4 Compound Interest Problems

 4 Compound Interest Problems with detailed step-by-step solutions and clear explanations for each. These problems cover various aspects of compound interest, including calculating interest, amount, principal, rate, time, and related scenarios. Each problem includes the problem statement, solution procedure, final answer, and an explanation of the underlying concepts. The compound interest formulas used are:

  1. Amount (A):
    A = P × (1 + R/100)^n
    Where:
    • P = Principal (initial amount)
    • R = Annual interest rate (%)
    • n = Number of years
    • (Note: For half-yearly compounding, R = R/2, n = 2n; for quarterly, R = R/4, n = 4n)
  2. Compound Interest (CI):
    CI = A - P
  3. Effective Rate for Successive Years:
    For two years at rates R1% and R2%:
    Equivalent single rate = R1 + R2 + (R1 × R2)/100
Problem 1: Calculate Compound Interest
Problem: Find the compound interest on $5000 at 5% per annum for 2 years, compounded annually.
Solution:
  1. Identify the given values:
    • Principal (P) = $5000
    • Rate (R) = 5%
    • Time (n) = 2 years
  2. Calculate Amount using the formula:
    • A = P × (1 + R/100)^n
    • A = 5000 × (1 + 5/100)^2
    • A = 5000 × (1 + 0.05)^2
    • A = 5000 × (1.05)^2
    • A = 5000 × 1.1025 = $5512.50
  3. Calculate Compound Interest:
    • CI = A - P
    • CI = 5512.50 - 5000 = $512.50
Answer:
  • Compound Interest = $512.50
Explanation:
  • The amount is calculated by applying the compound interest formula, which accounts for interest on both principal and accumulated interest.
  • CI is the difference between the final amount and the initial principal.
Problem 2: Calculate Amount with Half-Yearly Compounding
Problem: Find the amount on $8000 at 6% per annum for 1.5 years, compounded half-yearly.
Solution:
  1. Identify the given values:
    • P = $8000
    • Annual Rate (R) = 6%
    • Time = 1.5 years
    • For half-yearly compounding:
      • Rate per half-year = R/2 = 6/2 = 3%
      • Number of half-years = n × 2 = 1.5 × 2 = 3
  2. Calculate Amount:
    • A = P × (1 + R/100)^n
    • A = 8000 × (1 + 3/100)^3
    • A = 8000 × (1.03)^3
    • A = 8000 × 1.092727 = $8741.82 (approx.)
Answer:
  • Amount = $8741.82
Explanation:
  • For half-yearly compounding, the interest rate is halved, and the number of compounding periods is doubled.
  • The formula adjusts to reflect the more frequent compounding.
Problem 3: Find Principal
Problem: The compound interest on a certain sum at 4% per annum for 2 years is $408. Find the principal.
Solution:
  1. Identify the given values:
    • CI = $408
    • R = 4%
    • n = 2 years
  2. Relate Amount and Principal:
    • A = P × (1 + R/100)^n
    • CI = A - P
    • A = P + CI
    • A = P × (1 + 4/100)^2
    • A = P × (1.04)^2
    • A = P × 1.0816
    • CI = A - P = P × 1.0816 - P = P × (1.0816 - 1) = P × 0.0816
    • 408 = P × 0.0816
  3. Solve for P:
    • P = 408 / 0.0816
    • P = $5000
Answer:
  • Principal = $5000
Explanation:
  • CI is the difference between A and P. Use the CI formula to express CI in terms of P and solve.
  • Alternatively, compute A = P × (1 + R/100)^n and use CI = A - P to find P.
Problem 4: Find Rate of Interest
Problem: A sum of $10000 amounts to $12100 in 2 years at a certain rate of compound interest, compounded annually. Find the rate.
Solution:
  1. Identify the given values:
    • P = $10000
    • A = $12100
    • n = 2 years
  2. Use the Amount formula:
    • A = P × (1 + R/100)^n
    • 12100 = 10000 × (1 + R/100)^2
    • (1 + R/100)^2 = 12100 / 10000 = 1.21
    • 1 + R/100 = √1.21 = 1.1
    • R/100 = 1.1 - 1 = 0.1
    • R = 0.1 × 100 = 10%
Answer:
  • Rate of Interest = 10% per annum
Explanation:
  • Divide A by P to isolate (1 + R/100)^n, then take the nth root to solve for R.
  • The rate is expressed as a percentage per annum.
Problem 5: Find Time Period
Problem: A sum of $5000 becomes $7320.50 at 8% per annum, compounded annually. Find the time period.
Solution:
  1. Identify the given values:
    • P = $5000
    • A = $7320.50
    • R = 8%
  2. Use the Amount formula:
    • A = P × (1 + R/100)^n
    • 7320.50 = 5000 × (1 + 8/100)^n
    • 7320.50 / 5000 = (1.08)^n
    • 1.4641 = (1.08)^n
    • Estimate n by testing powers of 1.08:
      • (1.08)^2 = 1.1664
      • (1.08)^3 = 1.2597
      • (1.08)^4 = 1.3605
      • (1.08)^5 = 1.4693
    • 1.4641 is close to (1.08)^5 ≈ 1.4693, suggesting n ≈ 5 years
  3. Verify:
    • A = 5000 × (1.08)^5 = 5000 × 1.4693 ≈ $7346.50 (slightly higher, indicating n is slightly less than 5, but closest integer is 5 for practical purposes).
Answer:
  • Time ≈ 5 years (approximately)
Explanation:
  • Solve for n by isolating (1 + R/100)^n and estimating n through trial or logarithms.
  • Here, n is approximated as 5 years based on closest value.
Problem 6: Quarterly Compounding
Problem: Find the compound interest on $6000 at 12% per annum for 1 year, compounded quarterly.
Solution:
  1. Identify the given values:
    • P = $6000
    • Annual Rate (R) = 12%
    • Time = 1 year
    • For quarterly compounding:
      • Rate per quarter = R/4 = 12/4 = 3%
      • Number of quarters = n × 4 = 1 × 4 = 4
  2. Calculate Amount:
    • A = P × (1 + R/100)^n
    • A = 6000 × (1 + 3/100)^4
    • A = 6000 × (1.03)^4
    • A = 6000 × 1.125509 = $6753.05 (approx.)
  3. Calculate Compound Interest:
    • CI = A - P
    • CI = 6753.05 - 6000 = $753.05
Answer:
  • Compound Interest = $753.05
Explanation:
  • Quarterly compounding uses a quarter of the annual rate and multiplies the number of periods by 4.
  • The formula adjusts for more frequent compounding, increasing the interest earned.
Problem 7: Compare Simple and Compound Interest
Problem: A sum of $10000 is invested at 5% per annum for 3 years. Find the difference between compound interest and simple interest, compounded annually.
Solution:
  1. Calculate Compound Interest:
    • P = $10000, R = 5%, n = 3 years
    • A = P × (1 + R/100)^n
    • A = 10000 × (1 + 5/100)^3
    • A = 10000 × (1.05)^3
    • A = 10000 × 1.157625 = $11576.25
    • CI = A - P = 11576.25 - 10000 = $1576.25
  2. Calculate Simple Interest:
    • SI = (P × R × T) / 100
    • SI = (10000 × 5 × 3) / 100
    • SI = 150000 / 100 = $1500
  3. Find the Difference:
    • Difference = CI - SI
    • Difference = 1576.25 - 1500 = $76.25
Answer:
  • Difference = $76.25
Explanation:
  • CI includes interest on interest, making it higher than SI, which is linear.
  • The difference highlights the effect of compounding over multiple years.
Problem 8: Amount Doubles
Problem: At what rate of interest, compounded annually, will a sum double itself in 5 years?
Solution:
  1. Understand “Amount Doubles”:
    • Let Principal = P
    • Amount doubles: A = 2P
    • n = 5 years
  2. Use the Amount formula:
    • A = P × (1 + R/100)^n
    • 2P = P × (1 + R/100)^5
    • Divide by P (P ≠ 0):
    • 2 = (1 + R/100)^5
    • (1 + R/100) = (2)^(1/5)
    • (2)^(1/5) ≈ 1.1487 (using approximation or calculator)
    • 1 + R/100 = 1.1487
    • R/100 = 1.1487 - 1 = 0.1487
    • R = 0.1487 × 100 ≈ 14.87%
Answer:
  • Rate of Interest ≈ 14.87% per annum
Explanation:
  • When the amount doubles, A = 2P. Solve for R by taking the nth root of 2.
  • The rate ensures the principal doubles in the given time.
Problem 9: Successive Rates
Problem: A sum of $4000 is invested at 10% per annum for the first year and 12% per annum for the second year, compounded annually. Find the final amount.
Solution:
  1. Identify the given values:
    • P = $4000
    • Year 1: R1 = 10%
    • Year 2: R2 = 12%
  2. Calculate Amount Year by Year:
    • After Year 1:
      • A1 = P × (1 + R1/100)
      • A1 = 4000 × (1 + 10/100) = 4000 × 1.1 = $4400
    • After Year 2 (using A1 as principal):
      • A2 = A1 × (1 + R2/100)
      • A2 = 4400 × (1 + 12/100) = 4400 × 1.12 = $4928
Alternative Method:
  • Combine rates:
  • A = P × (1 + R1/100) × (1 + R2/100)
  • A = 4000 × (1 + 10/100) × (1 + 12/100)
  • A = 4000 × 1.1 × 1.12
  • A = 4000 × 1.232 = $4928
Answer:
  • Final Amount = $4928
Explanation:
  • Apply each year’s rate sequentially or combine them in one formula.
  • The result is the same, as compounding is multiplicative.
Problem 10: Effective Rate for Two Years
Problem: A loan of $10000 is taken at 8% per annum for the first year and 10% per annum for the second year, compounded annually. Find the effective annual rate for the two-year period.
Solution:
  1. Identify the given values:
    • P = $10000
    • Year 1: R1 = 8%
    • Year 2: R2 = 10%
  2. Calculate Final Amount:
    • A = P × (1 + R1/100) × (1 + R2/100)
    • A = 10000 × (1 + 8/100) × (1 + 10/100)
    • A = 10000 × 1.08 × 1.10
    • A = 10000 × 1.188 = $11880
  3. Find Effective Rate:
    • Effective rate for two years: A = P × (1 + R_eff/100)^2
    • 11880 = 10000 × (1 + R_eff/100)^2
    • (1 + R_eff/100)^2 = 11880 / 10000 = 1.188
    • 1 + R_eff/100 = √1.188 ≈ 1.0899
    • R_eff/100 = 1.0899 - 1 = 0.0899
    • R_eff ≈ 8.99%
Alternative Method:
  • Use successive rate formula:
  • Effective rate = R1 + R2 + (R1 × R2)/100
  • Effective rate = 8 + 10 + (8 × 10)/100
  • = 18 + 0.8 = 18.8% (for two years)
  • Annual effective rate = [(1 + 18.8/100)^(1/2) - 1] × 100
  • = (1.188)^(1/2) - 1 ≈ 0.0899 × 100 ≈ 8.99%
Answer:
  • Effective Annual Rate ≈ 8.99% per annum
Explanation:
  • The effective rate combines the effect of different rates over two years into a single equivalent annual rate.
  • Both methods yield the same result, confirming the calculation.
Key Concepts Summary:
  • Compound Interest Formula: A = P × (1 + R/100)^n; CI = A - P
  • Half-Yearly/Quarterly Compounding: Adjust R and n accordingly (R/2, 2n for half-yearly; R/4, 4n for quarterly).
  • Finding Principal: P = CI / [(1 + R/100)^n - 1]
  • Finding Rate: (1 + R/100) = (A/P)^(1/n)
  • Finding Time: n = log(A/P) / log(1 + R/100) or trial method.
  • Doubling Amount: Solve (1 + R/100)^n = 2.
  • Successive Rates: Multiply factors (1 + R1/100) × (1 + R2/100).
  • Effective Rate: Combines multiple rates into a single equivalent rate.

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