8.2: Simple and Compound Interest
- Page ID
- 84631
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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! We will refer to money in terms of present value P, which is an amount of money at the present time, and future value F, which is an amount of money in the future. Usually, if someone loans money to another person in present value, and are promised to be paid back in future value, then the person who loaned the money would like the future value to be more than the present value. That is because the value of money declines over time due to inflation. Therefore, when a person loans money, they will charge interest. They hope that the interest will be enough to beat inflation and make the future value more than the present value.
Simple Interest
Simple interest is interest that is only calculated on the initial amount of the loan (present value, P). This means you are paying the same amount of interest every year. An example of simple interest is when someone purchases a U.S. Treasury Bond.
To calculate simple interest only, \[I = Prt\]
where,
- \(P\) is the Present value
- \(r\) is the Annual percentage rate (APR) changed to a decimal
- \(t\) is the Number of years
To calculate the future value based on simple interest, \[F = P(1+rt)\]
where,
- \(F\) is the Future value
- \(P\) is the Present value
- \(r\) is the Annual percentage rate (APR) changed to a decimal
- \(t\) is the Number of years
Sue borrows $2000 at 5% annual simple interest from her bank. How much does she owe after five years?
Solution
|
Year |
Interest Earned |
Total Balance Owed |
|---|---|---|
|
1 |
$2000*.05 = $100 |
$2000 + $100 = $2100 |
|
2 |
$2000*.05 = $100 |
$2100 + $100 = $2200 |
|
3 |
$2000*.05 = $100 |
$2200 + $100 = $2300 |
|
4 |
$2000*.05 = $100 |
$2300 + $100 = $2400 |
|
5 |
$2000*.05 = $100 |
$2400 + $100 = $2500 |
After 5 years, Sue owes $2500.
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
P = $10,000, r = 0.08, t = 1
\[F = P(1+rt)\]
\[F = 10000(1+0.08(1)) = $10,800\]
Chad owes $10,800 after one year. He will pay $10800 - $10000 = $800 in interest.
Carlos deposits $20,000 into a savings account earning 7.25% annual simple interest. How much does he have in the account after 6 years? What was the total interest earned?
Solution
P = $20,000, r = 0.0725, t = 6
\[F = P(1+rt)\]
\[F = 20000(1+0.0725(6)) = $28,700\]
Carlos has $28,700 in his account after 6 years. He earned $28,700 - $20000 = $8,700 in interest.
Ben wants to buy a used car. He has $3000 but wants $3500 to spend. He invests his $3000 into an account earning 6% annual simple interest. How long will he need to leave his money in the account to accumulate the $3500 he wants?
Solution
F = $3500, P = $3000, r = 0.06
\[F = P(1+rt)\]
\[\begin{align*} 3500 &= 3000(1+0.06t) \\ \dfrac{3500}{3000} &= 1 + 0.06t \\ \dfrac{3500}{3000} - 1 &= 0.06t \\ \dfrac{\dfrac{3500}{3000}-1}{0.06} &= t \end{align*}\]
\[t\approx 2.8 \text{ years}\nonumber\]
Ben would need to invest his $3000 for about 2.8 years until he would have $3500 to spend on a used car.
Note: As shown above, wait to round your answer until the very last step so you get the most accurate answer.
Compound Interest
Compound interest is interest paid both on the original principal and on all interest that has been added to the original principal.
Most banks, loans, credit cards, etc. charge you compound interest, not simple interest. This is interest paid on the principal AND the interest accrued. Interest on a mortgage or auto loan is compounded monthly. Interest on a savings account can be compounded quarterly (four times a year). Interest on a credit card can be compounded weekly or daily!
|
Compounding type |
Number of compounding periods per year, m |
|---|---|
|
Annually |
1 |
|
Semiannually |
2 |
|
Quarterly |
4 |
|
Monthly |
12 |
|
Daily |
365 |
Suppose you invest $3000 into an account that pays you 7% interest per year for four years. Using compound interest, after the interest is calculated at the end of each year, then that amount is added to the total amount of the investment. Then the following year, the interest is calculated using the new total of the loan.
|
Year |
Interest Earned |
Total of Loan |
|---|---|---|
|
1 |
$3000*0.07 = $210 |
$3000 + $210 = $3210 |
|
2 |
$3210 *0.07 = $224.70 |
$3210 + $224.70 = $3434.70 |
|
3 |
$3434.70*0.07 = $240.43 |
$3434.70 + $240.43 = $3675.13 |
|
4 |
$3675.13 *0.07 = $257.26 |
$3675.13 + $257.26 = $3932.39 |
|
Total |
$932.39 |
So, after four years, you have earned $932.39 in interest for a total of $3932.39.
\[F=P\left(1+\frac{r}{m}\right)^{mt}\]
where
- F = Future value
- P = Present value
- r = Annual percentage rate (APR) changed into a decimal
- t = Number of years
- m = Number of compounding periods per year
Let’s compare a savings plan that pays 6% simple interest versus another plan that pays 6% annual interest compounded quarterly. If we deposit $8,000 into each savings account, how much money will we have in each account after three years?
Solution
6% Simple Interest: P = $8,000, r = 0.06, t = 3
\[F = P(1+rt)\]
\[F = 8000(1+0.06(3))\]
\[F = 9440\]
Thus, we have $9440.00 in the simple interest account after three years.
6% Interest Compounded Quarterly: P = $8,000, r = 0.06, t = 3, m=4
\[F=P\left(1+\frac{r}{m}\right)^{mt}\]
\[F=8000\left(1+\frac{0.06}{4}\right)^{4(3)}\]
\[F=8000\left(1+\frac{0.06}{4}\right)^{12}\]
\[F = 9564.95\]
So, we have $9564.95 in the compounded quarterly account after three years.
With simple interest we earn $1440.00 on our investment, while with compound interest we earn $1564.95.
In comparison with Example \(\PageIndex{6}\) consider another account with 6% interest compounded monthly. If we invest $8000 in this account, how much will there be in the account after three years?
Solution
P = $8,000, r = 0.06, t = 3, m = 12
\[F=P\left(1+\frac{r}{m}\right)^{mt}\]
\[F=8000\left(1+\frac{0.06}{12}\right)^{12(3)}\]
\[F=8000\left(1+\frac{0.06}{12}\right)^{36}\]
\[F = 9573.44\]
Thus, we will have $9573.44 in the compounded monthly account after three years.
Interest compounded monthly earns you $9573.44 - $9564.95 = $8.49 more than interest compounded quarterly.
Sophia’s grandparents bought her a savings bond for $200 when she was born. The interest rate was 3.28% compounded semiannually, and the bond would mature in 30 years. How much will Sophia’s bond be worth when she turns 30?
Solution
P = $200, r = 0.0328, t = 30, m = 2
\[F=P\left(1+\frac{r}{m}\right)^{mt}\]
\[F=200\left(1+\frac{0.0328}{2}\right)^{2(30)}\]
\[F=8000\left(1+\frac{0.0328}{2}\right)^{60}\]
\[F = 530.77\]
Sophia’s savings bond will be worth $530.77 after 30 years.
Suppose you know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?
Solution
F = $40,000, r = 0.04, t = 18, m = 4
In this case, we’re going to have to set up the equation, and solve for P.
\[40000=P\left(1+\frac{0.04}{4}\right)^{18(4)}\]
\[40000=P(1.01)^{72}\]
\[40000=P(2.0471)\]
\[P=\frac{40000}{2.0471}=\$ 19539.84\]
So you would need to deposit $19,539.84 now to have $40,000 in 18 years.