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A school purchased a printer for R250 000. The printer depreciates at 10% p.a. on a reducing balance method. The printer will be replaced after 5 years. The inflation rate during this period is 8% per annum. Calculate the book value of the printer at the end of the 5 year period.

Mathematics

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5.1.1 Calculate the book value of the printer at the end of the 5 year period.

  1. Use the reducing balance depreciation formula.

    The formula for depreciation on a reducing balance is A=P(1−i)nA = P(1 - i)^n, where AA is the book value, PP is the initial cost, ii is the depreciation rate, and nn is the number of years.

    A=R250 000(1−0.10)5A = R250\,000(1 - 0.10)^5
  2. Calculate the book value.

    A=R250 000(0.90)5A = R250\,000(0.90)^5 A=R250 000(0.59049)A = R250\,000(0.59049) A=R147 622.50A = R147\,622.50

    The book value of the printer at the end of 5 years is R147 622.50\boxed{\text{R147 622.50}}.

    5.1.2 Calculate the cost of a similar printer when the 5 year period expires.

  3. Use the compound interest formula for inflation.

    The formula for future value with inflation is A=P(1+i)nA = P(1 + i)^n, where AA is the future cost, PP is the current cost, ii is the inflation rate, and nn is the number of years.

    A=R250 000(1+0.08)5A = R250\,000(1 + 0.08)^5
  4. Calculate the future cost.

    A=R250 000(1.08)5A = R250\,000(1.08)^5 A=R250 000(1.4693280768)A = R250\,000(1.4693280768) A=R367 332.02A = R367\,332.02

    The cost of a similar printer after 5 years is R367 332.02\boxed{\text{R367 332.02}}.

    5.1.3 Calculate the amount of money that must be saved in order to have enough money to replace the printer in 5 years.

  5. Calculate the selling price of the used printer.

    The school sells the used printer for 95% of its book value (from 5.1.1).

    Selling Price=0.95×R147 622.50\text{Selling Price} = 0.95 \times R147\,622.50 Selling Price=R140 241.38\text{Selling Price} = R140\,241.38
  6. Calculate the amount to be saved.

    The amount to be saved is the cost of the new printer (from 5.1.2) minus the selling price of the old printer.

    Amount to be Saved=R367 332.02−R140 241.38\text{Amount to be Saved} = R367\,332.02 - R140\,241.38 Amount to be Saved=R227 090.64\text{Amount to be Saved} = R227\,090.64

    The amount of money that must be saved is R227 090.64\boxed{\text{R227 090.64}}.

    5.2.1 Determine which bank offers the best interest rate.

  7. State the effective interest rate for Bank A.

    Bank A offers an effective interest rate of 12%12\% per annum.

  8. Calculate the effective interest rate for Bank B.

    Bank B offers a nominal interest rate of 11.5%11.5\% p.a., compounded monthly.
    The formula for the effective annual rate is ieff=(1+inomm)m−1i_{eff} = \left(1 + \frac{i_{nom}}{m}\right)^m - 1.
    Here, inom=0.115i_{nom} = 0.115 and m=12m = 12.

    ieff=(1+0.11512)12−1i_{eff} = \left(1 + \frac{0.115}{12}\right)^{12} - 1 ieff=(1+0.0095833333)12−1i_{eff} = (1 + 0.0095833333)^{12} - 1 ieff=(1.0095833333)12−1i_{eff} = (1.0095833333)^{12} - 1 ieff=1.121339−1i_{eff} = 1.121339 - 1 ieff=0.121339i_{eff} = 0.121339 ieff=12.13%i_{eff} = 12.13\%
  9. Compare the effective interest rates.

    Bank A effective rate: 12%12\%
    Bank B effective rate: 12.13%12.13\%
    Since 12.13%>12%12.13\% > 12\%, Bank B offers a better interest rate.
    Bank B\boxed{\text{Bank B}} offers the best interest rate.

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