operations.
• Solve the equation. Be sure the solution can be explained in words including the units of measure.

Example: Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[/latex] and their sum is [latex]31[/latex]. Find the two numbers.

Let [latex]x[/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[/latex]. The sum of the two numbers is 31. We usually interpret the word is as an equal sign.

The two numbers are [latex]7[/latex] and [latex]24[/latex].

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[/latex], find the numbers.

Example: Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

• Write a linear equation that models the packages offered by both companies.
• If the average number of minutes used each month is 1,160, which company offers the better plan?
• If the average number of minutes used each month is 420, which company offers the better plan?
• How many minutes of talk-time would yield equal monthly statements from both companies?
• The model for Company A can be written as [latex]A=0.05x+34[/latex]. This includes the variable cost of [latex]0.05x[/latex] plus the monthly service charge of $34. Company B ’s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[/latex]. Company B ’s model can be written as [latex]B=0.04x+40[/latex].

Check the x- value in each equation.

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

Using Formulas to Solve Problems

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, [latex]A=LW[/latex]; the perimeter of a rectangle, [latex]P=2L+2W[/latex]; and the volume of a rectangular solid, [latex]V=LWH[/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Example: Solving an Application Using a Formula

It takes Andrew 30 minutes to drive to work in the morning. He drives home using the same route, but it takes 10 minutes longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

This is a distance problem, so we can use the formula [latex]d=rt[/latex], where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or [latex]\frac{1}{2}[/latex] h at rate [latex]r[/latex]. His drive home takes 40 min, or [latex]\frac{2}{3}[/latex] h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance [latex]d[/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

As both equations equal the same distance, we set them equal to each other and solve for r .

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.

The distance between home and work is 20 mi.

Analysis of the Solution

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[/latex].

On Saturday morning, it took Jennifer 3.6 hours to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

45 [latex]\frac{\text{mi}}{\text{h}}[/latex]

Example: Solving a Perimeter Problem

The perimeter of a rectangular outdoor patio is [latex]54[/latex] ft. The length is [latex]3[/latex] ft. greater than the width. What are the dimensions of the patio?

The perimeter formula is standard: [latex]P=2L+2W[/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.

Now we can solve for the width and then calculate the length.

The dimensions are [latex]L=15[/latex] ft and [latex]W=12[/latex] ft.

Find the dimensions of a rectangle given that the perimeter is [latex]110[/latex] cm. and the length is 1 cm. more than twice the width.

L = 37 cm, W = 18 cm

Example: Solving an Area Problem

The perimeter of a tablet of graph paper is 48 in 2 . The length is [latex]6[/latex] in. more than the width. Find the area of the graph paper.

The standard formula for area is [latex]A=LW[/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

Now, we find the area given the dimensions of [latex]L=15[/latex] in. and [latex]W=9[/latex] in.

The area is [latex]135[/latex] in 2 .

A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft 2 of new carpeting should be ordered?

Example: Solving a Volume Problem

Find the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[/latex] inches, and the volume is 1,600 in. 3 .

The formula for the volume of a box is given as [latex]V=LWH[/latex], the product of length, width, and height. We are given that [latex]L=2W[/latex], and [latex]H=8[/latex]. The volume is [latex]1,600[/latex] cubic inches.

The dimensions are [latex]L=20[/latex] in., [latex]W=10[/latex] in., and [latex]H=8[/latex] in.

Note that the square root of [latex]{W}^{2}[/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

Key Concepts

• A linear equation can be used to solve for an unknown in a number problem.
• Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities.
• There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the [latex]d=rt[/latex] formula.
• Many geometry problems are solved using the perimeter formula [latex]P=2L+2W[/latex], the area formula [latex]A=LW[/latex], or the volume formula [latex]V=LWH[/latex].
• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Question ID 52436. Authored by : Edward Wicks. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• Question ID 7647, 7679. Authored by : Tyler Wallace. License : CC BY: Attribution . License Terms : IMathAS Community License CC- BY + GPL
• Question ID 30987, 13665. Authored by : James Sousa. License : CC BY: Attribution . License Terms : IMathAS Community License CC- BY + GPL
• Question ID 92426. Authored by : Michael Jenck. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• Question ID 1688. Authored by : WebWork-Rochester. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL

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Topics: Implementation Support

Building Mathematical Thinkers, One Application Problem at a Time

by Carrie Thornton

every child is capable of greatness.

Posted in: Aha! Blog > Eureka Math Blog > Implementation Support > Building Mathematical Thinkers, One Application Problem at a Time

When am I ever going to use this in life?

My students asked me this question all the time. The truth is, I didn’t know what my students’ futures would be or what math they would need to use. We didn’t do math problems just so they would be able to do those same problems later in life. We did them so the students would learn to think mathematically and be able to solve the new problems they would encounter at each of life’s turns.

The Application Problems that we pose to students help them understand mathematical concepts by making the contexts more concrete. These problems are essential for students to build the skill to transfer mathematical learning to new and different contexts. That is the goal—educating students who know and can apply mathematical concepts to the world around them.

Help your students connect Application Problems and symbolic math.

Many of us learned how to solve math problems procedurally. For example, you probably learned to multiply fractions with the rule multiply the numerators, then multiply the denominators . But how many of us understood why we were doing that and when we needed to do it to solve a problem?

  Consider this symbolic math problem and a related Application Problem.

Use the Read–Draw–Write framework to help your students solve Application Problems.

Eureka Math ® emphasizes the Read–Draw–Write (RDW) process as an overarching framework to help students make sense of and solve Application Problems. This framework encourages students to make connections between the problem and the symbolic math they learn.

The RDW process is a flexible approach that can be applied to any word problem students encounter. It encourages students to make sense of the mathematical information instead of scanning for numbers and keywords, which can often point students in the wrong direction. More information about the RDW process can be found in the Module 1 Overview at each grade level in Kindergarten through Grade 5.

Use different modes of delivery for different purposes.

You and your students may engage with Application Problems in different ways, including modeling, guided application, and independent practice.

Think about how moving flexibly among these three modes of delivery can support students. Modeling is especially useful for situations where you might want students to try something new. Guided application might be more effective once students have some familiarity with the concepts or methods. With independent practice, students can engage in productive struggle when they have enough foundation to tackle the new aspects of the problem on their own. You will likely find yourself using a combination of these modes of delivery, depending on the math problem and your students’ needs. However, don’t neglect independent practice. Those moments of productive struggle can lead to the most student growth.

Eureka Math builds application throughout each topic and module instead of only addressing word problems in their own separate lesson. This allows students to gain experience with contextualizing and decontextualizing mathematics. The result is that your students will be more confident and flexible problem solvers. They will transfer their mathematical knowledge to new situations throughout the school year and develop the skills needed to independently solve new problems.

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Carrie Thornton

Carrie Thornton is a Eureka Math Implementation Leader. She is a former 4th grade teacher and new teacher mentor from Bethel School District in Washington state.

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Achievement Descriptors and Student Understanding

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Free Math Worksheets — Over 100k free practice problems on Khan Academy

Looking for free math worksheets.

You’ve found something even better!

That’s because Khan Academy has over 100,000 free practice questions. And they’re even better than traditional math worksheets – more instantaneous, more interactive, and more fun!

Just choose your grade level or topic to get access to 100% free practice questions:

Kindergarten, basic geometry, pre-algebra, algebra basics, high school geometry.

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High school statistics, ap®︎/college statistics, precalculus, differential calculus, integral calculus, ap®︎/college calculus ab, ap®︎/college calculus bc, multivariable calculus, differential equations, linear algebra.

• Addition and subtraction
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• Measurement and data
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• Place value
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• Add and subtract within 20
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• Intro to multiplication
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• Decimal place value
• Add decimals
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• Multi-digit multiplication and division
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• AP Calculus AB solved free response questions from past exams
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Frequently Asked Questions about Khan Academy and Math Worksheets

Why is khan academy even better than traditional math worksheets.

Khan Academy’s 100,000+ free practice questions give instant feedback, don’t need to be graded, and don’t require a printer.

What do Khan Academy’s interactive math worksheets look like?

Here’s an example:

What are teachers saying about Khan Academy’s interactive math worksheets?

“My students love Khan Academy because they can immediately learn from their mistakes, unlike traditional worksheets.”

Is Khan Academy free?

Khan Academy’s practice questions are 100% free—with no ads or subscriptions.

What do Khan Academy’s interactive math worksheets cover?

Our 100,000+ practice questions cover every math topic from arithmetic to calculus, as well as ELA, Science, Social Studies, and more.

Is Khan Academy a company?

Khan Academy is a nonprofit with a mission to provide a free, world-class education to anyone, anywhere.

Want to get even more out of Khan Academy?

Then be sure to check out our teacher tools . They’ll help you assign the perfect practice for each student from our full math curriculum and track your students’ progress across the year. Plus, they’re also 100% free — with no subscriptions and no ads.

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The 6 Best Homework Apps to Help Students (and Parents)

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Getting homework done can be difficult. While the information might make sense in the classroom , some students don't retain that knowledge. When your child needs extra help, there are apps to help with homework. These solutions help both students and parents succeed.

The apps in this article are free to download, and some may include in-app purchases .

Easy to use with over 30 languages to learn.

ESL courses to strengthen English skills.

The interface can be clunky at times.

It has a limited vocabulary.

This popular language-learning app sits at the top of both the iOS and Android app stores as an excellent solution to strengthen foreign language skills. Whether your child wants to learn outside of their curriculum or wishes to practice what they learned in class, Duolingo is an excellent addition to any digital device.

With over thirty languages to choose, your child can practice German, Italian, Spanish, French, or dozens of other languages. If your student is currently enrolled in ESL courses, they can strengthen their English skills from the bottom-up.

Unlike other language learning solutions that focus on memorizing vocabulary, Duolingo uses a combination of reading, writing, and speaking exercises to create a more natural learning experience.

Download For :

It's great for helping math students who fall behind.

The built-in calculator allows for smart, on-the-fly calculations and 2D graph plotting.

It's somewhat limited depth. Wrong answers don't provide much room for learning why an answer is incorrect.

Mathematics can be one of the most challenging courses for students, with complicated steps that are quickly forgotten after a long school day. Especially challenging is that many parents struggle to help their children with subject matter that they haven't been acquainted with for years. Photomath is an excellent solution for struggling mathematicians.

Children can scan complex or simple math problems, learning how to solve them with step-by-step instructions. A built-in calculator improves the experience, allowing for smart, on-the-fly calculations and 2D graph plotting abilities. Linear equations, logarithms, trigonometry, functions, and basic algebraic expressions are only a few of Photomath's vast capabilities.

Great music learning app with tools for learning and practicing guitar, bass, piano, and more.

Voice-overs and instructional design lack the human touch that helps when learning an instrument.

One school subject that is sadly overlooked is music. Music is an area of study that has been proven to increase a child's language and reasoning skills, fine-tune their motor skills, and decrease stress levels. So, if your child is trying to learn a musical instrument and struggling, consider investing in Yousician. This app allows students to practice guitar, bass, piano, or ukulele.

Students can practice their instruments along with the charts and diagrams that display on the screen. They'll receive real-time feedback when they miss a note or fall out of tune. Included step-by-step video tutorials are available to show your child how to be proficient in a specific skill set. Practicing an instrument has never seemed so natural. With different genres of music available, your child can quickly sort out their favorites.

Khan Academy

A range of subjects, from kindergarten mathematics to advanced placement physics.

More than 150,000 interactive exercises.

There's not much room for creativity, collaboration, or alternative teaching styles.

Ready to increase your knowledge in math, science, computing, history, economics, and more? Whether as an educational tool for your child, or an extra app for a parent to enjoy in their free time, everyone can use Khan Academy to unlock a world of education. Children can practice anything from kindergarten mathematics to advanced placement physics. Quickly access a collection of portable courses on-the-go or on your computer.

The Khan Academy app offers over 150,000 interactive exercises to strengthen old or new skills. Additionally, you can download content for offline study so that you have access to it no matter where you are. Parents wishing to learn something new can jump into advanced high school courses or enjoy courses on entrepreneurship and career-building. Khan Academy offers solutions for all ages and skill ranges.

Quizlet Flashcards

A digital flashcard mobile platform that's suitable for all ages.

The focus on memorization is ideal for studying.

The ad-supported platform may be distracting.

User-generated content means some content may be inaccurate.

Do you remember studying for exams and creating piles of flashcards to memorize facts and details? While flashcards can be an excellent method for analyzing new material, they use a large amount of paper that is eventually thrown in the trash. Learn new topics while also saving trees with the Quizlet Flashcards application. Study from existing flashcard sets or create your own.

Quizlet digital flashcards allow students of all ages to practice and excel at various topics using their smartphones. In addition to basic flashcards, Quizlet offers multiple modes to encourage different methods of memorization. If your child attends the same class as another student using Quizlet, the two can share flashcards. Those who use the app to study foreign languages can hear keywords spoken to them in over 18 languages.

PowerSchool

Allows parents and guardians to keep tabs on their child's education.

Access classroom handouts, attendance records, school bulletins, and more.

The complicated interface makes for a rather steep learning curve.

Here's an app that both children and parents can enjoy: PowerSchool Mobile. Many school districts use the PowerSchool system to manage grades and reports. If your child's school is a participant, you can use the mobile app to keep a close eye on your child's educational progress. Depending on how a teacher chooses to use the tool in their classroom, you may also have access to handouts, attendance records, school bulletins, and more.

While not every school offers support for the PowerSchool Mobile application, it is worth checking with your child's teacher to see if the option is available. Sorry kids, but it isn't possible to hide report cards with the PowerSchool app. Parents can manage what push and email notifications they receive from the child's school. You'll become more involved and aware of your child's school performance with this simple mobile app.

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ME 274: Basic Mechanics II

Homework h6.a - sp24.

NOTE : Please complete ONLY Part (a). Do not work on Part (b).

DISCUSSION THREAD

Any questions?? Please ask/answer questions regarding this homework problem through the "Leave a Comment" link.

Discussion and hints

The four-step plan:

• FBD : It is recommended that you draw individual free body diagrams of each disk and of bar AB. Note that the mass of AB is negligible and that forces are applied at only two points (i.e., a two-force member). What does this say about the direction of the forces on the disks at A and B?
• Kinetics : Recommended that you use an Euler equation about the contact point of each disk since those are no-slip points for whose accelerations are directed toward the center of mass of each disk. Note that both of these two equations will involve the load carried by AB. You can eliminate that load from these equations, resulting in a single equation.
• Kinematics : Note that rigid bar AB insures that the velocities of points A and B are the same. How can this be used to relate the angular accelerations of the two disks? Can you see this kinematic relationship between the angular accelerations in the animation above? Please note that the angular accelerations of the two disks are NOT the same, although their centers have the same accelerations.
• EOM: Combine your results from Steps 2 and 3 to produce a single differential equation of motion (EOM) in terms of the x -coordinate.

7 thoughts on “Homework H6.A - Sp24”

Page 388 of the textbook has some equations for natural frequency that may help with part b of this problem.

Please see the Note above stating that you are asked to do only Part (a) of this problem. Not Part (b).

For the two force member at A the reaction force will act in the direction of B correct?

Do the reaction forces have reactions in both x and y directions on both disks because they are pinned to the disk at those points?

For the analysis I don’t think the y component if it exists matters. If you are solving by taking the sum of the moments about the no slip point the y direction wouldn’t matter regardless.

As discussed above, AB is a two-force member - the reactions at A and B are aligned with the line connecting A and B. Therefore, the reaction forces on the disks WILL have both x- and y-components.

As Bradon points out, however, the actual directions of the reactions on the disks are not relevant if you take your moments about the centers of the disks.

I found it useful to look at lecture book example 6.A.2 as it required a similar process to this homework.

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The number of high school seniors who have filled out FAFSA is down from last year

Michel Martin

Elissa Nadworny

High school seniors aren't filling out a federal student aid application. This year's form is supposed to be simpler, but it's had problems. What does this mean for who goes to college and where?

MICHEL MARTIN, HOST:

Could fewer high school seniors end up in college next fall because of a form?

STEVE INSKEEP, HOST:

Apparently - the federal student aid form for college, known as the FAFSA. The Department of Education launched a new process to apply for financial aid this year, which should be simpler, except for all the delays and errors.

MARTIN: NPR higher education correspondent Elissa Nadworny is here to bring us up to date on how it's going. Good morning, Elissa.

ELISSA NADWORNY, BYLINE: Good morning, Michel.

MARTIN: So I take it it's going badly.

NADWORNY: Yes. It has been a very bumpy road for this new FAFSA. So the form didn't come out until about three months after it usually comes out. So essentially the starting line got pushed way back. And then there were the Education Department's miscalculations and missteps. Some students in mixed immigration status families are still having trouble filling out the online form, and it is April. The Education Department says they are working hard and fast to get this fixed and get that data out to colleges.

MARTIN: I understand in the meantime, though, that FAFSA submissions are down. How far down?

NADWORNY: Twenty-seven percent, Michel. That's for high school seniors. And that comes out to about half a million fewer students than the class of 2023. That's all according to the National College Attainment Network, which is using Department of Education data.

MARTIN: And what are the high schools - you know, there are also nonprofits that help students apply for college. What are they saying about all this?

NADWORNY: Well, they're in emergency mode. They are sounding the alarm. Here's Bill DeBaun with the National College Attainment Network.

BILL DEBAUN: In some parts of the country, we're less than 10 weeks from high school graduation. It's not that students can't complete the FAFSA after high school graduation. It's just that in general, they have less support to do so.

NADWORNY: So I've talked with some families who are planning on filling out the FAFSA. They were just kind of waiting until the chaos was over. And there have been steady gains over the last couple of weeks as more and more high school seniors fill it out. But we are, you know, a long way away from getting back up to the levels of the class of 2023.

MARTIN: Well, say more about what the experts think the implications are for all this. I mean, does this have implications for where people go to college or even if they go to college?

NADWORNY: Absolutely. I mean, that - high FAFSA completion numbers have historically meant higher college enrollment numbers come fall. The data shows that high schools with more resources have higher completion numbers. So it is an early indicator that college going rates might be in trouble, especially for students from more poorly funded schools.

MARTIN: And I understand that college enrollment has been declining, too.

NADWORNY: Exactly. Yes. So the pandemic saw about a million fewer students choose college. Now, last fall, the data showed the beginning of a recovery, but now this.

MARTIN: Is there still time to fill the FAFSA out?

NADWORNY: Absolutely. And high schools and college access nonprofits, even colleges, are trying to do whatever they can to get students to fill it out. I've seen pizza parties. You know, they're throwing these big community events after school, on weekends, offering students one-on-one help. Rocio Zamora runs a college access program at a high school in San Diego, and her staff has been putting on weekend FAFSA events.

ROCIO ZAMORA: The support is there, and the messages to complete it are there, but it's more of the fact that the application wasn't ready for them and that leading to frustrations and disappointment and discouragement and just really questioning their plans.

NADWORNY: And the thing is, low-income students need that financial aid offer to make a decision about if they can afford college. I've talked with students who have been accepted to college already, but it's not real until they get that financial aid document, which in many cases hasn't happened yet. And without the full financial picture, students may make preemptive decisions about where to go - to stay closer to home, to go to a less expensive community college, or to say, maybe I'm not going to go to college at all.

MARTIN: That's NPR's Elissa Nadworny. Elissa, thank you.

NADWORNY: You bet.

Copyright © 2024 NPR. All rights reserved. Visit our website terms of use and permissions pages at www.npr.org for further information.

NPR transcripts are created on a rush deadline by an NPR contractor. This text may not be in its final form and may be updated or revised in the future. Accuracy and availability may vary. The authoritative record of NPR’s programming is the audio record.

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6.5: Miscellaneous Application Problems

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• Rupinder Sekhon and Roberta Bloom
• De Anza College

Learning Objectives

In this section, you will learn to apply to concepts for compound interest for savings and annuities to:

• Find the outstanding balance, partway through the term of a loan, of the future payments still remaining on the loan.
• Perform financial calculations in situations involving several stages of savings and/or annuities.
• Find the fair market value of a bond.
• Construct an amortization schedule for a loan.

We have already developed the tools to solve most finance problems. Now we use these tools to solve some application problems.

OUTSTANDING BALANCE ON A LOAN

One of the most common problems deals with finding the balance owed at a given time during the life of a loan. Suppose a person buys a house and amortizes the loan over 30 years, but decides to sell the house a few years later. At the time of the sale, he is obligated to pay off his lender, therefore, he needs to know the balance he owes. Since most long term loans are paid off prematurely, we are often confronted with this problem.

To find the outstanding balance of a loan at a specified time, we need to find the present value \(\mathrm{P}\) of all future payments that have not yet been paid. In this case t does not represent the entire term of the loan. Instead:

• \(t\) represents the time that still remains on the loan
• \(nt\) represents the total number of future payments.

Example \(\PageIndex{1}\)

Mr. Jackson bought his house in 1995, and financed the loan for 30 years at an interest rate of 7.8%. His monthly payment was $1260. In 2015, Mr. Jackson decides to pay off the loan. Find the balance of the loan he still owes.

The reader should note that the original amount of the loan is not mentioned in the problem. That is because we don't need to know that to find the balance.

The original loan was for 30 years. 20 years have past so there are years still remaining. 12(10) = 120 payments still remain to be paid on this loan.

As for the bank or lender is concerned, Mr. Jackson is obligated to pay $1260 each month for 10 more years; he still owes a total of 120 payments. But since Mr. Jackson wants to pay it all off now, we need to find the present value \(\mathrm{P}\) at the time of repayment of the remaining 10 years of payments of $1260 each month. Using the formula we get for the present value of an annuity, we get

\[\begin{aligned} \mathrm{P}(1+.078 / 12)^{120} &=\frac{\left.\$ 1260\left[(1+.078 / 12)^{120}-1\right)\right]}{(.078 / 12)} \\ \mathrm{P}(2.17597) &=\$ 227957.85 \\ \mathrm{P} &=\$ 104761.48 \end{aligned} \nonumber \]

to find the outstanding balance of a loan

If a loan has a payment of \(m\) dollars made \(n\) times a year at an interest \(r\), then the outstanding value of the loan when there are \(t\) years still remaining on the loan is given by \(\mathrm{P}\):

\[\mathbf{P}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}=\frac{\mathbf{m}\left[(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}-\mathbf{1} |\right.}{\mathbf{r} / \mathbf{n}} \nonumber \]

IMPORTANT: Note that \(t\) is not the original term of the loan but instead \(t\) is the amount of time still remaining in the future \(nt\) is the number of payments still remaining in the future

If the problem does not directly state the amount of time still remaining in the term of the loan, then it must be calculated BEFORE using the above formula as \(t\) = original term of loan - time already passed since the start date of the loan.

Note that there are other methods to find the outstanding balance on a loan, but the method illustrated above is the easiest.

One alternate method would be to use an amortization schedule, as illustrated toward the end of this section. An amortization schedule shows the payments, interest, and outstanding balance step by step after each loan payment. An amortization schedule is tedious to calculate by hand but can be easily constructed using spreadsheet software.

Another way to find the outstanding balance, that we will not illustrate here, is to find the difference A - B, where

A = the original loan amount (principal) accumulated to the date on which we want to find the outstanding balance (using compound interest formula)

B = the accumulated value of all payments that have been made as of the date on which we want to find the outstanding balance (using formula for accumulated value of an annuity)

In this case we would need do a compound interest calculation and an annuity calculation; we then need to find the difference between them. Three calculations are needed instead of one.

It is a mathematically acceptable way to calculate the outstanding balance. However, it is very strongly recommended that students use the method explained in box above and illustrated in Example \(\PageIndex{1}\), as it is much simpler.

PROBLEMS INVOLVING MULTIPLE STAGES OF SAVINGS AND/OR ANNUITIES

Consider the following situations:

• Suppose a baby, Aisha, is born and her grandparents invest $5000 in a college fund. The money remains invested for 18 years until Aisha enters college, and then is withdrawn in equal semiannual payments over the 4 years that Aisha expects to need to finish college. The college investment fund earns 5% interest compounded semiannually. How much money can Aisha withdraw from the account every six months while she is in college?

Aisha graduates college and starts a job. She saves $1000 each quarter, depositing it into a retirement savings account. Suppose that Aisha saves for 30 years and then retires. At retirement she wants to withdraw money as an annuity that pays a constant amount every month for 25 years. During the savings phase, the retirement account earns 6% interest compounded quarterly. During the annuity payout phase, the retirement account earns 4.8% interest compounded monthly. Calculate Aisha’s monthly retirement annuity payout.

These problems appear complicated. But each can be broken down into two smaller problems involving compound interest on savings or involving annuities. Often the problem involves a savings period followed by an annuity period. ; the accumulated value from first part of the problem may become a present value in the second part. Read each problem carefully to determine what is needed.

Example \(\PageIndex{2}\)

Suppose a baby, Aisha, is born and her grandparents invest $8000 in a college fund. The money remains invested for 18 years until Aisha enters college, and then is withdrawn in equal semiannual payments over the 4 years that Aisha expects to attend college. The college investment fund earns 5% interest compounded semiannually. How much money can Aisha withdraw from the account every six months while she is in college?

Part 1: Accumulation of College Savings: Find the accumulated value at the end of 18 years of a sum of $8000 invested at 5% compounded semiannually.

\[\begin{array}{l} A=\$ 8000(1+.05 / 2)^{(2 \times 18)}=\$ 8000(1.025)^{36}=\$ 8000(2.432535) \\ A=\$ 19460.28 \end{array} \nonumber \]

Part 2: Seminannual annuity payout from savings to put toward college expenses. Find the amount of the semiannual payout for four years using the accumulated savings from part 1 of the problem with an interest rate of 5% compounded semiannually.

\(A\)= $19460.28 in Part 1 is the accumulated value at the end of the savings period. This becomes the present value \(P\)=$19460.28 when calculating the semiannual payments in Part 2.

\[\begin{array}{c} \$ 19460.28\left(1+\frac{.05}{2}\right)^{2 \times 4}=\frac{m\left[\left(1+\frac{05}{2}\right)^{2 \times 4}-1\right]}{(.05 / 2)} \\ \$ 23710.46=m(8.73612) \\ m=\$ 2714.07 \end{array} \nonumber \]

Aisha will be able to withdraw $2714.07 semiannually for her college expenses.

Example \(\PageIndex{3}\)

Part 1: Accumulation of Retirement Savings: Find the accumulated value at the end of 30 years of $1000 deposited at the end of each quarter into a retirement savings account earning 6% interest compounded quarterly.

\[\begin{array}{l} A=\frac{\$ 1000\left[(1+.06 / 4)^{4 \times 30}-1\right]}{(.06 / 4)} \\ A=\$ 331288.19 \end{array} \nonumber \]

Part 2: Monthly retirement annuity payout: Find the amount of the monthly annuity payments for 25 years using the accumulated savings from part 1 of the problem with an interest rate of 4.8% compounded monthly.

\(A\)= $331288.19 in Part 1 is the accumulated value at the end of the savings period. This amount will become the present value \(\mathrm{P}\) =$331288.19 when calculating the monthly retirement annuity payments in Part 2.

\[\begin{array}{l} \$ 331288.19(1+.048 / 12)^{12 \times 25}=\frac{m\left[(1+.048 / 12)^{12 \times 25}-1\right]}{(.048 / 12)} \\ \$ 1097285.90=m(578.04483) \\ m=\$ 1898.27 \end{array} \nonumber \]

Aisha will have a monthly retirement annuity income of $1898.27 when she retires.

FAIR MARKET VALUE OF A BOND

Whenever a business, and for that matter the U. S. government, needs to raise money it does it by selling bonds. A bond is a certificate of promise that states the terms of the agreement. Usually the business sells bonds for the face amount of $1,000 each for a stated term , a period of time ending at a specified maturity date.

The person who buys the bond, the bondholder , pays $1,000 to buy the bond.

The bondholder is promised two things: First that he will get his $1,000 back at the maturity date, and second that he will receive a fixed amount of interest every six months.

As the market interest rates change, the price of the bond starts to fluctuate. The bonds are bought and sold in the market at their fair market value .

The interest rate a bond pays is fixed, but if the market interest rate goes up, the value of the bond drops since the money invested in the bond could earn more if invested elsewhere. When the value of the bond drops, we say it is trading at a discount .

On the other hand, if the market interest rate drops, the value of the bond goes up since the bond now yields a higher return than the market interest rate, and we say it is trading at a premium .

Example \(\PageIndex{4}\)

The Orange Computer Company needs to raise money to expand. It issues a 10-year $1,000 bond that pays $30 every six months. If the current market interest rate is 7%, what is the fair market value of the bond?

The bond certificate promises us two things - An amount of $1,000 to be paid in 10 years, and a semi-annual payment of $30 for ten years. Therefore, to find the fair market value of the bond, we need to find the present value of the lump sum of $1,000 we are to receive in 10 years, as well as, the present value of the $30 semi-annual payments for the 10 years.

We will let P 1 = the present value of the face amount of $1,000

\[P_{1}(1+.07 / 2)^{20}=\$ 1,000 \nonumber \]

Since the interest is paid twice a year, the interest is compounded twice a year and \(nt\) = 2(10)=20

\[\begin{array}{l} P_{1}(1.9898)=\$ 1,000 \\ P_{1}=\$ 502.56 \end{array} \nonumber \]

We will let P 2 = the present value of the $30 semi-annual payments is

\[\begin{aligned} \mathrm{P}_{2}(1+.07 / 2)^{20} &=\frac{\$ 30\left[(1+.07 / 2)^{20}-1\right]}{(.07 / 2)} \\ \mathrm{P}_{2}(1.9898) &=848.39 \\ \mathrm{P}_{2} &=\$ 426.37 \end{aligned} \nonumber \]

The present value of the lump-sum $1,000 = $502.56

The present value of the $30 semi-annual payments = $426.37

The fair market value of the bond is P = P 1+ P 2 = $502.56 + $426.37 = $928.93 Note that because the market interest rate of 7% is higher than the bond’s implied interest rate of 6% implied by the semiannual payments, the bond is selling at a discount; its fair market value of $928.93 is less than its face value of $1000.

Example \(\PageIndex{5}\)

A state issues a 15 year $1000 bond that pays $25 every six months. If the current market interest rate is 4%, what is the fair market value of the bond?

The bond certificate promises two things - an amount of $1,000 to be paid in 15 years, and semi-annual payments of $25 for 15 years. To find the fair market value of the bond, we find the present value of the $1,000 face value we are to receive in 15 years and add it to the present value of the $25 semi-annual payments for the 15 years. In this example, \(nt = 2(15)=30\).

We will let P 1 = the present value of the lump-sum $1,000

\[\begin{array}{l} \mathrm{P}_{1}(1+.04 / 2)^{30}=\$ 1,000 \\ \mathrm{P}_{1}=\$ 552.07 \end{array} \nonumber \]

We will let P 2 = the present value of the $25 semi-annual payments is

\[\begin{array}{l} \mathrm{P}_{2}(1+.04 / 2)^{30}=\frac{\$ 25\left[(1+.04 / 2)^{30}-1\right]}{(.04 / 2)} \\ \mathrm{P}_{2}(1.18114)=\$ 1014.20 \\ \mathrm{P}_{2}=\$ 559.90 \end{array} \nonumber \]

The present value of the lump-sum $1,000 = $552.07

The present value of the $30 semi-annual payments = $559.90

Therefore, the fair market value of the bond is \[P=P_{1} + P_{2}=\$ 552.07+\$ 559.90=\$ 1111.97\nonumber \]

Because the market interest rate of 4% is lower than the interest rate of 5% implied by the semiannual payments, the bond is selling at a premium: the fair market value of $1,111.97 is more than the face value of $1,000.

To summarize:

to find the Fair Market Value of a Bond

Find the present value of the face amount \(\mathrm{A}\) that is payable at the maturity date:

\[\mathbf{A}=\mathbf{P}_{1}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}} ; \text { solve to find } \mathrm{P}_{1} \nonumber \]

Find the present value of the semiannually payments of $\(m\) over the term of the bond:

\[\mathbf{P}_{2}(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}=\frac{\mathbf{m}\left[(\mathbf{1}+\mathbf{r} / \mathbf{n})^{\mathbf{n} \mathbf{t}}-\mathbf{1}\right]}{\mathbf{r} / \mathbf{n}} \quad ; \text { solve to find } \mathbf{P}_{2} \nonumber \]

The fair market value (or present value or price or current value) of the bond is the sum of the present values calculated above:

\[\mathrm{P}=\mathbf{P}_{1}+\mathbf{P}_{2} \nonumber \]

AMORTIZATION SCHEDULE FOR A LOAN

An amortization schedule is a table that lists all payments on a loan, splits them into the portion devoted to interest and the portion that is applied to repay principal, and calculates the outstanding balance on the loan after each payment is made.

Example \(\PageIndex{6}\)

An amount of $500 is borrowed for 6 months at a rate of 12%. Make an amortization schedule showing the monthly payment, the monthly interest on the outstanding balance, the portion of the payment contributing toward reducing the debt, and the outstanding balance.

The reader can verify that the monthly payment is $86.27.

The first month, the outstanding balance is $500, and therefore, the monthly interest on the outstanding balance is

(outstanding balance)(the monthly interest rate) = ($500)(.12/12) = $5

This means, the first month, out of the $86.27 payment, $5 goes toward the interest and the remaining $81.27 toward the balance leaving a new balance of $500 - $81.27 = $418.73.

Similarly, the second month, the outstanding balance is $418.73, and the monthly interest on the outstanding balance is ($418.73)(.12/12) = $4.19. Again, out of the $86.27 payment, $4.19 goes toward the interest and the remaining $82.08 toward the balance leaving a new balance of $418.73 - $82.08 = $336.65. The process continues in the table below.

Note that the last balance of 3 cents is due to error in rounding off.

An amortization schedule is usually lengthy and tedious to calculate by hand. For example, an amortization schedule for a 30 year mortgage loan with monthly payments would have (12)(30)=360 rows of calculations in the amortization schedule table. A car loan with 5 years of monthly payments would have 12(5)=60 rows of calculations in the amortization schedule table. However it would be straightforward to use a spreadsheet application on a computer to do these repetitive calculations by inputting and copying formulas for the calculations into the cells.

Most of the other applications in this section's problem set are reasonably straightforward, and can be solved by taking a little extra care in interpreting them. And remember, there is often more than one way to solve a problem.