23. Interest calculations over time

23. Interest calculations over time - step by step solution procedure and detailed explanation for each problem.

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Below are 10 problems related to interest calculations over time, covering simple interest, compound interest, and continuous compounding. Each problem includes a step-by-step solution procedure and a detailed explanation to ensure clarity. The problems progress in complexity to provide a comprehensive understanding.
Problem 1: Simple Interest - Basic Calculation
Problem: Calculate the simple interest earned on a principal of $5,000 invested at an annual interest rate of 4% for 3 years.
Step-by-Step Solution:
  1. Identify the formula: Simple Interest (SI) =
    P \times R \times T
    , where:
    • ( P ) = Principal ($5,000)
    • ( R ) = Annual interest rate (4% = 0.04)
    • ( T ) = Time in years (3)
  2. Substitute values: SI =
    5,000 \times 0.04 \times 3
  3. Calculate: SI =
    5,000 \times 0.12 = 600
  4. State the result: The simple interest earned is $600.
Detailed Explanation: Simple interest is calculated as a fixed percentage of the principal, applied annually over the time period. Here, the interest rate is 4%, or 0.04 as a decimal. Multiplying the principal ($5,000) by the rate (0.04) and time (3 years) gives the total interest earned. The result, $600, represents the interest without any compounding, meaning the principal remains unchanged, and interest is not earned on previously accrued interest.
Problem 2: Simple Interest - Total Amount
Problem: Find the total amount after 5 years if $10,000 is invested at a simple interest rate of 5% per year.
Step-by-Step Solution:
  1. Calculate simple interest: SI =
    P \times R \times T
    • P = 10,000
      ,
      R = 5\% = 0.05
      ,
      T = 5
    • SI =
      10,000 \times 0.05 \times 5 = 2,500
  2. Calculate total amount: Total Amount =
    P + \text{SI} = 10,000 + 2,500
  3. State the result: Total Amount = $12,500.
Detailed Explanation: The total amount includes the original principal plus the interest earned. Simple interest is computed as in Problem 1, yielding $2,500 over 5 years. Adding this to the principal gives the final amount. This problem illustrates that simple interest calculations are straightforward, as the interest does not compound, making it linear over time.
Problem 3: Compound Interest - Annual Compounding
Problem: Calculate the future value of $2,000 invested at an annual interest rate of 6%, compounded annually, for 4 years.
Step-by-Step Solution:
  1. Identify the formula: Compound Interest Future Value =
    P \times (1 + R)^T
    • P = 2,000
      ,
      R = 6\% = 0.06
      ,
      T = 4
  2. Substitute values: FV =
    2,000 \times (1 + 0.06)^4
  3. Simplify inside parentheses:
    1 + 0.06 = 1.06
  4. Calculate the exponent:
    1.06^4 = 1.06 \times 1.06 \times 1.06 \times 1.06 \approx 1.26247696
  5. Compute future value: FV =
    2,000 \times 1.26247696 \approx 2,524.95
  6. State the result: The future value is approximately $2,524.95.
Detailed Explanation: Compound interest accounts for interest earned on both the principal and previously accumulated interest. The formula
P \times (1 + R)^T
reflects annual compounding, where the interest rate is applied once per year. Raising 1.06 to the 4th power accounts for the growth over 4 years. The result is higher than simple interest for the same parameters because compounding accelerates growth over time.
Problem 4: Compound Interest - Quarterly Compounding
Problem: Find the future value of $3,000 invested at 8% annual interest, compounded quarterly, for 2 years.
Step-by-Step Solution:
  1. Identify the formula: FV =
    P \times \left(1 + \frac{R}{n}\right)^{n \times T}
    • P = 3,000
      ,
      R = 8\% = 0.08
      ,
      n = 4
      (quarterly),
      T = 2
  2. Calculate the interest rate per period:
    \frac{R}{n} = \frac{0.08}{4} = 0.02
  3. Calculate the number of periods:
    n \times T = 4 \times 2 = 8
  4. Substitute values: FV =
    3,000 \times (1 + 0.02)^8
  5. Simplify:
    1 + 0.02 = 1.02
    , so FV =
    3,000 \times 1.02^8
  6. Calculate the exponent:
    1.02^8 \approx 1.17165938
  7. Compute future value: FV =
    3,000 \times 1.17165938 \approx 3,514.98
  8. State the result: The future value is approximately $3,514.98.
Detailed Explanation: Quarterly compounding divides the annual interest rate by 4 (since there are 4 quarters in a year) and applies it 4 times per year. The exponent
n \times T
reflects the total number of compounding periods (8 quarters over 2 years). More frequent compounding results in a higher future value compared to annual compounding, as interest is earned on interest more often.
Problem 5: Compound Interest - Interest Earned
Problem: How much interest is earned on $4,000 invested at 5% annual interest, compounded semi-annually, for 3 years?
Step-by-Step Solution:
  1. Use the compound interest formula: FV =
    P \times \left(1 + \frac{R}{n}\right)^{n \times T}
    • P = 4,000
      ,
      R = 5\% = 0.05
      ,
      n = 2
      (semi-annually),
      T = 3
  2. Calculate rate per period:
    \frac{R}{n} = \frac{0.05}{2} = 0.025
  3. Calculate periods:
    n \times T = 2 \times 3 = 6
  4. Substitute: FV =
    4,000 \times (1 + 0.025)^6 = 4,000 \times 1.025^6
  5. Calculate the exponent:
    1.025^6 \approx 1.159693
  6. Compute future value: FV =
    4,000 \times 1.159693 \approx 4,638.77
  7. Calculate interest earned: Interest = FV - P =
    4,638.77 - 4,000 = 638.77
  8. State the result: The interest earned is approximately $638.77.
Detailed Explanation: To find the interest earned, we calculate the future value and subtract the principal. Semi-annual compounding means the interest rate is halved, and the number of periods is doubled compared to annual compounding. The result shows how compounding increases the effective yield compared to simple interest.
Problem 6: Continuous Compounding
Problem: Calculate the future value of $1,500 invested at 7% annual interest, compounded continuously, for 5 years.
Step-by-Step Solution:
  1. Identify the formula: FV =
    P \times e^{R \times T}
    • P = 1,500
      ,
      R = 7\% = 0.07
      ,
      T = 5
      ,
      e \approx 2.71828
  2. Calculate the exponent:
    R \times T = 0.07 \times 5 = 0.35
  3. Compute the exponential:
    e^{0.35} \approx 1.419067
  4. Calculate future value: FV =
    1,500 \times 1.419067 \approx 2,128.60
  5. State the result: The future value is approximately $2,128.60.
Detailed Explanation: Continuous compounding assumes interest is added infinitely often, modeled by the exponential function
e^{R \times T}
. This results in the highest future value for a given rate and time compared to discrete compounding. The value of
e^{0.35}
is computed using a calculator or mathematical software for precision.
Problem 7: Finding the Interest Rate
Problem: A $2,000 investment grows to $2,500 in 3 years with annual compounding. Find the annual interest rate.
Step-by-Step Solution:
  1. Use the compound interest formula:
    FV = P \times (1 + R)^T
    • FV = 2,500
      ,
      P = 2,000
      ,
      T = 3
  2. Rearrange for the base:
    \frac{FV}{P} = (1 + R)^T
    • \frac{2,500}{2,000} = 1.25 = (1 + R)^3
  3. Solve for
    1 + R
    :
    (1 + R) = 1.25^{1/3}
  4. Calculate the cube root:
    1.25^{1/3} \approx 1.077217
  5. Solve for ( R ):
    R = 1.077217 - 1 = 0.077217
  6. Convert to percentage:
    R \approx 7.72\%
  7. State the result: The annual interest rate is approximately 7.72%.
Detailed Explanation: This problem requires solving for the unknown interest rate. Dividing the future value by the principal gives the growth factor, and taking the T-th root isolates
1 + R
. The cube root can be approximated using a calculator. The result is the rate that, when compounded annually, achieves the given growth.
Problem 8: Finding the Time Period
Problem: How long will it take for $5,000 to double at 6% annual interest, compounded annually?
Step-by-Step Solution:
  1. Set up the equation:
    FV = P \times (1 + R)^T
    • P = 5,000
      ,
      FV = 10,000
      ,
      R = 6\% = 0.06
    • 10,000 = 5,000 \times (1 + 0.06)^T
  2. Simplify:
    2 = 1.06^T
  3. Take the natural logarithm:
    \ln(2) = \ln(1.06^T)
  4. Apply logarithm properties:
    \ln(2) = T \times \ln(1.06)
  5. Solve for ( T ):
    T = \frac{\ln(2)}{\ln(1.06)}
  6. Calculate:
    \ln(2) \approx 0.693147
    ,
    \ln(1.06) \approx 0.058269
    • T \approx \frac{0.693147}{0.058269} \approx 11.896
  7. State the result: It takes approximately 11.9 years.
Detailed Explanation: To find the time required to double the investment, we set the future value to twice the principal and solve for ( T ). Logarithms are used to handle the exponent, allowing us to isolate ( T ). The result is slightly less than 12 years, reflecting the effect of compounding. For quick approximations, the Rule of 72 (72 ÷ 6 = 12) provides a close estimate.
Problem 9: Comparing Simple vs. Compound Interest
Problem: Compare the total amounts after 10 years for a $10,000 investment at 5% annual interest, using simple interest and annual compounding.
Step-by-Step Solution:
  1. Simple Interest:
    • SI =
      P \times R \times T = 10,000 \times 0.05 \times 10 = 5,000
    • Total Amount =
      P + \text{SI} = 10,000 + 5,000 = 15,000
  2. Compound Interest:
    • FV =
      P \times (1 + R)^T = 10,000 \times (1 + 0.05)^{10}
    • Calculate:
      1.05^{10} \approx 1.628894
    • FV =
      10,000 \times 1.628894 \approx 16,288.94
  3. Compare: Simple Interest = $15,000; Compound Interest = $16,288.94.
    • Difference =
      16,288.94 - 15,000 = 1,288.94
      .
Detailed Explanation: This problem highlights the difference between simple and compound interest over a long period. Simple interest grows linearly, adding $500 per year. Compound interest grows exponentially, as interest is earned on interest. Over 10 years, compounding yields an additional $1,288.94, demonstrating its power for long-term investments.
Problem 10: Effective Annual Rate
Problem: Find the effective annual rate (EAR) for a nominal interest rate of 12%, compounded monthly.
Step-by-Step Solution:
  1. Identify the formula: EAR =
    \left(1 + \frac{R}{n}\right)^n - 1
    • R = 12\% = 0.12
      ,
      n = 12
      (monthly)
  2. Calculate the rate per period:
    \frac{R}{n} = \frac{0.12}{12} = 0.01
  3. Compute the base:
    1 + 0.01 = 1.01
  4. Calculate the exponent:
    1.01^{12} \approx 1.126825
  5. Find EAR: EAR =
    1.126825 - 1 = 0.126825
  6. Convert to percentage: EAR
    \approx 12.6825\%
  7. State the result: The effective annual rate is approximately 12.68%.
Detailed Explanation: The effective annual rate converts a nominal rate with multiple compounding periods into an equivalent annual rate with one compounding period. Monthly compounding at 1% per month results in a higher effective rate than the nominal 12%, because interest is earned on interest within the year. This is useful for comparing investments with different compounding frequencies.
These 10 problems cover a range of interest calculation scenarios, from basic simple interest to advanced concepts like continuous compounding and effective rates. Each solution is broken down into clear steps, with explanations to build understanding of the underlying principles. Let me know if you need further clarification or additional problems!

 

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