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An investment of GHS 10,000 is being considered for two options. Option A offers an 8% annual interest rate compounded quarterly. Option B offers a 7.5% annual interest rate compounded monthly. Both i

Mathematics

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Step-by-step solution

5 steps
  1. Calculate the final amount for Option A.

    The formula for compound interest is A=P(1+rn)ntA = P \left(1 + \frac{r}{n}\right)^{nt}.
    For Option A:
    P=GHS 10,000P = \text{GHS } 10,000
    r=8%=0.08r = 8\% = 0.08
    n=4n = 4 (compounded quarterly)
    t=5t = 5 years

    AA=10000(1+0.084)4×5A_A = 10000 \left(1 + \frac{0.08}{4}\right)^{4 \times 5} AA=10000(1+0.02)20A_A = 10000 (1 + 0.02)^{20} AA=10000(1.02)20A_A = 10000 (1.02)^{20} AA=10000×1.4859473959A_A = 10000 \times 1.4859473959 AA=14859.47A_A = 14859.47

    The final amount for Option A is GHS 14859.47\boxed{\text{GHS } \textbf{14859.47}}.

  2. Calculate the final amount for Option B.

    For Option B:
    P=GHS 10,000P = \text{GHS } 10,000
    r=7.5%=0.075r = 7.5\% = 0.075
    n=12n = 12 (compounded monthly)
    t=5t = 5 years

    AB=10000(1+0.07512)12×5A_B = 10000 \left(1 + \frac{0.075}{12}\right)^{12 \times 5} AB=10000(1+0.00625)60A_B = 10000 (1 + 0.00625)^{60} AB=10000(1.00625)60A_B = 10000 (1.00625)^{60} AB=10000×1.449948408A_B = 10000 \times 1.449948408 AB=14499.48A_B = 14499.48

    The final amount for Option B is GHS 14499.48\boxed{\text{GHS } \textbf{14499.48}}.

  3. Determine which option yields more profit and by how much.

    Profit for Option A = Final Amount - Principal

    PA=14859.47−10000=4859.47P_A = 14859.47 - 10000 = 4859.47

    Profit for Option B = Final Amount - Principal

    PB=14499.48−10000=4499.48P_B = 14499.48 - 10000 = 4499.48

    Option A yields more profit.
    Difference in profit = PA−PB=4859.47−4499.48=359.99P_A - P_B = 4859.47 - 4499.48 = 359.99
    Option A yields more profit by GHS 359.99\boxed{\text{GHS } \textbf{359.99}}.

  4. Calculate the time for the investment to double under Option A.

    The formula given is t=ln⁡(2)nln⁡(1+rn)t = \frac{\ln(2)}{n \ln\left(1+\frac{r}{n}\right)}.
    For Option A:
    r=0.08r = 0.08
    n=4n = 4

    t=ln⁡(2)4ln⁡(1+0.084)t = \frac{\ln(2)}{4 \ln\left(1+\frac{0.08}{4}\right)} t=ln⁡(2)4ln⁡(1.02)t = \frac{\ln(2)}{4 \ln(1.02)} t=0.6931474×0.0198026t = \frac{0.693147}{4 \times 0.0198026} t=0.6931470.0792104t = \frac{0.693147}{0.0792104} t=8.7506t = 8.7506

    It would take approximately 8.75 years\boxed{\textbf{8.75 years}} for the investment to double.

  5. Discuss the relationship between compounding frequency and final amount.

    Generally, for a given principal, annual interest rate, and time period, a higher compounding frequency (more times per year) leads to a higher final amount. This is because interest is calculated and added to the principal more often, allowing subsequent interest calculations to be based on a larger principal, leading to greater overall growth.

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