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A farmer is buying a tractor for R 750 000. He can afford a 15% deposit and will finance the rest over 10 years at an annual interest rate of 8.6%, compounded monthly. Calculate the deposit amount and

Mathematics

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Step-by-step solution

5 steps

Here are the calculations for the deposit and monthly instalments:

1. Calculate the deposit amount.

  1. Identify the cost of the tractor and the deposit percentage.

    The tractor costs R 750 000.
    The deposit is 15% of the tractor cost.

  2. Calculate the deposit amount.

    Deposit amount=15%×R 750 000\text{Deposit amount} = 15\% \times \text{R } 750\,000 Deposit amount=0.15×750 000\text{Deposit amount} = 0.15 \times 750\,000 Deposit amount=R 112 500\text{Deposit amount} = \text{R } 112\,500

    The deposit amount is R 112 500\boxed{\text{R } 112\,500}.

    2. Determine his monthly instalments.

  3. Calculate the loan amount.

    The loan amount is the tractor cost minus the deposit.

    Loan amount=R 750 000−R 112 500\text{Loan amount} = \text{R } 750\,000 - \text{R } 112\,500 Loan amount=R 637 500\text{Loan amount} = \text{R } 637\,500
  4. Determine the monthly interest rate and total number of payments.

    Annual interest rate (rr) = 8.6% = 0.086
    Compounded monthly, so the monthly interest rate (ii) is:

    i=r12=0.08612i = \frac{r}{12} = \frac{0.086}{12}

    Loan period = 10 years
    Total number of payments (nn) = 10 years ×\times 12 months/year = 120 months

  5. Use the present value of an annuity formula to find the monthly instalment (xx).

    The formula is:

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

    Rearrange to solve for xx:

    x=P⋅i1−(1+i)−nx = \frac{P \cdot i}{1 - (1 + i)^{-n}}

    Substitute the values: P=R 637 500P = \text{R } 637\,500, i=0.08612i = \frac{0.086}{12}, n=120n = 120.

    x=637 500×0.086121−(1+0.08612)−120x = \frac{637\,500 \times \frac{0.086}{12}}{1 - \left(1 + \frac{0.086}{12}\right)^{-120}} x=637 500×0.00716666666...1−(1.00716666666...)−120x = \frac{637\,500 \times 0.00716666666...}{1 - (1.00716666666...)^{-120}} x=4562.51−0.420697902...x = \frac{4562.5}{1 - 0.420697902...} x=4562.50.579302098...x = \frac{4562.5}{0.579302098...} x≈7876.00000...x \approx 7876.00000...

    Rounding to two decimal places for currency:

    x=R 7876.00x = \text{R } 7876.00

    His monthly instalments will be R 7876.00\boxed{\text{R } 7876.00}.

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