5.1: Simple Interest
- Page ID
- 106253
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Money is not free to borrow! When you borrow money, the cost you pay to borrow that money is called interest. The amount of money borrowed is known as principal or present value, \(P\).
At some point in the future, you pay back the amount you borrowed originally as well as the interest, \(I\), that was charged. The total amount paid back is known as future value, \(F\). We can summarize this with the formula \(F=P+I\).
The same happens when you make an investment. You deposit a certain amount of money (the present value, \(P\)) that accumulates interest \(I\) over time. In the future you withdraw the accumulated amount -- both your original principal as well the as interest earned. Again, \(F=P+I\).
The interest you pay or earn can be calculated in different ways. In this section we consider simple interest. Simple interest is interest that is calculated only on the principal, \(P\). This means you are paying or earning the same amount of interest every period. An example of simple interest is when someone purchases a U.S. Treasury Bond.
Simple interest is interest that is only paid on the principal:
\(\text{simple interest} = \text{principal} \times \text{rate} \times \text{time}\)
Sue borrows $2,000 at 5% annual simple interest from her bank. How much does she owe after five years?
Solution
Each year, Sue will pay 5% (or 0.05) of the principal that she borrowed as the cost for borrowing the money. The balance that Sue owes grows constantly by $100 each year because each year Sue is paying interest only the original $2,000 she borrowed.
| Year | Interest Owed | Total Balance Owed |
|---|---|---|
| 1 | \($2000 \times 0.05 \times 1 = $100\) | \($2000 + $100 = $2100\) |
| 2 | \($2000 \times 0.05 \times 1 = $100\) | \($2100 + $100 = $2200\) |
| 3 | \($2000 \times 0.05 \times 1 = $100\) | \($2200 + $100 = $2300\) |
| 4 | \($2000 \times 0.05 \times 1 = $100\) | \($2300 + $100 = $2400\) |
| 5 | \($2000 \times 0.05 \times 1 = $100\) | \($2400 + $100 = $2500\) |
After 5 years, Sue owes $2,500. The total accumulated interest during these 5 years was \($2000 \times 0.05 \times 5 = $500\).
This leads us to the simple interest formulas.
\(I = Prt\) \(F = P + I\) \(F = P(1+rt)\)
where,
- \(I\) is the interest accumulated over \(t\) years
- \(P\) is the present value, or principal
- \(r\) is the annual percentage rate (APR) written as a decimal
- \(t\) is number of years
- \(F\) is the future value
Chad got a student loan for $10,000 at 8% annual simple interest. How much does he owe after one year? How much interest will he pay for that one year?
Solution
In this scenario, \(P = $10,000\), \(r = 0.08\), and \(t = 1 \text{ year}\). We substitute these values into the third formula to find the future value:
\(F = P(1+rt)\)
\(F = 10,000(1+0.08 \times 1) = $10,800\)
Chad owes $10,800 after one year. Using the second formula, he will pay \(F-P=$10,800 - $10,000 = $800\) in interest.
Notice that you could have also used the first formula to find the amount Chad paid in interest: \(I=Prt = 10,000 \times 0.08 \times 1= $800\).
Try it Now 1
You borrowed $6,000 from a family member at an APR of 6% for 5 years to start a business. How much do you need to repay after 5 years?
- Answer
- $7,800.00
The formulas for \(I\) and \(F\) only work if time \(t\) and interest rate \(r\) are expressed using the same time units. For example, if interest rate is expressed as an annual rate then time \(t\) must be expressed in years. But, if the interest rate is 2% per month then time \(t\) should be expressed in months.
Typically, interest rate is given as an annual rate unless otherwise stated. But, sometimes the interest rate can be given monthly or even daily.
If time \(t\) and interest rate \(r\) are expressed in different time units, convert them to the same time units. It is always easier to convert \(t\) to the units in which \(r\) is given. We all know that number of days varies from to month and from year to year. To simplify this matter, it is common to use the following correspondence in financial math calculations:
- 1 month = 30 days
- 1 year = 360 days
- 1 year = 52 weeks
- I year = 12 month
The following two examples illustrate conversion of time units.
Aysha borrowed $4,000 from the bank for a term of 6 months at a 6.5% annual interest rate. How much did Aysha have to pay for the use of the money? How much did she repay to the bank on the due date of the loan?
Solution
In this scenario, the 6.5% interest rate is given as an annual (yearly) rate and time is given as 6 months. Therefore, time must be converted to years to match the unit on the rate: \(t = 6 \text{ months} = \frac{6}{12} = 0.5 \text{ year}\).
Then, we can find the accumulated interest to answer the first question using \(P = $4,000\), \(r = 0.065\), and \(t = 0.5\):
\(I=Prt = 4,000 \times 0.065\times 0.5 = $130\)
Aysha repays the future value \(F = P + I = $4,000 + $130 = $4,130\).
Notice that you could also find the future value using the formula: \(F = 10,000(1+0.065 \times 0.5) = $4,130\).
Try it Now 2
Suppose a loan is taken where \(P = $587\), \(r = 6.05 \text{% annually}\), and \(t = 90 \text{ days}\). Find the accumulated interest \(I\).
- Answer
- \(I =$8.88 \)
Anna pawned her ring for $300. To get her ring back in 40 days she needs to pay back $355. What is the annual interest rate charged by the pawn shop?
Solution
In this scenario, the future value \(F\) is known, but the interest rate \(r\) is unknown. We have \(P =$300\), \(F = $355\), and \(t = 40\) days.
First, find the amount of accumulated interest that Anna will pay: \(I = F – P = $355 - $300 = $55\).
Because we are looking for the annual interest rate, convert time to years: \(t = 40\) days = \(= \frac{40}{360} = \frac{1}{9}\) year.
Solve the equation to find the rate \(r\):
\(I = Prt\)
\(55 = 300 \cdot r \cdot \frac{1}{9}\)
\(55 \cdot 9 = 300 \cdot r\)
\(495 = 300 \cdot r\)
\(r =\frac{495}{300} = 1.65 = 165\%\)
Try it Now 3
Find \(P\) when \(r = 2.1\% \text{ annually}\), \(t = 135 \text{ days}\), and \(I = $37.80\).
- Answer
- \(P=$4,800\)