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Convert an interest rate of 14% p.a. compounded monthly to an effective interest rate per annum.

Mathematics

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Part 2

Part 2.1· 3 steps

  1. Convert the nominal interest rate of 13% p.a.

    compounded quarterly to an equivalent nominal interest rate compounded monthly.
    We use the formula for equivalent interest rates:

    (1+inom1m1)m1=(1+inom2m2)m2\left(1 + \frac{i_{nom1}}{m_1}\right)^{m_1} = \left(1 + \frac{i_{nom2}}{m_2}\right)^{m_2}

    Given inom1=0.13i_{nom1} = 0.13 (13%), m1=4m_1 = 4 (quarterly). We want to find inom2i_{nom2} (monthly rate), with m2=12m_2 = 12.

    (1+0.134)4=(1+imonthly12)12\left(1 + \frac{0.13}{4}\right)^4 = \left(1 + \frac{i_{monthly}}{12}\right)^{12}

  2. Calculate the left side of the equation.

    (1+0.0325)4=(1.0325)4=1.13647990625(1 + 0.0325)^4 = (1.0325)^4 = 1.13647990625

  3. Solve for imonthlyi_{monthly}.

    1.13647990625=(1+imonthly12)121.13647990625 = \left(1 + \frac{i_{monthly}}{12}\right)^{12}

    Take the 12th root of both sides:

    (1.13647990625)1/12=1+imonthly12(1.13647990625)^{1/12} = 1 + \frac{i_{monthly}}{12} 1.010720333=1+imonthly121.010720333 = 1 + \frac{i_{monthly}}{12} 0.010720333=imonthly120.010720333 = \frac{i_{monthly}}{12} imonthly=0.010720333×12i_{monthly} = 0.010720333 \times 12 imonthly=0.128643996i_{monthly} = 0.128643996

    Expressed as a percentage, this is 12.8643996%≈12.86%12.8643996\% \approx 12.86\%.
    Thus, the effective interest rate is 12.86% p.a. compounded monthly\boxed{\text{12.86\% p.a. compounded monthly}}.

Part 2.2· 3 steps

  1. Identify the known values for the present value annuity formula.

    Loan amount (PP) = R600 000
    Monthly instalment (xx) = R9 000
    Monthly interest rate (ii) = 0.12864399612=0.010720333\frac{0.128643996}{12} = 0.010720333
    We need to find the number of payments (nn).
    The present value annuity formula is:

    P=x[1−(1+i)−n]iP = \frac{x[1 - (1 + i)^{-n}]}{i}

  2. Substitute the values into the formula.

    600000=9000[1−(1+0.010720333)−n]0.010720333600000 = \frac{9000[1 - (1 + 0.010720333)^{-n}]}{0.010720333}

  3. Solve for nn.

    600000×0.010720333=9000[1−(1.010720333)−n]600000 \times 0.010720333 = 9000[1 - (1.010720333)^{-n}] 6432.1998=9000[1−(1.010720333)−n]6432.1998 = 9000[1 - (1.010720333)^{-n}] 6432.19989000=1−(1.010720333)−n\frac{6432.1998}{9000} = 1 - (1.010720333)^{-n} 0.71468886=1−(1.010720333)−n0.71468886 = 1 - (1.010720333)^{-n} (1.010720333)−n=1−0.71468886(1.010720333)^{-n} = 1 - 0.71468886 (1.010720333)−n=0.28531114(1.010720333)^{-n} = 0.28531114

    Take the natural logarithm of both sides:

    −nln⁡(1.010720333)=ln⁡(0.28531114)-n \ln(1.010720333) = \ln(0.28531114) −n×0.01066310=−1.254790-n \times 0.01066310 = -1.254790 n=−1.254790−0.01066310n = \frac{-1.254790}{-0.01066310} n=117.674n = 117.674

    Since nn is not an integer, there will be 117 full instalments of R9 000.
    The number of instalments of R9 000 that must be paid is 117\boxed{\text{117}}.

Part 2.3· 2 steps

  1. Calculate the outstanding balance after 117 payments.

    The balance outstanding after kk payments (BkB_k) is given by:

    Bk=P(1+i)k−x(1+i)k−1iB_k = P(1+i)^k - x \frac{(1+i)^k - 1}{i}

    Using P=600000P = 600000, x=9000x = 9000, i=0.010720333i = 0.010720333, and k=117k = 117:

    B117=600000(1.010720333)117−9000(1.010720333)117−10.010720333B_{117} = 600000(1.010720333)^{117} - 9000 \frac{(1.010720333)^{117} - 1}{0.010720333}

    First, calculate (1.010720333)117≈3.437990(1.010720333)^{117} \approx 3.437990.

    B117=600000(3.437990)−90003.437990−10.010720333B_{117} = 600000(3.437990) - 9000 \frac{3.437990 - 1}{0.010720333} B117=2062794−90002.4379900.010720333B_{117} = 2062794 - 9000 \frac{2.437990}{0.010720333} B117=2062794−9000×227.4170B_{117} = 2062794 - 9000 \times 227.4170 B117=2062794−2046753B_{117} = 2062794 - 2046753 B117=16041B_{117} = 16041

  2. Calculate the final payment.

    The final payment will be this outstanding balance plus one month's interest. This payment occurs at the end of the 118th month.

    Final payment=B117(1+i)\text{Final payment} = B_{117}(1+i) Final payment=16041(1+0.010720333)\text{Final payment} = 16041(1 + 0.010720333) Final payment=16041(1.010720333)\text{Final payment} = 16041(1.010720333) Final payment=16213.38\text{Final payment} = 16213.38

    The final payment will be R16 213.38\boxed{\text{R16 213.38}}.

Part 2.4· 1 step

  1. Calculate the total amount paid.

    Total cost = (Number of full payments ×\times Instalment amount) + Final payment

    Total cost=(117×R9000)+R16213.38\text{Total cost} = (117 \times \text{R}9000) + \text{R}16213.38 Total cost=R1053000+R16213.38\text{Total cost} = \text{R}1053000 + \text{R}16213.38 Total cost=R1069213.38\text{Total cost} = \text{R}1069213.38

    The car cost Jake in total R1 069 213.38\boxed{\text{R1 069 213.38}}.

Part 3

Part 3.1· 3 steps

  1. Identify the nominal interest rate and compounding frequency.

    Nominal rate (inomi_{nom}) = 0.14 (14%)
    Number of compounding periods per year (mm) = 12 (monthly)

  2. Use the formula for effective annual interest rate.

    (1+ieff)=(1+inomm)m(1 + i_{eff}) = \left(1 + \frac{i_{nom}}{m}\right)^m (1+ieff)=(1+0.1412)12(1 + i_{eff}) = \left(1 + \frac{0.14}{12}\right)^{12} (1+ieff)=(1+0.01166666666)12(1 + i_{eff}) = (1 + 0.01166666666)^{12} (1+ieff)=(1.01166666666)12(1 + i_{eff}) = (1.01166666666)^{12} (1+ieff)=1.149342026(1 + i_{eff}) = 1.149342026

  3. Calculate the effective annual rate.

    ieff=1.149342026−1i_{eff} = 1.149342026 - 1 ieff=0.149342026i_{eff} = 0.149342026

    Expressed as a percentage, this is 14.9342026%≈14.93%14.9342026\% \approx 14.93\%.
    The effective interest rate per annum is 14.93%\boxed{\text{14.93\%}}.

Part 3.2· 3 steps

  1. Identify the known values for the future value annuity formula.

    Future Value (FVFV) = R7003007.22
    Monthly payment (xx) = R800
    Nominal interest rate = 16% p.a. compounded monthly.
    Monthly interest rate (ii) = 0.1612=0.01333333333\frac{0.16}{12} = 0.01333333333
    We need to find the number of payments (nn).
    The future value annuity formula is:

    FV=x[(1+i)n−1]iFV = \frac{x[(1 + i)^n - 1]}{i}

  2. Substitute the values into the formula.

    7003007.22=800[(1+0.01333333333)n−1]0.013333333337003007.22 = \frac{800[(1 + 0.01333333333)^n - 1]}{0.01333333333}

  3. Solve for nn.

    7003007.22×0.01333333333=800[(1.01333333333)n−1]7003007.22 \times 0.01333333333 = 800[(1.01333333333)^n - 1] 93373.4296=800[(1.01333333333)n−1]93373.4296 = 800[(1.01333333333)^n - 1] 93373.4296800=(1.01333333333)n−1\frac{93373.4296}{800} = (1.01333333333)^n - 1 116.716787=(1.01333333333)n−1116.716787 = (1.01333333333)^n - 1 117.716787=(1.01333333333)n117.716787 = (1.01333333333)^n

    Take the natural logarithm of both sides:

    ln⁡(117.716787)=nln⁡(1.01333333333)\ln(117.716787) = n \ln(1.01333333333) 4.768390=n×0.01324404.768390 = n \times 0.0132440 n=4.7683900.0132440n = \frac{4.768390}{0.0132440} n=360n = 360

    Skomota made 360\boxed{\text{360}} payments.

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