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The Basics of Loans

Describe various reasons for loans.

Learning Objectives

After completing this section, you should be able to:

  1. Describe various reasons for loans.
  2. Describe the terminology associated with loans.
  3. Understand how credit scoring works.
  4. Calculate the payment necessary to pay off a loan.
  5. Read an amortization table.
  6. Determine the cost to finance for a loan.

Reasons for Loans

Even if you want a new car because you need one, or if you need a new computer since your current one no longer runs as fast or smoothly as you would like, or you need a new chimney because the one on your house is crumbling, it’s likely you do not have that cost in cash. Those are very large purchases. How do you buy that if you don ’t have the cash?
You borrow the money.

And for helping you with your purchase, the company or bank charges you interest.

Loans are taken out to pay for goods or services when a person does not have the cash to pay for the goods or services. We are most familiar with loans for the big purchases in our lives, such as cars, homes, and a college education. Loans are also taken out to pay for repairs, smaller purchases, and home goods like furniture and computers.

Loans can come from a bank, or from the company selling the goods or providing the service. The borrower agrees to pay back more than the amount borrowed. So there is a cost to borrowing that should be considered when deciding on a purchase bought with credit or a borrowed money.
Even using a credit card is a form of a loan.

Essentially, a loan can be obtained for just about any purchase, large or small, that has a cost beyond a person’s cash on hand.

The Terminology of Loans

There are many words and acronyms that get used in relation to loans. A few are below.

APR is the annual percentage rate. It is the annual interest paid on the money that was borrowed. The principal is the total amount of the loan, or that has been financed. A fixed interest rate loan has an interest rate that does not change during the life of the loan. A variable interest rate loan has an interest rate that may change during the life of the loan. The term of the loan is how long the borrower has to pay the loan back. An installment loan is a loan with a fixed period, and the borrower pays a fixed amount per period until the loan is paid off. The periods are almost uniformly monthly. Loan amortization is the process used to calculate how much of each payment will be applied to principal and how much is applied to interest. Revolving credit, also known as open-end credit, is how most credit cards work but is also a kind of loan account. (We will learn about credit cards in Credit Cards) You can use up to some specified value, called the limit, any way you want, and as long as you pay the issuer of the credit according to their terms, you can keep borrowing from this account.

These and other terminologies can be researched further at Forbes.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Calculating Loan Payments

Loan payments are made up of two components. One component is the interest that accrued during the payment period. The other component is part of the principal. This should remind you of partial payments from Simple Interest.

Over the course of the loan, the amount of principal remaining to be paid decreases. The interest you pay in a month is based on the remaining principal, just as in the partial payments of Simple Interest.

Interest for a Monthly Payment of a Loan

Try it.

Find the interest to be paid for the period on loans with the following remaining principal and given annual interest rate. Each period is a month.

  1. Remaining principal is $13,450, interest rate is 6.75%
  2. Remaining principal is $8,460, interest rate is 5.99%
Solution
  1. Substituting $13,450 for the remaining principal \(P\), 0.0675 for \(r\), and \(n\) = 12 since the period is a month into the formula, we find that the interest to be paid this period is \(I=P\times \frac{r}{n}=\text{\$}13,450\times \frac{0.0675}{12}=\text{\$}75.66\).
  2. Substituting $8,460 for the remaining principal \(P\), 0.0599 for \(r\), and \(n\) = 12 since the period is a month into the formula, we find that the interest to be paid this period is \(I=P\times \frac{r}{n}=\text{\$}8,450\times \frac{0.0599}{12}=\text{\$}42.18\).

The payment of the loan has to be such that the principal of the loan is paid off with the last payment. In any period, the amount of interest is defined by the formula above, but changes from period to period since the principal is decreasing with each payment. The trick is knowing how much principal should be paid each payment so that the loan is paid off at the stated time. Fortunately, that is found using the following formula.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Reading Amortization Tables

An amortization table or amortization schedule is a table that provides the details of the periodic payments for a loan where the payments are applied to both the principal and the interest. The principal of the loan is paid down over the life of the loan. Typically, the payments each period are equal. Importantly, one of the columns will show how much of each payment is used for interest, another column shows how much is applied to the outstanding principal, and another column shows the remaining principal or balance .

Reading from an Amortization Table

Try it.

Using the partial amortization table (Figure 6.23), answer the following questions.

  1. What is the loan amount (principal), the interest rate, and the term of the loan?
  2. How much is the monthly payment?
  3. How much remaining balance is there after the payment in month 15?
  4. How much was the interest in payment 10?
  5. What is the total of the interest paid after payment 18?
  6. What happens to the amount paid in interest each month?
Solution
  1. Reading the values at the top of the table, we see the principal is $10,000, the interest rate is 4.75%, and has a term of 20 years.
  2. The monthly payment is listed below the term of the loan, and is $64.62.
  3. $9,613.83
  4. $38.68
  5. $697.01
  6. The amount paid to interest decreases each month.

Cost of Finance

There are often costs associated with a loan beyond the interest being paid. The cost of finance of a loan is the sum of all costs, fees, interest, and other charges paid over the life of the loan.

Cost of Financing a Personal Loan

Try it.

Irena signed for a loan of $15,000 at 6.33% for 5 years. When she took out the loan, Irena paid a $750 origination fee. Over the course of the loan, she pays $2,537.96 in interest. What was her cost to finance the loan?

Solution

The cost of finance is the sum is all interest and any fees paid for the loan. The fees paid were $750.00 and the interest was $2,537.96. Her cost of finance for this loan was $3,287.96.

Key Concepts

  • There are many reasons for a loan, but primarily it is taken out for a large expense when cash is not available.
  • Each payment for an installment loan consists of an interest portion and a principal portion.
  • There is a formula to calculate the payment necessary to pay off a loan in installments.
  • Amortization schedules, or tables, show how each payment is applied to principal and interest. It also includes other details such as remaining balance and total interest paid.
  • Loans often have other fees associated with them such as origination fees or application fees. The total of the interest paid and the fees is the cost of finance.

Practice (4)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Find the interest to be paid for the period on loans with the following remaining principal and given annual interest rate. Each period is a month.

    1. Remaining principal is $13,450, interest rate is 6.75%
    2. Remaining principal is $8,460, interest rate is 5.99%
    Jawaabta muuji
    1. Substituting $13,450 for the remaining principal \(P\), 0.0675 for \(r\), and \(n\) = 12 since the period is a month into the formula, we find that the interest to be paid this period is \(I=P\times \frac{r}{n}=\text{\$}13,450\times \frac{0.0675}{12}=\text{\$}75.66\).
    2. Substituting $8,460 for the remaining principal \(P\), 0.0599 for \(r\), and \(n\) = 12 since the period is a month into the formula, we find that the interest to be paid this period is \(I=P\times \frac{r}{n}=\text{\$}8,450\times \frac{0.0599}{12}=\text{\$}42.18\).
  2. In the following, calculate the payment necessary to pay off the loan with the given details. The payments are monthly.

    1. A car loan taken out for $28,500 at an annual interest rate of 3.99% for 5 years.
    2. A home loan taken out for $136,700 and an annual interest rate of 5.75% for 15 years.
    Jawaabta muuji
    1. The loan is for $28,500, which is the principal. The rate is 3.99%, so \(r\) = 0.0399. The term of the loan is 5 years, so \(t\) =5. Monthly payments means \(n\) = 12. Substituting these values for \(P\), \(r\), \(n\), and \(t\) into the formula \(pmt=\frac{P\times (r/n)\times {(1+r/n)}^{n\times t}}{{(1+r/n)}^{n\times t}-1}\) and calculating, we find the payment for the loan.
      \[\begin{array}{lll}pmt & = & \frac{P\times (r/n)\times {(1+r/n)}^{n\times t}}{{(1+r/n)}^{n\times t}-1} \\ & = & \frac{\text{\$}28,500\times (0.0399/12)\times {(1+0.399/12)}^{12\times 5}}{{(1+0.0399/12)}^{12\times 5}-1} \\ & = & \frac{\text{\$}28,500\times (0.003325)\times {(1.003325)}^{60}}{{(1.003325)}^{60}-1} \\ & = & \frac{\text{\$}115.6470437}{0.220388273} \\ & = & \text{\$}524.75\end{array}\]
      The monthly payment needed is $524.75.
    2. The loan is for $136,000, which is the principal \(P\). The rate is 5.75% so \(r\) = 0.0575. The term of the loan is 15 years, so \(n\) = 15. Monthly payments mean \(n\) =12. Substituting these values for \(P\), \(r\), \(n\), and \(t\) into the formula \(pmt=\frac{P\times (r/n)\times {(1+r/n)}^{n\times t}}{{(1+r/n)}^{n\times t}-1}\) and calculating, we find the payment for the loan.
      \[\begin{array}{lll}pmt & = & \frac{P\times (r/n)\times {(1+r/n)}^{n\times t}}{{(1+r/n)}^{n\times t}-1} \\ & = & \frac{\text{\$}136,700\times (0.0575/12)\times {(1+0.0575/12)}^{12\times 15}}{{(1+0.0575/12)}^{12\times 15}-1} \\ & = & \frac{\text{\$}28,500\times (0.0047917)\times {(1.0047917)}^{180}}{{(1.003325)}^{180}-1} \\ & = & \frac{\text{\$}1,548.600986}{1.364201118} \\ & = & \text{\$}1,135.18\end{array}\]
      The monthly payment needed is $1,135.18.
  3. Using the partial amortization table (Figure 6.23), answer the following questions.

    1. What is the loan amount (principal), the interest rate, and the term of the loan?
    2. How much is the monthly payment?
    3. How much remaining balance is there after the payment in month 15?
    4. How much was the interest in payment 10?
    5. What is the total of the interest paid after payment 18?
    6. What happens to the amount paid in interest each month?
    Jawaabta muuji
    1. Reading the values at the top of the table, we see the principal is $10,000, the interest rate is 4.75%, and has a term of 20 years.
    2. The monthly payment is listed below the term of the loan, and is $64.62.
    3. $9,613.83
    4. $38.68
    5. $697.01
    6. The amount paid to interest decreases each month.
  4. Irena signed for a loan of $15,000 at 6.33% for 5 years. When she took out the loan, Irena paid a $750 origination fee. Over the course of the loan, she pays $2,537.96 in interest. What was her cost to finance the loan?

    Jawaabta muuji

    The cost of finance is the sum is all interest and any fees paid for the loan. The fees paid were $750.00 and the interest was $2,537.96. Her cost of finance for this loan was $3,287.96.

Symbols used here

\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\approx
approximately equal
Equal to the precision shown, not exactly.
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
a \bmod n
remainder
What is left after dividing a by n.
\%
per cent
Per hundred: 15% = 15/100.
a : b,\ \frac{a}{b}
ratio, fraction
a for every b; a divided by b.

How to: The Basics of Loans

  1. Describe various reasons for loans.
  2. Describe the terminology associated with loans.
  3. Understand how credit scoring works.
  4. Calculate the payment necessary to pay off a loan.
  5. Read an amortization table.
  6. Determine the cost to finance for a loan.
  7. Remaining principal is $13,450, interest rate is 6.75%
  8. Remaining principal is $8,460, interest rate is 5.99%

Questions people ask

Why does the order of operations matter?

Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.

How do I check an arithmetic answer?

Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.

Why are fractions harder than decimals?

They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.

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Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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