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Dean bought a vehicle for R120 000 at the end of 2008. The value depreciated by 8% p.a. Dean invested R55 000 at 10.5% p.a. compounded yearly. Calculate the values of the vehicle and investment at the

Mathematics

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3 parts
a

R67 002\text{R}67\,002

b

R90 610\text{R}90\,610

c

2013\text{2013}

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Here's the solution to Question 3.

Part 3

Dean bought a vehicle for R120 000 at the end of the year 2008. The value of the vehicle depreciated by 8% p.a.
At the end of 2008 Dean invested R55 000 at 10.5% p.a. compounded yearly.

Formulae:
Investment: A=P(1+i)nA = P(1+i)^n
Depreciation: A=P(1−i)nA = P(1-i)^n

Given table:

At the end of:200920102011201220132014
Value of vehicleR 101 568R 93 443R 85 967R 79 090R 72 763(a)
Value of investmentR 55 000R 60 775R 67 157R 74 208R 82 000(b)
(c)

The question asks to use the given information and the table to answer the questions that follow. The table has missing values (a), (b), and (c). We need to calculate these values.

Part a

R67 002\text{R}67\,002

Value of vehicle at the end of 2014

  1. Identify the given values for depreciation.

    The initial value of the vehicle P=R120 000P = \text{R}120\,000 (at the end of 2008).
    The depreciation rate i=8%=0.08i = 8\% = 0.08.
    We need to find the value at the end of 2014. The number of years nn from the end of 2008 to the end of 2014 is 2014−2008=62014 - 2008 = 6 years.

  2. Apply the depreciation formula.

    The formula for depreciation is A=P(1−i)nA = P(1-i)^n.
    Substitute the values:

    A=120 000(1−0.08)6A = 120\,000(1 - 0.08)^6 A=120 000(0.92)6A = 120\,000(0.92)^6 A=120 000(0.60635506)A = 120\,000(0.60635506) A=72 762.6072A = 72\,762.6072

  3. Round the value to the nearest rand as per the note.

    A≈R72 763A \approx \text{R}72\,763

    However, looking at the table, the value for 2013 is R72 763. This implies that the value (a) for 2014 should be calculated from the 2013 value or from the original value for 6 years. Let's re-check the table values.
    The table shows the value of the vehicle at the end of 2013 as R72 763.
    To find the value at the end of 2014, we can depreciate the 2013 value for one more year.
    P2013=R72 763P_{2013} = \text{R}72\,763.
    n=1n = 1 year (from end of 2013 to end of 2014).

    A2014=P2013(1−i)1A_{2014} = P_{2013}(1-i)^1 A2014=72 763(1−0.08)A_{2014} = 72\,763(1 - 0.08) A2014=72 763(0.92)A_{2014} = 72\,763(0.92) A2014=67 002.00A_{2014} = 67\,002.00

  4. Round the value to the nearest rand.

    A2014≈R67 002A_{2014} \approx \text{R}67\,002

    So, the value for (a) is R67 002\boxed{\text{R}67\,002}.

Part b

R90 610\text{R}90\,610

Value of investment at the end of 2014

  1. Identify the given values for investment.

    The initial investment P=R55 000P = \text{R}55\,000 (at the end of 2008).
    The interest rate i=10.5%=0.105i = 10.5\% = 0.105.
    We need to find the value at the end of 2014. The number of years nn from the end of 2008 to the end of 2014 is 2014−2008=62014 - 2008 = 6 years.

  2. Apply the investment formula.

    The formula for investment is A=P(1+i)nA = P(1+i)^n.
    Substitute the values:

    A=55 000(1+0.105)6A = 55\,000(1 + 0.105)^6 A=55 000(1.105)6A = 55\,000(1.105)^6 A=55 000(1.8289006)A = 55\,000(1.8289006) A=100 590.033A = 100\,590.033

  3. Round the value to the nearest rand as per the note.

    A≈R100 590A \approx \text{R}100\,590

    However, looking at the table, the value for 2013 is R82 000.
    To find the value at the end of 2014, we can compound the 2013 value for one more year.
    P2013=R82 000P_{2013} = \text{R}82\,000.
    n=1n = 1 year (from end of 2013 to end of 2014).

    A2014=P2013(1+i)1A_{2014} = P_{2013}(1+i)^1 A2014=82 000(1+0.105)A_{2014} = 82\,000(1 + 0.105) A2014=82 000(1.105)A_{2014} = 82\,000(1.105) A2014=90 610A_{2014} = 90\,610

  4. Round the value to the nearest rand.

    A2014=R90 610A_{2014} = \text{R}90\,610

    So, the value for (b) is R90 610\boxed{\text{R}90\,610}.

Part c

2013\text{2013}

Determine in which year the value of the investment first exceeded the value of the vehicle.

  1. Compare the values in the table for each year.

    We need to find the first year where "Value of investment" > "Value of vehicle".

    • 2009:
      • Vehicle: R101 568
      • Investment: R55 000
      • Investment < Vehicle
    • 2010:
      • Vehicle: R93 443
      • Investment: R60 775
      • Investment < Vehicle
    • 2011:
      • Vehicle: R85 967
      • Investment: R67 157
      • Investment < Vehicle
    • 2012:
      • Vehicle: R79 090
      • Investment: R74 208
      • Investment < Vehicle
    • 2013:
      • Vehicle: R72 763
      • Investment: R82 000
      • Investment > Vehicle

  2. Identify the first year where the condition is met.

    The value of the investment first exceeded the value of the vehicle in the year 2013.

    So, the answer for (c) is 2013\boxed{\text{2013}}.

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