6.2 Compound Interest
Why does compound interest matter so much? It is one of the foundational ideas in finance. Savings accounts, certificates of deposit, retirement accounts, student loans, mortgages, credit cards, inflation, and long-term investing all depend on the fact that money can earn interest on past interest. Once you understand compound interest, you understand the basic engine behind how money grows—and how debt grows too.
- Find the future value of a lump-sum.
- Find the present value of a lump-sum.
- Find the effective interest rate.
Simple interest grows in a straight-line way because interest is earned only on the original principal. Compound interest grows faster because each new interest payment becomes part of the balance and starts earning interest too. That is why compound growth is so powerful in both investing and borrowing.
Introduction
In the last section, we examined problems involving simple interest. Simple interest is generally charged when the lending period is short and often less than a year. When money is loaned or borrowed for a longer time period, and interest is paid or charged not only on the principal but also on past interest, we say the interest is compounded.
Compound Interest
Compound interest means that interest is earned or charged on both the original principal and on interest that has already been added to the account.
Suppose we deposit \$200 in an account that pays \(8\%\) interest. At the end of one year, we will have
$$ 200 + 200(0.08) = 200(1 + 0.08) = 216. $$Now suppose we leave this amount, \$216, in the same account. After another year, we will have
$$ 216 + 216(0.08) = 216(1 + 0.08) = 233.28. $$So an initial deposit of \$200 has accumulated to \$233.28 in two years. Notice that under simple interest, this amount would have accumulated to only \$232. The reason the compound-interest amount is slightly higher is that the \$16 of interest earned during the first year was left in the account and itself earned interest during the second year:
$$ 16(0.08) = 1.28. $$So we earned interest on the principal as well as on past interest. That is exactly why we call it compound interest.
If we leave the amount \$233.28 in the bank for another year, the final amount becomes
$$ 233.28 + 233.28(0.08) = 233.28(1 + 0.08) = 251.94. $$Now let us look at the mathematical pattern behind this problem.
After one year, we had
$$ 200(1 + 0.08) = 216. $$After two years, we had
$$ 216(1 + 0.08). $$But since
$$ 216 = 200(1 + 0.08), $$the above expression becomes
$$ 200(1 + 0.08)(1 + 0.08) = 200(1 + 0.08)^2 = 233.28. $$After three years, we get
$$ 233.28(1 + 0.08) = 200(1 + 0.08)(1 + 0.08)(1 + 0.08), $$which can be written as
$$ 200(1 + 0.08)^3 = 251.94. $$If we want the total amount at the end of five years, we get
$$ 200(1 + 0.08)^5 = 293.87. $$We can summarize this pattern as follows:
| Description | Expression | Value |
|---|---|---|
| The original amount | \$200 | \$200 |
| The amount after one year | \$200\((1 + 0.08)\) | \$216 |
| The amount after two years | \$200\((1 + 0.08)^2\) | \$233.28 |
| The amount after three years | \$200\((1 + 0.08)^3\) | \$251.94 |
| The amount after five years | \$200\((1 + 0.08)^5\) | \$293.87 |
| The amount after \(t\) years | \$200\((1 + 0.08)^t\) |
This table is the heart of many financial models. Whether you are tracking a retirement account, a savings account, or the unpaid balance on a loan, the same pattern appears: each period multiplies the current amount by the same growth factor.
The original amount invested is called the principal. The amount in the account after interest has been added is called the future value or accumulated value.
A compound-interest formula is really a repeated multiplication formula. Each compounding period multiplies the balance by the same factor. That is why exponents appear naturally.
A deposit of \$500 earns \(6\%\) interest compounded annually. What is the future value after \(3\) years?
Try It Now 6.2.1 Solution
Use annual compounding, so the amount after 3 years is
$$ 500(1 + 0.06)^3 = 500(1.06)^3 \approx 595.51. $$So the future value is approximately \$595.51.
Compounding Periods
Banks often compound interest more than once a year. Consider a bank that pays \(8\%\) interest but compounds it four times a year, or quarterly. This means that every quarter the bank will pay interest equal to one-fourth of \(8\%\), or \(2\%\).
Now if we deposit \$200 in the bank, after one quarter we will have
$$ 200\left(1 + \frac{0.08}{4}\right) = 204. $$After two quarters, we will have
$$ 200\left(1 + \frac{0.08}{4}\right)^2 = 208.08. $$After one year, we will have
$$ 200\left(1 + \frac{0.08}{4}\right)^4 = 216.49. $$After three years, we will have
$$ 200\left(1 + \frac{0.08}{4}\right)^{12} = 253.65. $$The pattern is summarized below:
| Description | Expression | Value |
|---|---|---|
| The original amount | \$200 | \$200 |
| The amount after one quarter | \$200\(\left(1 + \frac{0.08}{4}\right)\) | \$204 |
| The amount after two quarters | \$200\(\left(1 + \frac{0.08}{4}\right)^2\) | \$208.08 |
| The amount after one year | \$200\(\left(1 + \frac{0.08}{4}\right)^4\) | \$216.49 |
| The amount after two years | \$200\(\left(1 + \frac{0.08}{4}\right)^8\) | \$234.31 |
| The amount after three years | \$200\(\left(1 + \frac{0.08}{4}\right)^{12}\) | \$253.65 |
| The amount after five years | \$200\(\left(1 + \frac{0.08}{4}\right)^{20}\) | \$297.19 |
| The amount after \(t\) years | \$200\(\left(1 + \frac{0.08}{4}\right)^{4t}\) |
Therefore, if we invest a lump-sum amount of \(P\) dollars at an interest rate \(r\), compounded \(n\) times a year, then after \(t\) years the final amount is given by the following formula.
Compound Interest Formula
If \(P\) dollars are invested at an annual interest rate \(r\), compounded \(n\) times per year, then the future value after \(t\) years is
$$ A = P\left(1 + \frac{r}{n}\right)^{nt}. $$Here:
- \(A\) = future value
- \(P\) = principal (present value)
- \(r\) = annual interest rate written as a decimal
- \(n\) = number of compounding periods per year
- \(t\) = number of years
This is one of the most important formulas in all of finance. It tells you how a one-time deposit grows. Later, this same idea will help you understand annuities, loans, retirement planning, inflation, and many other real-world financial decisions.
The following examples use the compound interest formula
$$ A = P\left(1 + \frac{r}{n}\right)^{nt}. $$If \$3500 is invested at \(9\%\) compounded monthly, what will the future value be in four years?
Example 6.2.1 Solution
Clearly, interest of \$0.09/12 is paid every month for four years. The interest is compounded
$$ 4 \times 12 = 48 $$times over the four-year period. Therefore,
$$ A = 3500\left(1 + \frac{0.09}{12}\right)^{48} = 3500(1.0075)^{48} \approx 5009.92. $$So \$3500 invested at \(9\%\) compounded monthly will accumulate to \$5009.92 in four years.
Present Value of a Lump Sum
Sometimes we know the future value and want to work backward to find how much should be invested now. That amount is the present value.
The present value is the amount that must be invested now in order to reach a desired future value later.
Starting with the compound-interest formula
$$ A = P\left(1 + \frac{r}{n}\right)^{nt}, $$we solve for \(P\) by dividing both sides by \(\left(1 + \frac{r}{n}\right)^{nt}\).
Present Value Formula
If you want a future value of \(A\) dollars after \(t\) years at an annual rate \(r\), compounded \(n\) times per year, then the amount to deposit now is
$$ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}. $$Present value is the "work backward" version of future value. Instead of asking, "What will my money grow to?" we ask, "How much do I need today to hit a target later?" This is how colleges, businesses, and investors plan ahead for future payments.
How much should be invested in an account paying \(9\%\) compounded daily for it to accumulate to \$5000 in five years?
Example 6.2.2 Solution
We know the future value, but we need to find the principal.
$$ 5000 = P\left(1 + \frac{0.09}{365}\right)^{365 \cdot 5}. $$Evaluating the growth factor gives
$$ 5000 = P(1.568225). $$Now solve for \(P\):
$$ P = \frac{5000}{1.568225} \approx 3188.32. $$So \$3188.32 invested in an account paying \(9\%\) compounded daily will accumulate to \$5000 in five years.
You want \$2000 in \(4\) years. How much should you deposit now if the account pays \(5\%\) compounded annually?
Try It Now 6.2.2 Solution
Use the present value formula:
$$ P = \frac{2000}{(1.05)^4} \approx 1645.41. $$You should deposit approximately \$1645.41 now.
Solving for Time
If the variable is in the exponent, we use logarithms.
If \$4000 is invested at \(4\%\) compounded annually, how long will it take to accumulate to \$6000?
Example 6.2.3 Solution
Here \(n = 1\) because annual compounding means compounding only once per year.
The model becomes
$$ 6000 = 4000(1 + 0.04)^t. $$Divide both sides by \$4000:
$$ \frac{6000}{4000} = 1.04^t $$so
$$ 1.5 = 1.04^t. $$We use logarithms because the variable \(t\) is in the exponent:
$$ t = \log_{1.04}(1.5). $$Using the change-of-base formula,
$$ t = \frac{\ln(1.5)}{\ln(1.04)} \approx 10.33 \text{ years}. $$It takes 10.33 years for \$4000 to accumulate to \$6000 if invested at \(4\%\) interest, compounded annually.
Solving for Interest Rate
If the variable is in the base and the exponent is known, we use roots.
If \$5000 is invested now for \(6\) years, what interest rate compounded quarterly is needed to obtain an accumulated value of \$8000?
Example 6.2.4 Solution
We have \(n = 4\) for quarterly compounding, so
$$ 8000 = 5000\left(1 + \frac{r}{4}\right)^{4 \cdot 6}. $$Divide both sides by \$5000:
$$ \frac{8000}{5000} = \left(1 + \frac{r}{4}\right)^{24} $$which simplifies to
$$ 1.6 = \left(1 + \frac{r}{4}\right)^{24}. $$Now take the \(24\)th root of both sides:
$$ \sqrt[24]{1.6} = 1 + \frac{r}{4}. $$Many calculators have a built-in \(n\)th-root function. Roots can also be written as fractional exponents:
$$ 1.6^{1/24} = 1 + \frac{r}{4}. $$Evaluating the left side gives
$$ 1.0197765 = 1 + \frac{r}{4}. $$So
$$ 0.0197765 = \frac{r}{4} $$and therefore
$$ r = 4(0.0197765) = 0.0791. $$An interest rate of \(7.91\%\) is needed in order for \$5000 invested now to accumulate to \$8000 at the end of \(6\) years, with interest compounded quarterly.
Effective Interest Rate
Banks are required to state their interest rate in terms of an effective yield or effective interest rate for comparison purposes. The effective rate is also called the Annual Percentage Yield (APY) or sometimes the Annual Percentage Rate (APR) in everyday financial language.
The effective rate is the annual rate that would produce the same one-year growth as the stated rate together with its compounding schedule.
The effective interest rate is the actual annual rate of growth produced when nominal interest is compounded during the year.
Effective Interest Rate Formula
If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is
$$ r_{\mathrm{EFF}} = \left(1 + \frac{r}{n}\right)^n - 1. $$To compare several investments, we find and compare the effective rate for each one.
If Bank A pays \(7.2\%\) interest compounded monthly, what is the effective interest rate? If Bank B pays \(7.25\%\) interest compounded semiannually, what is the effective interest rate? Which bank pays more interest?
Example 6.2.5 Solution
Bank A: Suppose we deposit \$1 in this bank and leave it for one year. Then we get
$$ 1\left(1 + \frac{0.072}{12}\right)^{12} \approx 1.0744. $$So
$$ r_{\mathrm{EFF}} = 1.0744 - 1 = 0.0744. $$We earned interest of
$$ 1.0744 - 1.00 = 0.0744 $$on an investment of \$1. Therefore, the effective interest rate is \(7.44\%\).
Bank B: The effective rate is
$$ r_{\mathrm{EFF}} = \left(1 + \frac{0.0725}{2}\right)^2 - 1 \approx 0.0738. $$So the effective interest rate is \(7.38\%\).
Therefore, Bank A pays slightly higher interest, with an effective rate of \(7.44\%\), compared to Bank B with an effective rate of \(7.38\%\).
This is why banks advertise APY so heavily. Two accounts may look almost the same based on the stated rate, but the one that compounds more frequently can actually earn more. The effective rate gives a fair comparison.
Find the effective interest rate for an account paying \(6\%\) compounded quarterly.
Try It Now 6.2.3 Solution
Use
$$ r_{\mathrm{EFF}} = \left(1 + \frac{0.06}{4}\right)^4 - 1 = (1.015)^4 - 1 \approx 0.0614. $$So the effective interest rate is approximately \(6.14\%\).
Continuous Compounding
Interest can be compounded yearly, semiannually, quarterly, monthly, and daily. Using the same calculation methods, we could compound every hour, every minute, and even every second. As the compounding period gets shorter and shorter, we move toward the concept of continuous compounding.
But what do we mean when we say the interest is compounded continuously, and how do we compute such amounts? When interest is compounded "infinitely many times," we say that the interest is compounded continuously.
Suppose we put \$1 in an account that pays \(100\%\) interest. If the interest is compounded once a year, the total amount after one year will be
$$ 1(1 + 1) = 2. $$If the interest is compounded semiannually, in one year we will have
$$ 1\left(1 + \frac{1}{2}\right)^2 = 2.25. $$If the interest is compounded quarterly, in one year we will have
$$ 1\left(1 + \frac{1}{4}\right)^4 \approx 2.44. $$If the interest is compounded monthly, in one year we will have
$$ 1\left(1 + \frac{1}{12}\right)^{12} \approx 2.61. $$If the interest is compounded daily, in one year we will have
$$ 1\left(1 + \frac{1}{365}\right)^{365} \approx 2.71. $$We show the results as follows:
| Frequency of compounding | Formula | Total amount |
|---|---|---|
| Annually | \(1(1 + 1)\) | \$2 |
| Semiannually | \(1\left(1 + \frac{1}{2}\right)^2\) | \$2.25 |
| Quarterly | \(1\left(1 + \frac{1}{4}\right)^4\) | \$2.44140625 |
| Monthly | \(1\left(1 + \frac{1}{12}\right)^{12}\) | \$2.61303529 |
| Daily | \(1\left(1 + \frac{1}{365}\right)^{365}\) | \$2.71456748 |
| Hourly | \(1\left(1 + \frac{1}{8760}\right)^{8760}\) | \$2.71812699 |
| Every minute | \(1\left(1 + \frac{1}{525600}\right)^{525600}\) | \$2.71827922 |
| Every second | \(1\left(1 + \frac{1}{31536000}\right)^{31536000}\) | \$2.71828247 |
| Continuously | \(1(e)\) | \$2.718281828\(\ldots\) |
We have noticed that the \$1 we invested does not grow without bound. It starts to stabilize near the irrational number
$$ 2.718281828\ldots $$which is given the name \(e\), after the great mathematician Euler.
In mathematics, we say that as \(n\) becomes infinitely large, the expression
$$ \left(1 + \frac{1}{n}\right)^n $$approaches \(e\).
Therefore, it is natural that the number \(e\) plays a part in continuous compounding.
It can be shown that as \(n\) becomes infinitely large,
$$ \left(1 + \frac{r}{n}\right)^{nt} \to e^{rt}. $$Therefore, if we invest \(P\) at an interest rate \(r\) per year, compounded continuously, after \(t\) years the final amount is given by the following formula.
Continuous Compounding Formula
If \(P\) dollars are invested at annual rate \(r\), compounded continuously, then the future value after \(t\) years is
$$ A = Pe^{rt}. $$Continuous compounding is an idealized model, but it is still extremely useful. It appears in finance, economics, population growth, radioactive decay, inflation modeling, and differential equations. Even when money is not literally compounded every instant, this formula often gives a very accurate model.
\$3500 is invested at \(9\%\) compounded continuously. Find the future value in \(4\) years.
Example 6.2.6 Solution
Using the continuous compounding formula,
$$ A = Pe^{rt}, $$we get
$$ A = 3500e^{0.09 \cdot 4} = 3500e^{0.36} \approx 5016.65. $$So the future value is \$5016.65.
If an amount is invested at \(7\%\) compounded continuously, what is the effective interest rate?
Example 6.2.7 Solution
If we deposit \$1 in the bank at \(7\%\) compounded continuously for one year, and subtract the original \$1 from the final amount, we get the effective interest rate in decimal form.
$$ r_{\mathrm{EFF}} = 1e^{0.07} - 1. $$So
$$ r_{\mathrm{EFF}} \approx 1.0725 - 1 = 0.0725. $$Therefore,
$$ r_{\mathrm{EFF}} \approx 0.0725 \text{ or } 7.25\%. $$Effective Interest Rate for Continuous Compounding
If an account pays annual rate \(r\) compounded continuously, then the effective interest rate is
$$ r_{\mathrm{EFF}} = e^r - 1. $$Find the future value of \$1000 invested at \(5\%\) compounded continuously for \(3\) years.
Try It Now 6.2.4 Solution
Use the formula \(A = Pe^{rt}\):
$$ A = 1000e^{0.05 \cdot 3} = 1000e^{0.15} \approx 1161.83. $$So the future value is approximately \$1161.83.
If an amount is invested at \(7\%\) compounded continuously, how long will it take to double?
We offer two solutions.
Solution 1 uses logarithms to calculate the exact answer, so it is preferred.
We already used this method in Example 6.2.3 to solve for the time needed for an investment to accumulate to a specified future value.
Solution 2 provides an estimated solution that is applicable only to doubling time, but not to other multiples. Students should find out from their instructor if there is a preference as to which method should be used for doubling-time problems.
Example 6.2.8 Solution
Solution 1: Calculating the answer exactly
Start with
$$ Pe^{0.07t} = A. $$We do not know the initial principal, but we do know that the accumulated value is double the principal. So
$$ Pe^{0.07t} = 2P. $$Divide both sides by \(P\):
$$ e^{0.07t} = 2. $$Now use natural logarithms:
$$ 0.07t = \ln(2) $$so
$$ t = \frac{\ln(2)}{0.07} \approx 9.9 \text{ years}. $$It takes 9.9 years for money to double if invested at \(7\%\) continuous interest.
Solution 2: Estimating the answer using the Law of 70
The Law of 70 is a useful tool for estimating the time needed for an investment to double in value. It is an approximation, not an exact rule, and it comes from the exact doubling-time formula.
From the previous work, we had
$$ t = \frac{\ln(2)}{r}. $$Since
$$ \ln(2) \approx 0.693, $$we get
$$ t \approx \frac{0.693}{r}. $$Multiplying numerator and denominator by \$100 gives
$$ t \approx \frac{69.3}{100r}. $$If we estimate \$69.3 by \$70 and state the interest rate as a percent instead of a decimal, we obtain the Law of 70.
Law of 70
The number of years required to double money is approximately
$$ 70 \div \text{interest rate} $$where the interest rate is written as a percent, not a decimal.
Using the Law of 70 gives us
$$ t \approx \frac{70}{7} = 10, $$which is close to, but not exactly, the value of \$9.9 years calculated in Solution 1.
Approximate doubling time in years as a function of interest rate:
| Annual interest rate | 1% | 2% | 3% | 4% | 5% | 6% | 7% | 8% | 9% | 10% |
|---|---|---|---|---|---|---|---|---|---|---|
| Number of years to double money | 70 | 35 | 23 | 18 | 14 | 12 | 10 | 9 | 8 | 7 |
The pattern in the table approximates the Law of 70.
With technology available to do calculations using logarithms, we would use the Law of 70 only for quick estimates of doubling times. It works only for doubling times, not for other multiples, so it is not a replacement for exact methods.
However, the Law of 70 is useful for mental estimates in compound-interest applications and in other applications involving exponential growth.
a. At the peak growth rate in the 1960s, the world's population had a doubling time of \(35\) years. At that time, approximately what was the growth rate?
b. As of 2015, the world population's annual growth rate was approximately \(1.14\%\). Based on that rate, find the approximate doubling time.
Example 6.2.9 Solution
a. According to the Law of 70,
$$ 35 \approx 70 \div r, $$so
$$ r \approx 2 $$expressed as a percent. Therefore, the world's population was growing at an approximate rate of \(2\%\) in the 1960s.
b. According to the Law of 70,
$$ t \approx 70 \div r = 70 \div 1.14 \approx 61 $$years.
If the world's population were to continue to grow at the annual growth rate of \(1.14\%\), it would take approximately 61 years for the population to double.
Use the Law of 70 to estimate how long it takes money to double at an annual interest rate of \(8\%\).
Try It Now 6.2.5 Solution
Use the interest rate as a percent:
$$ t \approx 70 \div 8 = 8.75. $$So the doubling time is about 8.75 years, or roughly 9 years.
Section 6.2 Summary
Below is a summary of the formulas developed in this section for calculations involving compound interest.
Compound Interest \(n\) Times per Year
-
If an amount \(P\) is invested for \(t\) years at an interest rate \(r\) per year, compounded \(n\) times a year, then the future value is given by
$$ A = P\left(1 + \frac{r}{n}\right)^{nt}. $$\(P\) is called the principal and is also called the present value.
-
If you know the future value \(A\) and want the amount to invest now, then the present value is
$$ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}. $$ -
If a bank pays an interest rate \(r\) per year, compounded \(n\) times a year, then the effective interest rate is given by
$$ r_{\mathrm{EFF}} = \left(1 + \frac{r}{n}\right)^n - 1. $$
Continuously Compounded Interest
-
If an amount \(P\) is invested for \(t\) years at an interest rate \(r\) per year, compounded continuously, then the future value is given by
$$ A = Pe^{rt}. $$ -
If an account pays an interest rate \(r\) per year, compounded continuously, then the effective interest rate is given by
$$ r_{\mathrm{EFF}} = e^r - 1. $$ -
The Law of 70 states that the number of years to double money is approximately
$$ 70 \div \text{interest rate}. $$The interest rate is written as a percent, not a decimal, when using the Law of 70.
Section 6.2 Problem Set: Compound Interest
Do the following compound-interest problems involving a lump-sum amount.
1) What will the final amount be in 4 years if \$8,000 is invested at \(9.2\%\) compounded monthly?
Problem 1 Solution
Step 1: Use the compound-interest formula
$$ A=P\left(1+\frac{r}{n}\right)^{nt}. $$Here,
$$ P=8000,\quad r=0.092,\quad n=12,\quad t=4. $$Step 2: Substitute the values.
$$ A=8000\left(1+\frac{0.092}{12}\right)^{12\cdot 4} =8000(1.007666\overline{6})^{48}. $$Step 3: Evaluate.
$$ A\approx 8000(1.4428156287)=11542.5250297. $$Step 4: Round to the nearest cent.
$$ A\approx \$11,542.53. $$Verification: Since the balance grows for 48 monthly compounding periods at a positive rate, the final amount should be greater than \$8000, which it is.
Answer: The final amount is \(\$11,542.53\).
2) How much should be invested at \(10.3\%\) for it to amount to \$10,000 in 6 years?
Problem 2 Solution
Step 1: Since no compounding period is stated, use annual compounding.
$$ A=P(1+r)^t. $$Here,
$$ A=10000,\quad r=0.103,\quad t=6. $$Step 2: Solve for the present value \(P\).
$$ P=\frac{A}{(1+r)^t}. $$Step 3: Substitute the given values.
$$ P=\frac{10000}{(1.103)^6}. $$Step 4: Evaluate.
$$ P\approx \frac{10000}{1.8007473070}=5553.246165. $$Step 5: Round to the nearest cent.
$$ P\approx \$5,553.25. $$Verification: Check by compounding forward:
$$ 5553.246165(1.103)^6\approx 10000. $$Answer: You should invest \(\$5,553.25\).
3) Lydia's aunt Rose left her \$5,000. Lydia spent \$1,000 on her wardrobe and deposited the rest in an account that pays \(6.9\%\) compounded daily. How much money will she have in 5 years?
Problem 3 Solution
Step 1: Find the amount Lydia deposits.
$$ 5000-1000=4000. $$So,
$$ P=4000. $$Step 2: Use the compound-interest formula with daily compounding.
$$ A=P\left(1+\frac{r}{n}\right)^{nt}, $$where
$$ r=0.069,\quad n=365,\quad t=5. $$Step 3: Substitute.
$$ A=4000\left(1+\frac{0.069}{365}\right)^{365\cdot 5}. $$Step 4: Evaluate.
$$ A\approx 4000(1.4119438818)=5647.7755272. $$Step 5: Round to the nearest cent.
$$ A\approx \$5,647.78. $$Verification: The account earned interest for 5 years, so the final amount should be larger than the deposited \$4000, which it is.
Answer: Lydia will have \(\$5,647.78\) in 5 years.
4) Thuy needs \$1,850 in eight months for her college tuition. How much money should she deposit lump sum in an account paying \(8.2\%\) compounded monthly to achieve that goal?
Problem 4 Solution
Step 1: Use the present-value formula for monthly compounding.
$$ P=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}. $$Here,
$$ A=1850,\quad r=0.082,\quad n=12. $$Eight months is
$$ t=\frac{8}{12}\text{ year}. $$Since
$$ nt=12\cdot \frac{8}{12}=8, $$there are 8 monthly compounding periods.
Step 2: Substitute.
$$ P=\frac{1850}{\left(1+\frac{0.082}{12}\right)^8}. $$Step 3: Evaluate.
$$ P\approx \frac{1850}{1.0562206616}=1751.9069908. $$Step 4: Round to the nearest cent.
$$ P\approx \$1,751.91. $$Verification: Compounding this amount for 8 months gives
$$ 1751.9069908\left(1+\frac{0.082}{12}\right)^8\approx 1850. $$Answer: Thuy should deposit \(\$1,751.91\) now.
5) Bank A pays \(5\%\) compounded daily, while Bank B pays \(5.12\%\) compounded monthly. Which bank pays more? Explain.
Problem 5 Solution
Step 1: Compare the effective annual rates.
Use
$$ r_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^n-1. $$Step 2: Find Bank A's effective rate.
For Bank A,
$$ r=0.05,\quad n=365. $$So
$$ r_{\mathrm{EFF}}=\left(1+\frac{0.05}{365}\right)^{365}-1\approx 0.0512674965. $$Thus Bank A's effective rate is about
$$ 5.1267\%. $$Step 3: Find Bank B's effective rate.
For Bank B,
$$ r=0.0512,\quad n=12. $$So
$$ r_{\mathrm{EFF}}=\left(1+\frac{0.0512}{12}\right)^{12}-1\approx 0.0524187464. $$Thus Bank B's effective rate is about
$$ 5.2419\%. $$Step 4: Compare the two rates.
$$ 5.2419\%>5.1267\%. $$Verification: Even though the rates are close, Bank B starts with a higher nominal rate, and its effective rate still ends up higher.
Answer: Bank B pays more, because its effective interest rate is about \(5.24\%\), compared to about \(5.13\%\) for Bank A.
6) EZ Photo Company needs five copying machines in \(2\tfrac{1}{2}\) years for a total cost of \$15,000. How much money should be deposited now to pay for these machines, if the interest rate is \(8\%\) compounded semiannually?
Problem 6 Solution
Step 1: Use the present-value formula.
$$ P=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}. $$Here,
$$ A=15000,\quad r=0.08,\quad n=2,\quad t=2.5. $$Step 2: Compute the number of compounding periods.
$$ nt=2(2.5)=5. $$Step 3: Substitute.
$$ P=\frac{15000}{\left(1+\frac{0.08}{2}\right)^5} =\frac{15000}{(1.04)^5}. $$Step 4: Evaluate.
$$ P\approx \frac{15000}{1.2166529024}=12328.9066014. $$Step 5: Round to the nearest cent.
$$ P\approx \$12,328.91. $$Verification: Depositing this amount now and compounding semiannually for 5 periods gives back about \$15,000.
Answer: EZ Photo Company should deposit \(\$12,328.91\) now.
7) Jon's grandfather was planning to give him \$12,000 in 10 years. Jon has convinced his grandfather to pay him \$6,000 now, instead. If Jon invests this \$6,000 at \(7.5\%\) compounded continuously, how much money will he have in 10 years?
Problem 7 Solution
Step 1: Use the continuous compounding formula.
$$ A=Pe^{rt}. $$Here,
$$ P=6000,\quad r=0.075,\quad t=10. $$Step 2: Substitute.
$$ A=6000e^{0.075\cdot 10}=6000e^{0.75}. $$Step 3: Evaluate.
$$ A\approx 6000(2.1170000166)=12702.0000997. $$Step 4: Round to the nearest cent.
$$ A\approx \$12,702.00. $$Verification: Since \(12702>12000\), Jon's choice gives him slightly more than the \$12,000 he would have received later.
Answer: Jon will have \(\$12,702.00\) in 10 years.
8) What will be the price of a \$20,000 car in 5 years if the inflation rate is \(6\%\)?
Problem 8 Solution
Step 1: Model inflation with annual compounding.
$$ A=P(1+r)^t. $$Here,
$$ P=20000,\quad r=0.06,\quad t=5. $$Step 2: Substitute.
$$ A=20000(1.06)^5. $$Step 3: Evaluate.
$$ A=20000(1.3382255776)=26764.511552. $$Step 4: Round to the nearest cent.
$$ A\approx \$26,764.51. $$Verification: Prices rising \(6\%\) per year for 5 years should make the car cost more than \$20,000, which matches the result.
Answer: The car will cost about \(\$26,764.51\) in 5 years.
Do the following compound-interest problems.
9) At an interest rate of \(8\%\) compounded continuously, how many years will it take to double your money?
Problem 9 Solution
Step 1: Use the continuous compounding formula.
$$ A=Pe^{rt}. $$To double, we have
$$ A=2P. $$So,
$$ 2P=Pe^{0.08t}. $$Step 2: Divide both sides by \(P\).
$$ 2=e^{0.08t}. $$Step 3: Take the natural logarithm of both sides.
$$ \ln(2)=0.08t. $$Step 4: Solve for \(t\).
$$ t=\frac{\ln(2)}{0.08}. $$Step 5: Evaluate.
$$ t\approx \frac{0.6931471806}{0.08}=8.664339757. $$Answer: It will take about \(8.66\) years to double the money.
10) If an investment earns \(10\%\) compounded continuously, in how many years will it triple?
Problem 10 Solution
Step 1: Use the continuous compounding formula.
$$ A=Pe^{rt}. $$To triple, we have
$$ A=3P. $$So,
$$ 3P=Pe^{0.10t}. $$Step 2: Divide both sides by \(P\).
$$ 3=e^{0.10t}. $$Step 3: Take natural logarithms.
$$ \ln(3)=0.10t. $$Step 4: Solve for \(t\).
$$ t=\frac{\ln(3)}{0.10}. $$Step 5: Evaluate.
$$ t\approx \frac{1.0986122887}{0.10}=10.986122887. $$Answer: It will take about \(10.99\) years to triple the investment.
11) The City Library ordered a new computer system costing \$158,000; it will be delivered in 6 months. The full amount will be due 30 days after delivery. How much must be deposited today into an account paying \(7.5\%\) compounded monthly to have \$158,000 in 7 months?
Problem 11 Solution
Step 1: The money is needed in 7 months total, so use the present-value formula with monthly compounding.
$$ P=\frac{A}{\left(1+\frac{r}{n}\right)^{nt}}. $$Here,
$$ A=158000,\quad r=0.075,\quad n=12,\quad t=\frac{7}{12}. $$Since
$$ nt=12\cdot \frac{7}{12}=7, $$there are 7 monthly compounding periods.
Step 2: Substitute.
$$ P=\frac{158000}{\left(1+\frac{0.075}{12}\right)^7}. $$Step 3: Evaluate.
$$ P\approx \frac{158000}{1.0445785421}=151257.1222067. $$Step 4: Round to the nearest cent.
$$ P\approx \$151,257.12. $$Verification: Compounding this deposit for 7 months at the stated rate returns approximately \$158,000.
Answer: The City Library must deposit \(\$151,257.12\) today.
12) Mr. and Mrs. Tran are expecting a baby girl in a few days. They want to put away money for her college education now. How much money should they deposit in an account paying \(10.2\%\) so they will have \$100,000 in 18 years to pay for their daughter's educational expenses?
Problem 12 Solution
Step 1: Since no compounding period is given, use annual compounding.
$$ P=\frac{A}{(1+r)^t}. $$Here,
$$ A=100000,\quad r=0.102,\quad t=18. $$Step 2: Substitute.
$$ P=\frac{100000}{(1.102)^{18}}. $$Step 3: Evaluate.
$$ P\approx \frac{100000}{5.7443000335}=17407.2953760. $$Step 4: Round to the nearest cent.
$$ P\approx \$17,407.30. $$Verification: If \$17,407.30 is invested and compounded annually for 18 years at \(10.2\%\), it grows to about \$100,000.
Answer: They should deposit \(\$17,407.30\) now.
13) Find the effective interest rate for an account paying \(7.2\%\) compounded quarterly.
Problem 13 Solution
Step 1: Use the effective interest rate formula.
$$ r_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^n-1. $$Here,
$$ r=0.072,\quad n=4. $$Step 2: Substitute.
$$ r_{\mathrm{EFF}}=\left(1+\frac{0.072}{4}\right)^4-1=(1.018)^4-1. $$Step 3: Evaluate.
$$ r_{\mathrm{EFF}}\approx 1.0739674330-1=0.0739674330. $$Step 4: Convert to a percent.
$$ 0.0739674330\approx 7.40\%. $$Answer: The effective interest rate is about \(7.40\%\).
14) If a bank pays \(5.75\%\) compounded monthly, what is the effective interest rate?
Problem 14 Solution
Step 1: Use the effective interest rate formula.
$$ r_{\mathrm{EFF}}=\left(1+\frac{r}{n}\right)^n-1. $$Here,
$$ r=0.0575,\quad n=12. $$Step 2: Substitute.
$$ r_{\mathrm{EFF}}=\left(1+\frac{0.0575}{12}\right)^{12}-1. $$Step 3: Evaluate.
$$ r_{\mathrm{EFF}}\approx 1.0590398313-1=0.0590398313. $$Step 4: Convert to a percent.
$$ 0.0590398313\approx 5.90\%. $$Answer: The effective interest rate is about \(5.90\%\).
15) The population of the African nation of Cameroon was 12 million people in the year 2015; it has been growing at the rate of \(2.5\%\) per year. If the population continues to grow at that rate, what will the population be in 2030? (http://databank.worldbank.org/data on 4/26/2016)
Problem 15 Solution
Step 1: Find the time interval.
$$ 2030-2015=15\text{ years}. $$Step 2: Use annual compound growth.
$$ A=P(1+r)^t. $$Here,
$$ P=12\text{ million},\quad r=0.025,\quad t=15. $$Step 3: Substitute.
$$ A=12(1.025)^{15}. $$Step 4: Evaluate.
$$ A\approx 12(1.4482981665)=17.379577998. $$Step 5: Interpret the result.
$$ A\approx 17.38\text{ million people}. $$Verification: Because the population is growing, the 2030 population should be larger than 12 million, which it is.
Answer: The population will be about \(17.38\) million people in 2030.
16) According to the Law of 70, if an amount grows at an annual rate of \(1\%\), then it doubles every seventy years. Suppose a bank pays \(5\%\) interest. How long will it take for you to double your money? How about at \(15\%\)?
Problem 16 Solution
Step 1: Use the Law of 70.
$$ t\approx \frac{70}{\text{interest rate as a percent}}. $$Step 2: For \(5\%\) interest,
$$ t\approx \frac{70}{5}=14. $$So the doubling time is about 14 years.
Step 3: For \(15\%\) interest,
$$ t\approx \frac{70}{15}=4.666\overline{6}. $$So the doubling time is about 4.67 years, or about 5 years.
Verification: A higher interest rate should produce a shorter doubling time, and \(4.67\) years is indeed less than \(14\) years.
Answer: At \(5\%\), the money doubles in about 14 years. At \(15\%\), it doubles in about 4.67 years.