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Explain the compound interest and depreciation formulas and apply them to calculate the value of a vehicle and an investment.

Mathematics

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6 parts
a

R67 002\text{R}67\,002

b

R90 610\text{R}90\,610

c

2013\text{2013}

3.6

2013\text{2013}

3.7

R47 237\text{R}47\,237

3.8

R35 610\text{R}35\,610

Step-by-step solution

12 steps

Here are the solutions to the questions:

Part 3.1

The formula Dean used to calculate the value of his vehicle during depreciation is:

A=P(1−i)nA = P(1-i)^n

Where:

  • AA is the future value
  • PP is the principal amount (initial value)
  • ii is the depreciation rate per period
  • nn is the number of periods (years)

Part 3.2

The formula Dean used to calculate the value of his investment is:

A=P(1+i)nA = P(1+i)^n

Where:

  • AA is the future value
  • PP is the principal amount (initial investment)
  • ii is the interest rate per period
  • nn is the number of periods (years)

Part 3.3

Part a

R67 002\text{R}67\,002

Value of vehicle at the end of 2014

  1. Identify the value of the vehicle at the end of 2013 and the depreciation rate.

    Value at end of 2013 (P2013P_{2013}) = R72 763
    Depreciation rate (ii) = 8% = 0.08
    Number of years (nn) = 1 (from end of 2013 to end of 2014)

  2. Apply the depreciation formula for one year.

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

  3. Round the value to the nearest rand.

    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 value of the investment at the end of 2013 and the interest rate.

    Value at end of 2013 (P2013P_{2013}) = R82 000
    Interest rate (ii) = 10.5% = 0.105
    Number of years (nn) = 1 (from end of 2013 to end of 2014)

  2. Apply the investment formula for one year.

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

  3. Round the value to the nearest rand.

    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 of the vehicle and investment for each year from the table.

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

  2. Identify the first year where the investment value is greater than the vehicle value.

    The investment value first exceeded the vehicle value at the end of 2013.
    The answer for (c) is 2013\boxed{\text{2013}}.

Part 3.4

I cannot draw a graph. However, here are the data points for the value of Dean's vehicle from 2008 to 2014, which you can use to plot the graph:

  • 2008: R120 000
  • 2009: R101 568
  • 2010: R93 443
  • 2011: R85 967
  • 2012: R79 090
  • 2013: R72 763
  • 2014: R67 002

Part 3.5

I cannot draw a graph. However, here are the data points for the value of Dean's investment from 2008 to 2014, which you can use to plot the graph on the same axes as the vehicle value:

  • 2008: R55 000
  • 2009: R60 775
  • 2010: R67 157
  • 2011: R74 208
  • 2012: R82 000
  • 2013: R90 610
  • 2014: R100 590 (calculated using 55000(1.105)655000(1.105)^6)

Part 3.6

2013\text{2013}

Assuming the question meant "In which year did Dean's investment first exceed the value of his vehicle?", based on the comparison in Question 3.3 (c):
The year in which Dean's investment first exceeded the value of his vehicle was 2013\boxed{\text{2013}}.

Part 3.7

R47 237\text{R}47\,237

  1. Identify the value of the vehicle at the end of 2008 and 2013.

    Value at end of 2008 = R120 000
    Value at end of 2013 = R72 763

  2. Calculate the depreciation.

    Depreciation = Value at end of 2008 - Value at end of 2013
    Depreciation = R120 000 - R72 763
    Depreciation = R47 237

    The vehicle depreciated by R47 237\boxed{\text{R}47\,237} between 2008 and 2013.

Part 3.8

R35 610\text{R}35\,610

  1. Identify the value of the investment at the end of 2008 and 2014.

    Value at end of 2008 = R55 000
    Value at end of 2014 = R90 610 (from Question 3.3 (b))

  2. Calculate the growth.

    Growth = Value at end of 2014 - Value at end of 2008
    Growth = R90 610 - R55 000
    Growth = R35 610

    Dean's investment grew by R35 610\boxed{\text{R}35\,610} between 2008 and 2014.

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