Compound Interest Problems with Detailed Solutions

A principal of \( \$2000 \) is placed in a savings account at \( 3\% \) per annum compounded annually. How much is in the account after one year, two years, and three years?

Solution:

For annual compounding, the amount after \(t\) years is given by \( A = P(1 + r)^t \).

• After 1 year: \( A = 2000(1.03)^1 = 2060 \)
• After 2 years: \( A = 2000(1.03)^2 = 2121.80 \)
• After 3 years: \( A = 2000(1.03)^3 = 2185.45 \)

What would \( \$1000 \) become in a savings account at \( 3\% \) per year for 3 years when the interest is not compounded (simple interest)? What would the same amount become after 3 years if it is compounded annually?

Solution:

• Simple interest: \( A = P(1 + rt) = 1000(1 + 0.03 \times 3) = 1090 \)
• Compound interest: \( A = P(1 + r)^t = 1000(1.03)^3 = 1092.73 \)

The compounded account yields a higher return.

\( \$100 \) is deposited in a \( 5\% \) savings account with simple interest. The same amount is placed in a \( 5\% \) savings account compounded annually. Find the total amount \(A\) after \(t\) years in each plan and graph both. Use the graphs to estimate how long each plan takes to double the initial amount.

Solution:

• Simple interest amount: \( A = 100(1 + 0.05t) \)
• Compound interest amount: \( A = 100(1.05)^t \)

The compound interest investment doubles in about 14 years, while the simple interest investment takes about 20 years.

If \( \$3000 \) is placed in an account at \( 5\% \) compounded quarterly for 5 years, how much is in the account at the end of the 5 years?

Solution:

When interest is compounded \(n\) times per year for \(t\) years, the amount is \( A = P\left(1 + \frac{r}{n}\right)^{nt} \). Quarterly compounding gives \( n = 4 \):

\[ A = 3000\left(1 + \frac{0.05}{4}\right)^{4 \times 5} = 3000(1.0125)^{20} \approx 3846.11 \]

\( \$1200 \) is placed in an account at \( 4\% \) compounded annually for 2 years. It is then withdrawn and placed in another bank at \( 5\% \) compounded annually for 4 years. What is the final balance?

Solution:

• First account (4% annually for 2 years): \[ A = 1200(1.04)^2 = 1297.92 \]
• Second account (5% annually for 4 years): \[ A = 1297.92(1.05)^4 = 1577.63 \]

\( \$1200 \) is placed in an account at \( 4\% \) compounded daily for 2 years. It is then withdrawn and placed in another bank at \( 5\% \) compounded daily for 4 years. What is the final balance? (Compare with the previous problem.)

Solution:

Assuming 365 days per year:

• First account (4% daily for 2 years): \[ A = 1200\left(1 + \frac{0.04}{365}\right)^{365 \times 2} = 1299.94 \]
• Second account (5% daily for 4 years): \[ A = 1299.94\left(1 + \frac{0.05}{365}\right)^{365 \times 4} = 1587.73 \]

The final balance is higher than with annual compounding due to more frequent compounding periods.

\( \$1200 \) is placed in an account at \( 4\% \) compounded continuously for 2 years. It is then withdrawn and placed in another bank at \( 5\% \) compounded continuously for 4 years. What is the final balance?

Solution:

For continuous compounding, \( A = Pe^{rt} \):

• First account (4% for 2 years): \[ A = 1200e^{0.04 \times 2} = 1299.94 \]
• Second account (5% for 4 years): \[ A = 1299.94e^{0.05 \times 4} = 1587.75 \]

This result is essentially identical to daily compounding.

What principal must be deposited in a \( 4.5\% \) savings account compounded monthly to obtain \( \$10,000 \) after 8 years?

Solution:

Given \( A = 10,000 \), \( r = 0.045 \), \( n = 12 \), and \( t = 8 \):

\[ 10{,}000 = P\left(1 + \frac{0.045}{12}\right)^{12 \times 8} = P(1.00375)^{96} \] \[ P = \frac{10{,}000}{(1.00375)^{96}} = 6981.46 \]

\( \$120 \) is deposited in a \( 7\% \) account compounded continuously, while \( \$150 \) is deposited in a \( 5\% \) account compounded annually. How long does it take for the two balances to be equal?

Solution:

• Continuous account: \( A_1 = 120e^{0.07t} \)
• Annual account: \( A_2 = 150(1.05)^t \)

Set them equal: \( 120e^{0.07t} = 150(1.05)^t \)

Taking natural logarithms:

\[ \ln(120) + 0.07t = \ln(150) + t\ln(1.05) \implies t = \frac{\ln(150) - \ln(120)}{0.07 - \ln(1.05)} \approx 10.5 \text{ years} \]

One savings account pays \( 5\% \) compounded annually. Another pays \( 5\% \) compounded continuously. Which investment is better in the long term?

Solution:

• Annual compounding: \( A_1 = 100(1.05)^t \)
• Continuous compounding: \( A_2 = 100e^{0.05t} \)

Continuous compounding yields a higher effective annual rate (\( e^{0.05} - 1 \approx 5.13\% \) versus \( 5\% \)), making continuous compounding superior in the long term.

What annual interest rate is needed for \( \$4000 \) to grow to \( \$4500 \) in 10 years (compounded annually)?

Solution:

\[ 4000(1 + r)^{10} = 4500 \implies (1 + r)^{10} = \frac{4500}{4000} = 1.125 \] \[ 1 + r = (1.125)^{0.1} \approx 1.01177 \implies r \approx 0.0118 \text{ or } 1.18\% \]

\( \$1000 \) is deposited at \( 2\% \) compounded continuously. \( \$500 \) is deposited at \( 8\% \) compounded continuously. When will the two balances be equal? When will the second account be 50% greater than the first?

Solution:

• Balances: \( A_1 = 1000e^{0.02t} \) and \( A_2 = 500e^{0.08t} \)
• When equal: \( 1000e^{0.02t} = 500e^{0.08t} \implies 2 = e^{0.06t} \implies t = \frac{\ln 2}{0.06} \approx 11.5 \text{ years} \)
• When \( A_2 = 1.5A_1 \): \[ 500e^{0.08t} = 1.5(1000e^{0.02t}) \implies 500e^{0.08t} = 1500e^{0.02t} \implies e^{0.06t} = 3 \implies t = \frac{\ln 3}{0.06} \approx 18.5 \text{ years} \]

A bank offers \( 4\% \) interest compounded quarterly. A customer deposits \( \$200 \) at the beginning of each quarter for one year. What is the total amount at the end of the year?

Solution:

Interest rate per quarter = \( 4\% / 4 = 1\% = 0.01 \). Each deposit grows for a different duration:

• 1st deposit (grows 4 quarters): \( 200(1.01)^4 = 208.12 \)
• 2nd deposit (grows 3 quarters): \( 200(1.01)^3 = 206.06 \)
• 3rd deposit (grows 2 quarters): \( 200(1.01)^2 = 204.02 \)
• 4th deposit (grows 1 quarter): \( 200(1.01)^1 = 202.00 \)

Total amount = \( 208.12 + 206.06 + 204.02 + 202.00 = 820.20 \)

\( \$1500 \) is invested for 5 years at \( 2\% \) for the first two years, \( 5\% \) for the third year, and \( 6\% \) for the final two years, all compounded continuously. What is the final amount?

Solution:

The investment grows continuously through changing rates over the time periods:

\[ A = 1500e^{(0.02 \times 2 + 0.05 \times 1 + 0.06 \times 2)} = 1500e^{0.21} \approx 1850.51 \]