6.4 Amortized Loans

Robert Foth

6.4 Amortized Loans

6.4: Amortized Loans

Learning Objectives

Upon completion of this section, you should be able to

Compute the monthly payment and interest costs for amortized loans.
Find the amount that can be borrowed with a known payment amount for an amortized loan.
Calculate the loan balance for an amortized loan.
Create an amortized loan schedule.

Amortized Loan Formula

In this section, we’ll explore amortized loans, a common type of installment loan used for major purchases such as homes and vehicles where there is a gradual reduction in the loan amount through periodic repayments (typically of equal amounts) over a period of time. In an amortized loan, each payment you make goes towards paying both the interest accrued and a portion of the principal. As you continue making payments, a larger percentage goes towards the principal, accelerating the loan payoff.

Common examples of amortized loans include:

Mortgage loans for home purchases
Auto loans for vehicle financing

Amortized Loan Formula

The monthly payment for an amortized loan can be calculated using the following formula:

pymt = \frac{P \cdot (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{(- n \cdot t)})}

The loan principal for an amortized loan when payments are known can be calculated using the following formula:

$P = \frac{pymt \cdot (1 - {(1 + \frac{r}{n})}^{(- n \cdot t)})}{(\frac{r}{n})}$

$P$ is the principal balance or loan proceeds (how much is being borrowed).
$pymt$ is your loan payment (your monthly payment, annual payment, etc).
$r$ is the annual interest rate (in decimal form).
$n$ is the frequency of payments in a year (typically 12 for monthly payments).
$t$ is the length of the loan, in years.

Example 1

You can afford $200 per month as a car payment. If you can get an auto loan through a local credit union at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month? In addition how much interest is paid over the life of the loan (what is the cost of the loan)?

Solution

In this example we are given the following:

$\begin{array}{l} pymt & = $ 200 & the monthly loan payment \\ r & = 0.03 & 3% annual rate \\ n & = 12 & since we’re doing monthly payments \\ t & = 5 & since we’re making monthly payments for 5 years \end{array}$

We’re looking for $P$ , the starting principal amount of the loan.

$\begin{array}{l} P = \frac{200 \cdot (1 - {(1 + \frac{0.03}{12})}^{(- 12 \cdot 5)})}{(\frac{0.03}{12})} \\ P \approx $ 11, 130.47 \end{array}$

You can afford a $11,130 loan.

You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $12,000-$11,130 = $870 interest total.

Now in the above example we did the complete evaluation of the loan formula all at once to avoid any rounding errors introduced while trying to evaluate the principal amount. If you had to do the calculations in steps instead and rounded to four places the principal amount you could borrow would be slightly off due to rounding errors. See below for the result of rounding during the steps.

$\begin{array}{l} P = \frac{200 \cdot (1 - {(1 + \frac{0.03}{12})}^{(- 12 \cdot 5)})}{(\frac{0.03}{12})} \\ P = \frac{200 \cdot (1 - {(1 + 0.0025)}^{(- 60)})}{(0.0025)} \\ P \approx \frac{200 \cdot (1 - 0.8609)}{(0.0025)} \\ P \approx \frac{200 \cdot (0.1391)}{(0.0025)} \\ P \approx \frac{27.82}{(0.0025)} \\ P \approx $ 11, 128 \end{array}$

A difference of about $2 dollars, but would be amplified further if we had rounded only to three or two decimal places in the work. To avoid the rounding errors it is best to do the full evaluation of a formula at once on a calculator. When that is not possible it is recommended you round values to 8 decimal places until the very end to reduce the amount of rounding error that may end up in your final answer.

Example 2

You want to take out a $140,000 mortgage (amortized loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?

Solution

In this example we’re looking for the amount of the payment, $pymt$ .

$\begin{array}{l} P & = $ 140, 000 & the starting loan amount \\ r & = 0.06 & 6% annual rate \\ n & = 12 & since we’re doing monthly payments \\ t & = 30 & since we’re making monthly payments for 30 years \end{array}$

Use the formula where pymt was already solved for

$\begin{array}{l} pymt = \frac{140, 000 \cdot (\frac{0.06}{12})}{(1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 30)})} \\ pymt \approx $ 839.37 \end{array}$

You will make payments of $839.37 per month for 30 years.

If you add all the payments up for 30 years you are paying a total of $302,173.20 to the loan company. This means the cost of the loan (interest) is:

$$ 302, 173.20 - $ 140, 000 = $ 162, 173.20$

This is not unusual to see the interest being paid over the life of a 30 year loan to be about the same as the amount borrowed for interest rates around 6%.

Try it Now 1

Lucía bought a used car for $8,500 (including fees and taxes). She needed a loan for the purchase and will be financing the whole amount. The bank offers an amortized loan with an annual rate of 6.9% with monthly payments over the next 3 years. Calculate the monthly payments, the total cost of the car purchase, and the total amount of interest paid.

Hint 1 (click to Show/Hide)

In this scenario we are finding the amount of payments on the loan for Lucía. Pick the second loan formula and plug in the values that were given.

$\begin{array}{l} P & = $ 8, 500 & the starting loan amount \\ r & = 0.069 & 6.9% annual rate \\ n & = 12 & since we’re doing monthly payments \\ t & = 3 & since we’re taking 3 years to repay \end{array}$

Answer (click to Show/Hide)

Start with solving for the payment amount, $pymt$

$\begin{array}{l} pymt = \frac{P \cdot (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{(- n \cdot t)})} \\ pymt = \frac{8500 \cdot (\frac{0.069}{12})}{(1 - {(1 + \frac{0.069}{12})}^{(- 12 \cdot 3)})} \\ pymt = $ 262.07 \end{array}$

With the payments being $262.07 each month for the next three years we have a total cost for the purchase of the car as:

$$ 262.07 \cdot 12 \cdot 3 = $ 9, 434.52$

The interest is then $$ 9, 434.52 - $ 8, 500 = $ 934.52$ .

Calculating Remaining Loan Balance

With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.

Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will not have paid off $12,000 of the loan balance.

To determine the remaining balance on an amortized loan, we will answer this question as our strategy: “What loan amount would the remaining payments fully pay off at the current interest rate?”

This strategy works because the remaining balance is equivalent to a new loan that would be fully amortized by the remaining scheduled payments. In other words:

We consider the remaining payments as if they were payments on a new loan.
We calculate what loan amount these payments would fully cover.
This amount represents the remaining balance on the original loan.

How to determine the remaining balance on an amortized loan

Identify the number of remaining payments or the remaining time of the loan.
Use the current interest rate, payment amount, frequency, and remaining time you just found in the amortized loan formula to find the principal balance.
This new principal balance is the amount remaining on the loan.

Why this works

In an amortized loan, each payment is calculated to cover interest and reduce the principal so that the loan is paid off exactly at the end of the term. At any point, the remaining balance is precisely the amount that would be paid off by the remaining payments at the given interest rate.

This approach allows us to calculate the remaining balance without needing to track every previous payment or perform complex interest calculations from the loan’s start date.

Example 3

Going back to the second example. You take out a loan with an initial principal of $140,000. The interest rate on the loan is 6% with a term of 30 years. The monthly payments were found to be $839.37. Determine the balance of the loan at 5, 10, 15, 20, and 25 years into the loan.

Was half of the balance paid at the half way mark of 15 years?

Solution

In the table below the first column time in years represents the amount of time after the loan has begun. When looking in the formula in the second column we see the time remaining is used to find the principal amount on a loan with the given time remaining. This will represent the balance of the loan. For instance 5 years after the loan was started there was 25 years remaining (left for the loan). The balance of the loan after 5 years was found to be $130,275.99.

Time in Years	Years Remaining	Balance of Loan
$5$	$25$	$\frac{839 .37 \cdot (1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 25)})}{(\frac{0.06}{12})} \approx $ 130, 275.99$
$10$	$20$	$\frac{839 .37 \cdot (1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 20)})}{(\frac{0.06}{12})} \approx $ 117, 159.91$
$15$	$15$	$\frac{839 .37 \cdot (1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 15)})}{(\frac{0.06}{12})} \approx $ 99, 468.30$
$20$	$10$	$\frac{839 .37 \cdot (1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 10)})}{(\frac{0.06}{12})} \approx $ 75, 609.95$
$25$	$5$	$\frac{839 .37 \cdot (1 - {(1 + \frac{0.06}{12})}^{(- 12 \cdot 5)})}{(\frac{0.06}{12})} \approx $ 43416.88$

Looking at the table we see that after half the time for the term of the loan has passed (15 years) there is still more than half of the balance remaining. We don’t actually see the balance dip below half ($70,000) until after 20 years has passed.

If we plot the data for the balance of the loan after a given number of years we get the graph as shown below. Notice how slowly the balance is reduced at the beginning of the loan term, but quickly accelerates to zero as you get closer to the end of the term.

Graph produced on Desmos. Go to Desmos to explore this data.

Often times answering remaining balance questions requires two steps:

Calculating the monthly payments on the loan
Calculating the remaining loan balance based on the remaining time on the loan

Example 4

A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?

Solution

First we will calculate their monthly payments as that will be needed to find the remaining balance. We will start with listing out what is known:

$\begin{array}{l} P & = $ 180, 000 & the starting loan amount \\ r & = 0.04 & 4% annual rate \\ n & = 12 & since they’re paying monthly \\ t & = 30 & since they’re making monthly payments for 30 more years \end{array}$

Now solve for $pymt$ :

$\begin{array}{l} pymt = \frac{P \cdot (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{(- n \cdot t)})} \\ pymt = \frac{180000 \cdot (\frac{0.04}{12})}{(1 - {(1 + \frac{0.04}{12})}^{(- 12 \cdot 30)})} \\ pymt = $ 859.35 \end{array}$

Now that we know the monthly payments, we can determine the remaining balance after 5 years (or 25 years left on the loan). Calculate a loan balance that will be paid off with the monthly payments of $859.35 over 25 years:

$\begin{array}{l} pymt & = $ 859.35 & the monthly loan payment we calculated above \\ r & = 0.04 & 4% annual rate \\ n & = 12 & since they’re paying monthly \\ t & = 25 & since they’d be making monthly payments for 25 more years \end{array}$

Now solve for $P$ in the loan formula:

$\begin{array}{l} P = \frac{pymt \cdot (1 - {(1 + \frac{r}{n})}^{(- n \cdot t)})}{(\frac{r}{n})} \\ P = \frac{859 .35 \cdot (1 - {(1 + \frac{0.04}{12})}^{(- 12 \cdot 25)})}{(\frac{0.04}{12})} \\ P \approx $ 162, 805.99 \end{array}$

The loan balance after 5 years, with 25 years remaining on the loan, will be $162,805.99

Over that 5 years, the couple has paid off $$ 180, 000 - $ 162, 805.99 = $ 17, 194.01$ of the loan balance.

If we were interested in finding out how much interest was paid over those five yeras we first find the total that was paid in payments over the five years and then subtract off the amount paid to the loan over the five years we just saw from above.

They have paid a total of $859.35 a month for 5 years (60 months), for a total of $51,561.00, so $$ 51561.00 - $ 17, 194.01 = $ 34, 366.99$ of what they have paid so far has been interest.

Amortized Loan Schedule

To better understand how the principal is paid down on an amortized loan we can build an 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.

Amortization Schedule

An amortized loan schedule, also known as an amortization schedule, is a detailed breakdown of each payment over the life of a loan. It shows how each payment is applied to both the principal balance and the interest, and how the loan balance changes over time.

The table below shows the titles for each column and the calculation done for each cell on an amortized loan paid monthly with principal $P$ , annual rate $r$ , time $t$ , and payment $pymt$ . Let $B$ be the balance from previous payment or the principal if it is the first payment.

Payment #

Payment

Interest

Principal Paid

Extra Payment

Balance After Payment

1

pymt

P \cdot (\frac{r}{12})

pymt minus interest

B = P minus (principal plus extra payment)

2

pymt

B \cdot (\frac{r}{12})

pymt minus interest

New B = P minus (principal plus extra payment)

The table would continue until you reach the last payment for the given amount of time for the loan.

In the example below we will show how to build out an amortization schedule for a fairly short loan so you can see how all the payments are applied to the loan.

Example 5

An amount of $500 is borrowed with an amortized loan for 6 months at a rate of 12%. This loan would have a payment of $86.27. 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.

Solution

With the first payment we use the entire principal to determine the amount of interest paid from the payment:

$(principal) (monthly interest rate) = (500) (\frac{0.12}{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 principal (or balance of the loan) leaving a new balance of:

$$ 500 - $ 81.27 = $ 418.73$

Let us fill that out in the first row of the table:

Payment #	Payment	Interest	Principal Paid	Balance After Payment
1	$86.27	$5	$81.27	$418.73
2	$86.27
3	$86.27
4	$86.27
5	$86.27
6	$86.27

Now in the second month we see the outstanding balance is $418.73. Calculate the monthly interest on the outstanding balance:

$(418.73) (\frac{0.12}{12}) = $ 4.19$

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 for the other months as well:

Payment #	Payment	Interest	Principal Paid	Balance After Payment
1	$86.27	$5	$81.27	$418.73
2	$86.27	$4.19	$82.08	$336.65
3	$86.27	$3.37	$82.90	$253.75
4	$86.27	$2.54	$83.73	$170.02
5	$86.27	$1.70	$84.57	$85.45
6	$86.27	$0.85	$85.42	$0.03

Note that the last balance of 3 cents is due to error in rounding off. This round off value is typically added to the last payment for the loan, so instead of paying $86.27 the borrow would have paid $86.30.

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 table. A car loan with 5 years of monthly payments would have $(12) (5) = 60$ rows of calculations. It would be straightforward to use a spreadsheet application to do these repetitive calculations by inputting and copying formulas for the calculations into the cells. In the next example we will show what that will look like along with how we can use the table to answer questions about the loan.

Example 6

Abigail purchases a home for $160,000 and the loan is amortized over 30 years at an interest rate of 4.4%. Complete an amortized loan schedule in a spread sheet and answer the following questions.

Find the balance after the 100th payment.
Find the balance owed after 20 years
Determine the cost of the loan (interest).
If an extra $200 a month is paid towards the balance find the new number of payments and the total cost of the loan.
If an extra $400 a month is paid towards the balance find the new number of payments and the total cost of the loan.

Solution

To complete the amortization schedule we need to start by finding the monthly payments

$\begin{array}{l} pymt = \frac{160, 000 \cdot (\frac{0.044}{12})}{(1 - {(1 + \frac{0.044}{12})}^{(- 12 \cdot 30)})} \\ pymt \approx $ 801.22 \end{array}$

Below represents the first three payments in the amortization schedule. View the full schedule in Google Sheets.

Payment #	Payment	Interest	Principal Paid	Balance After Payment
1	$801.22	$587.00	$214.55	$159,785.45
2	$801.22	$585.88	$215.34	$159,570.11
3	$801.22	$585.09	$216.13	$159,353.98

From the schedule we see the balance after the 100th payment is $134,139.52.

Payment # Payment Interest Principal Paid Extra Payment Balance After Payment

100 $801.22 $492.98 $308.24 $134,139.52
After 20 years is the same as after the 20*12=240th payment. This balance is $77,668.70

Payment # Payment Interest Principal Paid Extra Payment Balance After Payment

240 $801.22 $286.67 $514.55 $77,668.70
The cost of the loan can be found by adding up all the interest payments in the schedule. This total was found to be $128,437.30 when adding the interest payments found in the google sheets document. It will be slightly different due to rounding from taking the sum of all payments and subtracting the principal of the loan (that value would be $128.439.20).
In the second tab of the Google Sheets file we can see the extra $200 per month added to each payment that goes towards reducing the balance in addition to the principal portion of a payment. The loan would now finish at the 241 payment as that is where the balance is now a negative value. This is 20 years plus 1 month. The cost of the loan is found by adding all the interest payments and we find the cost is $81,218.
By adding an extra $200 per month for each payment the borrower pays approximately $47,000 less in interest over the life of the loan.
In the third tab of the document we see what happens when an extra $400 is applied each month. There is a total of 184 payments yielding a cost for the loan of $59,963.

Exercises

You can afford a $700 per month mortgage payment. You’ve found a 30 year loan with an annual rate of 5% paid monthly.
1. How big of a loan can you afford?
2. How much total money will you pay the loan company?
3. How much of that money is interest?
Answer (click to Show/Hide)
This problem uses the loan payment formula to find the affordable loan amount, and then basic calculations to find the total paid and total interest.
1. How big of a loan can you afford?
  We need to find the loan principal ( $P$ ) using the loan payment formula.
  
  Step 1: Identify the given values.
  - Monthly Payment ( $pymt$ ): $700
  - Interest Rate ( $r$ ): 5% or 0.05
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 30 years
  Step 2: Use the loan payment formula and solve.
  The formula is $P = \frac{pymt (1 - {(1 + \frac{r}{n})}^{- n t})}{(\frac{r}{n})}$ .
  
  $P = \frac{700 (1 - {(1 + \frac{0.05}{12})}^{(- 12 \times 30)})}{(\frac{0.05}{12})} \approx 130,397.13$
  
  Answer (a): You can afford a loan of approximately $130,397.13.
2. How much total money will you pay the loan company?
  This is the total of all monthly payments over the 30 years.
  
  Monthly Payment: $700
  
  Total Payments: 360
  
  $Total Paid = 700 \times 360 = 252000$
  
  Answer (b): You will pay a total of $252,000 to the loan company.
3. How much of that money is interest?
  The total interest is the difference between the total amount paid and the original loan amount.
  
  Total Paid: $252,000
  
  Loan Amount: $130,397.15
  
  $Total Interest = 252000 - 130397.13 = 121602.87$
  
  Answer (c): You will pay approximately $121,602.87 in interest.
Marie can afford a $250 per month car payment. She’s found a 5 year loan with an annual rate of 7% paid monthly.
1. How expensive of a car can she afford?
2. How much total money will she pay the loan company?
3. How much of that money is interest?
Answer (click to Show/Hide)
This problem uses the loan payment formula to find the affordable loan amount, followed by calculations for the total paid and total interest.
1. How expensive of a car can she afford?
  We need to find the loan principal ( $P$ ) using the loan payment formula.
  
  Step 1: Identify the given values.
  - Monthly Payment ( $pymt$ ): $250
  - Interest Rate ( $r$ ): 7% or 0.07
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 5 years
  Step 2: Use the loan payment formula and solve.
  The formula is $P = \frac{pymt (1 - {(1 + \frac{r}{n})}^{- n t})}{(\frac{r}{n})}$ .
  
  Now substitute the values into the formula and evaluate.
  $P = \frac{250 (1 - {(1 + \frac{0.07}{12})}^{(- 12 \times 5)})}{(\frac{0.07}{12})} \approx 12,625.50$
  
  Answer (a): She can afford a car that costs approximately $12,625.50.
2. How much total money will she pay the loan company?
  This is the total of all monthly payments over the 5 years.
  
  Monthly Payment: $250n
  
  Total Payments: 60
  
  $Total Paid = 250 \times 60 = 15000$
  
  Answer (b): She will pay a total of $15,000 to the loan company.
3. How much of that money is interest?
  The total interest is the difference between the total amount paid and the original loan amount.
  
  Total Paid: $15,000
  
  Loan Amount: $12,625.50
  
  $Total Interest = 15000 - 12625.50 = 2374.5$
  
  Answer (c): Marie will pay approximately $2,374.50 in interest.
You want to buy a $25,000 car. The company is offering a loan with a 2% annual rate paid monthly for 48 months (4 years). What will your monthly payments be?
Answer (click to Show/Hide)
This problem requires us to solve the loan payment formula for the monthly payment ( $pymt$ ).

Step 1: Identify the given values.
- Loan Principal ( $P$ ): $25,000
- Interest Rate ( $r$ ): 2% or 0.02
- Frequency of payments per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 4 years
Step 2: Use the loan payment formula and solve for pymt.
The formula, solved for the payment, is $pymt = \frac{P \times (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{- n t})}$ .

Now, substitute the values into the formula.
$pymt = \frac{25000 \times (\frac{0.02}{12})}{(1 - {(1 + \frac{0.02}{12})}^{(- 12 \times 4)})} \approx 542.38$

Answer: Your monthly payments will be approximately $542.38.
You decide finance a $12,000 car with an annual rate of 3% paid monthly for 4 years.
1. What will your monthly payments be?
2. How much interest will you pay over the life of the loan?
Answer (click to Show/Hide)
1. What will your monthly payments be?
  This problem requires us to find the loan payment ( $pymt$ ) given we know the loan amount.
  
  Step 1: Identify the given values.
  - Loan Principal ( $P$ ): $12,000
  - Interest Rate ( $r$ ): 3% or 0.03
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 4 years
  Step 2: Use the loan payment formula and solve for pymt.
  The formula, solved for the payment, is $pymt = \frac{P \times (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{- n t})}$ .
  
  Now, substitute the values into the formula.
  $pymt = \frac{12000 \times (\frac{0.03}{12})}{(1 - {(1 + \frac{0.03}{12})}^{(- 12 \times 4)})} \approx 265.61$
  
  Answer (a): Your monthly payments will be approximately $265.61.
2. How much interest will you pay over the life of the loan?
  Step 1: Calculate the total amount paid.
  Total Paid = Monthly Payment × Total Number of Payments
  $Total Paid = 265.61 \times (12 \times 4) = 265.61 \times 48 = 12749.28$
  
  Step 2: Calculate the total interest paid.
  Total Interest = Total Paid – Loan Principal
  $Total Interest = 12749.28 - 12000 = 749.28$
  
  Answer (b): You will pay $749.28 in interest over the life of the loan.
You want to buy a $200,000 home. You plan to pay 10% as a down payment, and take out a 30 year loan for the rest.
1. How much is the loan amount going to be?
2. What will your monthly payments be if the annual interest rate is 5% monthly?
3. What will your monthly payments be if the annual interest rate is 6% monthly?
Answer (click to Show/Hide)
1. How much is the loan amount going to be?
  Step 1: Calculate the down payment.
  Down Payment = 10% of $200,000
  $Down Payment = 0.10 \times 200000 = 20000$
  
  Step 2: Calculate the loan amount.
  Loan Amount = Total Price – Down Payment
  $Loan Amount = 200000 - 20000 = 180000$
  
  Answer (a): The loan amount will be $180,000.
2. What will your monthly payments be if the annual interest rate is 5%?
  Step 1: Identify the given values.
  - Loan Principal ( $P$ ): $180,000
  - Interest Rate ( $r$ ): 5% or 0.05
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 30 years
  Step 2: Use the loan payment formula and solve for pymt.
  $pymt = \frac{180000 \times (\frac{0.05}{12})}{(1 - {(1 + \frac{0.05}{12})}^{(- 12 \times 30)})} \approx 966.28$
  
  Answer (b): Your monthly payments will be approximately $966.28.
3. What will your monthly payments be if the annual interest rate is 6%?
  Step 1: Identify the given values.
  - Loan Principal ( $P$ ): $180,000
  - Interest Rate ( $r$ ): 6% or 0.06
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 30 years
  Step 2: Use the loan payment formula and solve for pymt.
  $pymt = \frac{180000 \times (\frac{0.06}{12})}{(1 - {(1 + \frac{0.06}{12})}^{(- 12 \times 30)})} \approx 1079.19$
  
  Answer (c): Your monthly payments will be approximately $1079.19.
Lynn bought a $300,000 house, paying 10% down, and financing the rest with an annual rate of 6% paid monthly for 30 years.
1. Find her monthly payments.
2. How much interest will she pay over the life of the loan?
Answer (click to Show/Hide)
1. Find her monthly payments.
  Step 1: Calculate the down payment and loan amount.
  - Down Payment = 10% of $300,000 = $30,000
  - Loan Principal ( $P$ ) = $300,000 – $30,000 = $270,000
  Step 2: Identify the given values for the loan.
  - Loan Principal ( $P$ ): $270,000
  - Interest Rate ( $r$ ): 6% or 0.06
  - Frequency of payments per year ( $n$ ): 12 (monthly)
  - Time ( $t$ ): 30 years
  Step 3: Use the loan payment formula and solve for pymt.
  The formula, solved for the payment, is $pymt = \frac{P \times (\frac{r}{n})}{(1 - {(1 + \frac{r}{n})}^{- n t})}$ .
  
  Now, substitute the values into the formula.
  $pymt = \frac{270000 \times (\frac{0.06}{12})}{(1 - {(1 + \frac{0.06}{12})}^{(- 12 \times 30)})} \approx 1618.79$
  
  Answer (a): Her monthly payments will be approximately $1,618.79.
2. How much interest will she pay over the life of the loan?
  Step 1: Calculate the total amount paid.
  Total Paid = Monthly Payment × Total Number of Payments
  $Total Paid = 1618.79 \times (12 \times 30) = 1618.79 \times 360 = 582764.40$
  
  Step 2: Calculate the total interest paid.
  Total Interest = Total Paid – Loan Principal
  $Total Interest = 582764.40 - 270000 = 312764.40$
  
  Answer (b): She will pay $312,764.40 in interest over the life of the loan.
Emile bought a car for $24,000 three years ago. The loan had a 5 year term with an annual rate of 4.8% paid monthly. How much does is still owed on the car?
Answer (click to Show/Hide)
To find the remaining balance on the loan, we first need to calculate the original monthly payment. Then, we can determine the present value of the remaining payments.

Part 1: Find the original monthly payment.

Step 1: Identify the given values for the original loan.
- Loan Principal ( $P$ ): $24,000
- Interest Rate ( $r$ ): 4.8% or 0.048
- Frequency of payments per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 5 years
Step 2: Use the loan payment formula to find the monthly payment.
$pymt = \frac{24000 \times (\frac{0.048}{12})}{(1 - {(1 + \frac{0.048}{12})}^{(- 12 \times 5)})} \approx 450.71$
The original monthly payment was $450.71.

Part 2: Find the remaining balance.

The remaining balance is the present value of the remaining loan payments.

Step 1: Identify the values for the remaining portion of the loan.
- Monthly Payment ( $pymt$ ): $450.71
- Interest Rate ( $r$ ): 4.8% or 0.048
- Frequency of payments per year ( $n$ ): 12 (monthly)
- Remaining Time ( $t$ ): The original term was 5 years, and 3 years have passed, so there are 2 years remaining.
Step 2: Use the loan principal formula to find the remaining balance.
$P = \frac{450.71 (1 - {(1 + \frac{0.048}{12})}^{- 12 \times 2})}{(\frac{0.048}{12})} \approx 10,294.44$

Answer: Emile still owes approximately $10,294.44 on the car.
A friend bought a house 15 years ago, taking out a $120,000 mortgage with an annual rate of 6% paid monthly for 30 years. How much does she still owe on the mortgage?
Answer (click to Show/Hide)
To find the remaining balance on the mortgage, we first need to calculate the original monthly payment. Then, we can find the present value of the remaining payments.

Part 1: Find the original monthly payment.

Step 1: Identify the given values for the original loan.
- Loan Principal ( $P$ ): $120,000
- Interest Rate ( $r$ ): 6% or 0.06
- Frequency of payments per year ( $n$ ): 12 (monthly)
- Time ( $t$ ): 30 years
Step 2: Use the loan payment formula to find the monthly payment.
$pymt = \frac{120000 \times (\frac{0.06}{12})}{(1 - {(1 + \frac{0.06}{12})}^{(- 12 \times 30)})} \approx 719.46$
The original monthly payment was $719.46.

Part 2: Find the remaining balance.

The remaining balance is the present value of the remaining loan payments.

Step 1: Identify the values for the remaining portion of the loan.
- Monthly Payment ( $pymt$ ): $719.46
- Interest Rate ( $r$ ): 6% or 0.06
- Frequency of payments per year ( $n$ ): 12 (monthly)
- Remaining Time ( $t$ ): The original term was 30 years, and 15 years have passed, so there are 15 years remaining.
Step 2: Use the loan principal formula to find the remaining balance.
$P = \frac{719.46 (1 - {(1 + \frac{0.06}{12})}^{- 12 \times 15})}{(\frac{0.06}{12})} \approx 85,258.54$

Answer: She still owes approximately $85,258.54 on the mortgage.
Taye is purchasing a used car for $9700 from a dealer. They offer an amortized loan that has an annual rate of 5.17% paid monthly. Determine the payments and cost of the loan for each of the following terms on the loan (2 years, 4 years, 6 years).
Answer (click to Show/Hide)
We will calculate the monthly payment and the cost of the loan for each of the three different loan terms.

Given Values for all Scenarios:
- Loan Principal ( $P$ ): $9,700
- Interest Rate ( $r$ ): 5.17% or 0.0517
- Frequency of payments per year ( $n$ ): 12 (monthly)
1. Term of 2 years
  Step 1: Calculate the monthly payment.
  Using the loan payment formula with $t = 2$ :
  $pymt = \frac{9700 \times (\frac{0.0517}{12})}{(1 - {(1 + \frac{0.0517}{12})}^{(- 12 \times 2)})} \approx 427.53$
  The monthly payment is approximately $427.53.
  
  Step 2: Calculate the cost of the loan.
  
  Total Paid = $427.53 × (12 × 2) = $10,260.72
  
  Cost of Loan = $10,260.72 – $9,700 = $560.72
  
  Answer (2 years): The monthly payment is $427.53 and the cost of the loan is $560.72.
2. Term of 4 years
  Step 1: Calculate the monthly payment.
  Using the loan payment formula with $t = 4$ :
  $pymt = \frac{9700 \times (\frac{0.0517}{12})}{(1 - {(1 + \frac{0.0517}{12})}^{(- 12 \times 4)})} \approx 224.75$
  The monthly payment is approximately $224.75.
  
  Step 2: Calculate the cost of the loan.
  
  Total Paid = $224.75 × (12 × 4) = $10,788.00
  
  Cost of Loan = $10,788.00 – $9,700 = $1,088.00
  
  Answer (4 years): The monthly payment is $224.75 and the cost of the loan is $1,088.00.
3. Term of 6 years
  Step 1: Calculate the monthly payment.
  Using the loan payment formula with $t = 6$ :
  $pymt = \frac{9700 \times (\frac{0.0517}{12})}{(1 - {(1 + \frac{0.0517}{12})}^{(- 12 \times 6)})} \approx 156.64$
  The monthly payment is approximately $156.64.
  
  Step 2: Calculate the cost of the loan.
  
  Total Paid = $156.64 × (12 × 6) = $11,278.08
  
  Cost of Loan = $11,278.08 – $9,700 = $1,578.08
  
  Answer (6 years): The monthly payment is $156.64 and the cost of the loan is $1,578.08.

A home is purchased for $230,000 with an amortized loan over 15 years at an interest rate of 4.4%. Complete an amortized loan schedule in a spread sheet and answer the following questions.

Find the balance owed after 5 years
Determine the cost of the loan (interest).
If an extra $100 a month is paid towards the balance find the new number of payments and the total cost of the loan.
If the interest rate of the loan is doubled to 8.8% does the cost of the loan double? Explain
If the principal of the loan is doubled to $460,000 does the cost of the loan double? Explain
If the term of the loan is doubled to 30 years does the cost of the loan double? Explain

Answer (click to Show/Hide)

First, we need to calculate the original monthly payment for the loan.

Original Loan Details:

Loan Principal ( $P$ ): $230,000
Interest Rate ( $r$ ): 4.4% or 0.044
Frequency of payments per year ( $n$ ): 12 (monthly)
Time ( $t$ ): 15 years

$Original Payment = \frac{230000 \times (\frac{0.044}{12})}{(1 - {(1 + \frac{0.044}{12})}^{(- 12 \times 15)})} \approx 1747.75$

Build the amortized loan schedule. Below is the first three rows to check your work:

Payment #	Payment	Interest	Principal Paid	Balance After Payment
1	$1,747.75	$843	$904.42	$229,095.58
2	$1,747.75	$840.02	$907.73	$228,187.85
3	$1,747.75	$836.69	$911.06	$227,276.79

Find the balance owed after 5 years
On the table we can find the balance remaining after five years by looking at the row for the 60th payment. Below is that row from the table:

Payment # Payment Interest Principal Paid Extra Payment Balance After Payment

60 $1,747.75 $625.35 $1122.4 $169,426.58

Answer (a): The balance after 5 years is approximately $169,426.58.
Determine the cost of the loan (interest).
Total Paid = $1747.75 × (12 × 15) = $314,595.00
Cost of Loan = $314,595 – $230,000 = $84,595
Alternatively the Cost of the Loan can be found by adding all interest payments. In the spread sheet this would add to be $84,596 (differnce due to rounding).
Answer (b): The total cost of the loan is $84,595.
If an extra $100 a month is paid…
The new payment would be $1847.75. This would reduce the loan term to approximately 167 months (13.9 years) instead of 180. In the 167th month a full payment is not needed as the balance is $1011.00, so we have 166 payments at $1847.75 and the last payment as the balance plus the interest for that one month: $1011.00 + .044/12*($1011.00)=$1014.71
New Total Paid ≈ $1847.75 × 166 + $1014.71 = $307,741.21

New Cost of Loan ≈ $307,741.21 – $230,000 = $77,741.21

This new cost of the loan matches up with adding all interest payments in the new amortized loan schedule when you add in the extra $100 payment.
Answer (c): The loan would be paid off faster, and the new cost of the loan would be approximately $77,741.21, a savings of about $7,000.
If the interest rate is doubled to 8.8% we will first need to find the new payment amount and change the interest rate in the amortized loan schedule.
New Payment ≈ $2,305.53

New Total Paid = $2,305.53 × 180 = $414,995.40

New Cost of Loan = $414,995.40 – $230,000 = $184,995.40
In the new amortized loan schedule if we add up all the interest we get a value of $184,995. The difference is due to slight rounding errors.
Answer (d): No, the cost of the loan more than doubles. It increases from ~$84.6k to ~$185k because interest compounds on a larger balance for a longer time.
If the principal is doubled to $460,000 we will go back to using the original interest rate and find the new monthly payments.
New Payment ≈ $3,495.51 (double the original payment found in part a)

New Total Paid = $3,495.51 × 180 = $629,191.80

New Cost of Loan = $629,191.80 – $460,000 = $169,191.80
Answer (e): Yes, the cost of the loan doubles ($84,595 × 2 = $169,190). This is because the payment amount scales linearly with the principal, and so does the total interest.
If the term is doubled to 30 years we first find the new monthly payment.
New Payment ≈ $1,151.75

New Total Paid = $1,151.475 × 360 = $414,630.00

New Cost of Loan = $414,630 – $230,000 = $184,630.00
Answer (f): No, the cost of the loan more than doubles when doubling the time (Original loan cost x 2 = $84,595 x 2 = $169,190). The longer term means interest accrues for many more years, significantly increasing the total interest paid, even though the monthly payment is lower.

You take out a new amortized loan for a home in the amount of $178,000 for 30 years at an annual rate of 5.15%. If you make an additional $50 in payments each month, then how much sooner do you pay off the loan?

Answer (click to Show/Hide)

First, we need to calculate the original monthly payment for the loan.

Original Loan Details:

Loan Principal ( $P$ ): $178,000
Interest Rate ( $r$ ): 5.15% or 0.0515
Frequency of payments per year ( $n$ ): 12 (monthly)
Time ( $t$ ): 30 years

$Original Payment = \frac{178000 \times (\frac{0.0515}{12})}{(1 - {(1 + \frac{0.0515}{12})}^{(- 12 \times 30)})} \approx 971.93$

Build the amortized loan schedule. Below is the first three rows to check your work:

Payment #	Payment	Interest	Principal Paid	Extra Payment	Balance After Payment
1	$971.93	$764	$208.01	$50	$177,741.99
2	$971.93	$762.81	$209.12	$50	$$177,482.87
3	$971.93	$761.70	$210.23	$50	$177,222.63

When you contiue to look down the table the last payment will be the 322nd payment. This means the loan is paid off in that 322nd month or 26 years and 10 months.

Answer: By paying an extra $50 each month, the loan would be paid off 38 months (3 years and 2 months) sooner.

Compare an Add-On Loan cost to an amortized loan cost for a $10,000 loan with an annual rate of 5% and a 10 year term.
Answer (click to Show/Hide)
To compare the costs, we need to calculate the total interest paid for both a simple interest add-on loan and a standard amortized loan.

Part 1: Simple Interest Add-On Loan

In an add-on loan, the total interest is calculated upfront using the simple interest formula and added to the principal.

Step 1: Identify the given values.
- Principal ( $P$ ): $10,000
- Interest Rate ( $r$ ): 5% or 0.05
- Time ( $t$ ): 10 years
Step 2: Calculate the total interest (the cost of the loan).
We use the simple interest formula $I = P r t$ .
$Interest = 10000 \times 0.05 \times 10 = 5000$
The cost of the add-on loan is $5,000.

Part 2: Amortized Loan

In an amortized loan, interest is calculated on the remaining balance with each payment.

Step 1: Calculate the monthly payment.
Using the same values as above, with $n = 12$ (monthly payments).
$pymt = \frac{10000 \times (\frac{0.05}{12})}{(1 - {(1 + \frac{0.05}{12})}^{(- 12 \times 10)})} \approx 106.07$
The monthly payment is approximately $106.07.

Step 2: Calculate the total amount paid.
The total number of payments is 10 years × 12 months/year = 120 payments.
$Total Paid = 106.07 \times 120 = 12728.40$

Step 3: Calculate the cost of the loan (total interest).
$Interest = 12728.40 - 10000 = 2728.40$
The cost of the amortized loan is approximately $2,728.40.

Conclusion:

The add-on loan costs $5,000 in interest, while the amortized loan costs only $2,728.40. The amortized loan is significantly cheaper because interest is calculated on a decreasing balance as the loan is paid down. In contrast, the add-on loan calculates interest on the full original principal for the entire 10-year term, making it much more expensive for the borrower.

Attributions

This page contains modified content from David Lippman, “Math In Society, 2nd Edition.” Licensed under CC BY-SA 4.0.
This page contains content by Robert Foth, Math Faculty, Pima Community College, 2021. Licensed under CC BY 4.0.
Portions of the exercise solutions in this answer key were generated with the assistance of Gemini, a large language model from Google.

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Amortized Loan Formula

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Calculating Remaining Loan Balance

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Amortized Loan Schedule

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