6.3.0: Exercises
- Page ID
- 171715
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)If $1,500.00 is invested in an account bearing 3.5% interest, what is the principal?
If $1,500 is invested in an account bearing 3.5% interest, what is the interest rate?
What is simple interest?
What is the present value of an investment?
What is the future value of an investment?
What is a partial payment on a loan?
In the following exercises, calculate the simple interest and payoff for the loan with the given principal, simple interest rate, and time.
Principal \(P\) = $5,000, annual interest rate \(r\) = 6.5%, and number of years \(t\) = 6
Principal \(P\) = $3,500, annual interest rate \(r\) = 12%, and number of years \(t\) = 7
Principal \(P\) = $7,800, annual interest rate \(r\) = 11.5%, and number of years \(t\) = 10
Principal \(P\) = $62,500, annual interest rate \(r\) = 4.88%, and number of years \(t\) = 4
Principal \(P\) = $4,600, annual interest rate \(r\) = 9.9%, for 18 months
Principal \(P\) = $19,000, annual interest rate \(r\) = 16.9%, for 14 months
Principal \(P\) = $8,500, annual interest rate \(r\) = 10.66%, for 6 months
Principal \(P\) = $17,600, annual interest rate \(r\) = 17.9%, for 20 months
Principal \(P\) = $4,000, annual interest rate \(r\) = 8.5%, for 130 days
Principal \(P\) = $9,900, annual interest rate \(r\) = 15.9%, for 90 days
Principal \(P\) = $600, annual interest rate \(r\) = 16.8%, for 25 days
Principal \(P\) = $890, annual interest rate \(r\) = 9.75%, for 200 days
In the following exercises, find the future value of the investment with the given principal, simple interest rate, and time.
Principal is $5,300, annual interest rate is 2.07%, and time is 18 years.
Principal is $14,700, annual interest rate is 3.11%, and time is 10 years.
Principal is $5,600, annual interest rate is 2.55%, for 30 months.
Principal is $10,000, annual interest rate is 1.99%, for 15 months.
Principal is $2,000, annual interest rate is 3.22%, for 100 days.
Principal is $900, annual interest rate is 3.75%, for 175 days.
In the following exercises, determine the amount applied to principal for the indicated partial payment on the loan with the given principal, interest rate, and time when the partial payment was made.
A simple interest loan for $2,700 is taken out at 11.6% annual percentage rate. A partial payment of $1,500 is made 28 days into the loan period.
A simple interest loan for $900 is taken out at 18.9% annual percentage rate. A partial payment of $400 is made 30 days into the loan period.
A simple interest loan for $13,500 is taken out at 14.8% annual percentage rate. A partial payment of $8,000 is made 75 days into the loan period.
A simple interest loan for $9,900 is taken out at 9.875% annual percentage rate. A partial payment of $4,000 is made 65 days into the loan period.
In the following exercises, determine the remaining principal for the indicated partial payment on the loan with the given principal, interest rate, and time when the partial payment was made.
A simple interest loan for $2,700 is taken out at 11.6% annual percentage rate. A partial payment of $1,500 is made 28 days into the loan period.
A simple interest loan for $900 is taken out at 18.9% annual percentage rate. A partial payment of $400 is made 30 days into the loan period
A simple interest loan for $13,500 is taken out at 14.8% annual percentage rate. A partial payment of $8,000 is made 75 days into the loan period.
A simple interest loan for $9,900 is taken out at 9.875% annual percentage rate. A partial payment of $4,000 is made 65 days into the loan period.
In the following exercises, find the payoff value of the loan with the given principal, annual simple interest rate, term, partial payment, and time at which the partial payment was made.
Principal = $1,500, rate = 6.99%, term is 5 years, partial payment of $900 made 2 years into the loan.
Principal = $21,500, rate = 7.44%, term is 10 years, partial payment of 15,000 made after 6 years.
Principal = $6,800, rate = 11.9%, term is 200 days, partial payment of $4,000 made after 100 days.
Principal = $800, rate = 13.99%, term is 150 days, partial payment of $525 made after 50 days.
In the following exercises, find the monthly payment for a loan with the given principal, annual simple interest rate and number of years.
Principal = $4,500, rate = 8.75%, years = 3
Principal = $2,700, rate = 15.9%, years = 5
Principal = $13,980, rate = 10.5%, years = 4
Principal = $8,750, rate = 9.9%, years = 10
In the following exercises, find the present value for the given future value, \(FV\), annual simple interest rate \(r\), and number of years \(t\).
\(FV\) = $25,000, \(t\) = 15 years, annual simple interest rate of 6.5%
\(FV\) = $12,000, \(t\) = 10 years, annual simple interest rate of 4.5%
\(FV\) = $15,000, \(t\) = 16 years, annual simple interest rate of 3.5%
\(FV\) = $100,000, \(t\) = 30 years, annual simple interest rate of 5.5%
Rita takes out a simple interest loan for $4,000 for 5 years. Her interest rate is 7.88%. How much will Rita pay when the loan is due?
Humberto runs a private computer networking company, and needs a loan of $31,500 for new equipment. He shops around for the lowest interest rate he can find. He finds a rate of 8.9% interest for a 10-year term. How much will Humberto’s payoff be at the end of the 10 years?
Jaye needs a short-term loan of $3,500. They find a 75-day loan that charges 14.9% interest. What is Jaye’s payoff?
Theethat’s car needs new struts, which cost $1,189.50 installed, but he doesn’t have the money to do so. He asks the repair shop if they offer any sort of financing. It offers him a short-term loan at 18.9% interest for 60 days. What is Theethat’s payoff for the struts?
Michelle opens a gaming shop in her small town. She takes out an $8,500 loan to get started. The loan is at 9.5% interest and has a term of 5 years. Michelle decides to make a partial payment of $4,700 after 3 years. What will Michelle pay when the loan is due?
A small retailer borrows $3,750 for a repair. The loan has a term of 100 days at 13.55% interest. If the retailer pays a partial payment of $2,000 after 30 days, what will the loan payoff be when the loan is due?
Sharon invests $2,500 in a CD for her granddaughter. The CD has a term of 5 years and has a simple interest rate of 3.11%. After that 5-year period, how much will the CD be worth?
Jen and Fred have a baby, and deposit $1,500 in a savings account bearing 1.76% simple interest. How much will the account be worth in 18 years?
Yasmin decides to buy a used car. Her credit union offers 7.9% interest for 5-year loans on used cars. The cost of the car, including taxes and fees, is $11,209.50. How much will Yasmin’s monthly payment be?
Cleo runs her own silk-screening company. She needs new silk-screening printing machines, and finds two that will cost her, in total, $5,489.00. She takes out a 3-year loan at 8.9% interest. What will her monthly payments be for the loan?
Kylie wants to invest some money in an account that yields 4.66% simple interest. Her goal is to have $20,000 in 15 years. How much should Kylie invest to reach that goal?
Ishraq wants to deposit money in an account that yields 3.5% simple interest for 10 years, to help with a down payment for a home. Her goal is to have $25,000 for the down payment. How much does Ishraq need to deposit to reach that goal?
In the following exercises, use the Cost of Financing. The difference between the total paid for a loan, along with all other charges paid to obtain the loan, and the original principal of the loan is the cost of financing. It measures how much more you paid for an item than the original price. In order to find the cost of financing, find the total paid over the life of the loan. Add to that any fees paid for the loan. Then subtract the principal.
\({\text{Cost of Financing }} = {\text{ Total of payments }} + {\text{ fees }} - {\text{ principal}}.\)
Yasmin decides to buy a used car. Her credit union offers 7.9% interest for 5-year loans on used cars. The cost of the car, including taxes and fees, is $11,209.50. How much did she pay the credit union over the 5 years? What was the cost of financing for Yasmin?
Cleo runs her own silk-screening company. She needs new silk-screening printing machines, and finds two that will cost her, in total, $5,489.00. She takes out a 3-year loan at 8.9% interest. What was the cost of financing for Cleo?