Compound Interest

Practice (10)

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

1. Abena invests $1,000 in a CD (certificate of deposit) earning 4% compounded annually. How much will Abena’s CD be worth after 3 years?

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   Since the interest is compounded annually, the interest will be computed at the end of each year and added to the CD’s value. The interest at the end of the following year will be based on the value found form the previous year.

   Step 1: After the first year, the interest in Abena’s CD is computed using the interest formula \(I=P\times r\times t\). The principal is \(P\) = 1,000, the rate, as a decimal, is 0.04, and the time is one year, so \(t\) = 1. Using that, the interest earned in the first year is \(I=P\times r\times t=1,000\times 0.04\times 1=40\), so the interest earned in the first year was $40.00. This is added to the value of the CD, making the CD worth \(\text{\$}1,000+\text{\$}40=\text{\$}1,040\).

   Step 2: At the end of the second year, interest is again computed, but is computed based on the CD’s new value, $1,040. Using this new value and the interest formula (\(r\) and \(t\) are still 0.04 and 1, respectively), we see that the CD earned \(I=P\times r\times t=1,040\times 0.04\times 1=41.6\), or $41.60. This is added to the value of the CD, making the CD now worth \(\text{\$}1,040.00+\text{\$}41.60=\text{\$}1,081.60\).

   Step 3: At the end of the third year, interest is again computed, but is computed based on Abena’s CD’s new value, $1,081.60. Using this value and the interest formula (\(r\) and \(t\) are still 0.04 and 1, respectively), we see that the CD earned \(I=P\times r\times t=1,081.60\times 0.04\times 1=43.264\), or $43.26 (remember to round down). This is added to the value of the CD, making the CD now worth \(\text{\$}1,081.60+\text{\$}43.26=\text{\$}1,124.86\).

   After 3 years, Abena’s CD is worth $1,124.86.

2. Abena invested $1,000 in a CD that earned 4% compounded annually, and the CD was worth $1,124.86 after 3 years. Had Abena invested in a CD with simple interest, how much would the CD have been worth after 3 years? How much more did Abena earn using compound interest?

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   Had Abena invested $1,000 in a 4% simple interest CD for 3 years, her CD would have been worth \(P+P\times r\times t=1,000+1,000\times 0.04\times 3=1,120\), or $1,120.00. With interest compounded annually, Abena’s CD was worth $1,124.86. The difference between compound and simple interest is \(\text{\$}1,124.86-\text{\$}1,120.00=\text{\$}4.86\). So compound interest earned Abena $4.86 more than the simple interest did.

3. In the following, compute the future value of the investment with the given conditions.

   1. Principal is $5,000, annual interest rate is 3.8%, compounded monthly, for 5 years.
   2. Principal is $18,500, annual interest rate is 6.25%, compounded quarterly, for 17 years.
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   1. The principal is \(P\) = $5,000, interest rate, in decimal form, \(r\) = 0.038, compounded monthly so \(n\) = 12, and for \(t\) = 5 years. Substituting these values into the formula, we find \[\begin{array}{lll}A & = & P{(1+\frac{r}{n})}^{nt}=5,000{(1+\frac{0.038}{12})}^{12\times 5} \\ & = & 5,000{(1+0.0031\overset{\bar}{6})}^{60} \\ & = & 5,000{(1.0031\overset{\bar}{6})}^{40} \\ & = & 5,000\times 1.20888663572 \\ & = & 6,044.4332\end{array}\]
      The future value of the investment is $6,044.43.
      

   1. The principal is \(P\) = $18,500, interest rate, in decimal form, \(r\) = 0.0625, compounded quarterly so \(n\) = 4, and for \(t\) = 17 years. Substituting these values into the formula, we \[\begin{array}{lll}A & = & P{(1+\frac{r}{n})}^{nt}=18,500{(1+\frac{0.0625}{4})}^{4\times 17} \\ & = & 18,500{(1+0.015625)}^{68} \\ & = & 18,500{(1.015625)}^{68} \\ & = & 18,500\times 2.86992151999 \\ & = & 53,093.5481\end{array}\] The future value of the investment is $53,093.54.

4. Cody invests $7,500 in an account that earns 4.5% interest compounded quarterly (4 times per year). Determine the value of Cody’s investment after 10 years.

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   Cody’s initial investment is $7,500, so \(P\) = $7,500. The annual interest rate is 4.5%, which is 0.045 in decimal form. Compounding quarterly means there are four periods in a year, so \(n\) = 4. He invests the money for 10 years. Substituting those values into the formula, we calculate

   \[\begin{array}{lll}A & = & P{(1+\frac{r}{n})}^{nt}=7,500{(1+\frac{0.045}{4})}^{4\times 10} \\ & = & 7,500{(1+0.01125)}^{40} \\ & = & 7,500{(1.01125)}^{40} \\ & = & 7,500\times 1.564376865391 \\ & = & 11,732.8265\end{array}\]

   After 10 years, Cody’s initial investment of $7,500 is worth $11,732.82.

5. Kathy invests $10,000 in an account that yields 5.6% compounded daily. How much money will be in her account after 20 years?

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   Kathy’s initial investment is $10,000, so \(P\) = $10,000. The annual interest rate is 5.6%, which is 0.056 in decimal form. Compounding daily means there are 364 periods in a year, so \(n\) = 365. She invests the money for 20 years, so \(t\) = 20. Substituting those values into the formula, we calculate

   \[\begin{array}{lll}A & = & P{(1+\frac{r}{n})}^{nt}=10,000{(1+\frac{0.056}{365})}^{365\times 20} \\ & = & 10,000{(1+0.000153424657534)}^{7300} \\ & = & 10,000{(1.000153424657534)}^{7300} \\ & = & 10,000\times 3.06459091598 \\ & = & 30,645.909\end{array}\]

   After 20 years, Kathy’s initial investment of $10,000 is worth $30,645.90.

6. Find the present value of the accounts under the following conditions.

   1. \(A\) = $250,000, invested at 6.75 interest, compounded monthly, for 30 years.
   2. \(A\) = $500,000, invested at 7.1% interest, compounded quarterly, for 40 years.
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   1. To reach a final account value of \(A\) = $250,000, invested at 6.75% interest, in decimal form \(r\) = 0.0675 (decimal form!), compounded monthly, so \(n\) = 12, for 30 years, substitute those values into the formula for present value. Calculating, we find the present value of the $250,000.
      \[\begin{array}{lll}PV & = & \frac{A}{{(1+\frac{r}{n})}^{n\times t}}=\frac{250,000}{{(1+\frac{0.0675}{12})}^{12\times 30}} \\ & = & \frac{250,000}{{(1+0.005625)}^{360}} \\ & = & \frac{250,000}{{(1.005625)}^{360}} \\ & = & \frac{250,000}{7.5332454772} \\ & = & 33,186.2277\end{array}\]
      In order for this account to reach $250,000 after 30 years, $33,186.23 needs to be invested.
      

   1. To reach a final account value of \(A\) = $500,000, invested at 7.1% interest, in decimal form \(r\) = 0.071, compounded quarterly, so \(n\) = 4, for 40 years, substitute those values into the formula for present value. Calculating, we find the present value of the $500,000.
      \[\begin{array}{lll}PV & = & \frac{A}{{(1+\frac{r}{n})}^{n\times t}}=\frac{500,000}{{(1+\frac{0.071}{4})}^{4\times 40}} \\ & = & \frac{500,000}{{(1+0.01775)}^{160}} \\ & = & \frac{500,000}{{(1.01775)}^{160}} \\ & = & \frac{500,000}{16.6946672846} \\ & = & 29,949.6834\end{array}\]
      In order for this account to reach $500,000 after 40 years, $29,949.69 needs to be invested.

7. Pilar plans early for retirement, believing she will need $1,500,000 to live comfortably after the age of 67. How much will she need to deposit at age 23 in an account bearing 6.35% annual interest compounded monthly?

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   Knowing how much to deposit at age 23 to reach a certain value later is a present value question. The target value for Pilar is $1,500,000. The interest rate is 6.35%, which in decimal form is 0.0635. Compounded monthly means \(n\) = 12. She’s 23 and will leave the money in the account until the age of 67, which is 44 years, making \(t\) = 44. Using this information and substituting in the formula for present value, we calculate

   \[\begin{array}{lll}PV & = & \frac{A}{{(1+\frac{r}{n})}^{n\times t}}=\frac{1,500,000}{{(1+\frac{0.0635}{12})}^{12\times 44}} \\ & = & \frac{1,500,000}{{(1+0.005291\overset{\bar}{6})}^{528}} \\ & = & \frac{1,500,000}{{(1.005291\overset{\bar}{6})}^{528}} \\ & = & \frac{1,500,000}{16.226302189} \\ & = & 92,442.5037\end{array}\]

   Pilar will need to invest $92,442,51 in this account to have $1,500,000 at age 67.

8. Suppose you have an investment paying a rate of 6% compounded quarterly. Determine and interpret that effective annual yield of the investment.

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   Here, \(n\) = 4 (quarterly) and \(r\) = 0.06 (decimal form). Substituting into the formula we find that the effective annual yield is

   \[\begin{array}{lll}Y & = & {(1+\frac{0.06}{4})}^{4}-1 \\ & = & {(1.015)}^{4}-1 \\ & = & 1.06136-1 \\ & = & 0.0614 \\ & = & 6.14\%\end{array}\]

   Therefore, a rate of 6% compounded quarterly is equivalent to a simple interest rate of 6.14%.

9. Calculate and interpret the effective annual yield on a deposit earning interest at a rate of 5% compounded daily.

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   In this case, the rate is \(r\) = 0.05 and \(n\) = 365 (daily). Using the formula \(Y={(1+\frac{r}{n})}^{n}-1\), we have

   \[\begin{array}{lll}Y & = & {(1+\frac{0.05}{365})}^{365}-1 \\ & = & {(1.0001369863)}^{365}-1 \\ & = & 1.051267-1 \\ & = & 0.0513\end{array}\]

   This tells us that an account earning 5% compounded daily is equivalent to earning 5.13% as simple interest.

10. Minh has a choice of banks in which he will open a savings account. He will deposit $3,200 and he wants to get the best interest he can. The banks advertise as follows:

    Bank	Interest Rate
    ABC Bank	2.08% compounded monthly
    123 Bank	2.09% compounded annually
    XYZ Bank	2.05% compounded daily

    Which bank offers the best interest?

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    To compare these directly, Minh could change each interest rate to its effective annual yield, which would allow direct comparison between the rates. Computing the effective annual yield for all three choices gives:

    ABC Bank: \(Y={(1+\frac{0.0208}{12})}^{12}-1=0.0210=2.10\%\)

    123 Bank: \(Y={(1+\frac{0.0209}{1})}^{1}-1=0.0209=2.09\%\)

    XYZ Bank: \(Y={(1+\frac{0.0205}{365})}^{365}-1=0.0207=2.07\%\)

    ABC Bank has the highest effective annual yield, so Minh should choose ABC bank.

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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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