# 5.11: Loans and Interest, Home Loans, Auto Loans, Insurance

- Page ID
- 6993

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Loans & Interest, Home Loans, Auto Loans, Insurance

Many people buy cars and houses they cannot afford. This causes major problems for both the banks and the people who have their property taken. In order to make wise choices when you purchase a car or a house, it is important to know how much you can afford to pay each period and calculate a maximum loan amount.

### Universal Generalizations

- Banks are essential to the economy, they help people save and borrow money.
- Financial institutions charge interest rates to make money on the money they lend.
- When people borrow money they will have certain responsibilities and obligations to ensure the repayment of that loan.
- A person’s credit score can impact how much they can borrow and at what interest rate they will have to pay to use someone else’s money.
- There are economic benefits to owning your own home.
- People who are knowledgeable about borrowing can make decisions in their best interest.
- There are different types of insurance depending on a person’s needs over time.
- As a consumer, the rule is “buyer beware”.

### Guiding Questions

- What must a bank consider before it decides to lend money?
- What types of questions should a borrower ask when shopping for a loan?
- How does a person get and maintain a good credit score?
- What costs are associated with purchasing and maintaining a car?
- Which types of insurance are required by law? Which ones will you need?
- How can a person save money on auto insurance? Health insurance?

### Buying a House

Joanna knows she can afford to pay $12,000 a year for a house loan. Interest rates are 4.2% annually and most house loans go for 30 years. What is the maximum loan she can afford? What will she end up paying after 30 years?

### Video: How to Determine the Monthly Loan Payment for a Mortgage

### Video: How to Determine the Payment for a Fixed Installment Loan

### Formulas with Examples

The present value can be found from the future value using the regular compound growth formula:

\(\ \begin{aligned} P V(1+i)^{n} &=F V \\ P V &=\frac{F V}{(1+i)^{n}} \end{aligned}\)

You also know the future value of an annuity:

\(\ F V=R \cdot \frac{(1+i)^{n}-1}{i}\)

So by substitution, the formula for the present value of an annuity is:

\(\ P V=R \cdot \frac{(1+i)^{n}-1}{i} \cdot \frac{1}{(1+i)^{n}}=R \cdot \frac{(1+i)^{n}-1}{i(1+i)^{n}}=R \cdot \frac{1-(1+i)^{-n}}{i}\)

The present value of a series of equal payments *R* with interest rate *i* per period for *n* periods is:

\(\ P V=R \cdot \frac{1-(1+i)^{-n}}{i}\)

What is the monthly payment of a $1,000,000 house loan over 30 years with a nominal interest rate of 6% convertible monthly?

**Solution**

\(\ \begin{array}{center} P V=\$ 1,000,000, R=?, i=0.005, n =360 \\ P V =R \cdot \frac{1-(1+i)^{-n}}{i} \\ 1,000,000 =R \cdot \frac{1-(1+0.005)^{-360}}{0.005} \\ R =\frac{1,000,000 \cdot 0.005}{1-(1+0.005)^{-360}} \approx 5995.51 \end{array}\)

It is remarkable that in order to pay off a $1,000,000 loan you will have to pay $5,995.51 a month, every month, for thirty years. After 30 years, you will have made 360 payments of $5995.51, and therefore will have paid the bank more than $2.1 million, more than twice the original loan amount. It is no wonder that people can get into trouble taking on more debt than they can afford.

How long will it take to pay off a $20,000 car loan with a 6% annual interest rate convertible monthly if you pay it off in monthly installments of $500? What about if you tried to pay it off in monthly installments of $100?

**Solution**

\(\ \begin{array}{center}P V=\$ 20,000, R=\$ 500, i=\frac{0.06}{12}=0.005, n=?\\

P V=R \cdot \frac{1-(1+i)^{-n}}{i}\\

20,000=500 \cdot \frac{1-(1+0.005)^{-n}}{0.005}\\

0.2=1-(1+0.005)^{-n}\\

(1+0.005)^{-n} =0.8 \\ n =-\frac{\ln 0.8}{\ln 1.005} \approx 44.74 \text { months} \end{array}

\)

For the $100 case, if you try to set up an equation and solve, there will be an error. This is because the interest on $20,000 is exactly $100 and so every month the payment will go to only paying off the interest. If someone tries to pay off less than $100, then the debt will grow.

It saves money to pay off debt faster in order to save money on interest. As shown in Example A, interest can more than double the cost of a 30 year mortgage. This example shows how much money can be saved by paying off more than the minimum.

Suppose a $300,000 loan has 6% interest convertible monthly with monthly payments over 30 years. What are the monthly payments? How much time and money would be saved if the monthly payments were larger by a fraction of \(\ \frac{13}{12}\)? This is like making 13 payments a year instead of just 12. First you will calculate the monthly payments if 12 payments a year are made.

\(\ \begin{aligned} P V &=R \cdot \frac{1-(1+i)^{-n}}{i} \\ 300,000 &=R \cdot \frac{1-(1+0.005)^{-360}}{0.005} \\ R &=\$ 1,798.65 \end{aligned}\)

After 30 years, you will have paid $647,514.57, more than twice the original loan amount.

If instead, the monthly payment was \(\ \frac{13}{12} \cdot 1798.65=1948.54\), you would pay off the loan faster. In order to find out how much faster, you will make your unknown.

\(\ \begin{aligned} P V &=R \cdot \frac{1-(1+i)^{-n}}{i} \\ 300,000 &=1948.54 \cdot \frac{1-(1+0.005)^{-n}}{0.005} \\ 0.7698 &=1-(1+0.005)^{-n} \\(1+0.005)^{-n} &=0.23019 \\ n &=-\frac{\ln 0.23019}{\ln 1.005} \approx 294.5 \text { months } \end{aligned}\)

294.5 months is about 24.5 years. Paying fractionally more each month saved more than 5 years of payments.

\(\ 294.5\ months \cdot \$ 1,948.54=\$ 573,847.99\)

The loan ends up costing $573,847.99, which saves you more than $73,000 over the total cost if you had paid over 30 years.

**Concept Problem Revisited**

Joanna knows she can afford to pay $12,000 a year to pay for a house loan. Interest rates are 4.2% annually and most house loans go for 30 years. What is the maximum loan she can afford? What does she end up paying after 30 years? You can use the present value formula to calculate the maximum loan:

\(\ P V=12,000 \cdot \frac{1-(1+0.042)^{-30}}{0.042} \approx \$ 202,556.98\)

For 30 years she will pay $12,000 a year. At the end of the 30 years she will have paid $12,000⋅30=$360,000 total

**Guided Practice**

- Mackenzie obtains a 15 year student loan for $160,000 with 6.8% interest. What will her yearly payments be?
- How long will it take Francisco to pay off a $16,000 credit card bill with 19.9% APR if he pays $800 per month?
*Note: APR in this case means nominal rate convertible monthly*. - What will the monthly payments be on a credit card debt of $8,000 with 34.99% APR if it is paid off over 3 years?

*After attempting to solve the Guided Practice Questions, check below for the correct solutions.*

**Solution**

1.

\(\ \begin{array}{center}P V=\$ 160,000, R=?, n=15, i=0.068\\

160,000=R \cdot \frac{1-(1+0.068)^{-15}}{0.068}\\

R \approx \$ 17,345.88\end{array}\)

2.

\(\ \begin{array}{center}P V=\$ 16,000, R=\$ 600, n=?, i=\frac{0.199}{12}\\

16,000=600 \cdot \frac{1-\left(1+\frac{0.199}{12}\right)^{-n}}{\frac{0.199}{12}}\\

n=24.50 \ months\end{array}\)

3.

\(\ \begin{array}{center}P V=\$ 8,000, R=?, n=36, i=\frac{0.3499}{12}\\

8,000=R \cdot \frac{1-\left(1+\frac{0.3499}{12}\right)^{-36}}{\frac{0.3499}{12}}\\

R=\$ 361.84\end{array}\)

### Activity

#### Independent Practice

For problems 1-10, find the missing value in each row using the present value for annuities formula.

Problem Number | PV |
R |
n (years) | i (annual) | Periods per year |

1. | $4,000 | 7 | 1.5% | 1 | |

2. | $15,575 | 5 | 5% | 4 | |

3. | $4,500 | $300 | 3% | 12 | |

4. | $1,000 | 12 | 2% | 1 | |

5. | $16,670 | 10 | 10% | 4 | |

6. | $400 | 4 | 2% | 12 | |

7. | $400 | 4 | 2% | 12 | |

7. | $315,000 | $1,800 | 5% | 12 | |

8. | $500 | 30 | 8% | 12 | |

9. | $1,000 | 40 | 6% | 4 | |

10. | $10,000 | 6 | 7% | 12 |

11. Charese obtains a 15 year student loan for $200,000 with 6.8% interest. What will her yearly payments be?

12. How long will it take Tyler to pay off a $5,000 credit card bill with 21.9% APR if he pays $300 per month? *Note: APR in this case means nominal rate convertible monthly*.

13. What will the monthly payments be on a credit card debt of $5,000 with 24.99% APR if it is paid off over 3 years?

14. What is the monthly payment of a $300,000 house loan over 30 years with a nominal interest rate of 2% convertible monthly?

15. What is the monthly payment of a $270,000 house loan over 30 years with a nominal interest rate of 3% convertible monthly?

### Learning Extension

Visit the lesson titled **Show Me the Money** on the EconEdLink website. Follow your teachers's instructions for completing this activity.