Unit outcomes

By the end of this unit you will be able to:

• Calculate straight-line depreciation.
• Calculate reducing-balance depreciation.
• Calculate the values of [latex]\scriptsize A,~P,~i~[/latex] and [latex]\scriptsize n[/latex].

What you should know

Before you start this unit, make sure you can:

• calculate simple and compound interest using the correct formulae. To revise simple and compound interest you can go over level 3 subject outcome 5.2 and unit 1 of this subject outcome.

Introduction

When the value of an asset such as a vehicle, computer or appliance decreases due to usage, we say it has depreciated. In other words, the asset has lost value over time. An investment or shares in the stock exchange can also depreciate.

The importance of calculating depreciation is that it affects tax calculations for businesses. Businesses treat the depreciation amounts as an expense, and thereby reduce their taxable income. A lower taxable income means that the business will pay less tax to the Revenue Service.

There are two methods to calculate depreciation: simple or straight-line depreciation, and compound or reducing-balance depreciation.

The following terminology is often used when discussing depreciation.

Decay: Another word to describe depreciation.

Book value: The value of an asset after depreciation is taken into account.

Scrap value: The book value of an asset at the end of its useful life. Also called the salvage value.

Straight-line depreciation

In straight-line depreciation the value of an asset depreciates by a constant amount each year. The amount of depreciation is calculated each year as a percentage of the original or principal value of the asset. The asset is then reduced by that amount every year. Straight-line depreciation can be represented by a straight line graph.

Example 3.1

Linda’s mum buys her a new car, which costs [latex]\scriptsize \displaystyle \text{R20}0\text{ }000[/latex], for her [latex]\scriptsize \displaystyle 18\text{th}[/latex] birthday. The car depreciates in value at a rate of [latex]\scriptsize \displaystyle 10\%[/latex] per annum simple depreciation.

1. How much will the car be worth on Linda’s [latex]\scriptsize \displaystyle 21\text{st}[/latex] birthday?
2. Calculate the amount of depreciation each year.

Solutions

1. Write down the simple depreciation formula, list the values that you are given and then solve for the unknown value.
   .
   [latex]\scriptsize \begin{align*}A&=P(1-in)\end{align*}[/latex]
   .
   [latex]\scriptsize \begin{align*}P&=\text{R}200~000\\i&=0.1\\n&=21-18\\&=3\\A&=?\end{align*}[/latex]
   .
   [latex]\scriptsize \begin{align*}A&=200~000(1-(0.1)\cdot 3)\\&=\text{R}140~000\end{align*}[/latex]
2. In three years’ time, Linda’s car will be worth [latex]\scriptsize \displaystyle \text{R}140\text{ }000[/latex]. That is a decrease in total value of [latex]\scriptsize \displaystyle \text{R60 }000[/latex].
   Depreciation per year:
   [latex]\scriptsize \displaystyle \frac{{\text{R}60\text{ }000}}{3}=\text{R}20\text{ }000[/latex].

Example 3.2

A computer was valued at [latex]\scriptsize \displaystyle \text{R}16\text{ }500[/latex] when it was bought. Four years later, its value had depreciated to an amount of [latex]\scriptsize \text{R}10\text{ }200[/latex]. Determine the rate at which the value depreciated using straight-line depreciation.

Solution

Write down the simple depreciation formula and list the values that you are given.

[latex]\scriptsize \displaystyle \begin{align*}A&=10\text{ }200\\P&=16\text{ }500\\n&=4\\i&=?\end{align*}[/latex]

[latex]\scriptsize \displaystyle \begin{align*}10\text{ }200&=16\text{ }500(1-i(4))\\i&=\left( {\displaystyle \frac{{10\text{ }200}}{{16\text{ }500}}-1} \right)\div -4\\&=0.0954...\end{align*}[/latex]

Rate of depreciation:

[latex]\scriptsize \begin{align*}&0.09545...\times 100\\&=9.55\%\end{align*}[/latex]

Exercise 3.1

1. A new smartphone costs [latex]\scriptsize \text{R}19\text{ }000[/latex] and depreciates at [latex]\scriptsize \displaystyle 22\%[/latex] p.a. on a straight-line basis. Determine the book value of the smartphone at the end of each year over a three-year period.
2. A car is valued at [latex]\scriptsize \displaystyle \text{R35}0\text{ }000[/latex]. If it depreciates at [latex]\scriptsize \displaystyle 15\%[/latex] p.a. using straight-line depreciation, calculate the value of the car after five years.
3. Seven years ago, Rocco’s drum kit cost him [latex]\scriptsize \displaystyle \text{R}12\text{ }500[/latex]. It has now been valued at [latex]\scriptsize \displaystyle \text{R}2\text{ }300[/latex]. What rate of simple depreciation does this represent?

The full solutions are at the end of the unit.

Reducing-balance depreciation

Reducing-balance depreciation or compound depreciation results in a higher depreciation expense in the earlier years of ownership of an asset. Compound depreciation is based on an asset’s previous value. Every period, the depreciation will be a percentage of the reduced value (reducing-balance) of the asset. The value of the asset will depreciate by smaller amounts each period. At the end of the period the asset will still have some value; its value will never depreciate to zero.

Compound depreciation can be represented by a decreasing exponential graph.

Example 3.3

Linda’s mum buys her a new car, which costs [latex]\scriptsize \displaystyle \text{R20}0\text{ }000[/latex], for her [latex]\scriptsize \displaystyle 18\text{th}[/latex] birthday. The car depreciates in value at a rate of [latex]\scriptsize \displaystyle 10\%[/latex] per annum using reducing-balance depreciation.

1. How much will the car be worth on Linda’s [latex]\scriptsize \displaystyle 21\text{st}[/latex] birthday?
2. Calculate the amount of depreciation each year.

Solutions

1. Write down the reducing-balance depreciation formula and list the values that you are given; then solve for the unknown value.
   [latex]\scriptsize A=P{{(1-i)}^{n}}[/latex]
   [latex]\scriptsize \begin{align*}P&=200\text{ }000\\i&=0.1\\n&=3\\A&=?\end{align*}[/latex]
   [latex]\scriptsize \begin{align*}A&=200\text{ }000{{(1-0.1)}^{3}}\\&=\text{R}145\text{ 800}\end{align*}[/latex]
2. In three years’ time, Linda’s car will be worth [latex]\scriptsize \text{R}145\text{ 800}[/latex]. That is a decrease in total value of [latex]\scriptsize \displaystyle \text{R54 }200[/latex].
   Depreciation at the end of year [latex]\scriptsize \displaystyle ~1[/latex]: [latex]\scriptsize 0.1\times \text{R}200\text{ }000=\text{R}20\text{ }000[/latex]
   Depreciation at the end of year [latex]\scriptsize \displaystyle 2[/latex]:
   The car is now worth [latex]\scriptsize \text{R}200\text{ }000-\text{R}20\text{ }000=\text{R}180\text{ }000[/latex], so we must calculate the depreciation on the reduced value of the car.
   [latex]\scriptsize 0.1\times \text{R18}0\text{ }000=\text{R18 }000[/latex]
   Depreciation at the end of year [latex]\scriptsize 3[/latex]:
   [latex]\scriptsize 0.1\times \text{R162 }000=\text{R16 2}00[/latex]
   .
   Compared to straight-line depreciation, we can see that in reducing-balance depreciation the depreciation amount decreases each year as the value of the car decreases.

Example 3.4

Simon bought a washing machine two years ago for [latex]\scriptsize \displaystyle \text{R9 }999[/latex] and sold it now for [latex]\scriptsize \displaystyle \text{R}5\text{ }500[/latex]. At what rate did the value of the washing machine depreciate on a reducing-balance method? Give your answer correct to two decimal places.

Solution

Write down known variables and the compound decay formula.

[latex]\scriptsize \begin{align*}P=9\text{ 999}\\i=?\\n=2\\A=5\text{ }500\end{align*}[/latex]
[latex]\scriptsize A=P{{(1-i)}^{n}}[/latex]
Substitute the values and solve for [latex]\scriptsize i[/latex].
[latex]\scriptsize \displaystyle \begin{align*}5\text{ }500=9\text{ }999{{(1-i)}^{2}}\\{{(1-i)}^{2}}=\displaystyle \frac{{5\text{ }500}}{{\text{9 999}}}\\1-i=\sqrt{{\displaystyle \frac{{5\text{ }500}}{{\text{9 999}}}}}\\i=1-\sqrt{{\displaystyle \frac{{5\text{ }500}}{{\text{9 999}}}}}\\=0.258...\end{align*}[/latex]

Rate of depreciation:

[latex]\scriptsize \begin{align*}0.2583...\times 100\\=25.83\%\end{align*}[/latex]

The next example is for enrichment only and is not part of the curriculum.

Example 3.5

Sam’s car cost [latex]\scriptsize \displaystyle \text{R}210\text{ }000[/latex]. After how many years will it be valued at [latex]\scriptsize \text{R}80\text{ }000[/latex] assuming a reducing-balance rate of depreciation of [latex]\scriptsize 15\%[/latex].

Note

Although solving for [latex]\scriptsize n[/latex] (the period of depreciation) is not required in the Assessment Guidelines for this unit, questions of this type have appeared in some past examination papers. For some explanation of how logs (logarithms) work, you could use the internet to read about “Exponents and Logarithms“.

Solution

Write down known variables and compound decay formula.

[latex]\scriptsize \begin{align*}P&=210\text{ 000}\\i&=0.15\\n&=?\\A&=80\text{ 0}00\end{align*}[/latex]
[latex]\scriptsize A=P{{(1-i)}^{n}}[/latex]

Substitute the known values and solve for [latex]\scriptsize n[/latex].

[latex]\scriptsize 80\text{ }000=210\text{ }000{{(1-0.15)}^{n}}[/latex]

We see that [latex]\scriptsize n[/latex] is a power and we cannot make the bases on either side of the equal sign the same. So we must use logs to solve for [latex]\scriptsize n[/latex].

[latex]\scriptsize \displaystyle \begin{align*}{{(0.85)}^{n}}&=\displaystyle \frac{{80\text{ }000}}{{210\text{ }000}}\\\log {{(0.85)}^{n}}&=\log (\displaystyle \frac{8}{{21}})\text{ }\\n\log 0.85&=\log (\displaystyle \frac{8}{{21}})\text{ since }\log {{x}^{n}}=n\log x\\n&=\log (\displaystyle \frac{8}{{21}})\div \log 0.85\end{align*}[/latex]

Use your calculator for this calculation, as follows:

So [latex]\scriptsize n=5.938=6\text{ years}[/latex]

Exercise 3.2

1. The number of pelicans at the Berg River mouth is decreasing at a compound rate of [latex]\scriptsize \displaystyle 12\%[/latex] p.a. If there are currently [latex]\scriptsize \displaystyle 3\text{ }200[/latex] pelicans in the wetlands of the Berg River mouth, what will the population be in five years?
2. The population of Bonduel decreases at a reducing-balance rate of [latex]\scriptsize \displaystyle 9.5\%[/latex] per annum as people migrate to the cities. Calculate the decrease in population over a period of five years if the initial population was [latex]\scriptsize \displaystyle 2\text{ }178\text{ }000[/latex].
3. After [latex]\scriptsize 15[/latex] years, an aeroplane is worth [latex]\scriptsize \displaystyle \frac{1}{6}[/latex] of its original value. What is the annual rate of depreciation?

Question 4 is for enrichment only

4. Andy’s car cost [latex]\scriptsize \displaystyle \text{R30}0\text{ }000[/latex]. After how many years will it be valued at [latex]\scriptsize \text{R12}0\text{ }000[/latex] assuming a reducing-balance rate of depreciation of [latex]\scriptsize 12\%[/latex].

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to calculate straight-line depreciation.
• How to calculate reducing-balance depreciation.
• How to calculate the unknown variables in depreciation questions.

Unit 3: Assessment

Suggested time to complete: 20 minutes

1. Fiona buys a DStv satellite dish for [latex]\scriptsize \displaystyle \text{R}3\text{ }000[/latex]. Due to weathering, its value depreciates simply at [latex]\scriptsize \displaystyle 15\%[/latex] per annum. After how long will the satellite dish have a book value of zero?
2. Harry’s grandpa is very fashionable. Harry wants to buy his grandpa’s jacket for [latex]\scriptsize \displaystyle \text{R}1\text{ }000[/latex]. His grandpa is quite pleased with the offer, seeing that it only depreciated at a rate of [latex]\scriptsize \displaystyle 3\%[/latex] per year using the straight-line method. Grandpa bought the jacket five years ago. What did grandpa pay for the jacket then?
3. Steven invested [latex]\scriptsize \displaystyle \text{R}50\text{ }000[/latex] in an investment scheme. His investment did not perform well and depreciated on a reducing balance basis at the rate of [latex]\scriptsize 6\%[/latex] per annum each year for the first five years. At the end of the five years he withdrew [latex]\scriptsize \text{R}10\text{ }000[/latex] for personal reasons. After that his investment grew at a rate of [latex]\scriptsize \displaystyle 12\%[/latex] p.a. compounded quarterly.
   1. Determine the value of Steven’s investment at the end of five years, before he withdrew the [latex]\scriptsize \text{R}10\text{ }000[/latex].
   2. Determine the value of his investment at the end of [latex]\scriptsize 10[/latex] years.
4. A [latex]\scriptsize \displaystyle 20\text{ kg}[/latex] watermelon consists of [latex]\scriptsize \displaystyle 98\%[/latex] water. If it is left outside in the sun it loses [latex]\scriptsize \displaystyle 3\%[/latex] of its water each day. How much does it weigh after a month of [latex]\scriptsize \displaystyle 31[/latex] days?

The full solutions are at the end of the unit.

Unit 3: Solutions

Exercise 3.1

1. Calculate depreciation amount first.
   Depreciation:
   [latex]\scriptsize 19\text{ }000\times \displaystyle \frac{{22}}{{100}}=4\text{ }180[/latex]
   Therefore the smartphone depreciates by [latex]\scriptsize \text{R}4\text{ }180[/latex] every year.
   Then find the value of the phone at the end of each year.
   Book value at end of year [latex]\scriptsize 1[/latex]: [latex]\scriptsize \text{R19 00}0-\text{R}4\text{ }180=\text{R1}4\text{ }820[/latex]
   Book value at end of year [latex]\scriptsize 2[/latex]: [latex]\scriptsize \text{R1}4\text{ }820-\text{R}4\text{ 18}0=\text{R10 64}0[/latex]
   Book value at end of year [latex]\scriptsize 3[/latex]: [latex]\scriptsize \text{R10 64}0-\text{R4 18}0=\text{R6 46}0[/latex]
2. .
   [latex]\scriptsize \begin{align*}A&=35\text{0 00}0(1-0.15\times 5)\\&=\text{R87 50}0\end{align*}[/latex]
3. .
   [latex]\scriptsize \begin{align*}2\text{ }300&=12\text{ 50}0(1-7i)\\\displaystyle \frac{{2\text{ 300}}}{{12\text{ 500}}}-1&=-7i\\i&=0.116...\end{align*}[/latex]
   Rate of depreciation is [latex]\scriptsize 11.66\%[/latex]

Back to Exercise 3.1

Exercise 3.2

1. .
   [latex]\scriptsize \begin{align*}A&=3\text{ }200{{(1-0.12)}^{5}}\\&\approx 1689\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}A&=2\text{ }178\text{ }000{{(1-0.095)}^{5}}\\&\approx 1\text{ }322\text{ }211\end{align*}[/latex]
3. .
   [latex]\scriptsize \begin{align*}\displaystyle \frac{1}{6}P&=P{{(1-i)}^{{15}}}\\i&=0.112...\end{align*}[/latex]
   Annual rate of depreciation was [latex]\scriptsize 11.26\%[/latex] .
4. .
   [latex]\scriptsize \displaystyle \begin{align*}{{(1-0.12)}^{n}}&=\displaystyle \frac{{120\text{ }000}}{{300\text{ }000}}\\\log {{(0.88)}^{n}}&=\log (\displaystyle \frac{2}{5})\text{ }\\n\log 0.88&=\log (\displaystyle \frac{2}{5})\text{ since }\log {{x}^{n}}=n\log x\\n&=\log (\displaystyle \frac{2}{5})\div \log 0.88\\&=7.167...\\\therefore n&\approx 7\displaystyle \frac{1}{6}\text{ years}\end{align*}[/latex]
   Note: You cannot round down to seven years as the car will be valued at over [latex]\scriptsize \text{R}120\text{ }000[/latex] at seven years. So you need to include the fractional part of the year into your answer. [latex]\scriptsize 0.167...\times 12=2[/latex] months and [latex]\scriptsize \displaystyle \frac{2}{{12}}=\displaystyle \frac{1}{6}[/latex] of a year.

Back to Exercise 3.2

Unit 3: Assessment

1. .
   [latex]\scriptsize \begin{align*}A&=P(1-in)\\0&=3\text{ }000(1-0.15(n))\\0&=3\text{ }000-450n\\n&=\displaystyle \frac{{20}}{3}\\&=6\displaystyle \frac{2}{3}\\&=6\text{ years and }8\text{ months}\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}A&=P(1-in)\\P(1-0.03(5))&=1\text{ }000\\P&=\text{R}1\text{ }176.47\end{align*}[/latex]
3. .
   1. The value of Steven’s investment at the end of five years, before he withdrew the [latex]\scriptsize \text{R}10\text{ }000[/latex]:
      [latex]\scriptsize \begin{align*}A&=P{{(1-i)}^{n}}\\&=50\text{ }000{{(1-0.06)}^{5}}\\&=\text{R}36\text{ }695.20\end{align*}[/latex]
   2. At the end of [latex]\scriptsize 10[/latex] years:
      New principal amount is [latex]\scriptsize \text{R}36\text{ }695.20-\text{R}10\text{ }000=\text{R}26\text{ }695.20[/latex]
      [latex]\scriptsize \begin{align*}A&=26\text{ }695.20{{(1+\displaystyle \frac{{0.12}}{4})}^{{5\times 4}}}\\&=\text{R}48\text{ 214}\text{.5}0\end{align*}[/latex]
4. The amount of water at the start: [latex]\scriptsize 0.98\times 20=19.6[/latex]. Therefore, [latex]\scriptsize 0.4\text{ kg}[/latex] makes up the remainder of the watermelon weight.
   At the end of [latex]\scriptsize \displaystyle 31[/latex] days the amount of water:
   [latex]\scriptsize \begin{align*}A&=P{{(1-i)}^{n}}\\&=19.6{{(1-0.03)}^{{31}}}\\&=7.62\text{ kg}\end{align*}[/latex]
   Watermelon in total weighs:
   [latex]\scriptsize 7.6\text{ kg}+0.4\text{ kg}=8.02\text{ kg}[/latex]

Back to Unit 3: Assessment