Growth and decay. VCEcoverage Area of study. Units 3 & 4 Business related mathematics

Transcription

1 Growth and decay VCEcoverage Area of study Units 3 & Business related mathematics In this cha chapter A Growth and decay functions B Compound interest formula C Finding time in compound interest using trial and error D Finding time in compound interest using logarithms E Flat rate depreciation F Reducing balance depreciation G Unit cost depreciation H Inflation

2 66 Further Mathematics Introduction to growth and decay Certain quantities in (for example) nature and business may change in a uniform (constant) way over time. A change may be an increase, as in the case of the value of an investment such as a house, or it may be a decrease, like the fall in the population of an endangered species. Growth and decay can often be modelled by equations or graphs. These in turn can be used to analyse the situations being modelled and to make predictions about them. For instance, if we know an equation that relates the falling value of a car to time, the future value of the car can be determined. (Falling values of this type are called depreciation.) This chapter investigates the general principles of growth and decay and takes a more detailed look at specific examples in the business world, such as compound interest, depreciation and inflation. Growth and decay functions Straight line and exponential growth If a quantity increases in size over a period of time, it is growing. This growth process may be straight line growth or exponential growth. Straight line growth A quantity may increase by a fixed amount for each time unit; that is, a fixed amount is added. Hence, the relationship is linear. A stamp collection worth $5000, for instance, may increase in value by $00 each year. This growth relationship can be written as the equation V = T where V = value of collection T = time in years The general form of linear growth function is y = a + bx where y is the dependent variable x is the independent variable (usually time) a is the initial or starting value of y b is the rate of growth. A graph of the equation could be drawn to represent this situation, like the one shown. By using the graph or the equation we can analyse the situation, for example to find the collection s value at any future time. V T Time (years) Value ($)

3 Chapter Growth and decay 67 WORKED Example Write an equation which describes the relationship between the number, N, of bees and time, T, if the population of 500 bees is known to be increasing by 0 each month. Find: a the size of the population in a year s time b when the population will have doubled. Using the equation a Let population be P and time in months be T. Substitute T = ( months = year). 3 Write your answer. b Substitute P = 000 and solve. Write your answer. WRITE/DISPLAY P = T When T =, P = P = 70 bees in one year s time When P = 000, 000 = T T = 5 months The population will double after years and month. Using the graph Refer to chapter Bivariate data (worked examples 8 and 0) and chapter 3 Introduction to regression (worked example ) for further details on using a graphics calculator for plotting relationships between variables. Draw a graph of population against time by plotting points. Set up tables in then graph using [STAT PLOT]. STAT nd and T P Continued over page

4 68 Further Mathematics WRITE/DISPLAY Find T = and read from the graph. Use STAT, select CALC to fit a straight line and then use TRACE to move along the function to T(x) = and read P(y) = 70. Number of bees N Time (months) T 3 Find N = 000 and read from the graph. Use TRACE to move to P(y) = 000 and read T(x) = 5. From graph P = 70 bees From graph T = 5 months Exponential growth Growth is exponential when the quantity present is multiplied by a constant for each unit time interval. This constant is called the growth or compounding factor and is greater than one. Consider the situation represented in the graph at right where the changes to a certain population of bacteria over a period of time are displayed. It can be seen that the population is increasing but also that the rate of growth is increasing; that is, the graph is getting steeper. Number of bacteria ('000) N T Time (hours)

5 Chapter Growth and decay 69 The readings taken from the graph are: Time, T Number, N The population is doubling every hour, so the growth or compounding factor is. The relationship could be written as N = 500 T. This means that as T increases by, the number of bacteria increases by a factor of. In general, the exponential growth function has an equation of the form: y = ka x, where k and a are constants a > is the growth or compounding factor k is the initial value of y (when x = 0). The number of bacteria in a group was recorded over a 6-hour period and a graph of the data is shown at right. a Determine whether the population growth is exponential. b If it is exponential, write an approximate equation for the relationship. WRITE a WORKED Example 3 5 a From the graph, N 0 = 000, N = 3000 N (T = ), and evaluate the ratio, N. N N = N =.5 Find the initial population of bacteria, N 0 (T = 0), and population after hour, Find N. For exponential growth,.5 is the growth or compounding factor. That is, N = N.5. N = 3000 N = = 500 Find N 3. It should be that N 3 = N.5. N 3 = = 6750 Repeat the process. N = = 0 5 N 5 = = 5 50 By using.5 as the growth or compounding factor, the calculated values compare favourably with values from the graph. Number of bacteria ('000) N T Time (hours) Growth or compounding factor =.5. Growth is exponential. b In the equation the initial value is multiplied by the growth or compounding factor,.5, raised to the power of unit time interval, that is, y = ka x. b N T = 000(.5) T Once an equation has been determined for a relationship it can be used to analyse the situation.

6 650 Further Mathematics WORKED Example The cost, C ($), of a deluxe puff pastry after time, T (years), is given by the equation C = 0.8(.6) T. Use the equation to complete the table below and plot a graph of cost against time. Time, T Cost, C WRITE Write the equation and substitute T = 0. C = 0.8(.6) T When T = 0, C = 0.8(.6) 0 = 0.8 = 0.8 Substitute T = and evaluate. When T =, C = 0.8(.6) =.8 Substitute T = and evaluate. When T =, C = 0.8(.6) Repeat for T = 3 to 5 and complete the table of T and C values. Draw the graph, joining the points with a smooth curve. =.05 T C Note: x y is the power function on a scientific calculator. It may be represented as y x or a x. On a graphics calculator, use the power button ^, for example, 5 ^ 3 = 5 3 = 5. The other alternative to that outlined above when finding a set of continuous points is simply to multiply by the growth or compounding factor each time. Cost ($) C T Time (years)

7 Chapter Growth and decay 65 Alternative Using a graphics calculator Refer to chapter 3 Introduction to regression, worked example, for further details. DISPLAY Enter the equation in Y=. Set up WINDOW of x are plotted. so that the integer values 3 GRAPH the function and use TRACE investigate values of C. to Complete the table or use the TABLE feature of the calculator.

8 65 Further Mathematics Meaning of growth or compounding factor The growth or compounding factor takes into account the quantity that we start with as well as the amount of the increase for the unit time interval. If the growth or compounding factor is.5 then the accounts for the initial quantity and the 0.5 accounts for the increase. That is:.5 = = = 00% + 5% = 5% So a growth or compounding factor of.5 means an increase each unit time interval of 5% of the previous value. For example, if the cost of an article increases by 5% per annum (p.a.) then the growth or compounding factor would be.05 (or 00% + 5% = 05%) each year. The cost, C ($), of a $6.50 cricket ball increases by 3% each year. Write the equation for the relationship between the cost and time, T (years), and use it to find the cost of the ball after 8 years. WRITE 3 WORKED Example Find the growth or compounding factor per year. Write the equation in the form y = ka x, where k is the initial value. Substitute T = 8 and evaluate. When T = 8, C = 6.5(.03) 8 Growth factor = original amount + increase = 00% + 3% = 03% =.03 C = 6.5(.03) T C = $8.3 Write a summary statement. Cost of ball after 8 years is $8.3. Decay If a quantity decreases in size over a period of time, it is decaying. This decay process may also be linear in nature or exponential, as was the case with growth. Straight line decay In this situation a quantity decreases by a fixed amount for each time unit interval, that is, a fixed amount is subtracted. Since the quantity is decreasing over time the slope of the straight line is negative. Suppose the number of starfish on a reef is 8000 at present but the population is decreasing by 50 each year. The decay relationship can be written as the equation: N = T where N = number of starfish T = time in years. A graph of the equation like the one shown could be drawn to represent this situation. The graph or the equation can be used in the same way they were used for growth situations (see worked example ). Number of starfish ($) N T Time (years)

9 Chapter Growth and decay 653 Exponential decay Decay can be exponential in a similar way to growth. That is, the quantity present can be multiplied by the growth or compounding factor for each unit time interval. It may seem to be a contradiction to say that there is a growth or compounding factor associated with decay; however, for decay the growth or compounding factor, a, is less than one, whereas for exponential growth the growth or compounding factor is greater than one. In comparison, the general equation for growth is: exponential y = ka x where k, a are constants, a > the general equation for decay is: exponential y = ka x where k, a are constants, a < The procedure that was adopted to analyse situations of exponential growth can be applied in the same way to exponential decay. WORKED Example The graph shows how the value of a car decreases over a 5-year period. a Determine whether the decay is exponential. b If it is exponential, write an approximate equation for the relationship between value, v, and time, T. a Find v 0 and v. Evaluate a From the graph, v 0 = 0 000, v 0 v = 000 v = v = Find v. If exponential decay then 0.6 is growth or compounding factor. So, v 0.6 = v. v = 000 v = = 700 Find v 3. It should be that v 0.6 = v 3. v 3 = = 30 Find v. It should be that v = v. v = = 59 5 By using 0.6 as the growth or compounding factor, the calculated values compare favourably with the values from the graph. b The decay equation is of the form y = ka x where a <. That is, the initial value, v 0, is multiplied by the growth or compounding factor, 0.6, once for each unit time interval. v 5 Value ($) T Time (years) WRITE Decay factor is 0.6. Decay is exponential. b v v = 0 000(0.6) T

10 65 Further Mathematics Meaning of growth or compounding factor As previously mentioned, the growth or compounding factor takes into account the initial quantity and, in the case of decay, also takes into account the decrease for the unit time interval. Consider a growth or compounding factor of = 0.5 = = 00% 5% = 85% That is, represents the initial quantity and 0.5 represents the actual decrease per unit time interval. So a growth or compounding factor of 0.85 means a decrease each unit time interval of 5% of the previous value. WORKED Example 6 The number, N, of tigers in a certain population is decreasing by 6% each year from an initial population of 5. Write an equation for the relationship between the number and time, T (years), and use it to find how many tigers there will be after 8 years. WRITE 3 Find the compounding factor per year. Write the equation in the form y = ka x, where k is the initial value. Compounding factor = original amount decrease = 00% 6% = 0.9 N = 5(0.9) T Substitute T = 8 and evaluate. When T = 8, N = 5(0.9) 8 Write a summary statement. N = 59 After 8 years there will be 59 tigers. Radioactive decay Radioactive decay is an example of exponential decay. Nuclear radiation is emitted from many different chemical elements. Uranium and plutonium are probably the best known radioactive elements. The original element decays to a different element over a period of time, which means that the amount of the radioactive element decreases. For example, uranium-38 decays to thorium-3 and an alpha particle is emitted. For the purposes of this work on growth and decay, all we need to remember is that, as with other cases of exponential decay, the rate of decrease (in mass or amount of the element in this case) is determined by the growth or compounding factor, which, of course, is less than one. The compounding factor is unique to each radioactive element.

11 remember remember Chapter Growth and decay 655. Growth and decay can be linear or exponential.. Linear growth and decay can be represented by the equation y = a + bx where y is the dependent variable x is the independent variable (usually time) a is the initial or starting value of y b is the rate of growth or decay. 3. Exponential growth and decay means an initial value multiplied by a growth or compounding factor for each unit time interval.. Exponential growth and decay can be represented by the equation y = ka x, where a = growth or compounding factor a > for growth, a < for decay k = the initial or starting value of y. A Growth and decay functions WORKED Example Write an equation that describes the relationship between the variables in each case, then solve each of the following problems. a The value of a Beatles record, currently worth $50, will increase by $0 every year. Assuming this relationship will continue indefinitely: i what will it be worth in 5 years time? ii when will its value have doubled? b From an initial population of 600 ants in a nest, the number grows by 30 each month indefinitely. i What will be the population in years time? ii When will the population reach 900? c If $00 is invested for 0 years and earns simple interest of $0 each year: i what will be the amount altogether after 8 years? ii when will the total amount ($00 + interest) be $800? d A coin collection, currently valued at $560, will increase in value by 5% of the current value each year. Assuming the trend will continue: i what will its value be in 8 -- years? ii when will its value reach $000? Verify your answers to question by graphing each relationship. EXCEL Spreadsheet SkillSHEET. Mathcad WORKED Example a 3 Determine whether the situations described below represent exponential growth. a The price of a certain food item over a 5-year period is detailed below. Year 3 5 Price ($) Function grapher

12 656 Further Mathematics b The value of a coin collection over a period of 5 years is shown below. Year 3 5 Value ($) c The number of rabbits in a population over a 3-month period is shown in the graph. Number of rabbits N T Time (months) d The amount in an investment account after each of 6 years is shown below. Amount in account, A ($) Year WORKED Example b Using T to represent time in hours, write an equation to describe the increase in number, N, of a population of bacteria if initially there are: a 000 bacteria and the number increases by a factor of each hour b 000 bacteria and the number increases by a factor of. each hour c 860 bacteria and the number increases by a factor of.5 each hour d 50 bacteria and the number increases by 50% each hour e 300 bacteria and the number increases by 00% each hour. 5 Using T to represent time in years, write an equation to describe the increase in value, V ($), of a painting if it was bought for: a $700 and its value increased by a factor of. each year b $00 and its value increased by a factor of.05 each year c $5000 and its value increased by a factor of.6 each year d $750 and its value increased to 0% each year e $380 and its value increased to 08% each year. 6 Using T to represent time in years, write an equation to describe the increase in cost, C ($), of buying: a a $5 000 new car if the cost of purchase grew by a factor of.07 each year b a $.50 loaf of bread if its cost grew by a factor of.03 each year c a $50 bike if its cost grew by a factor of.05 each year d a new $9.95 CD if its cost grew to 06% each year e a $7.0 packet of cigarettes if its cost grew to 0% each year.

13 Chapter Growth and decay 657 WORKED Example 3 7 Using T to represent time in years, write an equation to describe the increase in the amount, A ($), in an investment account if it was initially: a $000 and it increased by a factor of.6 each year b $850 and it increased by a factor of. each year c $900 and it increased by a factor of.06 each year d $5 000 and it increased to 09% each year e $ 600 and it increased to 5% each year. 8 a The amount, A ($), in an investment account after time, T (years), is given by the equation, A = 500(.08) T. Using this equation, copy and complete the table below and plot a graph of amount against time. b T A The value, V ($), of an antique chair over time, T (years), is given by the equation V = 850(.06) T. Use the equation to complete the table below and plot a graph of value against time. T V EXCEL Plotting relations Growth and decay Exponent Spreadsheet GC Mathcad program c The number, N, of possums in a National Park over time, T (months), is given by the equation, N = 00(.0) T. Use the equation to find the values of N for T values from 0 7 and plot a graph of number against time. WORKED Example d If the cost, C ($), of a new car is given by C = 7 000(.) T, where T is the time in years, plot a graph of cost against time for 0 5 years. 9 Given the exponential growth equations below, solve the problems provided. a If N = 650(.59) T, find N if T = 5. b If C = 0(.5) T, find C if T = 9. c If A = 3600(.09) T, find A ($) if T = 7 years. d If V = 050(.0) T, find V ($) if T = 0 years. e If N = 500(.85) T, find N (number of bacteria) if T = hours. 0 The amount, A ($), in an investment account is initially $000 and increases by 8% p.a. a Write the equation for the relationship between the amount and time, T (years). b Use the equation to find the amount in the account after 6 years. The value, V ($), of a piece of land bought for $ 000 increases by % each year. a Write an equation for the relationship between the value and time, T (years). b Use the equation to find how much the land is worth after 0 years. SkillSHEET. multiple choice The number, N, of bacteria in a colony increases by 0% per hour from an initial colony of 00. After 5 hours the number of bacteria present would be closest to: A 50 B 30 C 350 D 3000 E 500

14 658 Further Mathematics 3 multiple choice The cost, C ($), of a can of soft drink, which currently is $.5, increases by 5.5% p.a. The cost of the can in 0 years time will lie between: A 30 and 50 cents B $.0 and $.30 C $.30 and $.50 D $3.30 and $3.50 E $3.50 and $3.70 multiple choice The equation representing an increase of 9% p.a. in the number of animals, N, in a certain population which initially numbered 3600 is: A N = 3600T + 9 B N = 3600T 9 C N = 3600(.09) T D N = 3600(0.9) T E N = 9(3600) T 5 By using the equation that exists between the variables in each case below solve the given problems. a The number of elephants in a game reserve was initially 500, but the population is decreasing by 35 per year. If this trend continues: i what will be the population in years time? ii when will the population have halved? b The value of a car, currently worth $ 000, is decreasing by $800 per year. If this continues: i what will be its value in 6 years time? ii when will its value be $8800? c A cylindrical tank, containing water to a height of 3 m, has just been punctured and water is leaking out so that the water height is falling by 5 cm every minute. If this trend continues: i what will be the water height after 5 minutes? (Assume the puncture is below this height.) ii when will the height reach.65 m? d The number of wombats on an island is decreasing by a fixed number each year from an initial population of 366. If this trend continues and the population reaches 8 after 5 years: i what is the decay rate per year? ii what will be the population after 8 years? iii when will the population reach 6? e The value of a computer bought for $500 decreases by a fixed amount each year. If the value after years is $70 and this trend continues: i what is the decay rate? ii what will be the computer s value after 7 years? iii when will its value be $930? 6 Verify your answers to question 5 by graphing each relationship. WORKED Example 5a 7 Determine whether the situations described below represent exponential decay. a The value of a car over a 5-year period is detailed below. Year 3 5 Value ($)

15 Chapter Growth and decay 659 b The number in a colony of frogs over a -year period is described below. Year 3 Number c The value of a computer over a 6-year period is graphed below. d Value ($'000) V Year The mass of a radioactive element present over a time-unit interval is given at right. Does this situation represent exponential decay? T Mass of radioactive element (g) m Time units t WORKED Example 5b 8 In each case below, write an equation to represent the decrease in the given variable over time: a Amount, A, decreases from 300 by a compounding factor of -- every day. b c d e f Value, V, decreases from $5000 by a compounding factor of 0.75 every year. Number, N, decreases from 500 by a compounding factor of 0.95 each month. Mass, m, decreases from 900 g to 80% each minute. Value, V, decreases from $850 to 9% each year. Number, N, decreases from by a compounding factor of 75% each year. 9 multiple choice Which one of the equations below represents a decrease of 7% p.a. in the value, BV ($), of a boat after time, T (years), if it was bought for $7 800? A BV = 7 800(.7) T B BV = 7 800(0.83) T C BV = 7 800T 0.83 D BV = T E BV = T 0 Given the exponential decay equations below solve the given problems. a If N = 700(0.85) T, find N if T =. b If A = 0 000(0.9) T, find A if T = 0. c If V = 00(0.75) T, find V if T = 7. d If N = 60(0.96) T, find N if T = 5. e If A = 85.5(0.5) T, find A if T = 6.

16 660 Further Mathematics a The mass, m (g), of a radioactive element after time, T (days), is decreasing and is given by the equation, m = 80(0.76) T. Use this equation to complete the table below and plot a graph of mass against time. T m b The value, V ($), of a computer after time, T (years), is given by the equation, V = 600(0.80) T. Use this equation to complete the table below and plot a graph of value against time. T V WORKED Example 6 c d The number, N, in a population of seahorses after time, T (months), is given by the equation, N = 90(0.9) T. Use the equation to find the values of N for T values from 0 8 and plot a graph of number against time. The amount, A (grams), of a radioactive element after time, T (minutes), is decreasing and is given by the equation, A = 350(0.70) T. Use the equation to find the values of A for T values from 0 8 and plot a graph of amount against time. The number, N, of sulphur-crested cockatoos in a certain population is decreasing by % each month from an initial number of 00. a Write an equation for the relationship between the number and time, T (years). b Use the equation to find the number of cockatoos after year. 3 The value, V ($), of a washing machine bought for $899 decreases by 30% p.a. a Write an equation for the relationship between the value and time, T (years). b Use the equation to find how much the machine is worth after 5 years. 5 multiple choice The value, V ($), of a car bought for $7 500 decreases by 0% each year. The value of the car after 6 years would be closest to: A $7 300 B $ 00 C $700 D $5000 E $300 multiple choice The mass, m (g), of a radioactive sample of sodium decreases from an initial mass of 660 g by 5% each hour. The mass of the sodium left after hours would lie within the range: A grams B grams C grams D grams E grams 6 A 0-year research program is being carried out on two penguin rookery populations. Rookery A had 000 penguins at the start of the study and the population has been decreasing by 5% p.a. Rookery B had 3000 penguins initially and this population has been decreasing by 0% p.a. a On the same set of axes draw population against time graphs for the rookeries over the 0 years. b From your graphs, estimate when the two populations will be the same. c When the populations are the same, what is the population?

17 Chapter Growth and decay 66 Compound interest formula As you have seen in simple interest calculations, the amount present at the start does not change throughout the life of the investment. Interest is added at the end. In contrast, in the previous section graphical and algebraic methods were used to illustrate the concept of exponential growth whereby an amount increases at regular intervals over a period of time. This increasing factor was called the growth or compounding factor. Another example of exponential growth is compound interest. For investments, interest is added to the initial amount (principal) at the end of an interest-bearing period. Both the interest and the principal then earn further interest during the next period, which in turn is added to the balance. This process continues for the life of the investment. The interest is said to be compounded. The result is that the balance of the account increases at regular intervals and so too does the interest earned. Compound interest is illustrated in the next example. Consider $000 invested for years at an interest rate of % p.a. with interest compounded annually (added on each year). What will be the final balance of this account? Time period Starting principal, P ($) Interest ($) Balance ($) 000 % of 000 = % of 0 = % of 5.0 = % of 0.93 = So the balance after years is $ During the total period of an investment, interest may be compounded many times, so a formula has been derived to make calculations easier. In the above example the principal is increased by % each year. That is, the end of year balance = % or. of the start of the year balance. Now let us look at how this growth or compounding factor of. is applied in the example. Time period Balance ($) 0 = 000. = 000(.) 5.0 = 0. = = 000(.) = 5.0. = = 000(.) = = = 000(.) If this investment continued for n years the final balance would be: 000(.) n = 000( + 0.) n = 000( ) n 00

18 66 Further Mathematics The answer now is only in terms of information that was known at the start of the investment. From this pattern we are able to write a general formula that can be used to calculate compound interest. A = PR n where A = final or total amount ($) P = principal ($) R = growth or compounding factor = r = interest rate per period n = number of interest bearing periods r Note that the formula gives the total amount in an account, not just the interest earned as in the simple interest formula. To find the total interest compounded, I: I = A P where A = final or total amount ($) P = principal ($) Now let us consider how the formula is used. Find the amount in the account (balance) and the interest earned after $5000 is invested for years at 6.5% p.a., interest compounded annually. WRITE What is n? n = What is r? r = What is P? P = 5000 Write the compound interest (CI) formula. r A = P( ) n WORKED Example 7 Substitute known values into the formula. 6.5 = 5000( ) 00 Simplify. = 5000(.065) Evaluate (to decimal places). A = $63.33 Subtract the principal from the balance. I = A P = = $3.33 Write a summary statement. The amount of interest earned is $3.33 and the balance is $ In the last example interest was compounded annually. However, in many cases the interest is compounded more often than once a year, for example semi-annually (twice each year), quarterly (every 3 months), or weekly. In these situations n and r still have their usual meanings and we calculate them as follows. Number of interest periods, n = number of years number of interest periods per year nominal interest rate per annum Interest rate per period, r = number of interest periods per year Note: Nominal interest rate per annum is simply the annual interest rate advertised by a financial institution.

19 Chapter Growth and decay 663 WORKED Example 8 If $300 is invested for 5 years at 6% p.a., interest compounded quarterly, find: a the number of interest bearing periods, n b the interest rate per period, r c the balance of the account after 5 years. WRITE a Calculate n. a n = 5 (years) (quarters) = 0 b Convert % p.a. to % per quarter to match time over which interest is calculated. Divide r% p.a. by the number of compounding periods per year, namely. Write as a decimal. 6% p.a. b r% = =.5% per quarter r =.5 c What is P? c P = $300 Write the CI formula. r A = P( ) n 00 3 Substitute known values..5 = 300( ) 0 00 Simplify. = 300(.05) 0 5 Evaluate to decimal places. A = $309.9 Write a summary statement. Balance of account after 5 years is $ On occasions when interest is compounded monthly or more often, the value of r is a recurring decimal. Now the accuracy of the calculation should be maintained, so the value that has been determined for r is not to be approximated or truncated. Therefore, the order of the calculation on a calculator is changed slightly. Find the amount that accrues in an account which pays compound interest (compounded weekly) at a nominal rate of 5.6% p.a. if $50 is invested for 3 -- years. WRITE Calculate n. n = = Calculate r and retain on the calculator for r = step 5. = What is P? P = $50 3 WORKED Example 9 Continued over page

20 66 Further Mathematics WRITE r Write the CI formula and substitute. A = P( ) n = 50( ) 8 Evaluate to decimal places. A = $980.8 Subtract the principal from the balance. I = A P = = $530.8 Write a summary statement. Amount accrued after 3 -- years is $ Note: The order of calculation used here can be applied in all previous cases when finding A; see worked examples 7 and 8. The situation often arises where we require a certain amount of money by a future date. It may be to pay for a holiday or to finance the purchase of a car. It is then necessary to know what principal should be invested now in order that it will increase in value to the desired final balance within the time available. Find the principal that will grow to $000 in 6 years, if interest is added quarterly at 6.5% p.a. WRITE Calculate n. n = 6 = Calculate r. 6.5 r = =.65 3 What is A? A = $000 r Write the CI formula, substitute and A = P( ) n 00 simplify = P( ) WORKED Example = P(.065) Transpose to isolate P. 000 P = (.065) Evaluate to decimal places. P = $76.73 Write a summary statement. $76.73 would need to be invested. Sometimes we know how much we can afford to invest as well as the amount we want to have at a future date. Using the compound interest formula we can calculate the interest rate that is needed to increase the value of our investment to the amount we desire. This allows us to shop around various financial institutions for an account which provides the interest rate we want.

21 Chapter Growth and decay 665 We must first find the interest rate per period, r, and convert this to the corresponding nominal rate per annum. WORKED Example Find the interest rate per annum (to decimal places) that would enable an investment of $3000 to grow to $000 over years if interest is compounded quarterly. WRITE 3 What are A, P and n? For this example n needs to represent quarters of a year and therefore r will be evaluated in % per quarter. Write the CI formula and substitute. Divide A by P. A = $000 P = $3000 n = = 8 A = PR n 000 = 3000R = R Obtain R to the power of, that is, raise 8 both sides to the power of = ( ) = R R 8 = r Isolate r and evaluate. r = = r = 3.66 r% = 3.66% per quarter Multiply r by the number of interest periods per year to get the annual rate (to decimal places). Write a summary statement. -- Annual rate = r% per quarter = 3.66% per quarter =.65% per annum Interest rate of.65% p.a. is required. remember remember. Compound interest calculations can be made using the formula A = PR n where A = final amount ($) P = principal ($) R = growth or compounding factor = r = interest rate per period n = number of interest bearing periods.. For compound interest, I = A P. r

22 666 Further Mathematics B Compound interest formula EXCEL Spreadsheet WORKED Example 7 Mathcad Compound interest WORKED Example 8a Example 8b SkillSHEET.3 WORKED WORKED Example 8c Use the compound interest formula to find the amount, A, when: a P = $500, n =, r = 8 b P = $000, n =, r = 3 c P = $3600, n = 3, r = 7.5 d P = $95, n = 5, r = 5.5 e P = $850.0, n = 0, r = f P = $5, n =, r = 0.5 Find: i the balance, and ii the interest earned (interest compounded annually) after: a $000 is invested for 3 years at 8% p.a. b $7000 is invested for years at 6% p.a. c $6000 is invested for years at 5% p.a. d $900 is invested for 5 years at 0% p.a. e $35 is invested for 6 years at 7.5% p.a. f $96 is invested for 3 years at 6.5% p.a. g $650 is invested for 0 years at.95% p.a. h $505 is invested for 8 years at 7.% p.a. 3 Find the number of interest bearing periods, n, if interest is compounded: a annually for 5 years b quarterly for 5 years c semi-annually for years d monthly for 6 years e 6-monthly for -- years f quarterly for 3 years and 9 months g monthly for -- years h fortnightly for 3 -- years i daily for years j weekly for 3 -- years. Find the interest rate per period, r, if the annual rate is: a 6% and interest is compounded quarterly b % and interest is compounded half-yearly c % and interest is compounded 6-monthly d 8% and interest is compounded monthly e 7% and interest is compounded quarterly f 7.5% and interest is compounded quarterly g 8% and interest is compounded monthly h 5% and interest is compounded fortnightly i 9% and interest is compounded weekly j 8.5% and interest is compounded daily. 5 Find the balance of the account after: a years if $3000 is invested at 8% p.a., interest compounded quarterly b 5 years if $000 is invested at 6% p.a., interest compounded 6-monthly c years if $5000 is invested at % p.a., interest compounded monthly d 7 years if $500 is invested at % p.a., interest compounded quarterly e 3 years if $500 is invested at 7% p.a., interest compounded half-yearly f 6 years if $300 is invested at 9% p.a., interest compounded quarterly g years if $80 is invested at 9% p.a. (compounded monthly) h 3 -- years if $30 is invested at 6% p.a. (compounded quarterly) 3 i -- years if $965 is invested at 7.5% p.a. (compounded monthly) j 5 years months if $65 is invested at 5% p.a. (compounded monthly).

23 Chapter Growth and decay 667 WORKED Example 9 6 Find the amount that accrues in an account which pays compound interest at a nominal rate of: a 7% p.a. if $600 is invested for 3 years (compounded monthly) b 0% p.a. if $00 is invested for years (compounded monthly) c 8% p.a. if $3500 is invested for years (compounded monthly) d e f g h i j 5% p.a. if $850 is invested for 3 -- years (compounded fortnightly) % p.a. if $960 is invested for 5 -- years (compounded fortnightly) 6.6% p.a. if $90 is invested for 3 years (compounded fortnightly) 7.3% p.a. if $370 is invested for 5 years (compounded weekly) 8.% p.a. if $500 is invested for -- years (compounded weekly) 5.5% p.a. if $605 is invested for years (compounded daily).75% p.a. if $8970 is invested for 3 years (compounded daily) multiple choice The greatest return is likely to be made if interest is compounded: A annually B semi-annually C quarterly D monthly E fortnightly. multiple choice If $ 000 is invested for -- years at 6.75% p.a., compounded fortnightly, the amount of interest that would accrue would be closest to: A $3600 B $00 C $5000 D $ 00 E $ multiple choice Which account would provide the best investment opportunity if $500 was invested for one year? A simple interest at 6% p.a. (see simple interest topic) B interest compounded annually at 6% p.a. C interest compounded 6-monthly at 6.% p.a. D interest compounded quarterly at 6.% p.a. E interest compounded monthly at 6% p.a. 0 Peta wishes to invest $300 for 5 years. By comparing the interest earned, which of the following would be Peta s best investment option? a % p.a. simple interest b compound interest at 0.5% p.a., compounded annually c compound interest at 0% p.a., compounded monthly Cyril has just inherited $0 000 and after spending $90 on a holiday he would like to invest the balance for -- years. He is offered the investment opportunities detailed below. Which option should Cyril choose? a simple interest at 9.5% p.a. b compound interest at 9.% p.a., adjusted semi-annually c compound interest at 9.3% p.a., adjusted quarterly Agnes invests $050 for years. Interest is added quarterly. For the first years the rate is 7% p.a. and for the remaining years the rate rises to 9% p.a. What interest would accrue during this time?

24 668 Further Mathematics 3 Alex invests $685 for 3 years. Interest is added half-yearly. The rate starts at 8.5% p.a. for the first half of the investment period before it rises to 9.6% p.a. for the remaining time. What interest would Alex earn from this account? WORKED Example 0 Use the compound interest formula to find the principal, P, when: a A = $5000, r = 9, n = b A = $600, r = 8., n = 3 c A = $3550, r =.5, n = d A = $666.5, r = 0.8, n = 36 e A = $595.7, r =., n = 8 5 Find the principal that will grow to: 6 Find the interest accrued in each case in question a b c d e f g h i j $3000 in years, if interest is compounded 6-monthly at 9.5% p.a. $000 in 3 years, if interest is compounded quarterly at 9% p.a. $900 in 3 -- years, if interest is compounded quarterly at 0.6% p.a. $5600 in 5 -- years, if interest is compounded quarterly at 8.7% p.a. $0 000 in -- years, if interest is compounded monthly at 5% p.a. $8 000 in -- years, if interest is compounded monthly at 9% p.a. $7 500 in 6 years, if interest is compounded monthly at 0.5% p.a. $850 in years, if interest is compounded monthly at 8.5% p.a. $860 in -- years, if interest is compounded fortnightly at 6.5% p.a. $870 in -- years, if interest is compounded weekly at 7.5% p.a. multiple choice Lillian wishes to have $ 000 in a bank account after 6 years so that she can buy a new car. The account pays interest at 5.5% p.a. compounded quarterly. The amount (to the nearest dollar) that Lillian should deposit in the account now, if she is to reach her target, is: A $370 B $9637 C $0 09 D $ 7 E $ 30 multiple choice Peter has his heart set on a holiday in years time and it will cost him $700. His bank account pays interest at a rate of 0.5% p.a., compounded 6-monthly. The amount that Peter will need to deposit now into this account will lie between: A $000 and $050 B $50 and $00 C $30 and $360 D $360 and $0 E $500 and $550 9 Sarah needs $560 for a new stereo system which she is planning to buy in -- years time. Her bank offers a rate of 9.6% p.a. with interest compounded monthly. How much should she deposit now? 0 Glen s credit union offers an account which pays a rate of 8.% p.a. with interest compounded monthly. His house extension will cost him $5 000 in -- years time. How much should Glen invest in this account to be able to pay for his extension?

25 Chapter Growth and decay 669 WORKED Example Calculate the interest rate per year (to decimal places), given that the interest rate per period, r, is: a % and interest is compounded quarterly b.5% and interest is compounded quarterly c % and interest is compounded monthly d 3.5% and interest is compounded semi-annually e 0.65% and interest is compounded monthly f.65% and interest is compounded quarterly g 0.3% and interest is compounded fortnightly h 0.5% and interest is compounded weekly i 0.9% and interest is compounded weekly j 0.05% and interest is compounded daily. Find the interest rates per annum (to decimal places) that would enable investments of: a $000 to grow to $3000 over 3 years if interest is compounded 6-monthly b $8000 to grow to $9000 over years (interest compounded quarterly) c $ 000 to grow to $5 000 over years (interest compounded quarterly) d $5000 to grow to $7000 over 5 years (compounded semi-annually) e $650 to grow to $3750 over 3 -- years (compounded quarterly) f $5 000 to grow to $0 000 over -- years (compounded monthly) g $3 000 to grow to $ over -- years (compounded fortnightly) h $00 to grow to $950 over years (compounded weekly) i $960 to grow to $300 over -- years (compounded weekly) j $350 to grow to $00 over years (compounded daily). 3 If it takes years for $60 to grow to $00, find the annual interest rate (compounded semi-annually). After 3 -- years $950 has accumulated to $300. Find the annual interest rate (compounded quarterly). 5 6 multiple choice What is the minimum interest rate per annum (compounded quarterly) needed for $300 to grow to $300 in years time? A 6% p.a. B 7% p.a. C 8% p.a. D 9% p.a. E 0% p.a. multiple choice What will be the minimum annual rate of interest (compounded monthly) needed to enable $8 500 to accrue $000 interest in 8 months time? A.5% p.a. B 3.5% p.a. C.5% p.a. D 5.5% p.a. E 6.5% p.a. 7 Sophie has been told that if you invest $500 at 8% p.a. compounded annually then its value will double in 9 years. Is this true? 8 At an annual rate of 8.% (compounded half-yearly) $000 will double in value in years. Is this true? 9 a ii If you invested $000 now at % p.a. (compounded quarterly), how much would you have in 0 years time? ii At that time, what annual interest rate would give you $0 000 in a further 0 years (interest compounded quarterly)? b How would your answers to part a vary if interest was compounded fortnightly? WorkSHEET.

26 670 Further Mathematics Finding time in compound interest In the previous section we studied investments that compound over a period of time by manipulating the compound interest formula, A = PR n r, where R = Situations 00 where A, P and r were unknown were investigated. This section deals with two methods that can be used to find n, the number of interest-bearing periods, that is, to find the time for which an investment runs. The two methods are:. trial and error. logarithms. The value obtained for n may be a whole number, but it is more likely to be a decimal. That is, the time required will lie somewhere between two consecutive integers. The smaller of the two integers represents insufficient time for the investment to amount to the balance desired, while the second represents more than enough time. In practice, if this is the case, an investor may choose to: (a) withdraw funds as soon as the final balance is reached, in which case a fee may be imposed for early withdrawal (b) withdraw funds at the first integral value of n after the final balance is reached, that is, when the investment matures. In this section we will use the second option for calculating the time needed for an investment to run. Using trial and error to find time This process involves substituting reasonable values of n into a simplified form of the compound interest formula until we find the two integral values mentioned above. When deciding what values are reasonable we should primarily consider the frequency at which the interest is compounded. Generally, the greater the frequency, the greater the value of n. We can also consider by what proportion the principal grows, and the interest rate per annum. A large proportional increase and a small interest rate may indicate a large value of n, too. However, in the end the process is guesswork. WORKED Example Use the compound interest formula to find out how long it will take $000 to amount to $3500 at 8% p.a. (interest compounded annually). WRITE What are A, P and r? A = 3500, P = 000, r = 8 3 Substitute into the CI formula and simplify. Substitute a value for n which you think may give a value for.08 n equal to or greater than.75 and compare. Try n = 6. A = P r = 000(.08) n.08 n =.75 n Need.08 n =.587 too low

27 Chapter Growth and decay <.75 so repeat the process with a larger value, say n = 8. 5 Since.08 8 >.75, try n = 7. n = 7 represents insufficient time while n = 8 was just in excess of Write your answer. As the compounding period was annual, n is in years. WRITE.08 8 =.85 (possibly too high).08 7 =.7 (too small) Therefore n = 8 is sufficient for.08 n to exceed.75. It will take 8 years for $000 to amount to $3500. As we have seen, it may be that interest is compounded more often than once a year so r will need to be determined in the usual manner. In all these cases n has its usual meaning and could indicate the number of quarters, months, fortnights, weeks or days. It is then appropriate for the value of n obtained to be converted to time in more meaningful terms, for example years and months rather than just months. A Note: The decimal value for the ratio --- should be written to decimal places if P interest is compounded more often than monthly. Calculate the number of interest bearing periods, n, and hence the time it will take $3600 to amount to $500 at a rate of 7% p.a. (interest compounded quarterly). WRITE What are A, P and r? Note that n is the number of quarters. A = 500 P = 3600 r = =.75 Substitute into the formula and simplify. A = P r = 3600(.075) n.075 n =.7 Try n = 0 and compare with =.89 Try a much larger value for n, say n = =.5 (Remember n is in quarters.) <.7, so try n =..075 =.0 So quarters are needed. Change to time n = quarters in more meaningful units. Time = 5 -- years WORKED Example Write a summary statement It will take 5 -- years for $3600 to amount to $500. n

28 67 Further Mathematics Remember that r may be a recurring decimal (interest compounded monthly or more often) and although approximations are made to the calculated value of n in a problem, we should still maintain an accurate r value. This is because n could be very close to an integer and the accuracy consideration could determine whether n lies above or below that integer. This is especially important the greater the frequency of compounding. Once determined, store r in the calculator memory. The interest accrued, I, may be given rather than the final balance, A. As before, find A by using A = P + I. WORKED Example About how long would it take $800 to accrue $500 interest at 0% p.a., interest credited fortnightly? WRITE P and I are given, not A. A = P + I. What are A, P and r? r Store R = + in your calculator 00 memory A = P + I = = 300 P = 800 r = = Substitute and simplify. A = P r n 300 = 800( ) n n = Try n = 60 (remember n is in fortnights) and compare to = Try n = =.785 Try n = = fortnights are needed. Change to time in more meaningful terms. Write a summary statement. n = 6 fortnights Time = years and fortnights years and fortnights are needed for $800 to accrue $500 interest. remember remember. Finding time by trial and error involves educated guessing for the value of n.. Time = n number of times interest is compounded each year.

29 Chapter Growth and decay 673 C Finding time in compound interest using trial and error WORKED Example WORKED Example 3 Use the compound interest formula to find the amount, n, given: a A = $83.53, P = $3000, r = 0 b A = $5 03.5, P = $7000, r = 9 c A = $ , P = $6500, r = 6 d A = $75.0, P = $00, r = 7 e A = $37.50, P = $950, r =.5 f A = $ 7.0, P = $5 000, r = 8.5 g A = $773., P = $500, r =. h A = $39.8, P = $90, r = 0.5. Use the compound interest formula to find out how long it will take (with interest compounded annually) for: a $000 to amount to $335.0 at 9% p.a. b $7500 to amount to $ at % p.a. c $950 to amount to $ at 5% p.a. d $6 000 to amount to $ at 6% p.a. e $750 to amount to $000 at 8% p.a. f $95 to amount to $350 at % p.a. g $70 to amount to $600 at 0.5% p.a. h $ 000 to amount to $0 500 at.5% p.a. 3 Use your answers from question parts e to h (the number of years needed in each case) to calculate the actual balance after this time. For each of the following, calculate the number of interest bearing periods, n, and hence the time in more meaningful terms. a A = $00, P = $00, r = 3% per 6 months (compounded semi-annually) b A = $5000, P = $300, r = % per 6 months (compounded half-yearly) c A = $ 000, P = $8300, r =.5% per quarter (compounded quarterly) d A = $7300, P = $850, r =.5% per quarter e A = $5 600, P = $9600, r = % per month f A = $80, P = $990, r = 0.75% per month g A = $5 000, P = $6 750, r = 0.5% per fortnight h A = $ 050, P = $930, r = 0.% per fortnightly i A = $650, P = $380, r = 0.5% per week j A = $880, P = $6950, r = 0.03% per day 5 Calculate the number of interest bearing periods, n, and hence how long it will take for: a $8900 to amount to $ 000 at a rate of 8% p.a. (compounded quarterly) b $300 to amount to $5600 at a rate of 9% p.a. (compounded half-yearly) c $900 to amount to $00 at a rate of % p.a. (compounded quarterly) d $050 to amount to $800 at a rate of 0% p.a. (compounded 6-monthly) e $8 00 to amount to $8 600 at % p.a. (credited monthly) f $560 to amount to $930 at 5% p.a. (credited monthly) g $0 500 to amount to $5 000 at 9.5% p.a. (credited quarterly) h $7950 to amount to $7 00 at 9.6% p.a. (credited monthly). 6 Rachel has $50 to invest and she would like it to reach $000 by placing it in an account at a rate of 0.% p.a. with interest compounded quarterly. How long will this take? 7 Billy has invested $60 at a rate of 8.6% p.a., interest credited semi-annually. How long will this investment take to accumulate to $8000? Compound interest EXCEL Compound interest Mathcad Spreadsheet

30 67 Further Mathematics WORKED Example 8 Phillipa has invested $875 in an account which pays interest at a rate of.6% p.a., interest credited monthly. What time period is needed to enable this investment to grow to $8000? 9 Michael would like to have $3600 at some time in the near future. He has $60 that he can invest now and his bank offers him an account which pays 7.8% p.a., interest compounded quarterly. How long will it take him to achieve his target? 0 multiple choice What will be the least number of interest periods, n, required for $7590 to grow to at least $9000 in an account with interest paid at 6.5% p.a., interest compounded half-yearly? A 5 B 6 C 0 D E multiple choice What will be the minimum investment period required for $ 50 to result from an investment of $ 000 at a rate of 7.5% p.a., interest credited monthly? A 3 years 5 months B 3 -- years C 0 years months D 0 -- years E years multiple choice Sonya has $500 and she wants to be able to take a holiday to Spain, costing $60. Her credit union offers an account which pays compound interest at 9.5% p.a. (credited semi-annually). If Sonya invests her $500 in this account the first occasion that she can afford the holiday is after: A -- years B 5 years C 5 -- years D 6 years E years 3 About how long would it take for: a $500 to accrue $300 interest at 8% p.a., interest compounded monthly? b $300 to accrue $500 interest at 5% p.a., interest compounded monthly? c $9000 to accrue $00 interest at 9.6% p.a., interest credited fortnightly? d $ 60 to accrue $600 interest at 7.% p.a., interest credited fortnightly? e $80 to accrue $00 interest at 6.6% p.a., compounded weekly? f $875 to accrue $0 interest at 8.3% p.a., compounded weekly? g $ 0 to accrue $560 interest at 0.% p.a., credited daily? h $9 000 to accrue $ 000 interest at 9.5% p.a., credited daily? 5 multiple choice If, when you were born, your uncle invested $00 at % p.a. (credited monthly), by which of your birthdays would the $00 have amounted to at least $000? A 9th B 0th C st D 3st! E 3nd! multiple choice Consider question. If the initial investment that your uncle made was doubled, the birthday when the balance would reach at least $000 would be the: A th B 5th C 6th D 6st! E 6nd! 6 Vivian and Rick both want to save $6000 for a holiday. Vivian has chosen to invest $5000 with his bank at 5% p.a., compounded monthly. However, Rick has only $000 but he has found an account which pays interest at 9% p.a., compounded monthly. If they both invest their money at the same time: a who will get their holiday first? b how much longer will the other person have to wait for a holiday?

31 Chapter Growth and decay 675 Finding time in compound interest using logarithms The logarithm (log) is a special function just as sine and cosine are special functions in trigonometry. In the same way that we can find the sine of 30 (0.5) on a calculator, we can find the log of 0 (). Try this on your calculator so that you are comfortable with evaluating the log of a number. In the past you have used a set of rules to simplify indices. There is also a set of rules which applies to logs. (Your teacher may discuss with you the close relationship that exists between indices and logs and how the log rules are derived.) In using logs to find n in the CI formula we use one of the log rules: log x n = n log x. That is, the log of a number raised to a power equals the power multiplied by the log of the number to the power of one. Now let us see that this rule holds for a numerical example. Evaluate log 5. log 5 = 5 log (from log rule) log 5 = log 3 = (from calculator) = (from calculator) = You could try more of these to satisfy yourself that this log rule works. Now let us consider finding n in the CI formula. WORKED Example 5 Using the CI formula, find out how long it will take $000 to amount to $3500 at 8% p.a. with interest compounded annually. WRITE What are A, P and r? A = 3500 P = 000 r = 8 Substitute into the CI formula and simplify. Keep.75 on the calculator display for step 6. A = P r = 000(.08) n.08 n =.75 3 Take the log of both sides of the equation. log.08 n = log.75 Apply the log rule i.e. log x n = n log x. n log.08 = log.75 5 Isolate n by dividing both sides by log.75 log.08 n = log.08 Evaluate n. n = 7.7 years Interest is compounded annually, so n represents years. Raise n to the next whole year. Write a summary statement. n As the interest is compounded annually, n = 8 years It will take 8 years for $000 to amount to $3500.

32 676 Further Mathematics Calculate the number of interest bearing periods, n, required and hence the time it will take $3600 to amount to $500 at a rate of 7% p.a., with interest compounded quarterly. WRITE What are A, P and r? A = 500 P = r = -- =.75 r n Substitute into the formula and simplify. A = P Retain.6 66 on your calculator display for step WORKED Example = 3600(.075) n.075 n =.6 66 Take the log of both sides. log.075 n = log.6 66 Apply the log rule and isolate n. n log.075 = log.6 66 log.6 66 n = log.075 Evaluate n. n = 0.08 quarters As n represents quarters, raise n to the As the interest is compounded quarterly, next integer. n = quarters. Write the time in more meaningful terms. Time = years = 5 -- years. Write a summary statement. After 5 -- years, $3600 will amount to $500. WORKED Example 7 About how long would it take $800 to accrue $500 interest at 0% p.a., with interest credited fortnightly? WRITE P and I are given, not A. A = P + I r A= 300, P = What are A, P and r? Store R = in r = = the calculator memory Substitute and simplify. Retain A = P r n on the calculator display for step = 800( ) n ( ) n = Take the log of both sides. log ( ) n = log Apply the log rule and isolate n. n log = log log n = log Evaluate n. n = fortnights 6 Raise n to the next integer and write the time in more meaningful terms. As interest is credited fortnightly, n = 6 fortnights time = years fortnights 7 Write a summary statement. years fortnights are needed for $800 to accrue $500 interest.

33 remember remember. When using logs to find time, use the rule: log x n = n log x.. Change the n value to time in more meaningful terms. Chapter Growth and decay 677 D Finding time in compound interest using logarithms WORKED Example 5 Use the log rule to solve for n (to decimal places) in each case. a log 5 n = log 3 b log 3 n = log.5 c log n = log 7 d log. n = log e log. n = log.9 f log.05 n = log.5 g log.05 n = log.6 h log.05 n = log.53 Use the compound interest formula to find n, given: a A = $ 7.9, P = $8000, r = 0 b A = $ , P = $000, r = 9 c A = $ , P = $500, r = 6 d A = $950.73, P = $500, r = 7 e A = $53.53, P = $850, r =.5 f A = $ 80.55, P = $ 000, r = 8.5 g A = $937.68, P = $600, r =. h A = $63.75, P = $960, r = Use the CI formula to find n (to decimal places), given: a A = $6000, P = $000, r = 8 b A = $7500, P = $5500, r = 0 c A = $700, P = $600, r = 6 d A = $650, P = $90, r = 9 e A = $6 000, P = $3 300, r = 8.5 f A = $9 500, P = $6 50, r =.5 g A = $830, P = $750, r =.5 h A = $3680, P = $30, r =.7. Use the compound interest formula to find out how long it will take (with interest compounded annually) for: a $000 to amount to $ at 8% p.a. b $7500 to amount to $ at % p.a. c $950 to amount to $ at 6% p.a. d $6 000 to amount to $ at 9% p.a. e $850 to amount to $000 at 7% p.a. f $05 to amount to $350 at % p.a. g $50 to amount to $600 at.5% p.a. h $ 000 to amount to $0 500 at 3.5% p.a. SkillSHEET. Compound interest finding n Mathcad 5 Having determined the number of years needed in each case in question e h calculate the actual balance after this time.

34 678 Further Mathematics WORKED Example 6 6 Calculate the number of interest bearing periods, n, and hence the time in more meaningful terms when: a A = $00, P = $00, r = 3% per half-year b A = $000, P = $300, r = % per 6-month c A = $3 500, P = $8300, r =.5% per quarter d A = $800, P = $850, r =.5% per quarter e A = $6 900, P = $9600, r = % per month f A = $90, P = $990, r = 0.75% per month g A = $ 000, P = $6 750, r = 0.5% per fortnight h A = $3 050, P = $930, r = 0.% per fortnight i A = $550, P = $380, r = 0.5% per week j A = $900, P = $6950, r = 0.03% per day 7 Calculate the number of interest bearing periods, n, and hence how long it will take for: a b c d e f g h $7800 to amount to $0 000 at a rate of 8% p.a. (compounded quarterly) $500 to amount to $600 at a rate of 9% p.a. (compounded half-yearly) $800 to amount to $900 at a rate of % p.a. (compounded quarterly) $650 to amount to $800 at a rate of 0% p.a. (compounded 6-monthly) $0 00 to amount to $8 600 at % p.a. (credited monthly) $660 to amount to $0 30 at 5% p.a. (credited monthly) $ 00 to amount to $6 00 at 9.5% p.a. (credited quarterly) $0 500 to amount to $9 80 at 9.6% p.a. (credited monthly). 8 Wanda has invested $600 in an account at a rate of 0.% p.a., interest compounded quarterly. How long will it take to reach $00? 9 Baden has invested $5680 at a rate of 9.% p.a. with interest credited semi-annually. How long will this investment take to accumulate to $7000? 0 Stefan has invested $00 in an account which pays interest at a rate of.6% p.a., interest credited monthly. What time period is needed to enable this investment to grow to $7600? Don would like to have $3800 at some time in the near future. He has $00 that he can invest now and his bank offers him an account which pays 8.6% p.a., interest compounded quarterly. How long will it take him to achieve his goal? multiple choice What will be the least number of interest periods, n, required for $670 to grow to at least $9000 in an account with interest paid at 6.5% p.a. and compounded half-yearly? A 0 B C D 0 E 3 multiple choice What will be the minimum investment period required for $ 750 to result from an investment of $8000 at a rate of 6.9% p.a., interest credited monthly? A 6 months B 7 months C 6 years 9 months D 6 years 0 months E 7 years

35 Chapter Growth and decay 679 WORKED Example 7 multiple choice Rhiannon has $00 and she wants to be able to take a holiday to Thailand costing $3860. Her credit union offers an account which pays compound interest at 9.5% p.a. (credited semi-annually). If Rhiannon invests her $00 in this account, the first occasion that she can afford the holiday is after: A 3 years B -- years C 6 years D 6 -- years E 3 years 5 About how long would it take for: a $00 to accrue $300 interest at 8% p.a., interest compounded monthly? b $900 to accrue $500 interest at 5% p.a., interest compounded monthly? c $8000 to accrue $00 interest at 9.6% p.a., interest credited fortnightly? d $3 360 to accrue $600 interest at 7.% p.a., interest credited fortnightly? e $950 to accrue $300 interest at 6.6% p.a., compounded weekly? f $75 to accrue $0 interest at 8.3% p.a., compounded weekly? g $ 60 to accrue $560 interest at 0.% p.a., credited daily? h $8 000 to accrue $ 000 interest at 9.5% p.a., credited daily? 6 multiple choice If, when you were born, your aunt invested $00 at % p.a. (credited monthly), by which one of your birthdays would the $00 have amounted to at least $500? A 3th B th C 5th D 6st E 6nd 7 multiple choice Consider question 6. If the initial investment that your aunt made was doubled, the birthday when the balance would reach at least $500 would be the: A 8th B 9th C 0th D 9nd E 93rd 8 Jennifer and Dawn both want to save $5 000 for a car. Jennifer has $ 000 to invest in an account with her bank which pays 8% p.a., interest compounded quarterly. Dawn s credit union has offered her % p.a., interest compounded quarterly. a b How long will it take Jennifer to reach her target? How much will Dawn need to invest in order to reach her target at the same time as Jennifer? Assume their accounts were opened at the same time. WorkSHEET.

36 680 Further Mathematics Depreciation Many items such as antiques, jewellery or real estate increase in value (appreciate) with time. On the other hand, items such as computers, vehicles or machinery decrease in value (depreciate) with time as a result of wear and tear or a lack of demand for those specific items. The estimated loss in value of assets is called depreciation. Each financial year a business will set aside money equal to the depreciation of an item in order to cover the cost of the eventual replacement of that item. The estimated value of an item at any point in time is called its book value. When the book value becomes zero the item is said to be written off. At the end of an item s useful or effective life (as a contributor to a company s income) its book value is then called its scrap value. Book value = cost price total depreciation to that time When book value = $0, then the item is said to be written off. Scrap value is the book value of an item at the end of its useful life. There are 3 methods by which depreciation can be calculated. They are:. flat rate depreciation. reducing balance depreciation 3. unit cost depreciation. Flat rate (straight line) depreciation If an item depreciates by the flat rate method then its value decreases by a fixed amount each unit time interval, generally each year. This depreciation value may be expressed in dollars or as a percentage of the cost price. This method of depreciation may also be referred to as prime cost depreciation. Since the depreciation is the same for each unit time interval, the flat rate method is an example of straight line (linear) decay. The relationship can be represented by the linear equation: BV = P dt where P = cost price ($) BV = P dt where BV T = book value ($) after time, T BV = P dt where T = time since purchase (years) BV = P dt where d = rate of depreciation ($ per year) = fixed amount per year or = percentage of P per year We can use this relationship to analyse flat rate depreciation or we can use a depreciation schedule (table) which can then be used to draw a graph of book value against time. The schedule displays the book value after each unit time interval, that is: Time, T Depreciation, d Book value, V

37 Chapter Growth and decay 68 WORKED Example 8 Fast Word Printing Company bought a new printing press for $5 000 and chose to depreciate it by the flat rate method. The depreciation was 0% of the prime cost price each year and its useful life was years. a Find the annual depreciation. b Draw a depreciation schedule for the press useful life and use it to draw a graph of book value against time. c Find the relationship between the book value and time and use it to find the scrap value. WRITE a State the cost price. a P = $5 000 Find the depreciation rate as 0% of the prime cost price. d = 0% of $5 000 = $3000 per year 3 Write your answer. Annual depreciation is $3000. b Draw a depreciation schedule for b 0 years, using depreciation of $3000 Book each year and a starting value of Time, Depreciation, value, $ T (years) d ($) V ($) c 3 Draw a graph of the tabled values for book value against time. Set up the equation: BV = P dt State d and P. The press is scrapped after years so substitute T = into the equation. V Time (years) c d = 3000 P = 5000 BV T = T BV = () = BV = $3000 Write your answer. The scrap value is $3000. Book value ($) T Note: If finding time, T, or depreciation, d, simply substitute into the equation and solve. The depreciation schedule gives the scrap value, as can be seen in the previous example. So too does a graph of book value against time, since it is only drawn for the item s useful life and its end point is the scrap value.

38 68 Further Mathematics Businesses need to keep records of depreciation for tax purposes on a year-to-year basis. What if an individual wants to investigate the rate at which an item has depreciated over many years? An example is the rate at which a private car has depreciated. If a straight line depreciation model is chosen, then the following example demonstrates its application. WORKED Example 9 Jarrod bought his car 5 years ago for $ Its current market value is $7500. Assuming straight line depreciation, find: a the car s annual depreciation rate b the relationship between the book value and time, and use it to find when the car will have a value of $3000. WRITE a Find the total depreciation over the 5 years and thus the rate of depreciation. a Total depreciation = cost price current value = $5 000 $7500 = $7500 Rate of depreciation total depreciation = number of years $7500 = years = $500 per year Write your answer. The annual depreciation rate is $500. b Set up the book value equation. b BV = P dt BV T = T 3 Use the equation and substitute BV T = $3000 and transpose the equation to find T. Write your answer. When BV T = 3000, 3000 = T 500T = T = T = = 8 The depreciation equation for the car is BV T = T. The book value of $3000 for the car is expected to occur when it is 8 years old.

39 remember remember Chapter Growth and decay 683. To make calculations in flat rate depreciation, use the formula BV = P dt where BV = book value after T years ($) P = purchase or cost price ($) T = time (years of depreciation) d = rate of depreciation ($ per year) = fixed amount per year or = percentage of P per year.. Flat rate depreciation is also known as prime cost depreciation. 3. An item is written off when its book value becomes zero.. Scrap value is the item s book value when it is no longer used. 5. Total depreciation = cost price current value 6. Rate of depreciation = total depreciation number of years E Flat rate depreciation WORKED Example 8 A mining company bought a vehicle for $5 000 and chose to depreciate it by the flat rate method. The depreciation was 0% of the cost price each year and its useful life was years. Flat rate depreciation a Find the annual depreciation. b Draw a depreciation schedule for the item s useful life and draw a graph of book value against time. c Find the relationship between book value and time. Use it to find the scrap value. All Clean carpet cleaners bought a cleaner for $0 000 and chose to depreciate it by the flat rate method. The depreciation was 5% of the cost price each year and its useful life was 5 years. a Find the annual depreciation. b Draw a depreciation schedule for the item s useful life and draw a graph of book value against time. For the situations outlined in questions 3 and : a draw a depreciation schedule for the item s useful life and draw a graph of book value against time b find the relationship between book value and time. Use it to find the scrap value. 3 A farming company chose to depreciate a tractor by the prime cost method and the annual depreciation was $000. The tractor was purchased for $5 000 and its useful life was 0 years. Mathcad

40 68 Further Mathematics WORKED Example 9 A winery chose to depreciate a corking machine, costing $3 500 new, by the prime cost method. The annual depreciation was $000 and its useful life was 6 years. For the situations outlined in questions 5 and 6: a find the annual depreciation b draw a depreciation schedule for the item s useful life and draw a graph of book value against time c find the relationship between book value and time. Use it to find how long it will take for the item to reach its scrap value. 5 Machinery is bought for $7750 and depreciated by the flat rate method. The depreciation is 0% of the cost price each year and its scrap value is $ An excavation company buys a digger for $9 000 and depreciates it by the flat rate method. The depreciation is 5% of the cost price per year and its scrap value is $900. For the situations outlined in questions 7 and 8: a find the annual depreciation b draw a depreciation schedule for the item s useful life and draw a graph of book value against time c find the relationship between book value and time. Use it to find how long it will take for the item to be written off. 7 The owner of a rental property chooses to depreciate the carpets, which were purchased for $5000, by the prime cost method. The annual depreciation is 7% of the cost price per year. 8 A writer buys a set of books for $950 to help him with his work. He opts to depreciate this library by the prime cost method and depreciation is 7% of the cost price each year. 9 For the situations described below, and using a straight line depreciation model, find: i the annual rate of depreciation ii the relationship between the book value and time and use it to find at what age the item will be written off, that is, have a value of $0. a A car purchased for $ with a current value of $5 000 now it is 5 years old b A stereo unit bought for $850 seven years ago which has a current value of $50 c A refrigerator with a current value of $85 bought 0 years ago for $35 d An aeroplane with purchase price of $5 000 sold at market value of $0 50 when it was 7 years old 0 Each graph below represents the flat rate depreciation of four particular items. In each case determine: i the cost price of the item ii the annual depreciation iii the time taken for the item to reach its scrap value or to be written off. a b V c V d 800 V V T 0 3 T T T Time (years) Time (years) Time (years) Time (years) Book value ($ '000) Book value ($) Book value ($) Book value ($ '000)

41 Chapter Growth and decay 685 multiple choice A -tonne truck, bought for $3 000, was depreciated using the flat rate method. If the scrap value of $5000 was reached after 5 years, the annual depreciation would be: A $3 B $000 C $500 D $600 E $6 000 multiple choice A half-tonne truck, bought for $7 000, was depreciated using the flat rate method. If the scrap value of $3600 was reached after 9 years, the annual depreciation would be closest to: A $3 00 B $3500 C $3000 D $500 E $000 3 A business chose to depreciate its $600 photocopier by the prime cost method. What was the annual depreciation if the scrap value of $30 was reached in 6 years? multiple choice The depreciation of a piece of machinery is given by the equation, V = T. The machinery will have a book value of $00 after: A 7 years B 8 years C 9 years D 0 years E years 5 The depreciation of a computer is given by the equation, V = T. After how many years will the computer have a book value of $770? 6 multiple choice Listed below are the depreciation equations for 5 different items. Which item would be written off in the least amount of time? A V = 650T B V = 750T C V = 650T D V = 750T E V = 850T multiple choice Listed below are the depreciation equations for 5 different items. Which item would be written off in the greatest amount of time? A V = T B V = T C V = T D V = T E V = T 8 A business buys two different photocopiers at the same time. One costs $00 and is to be depreciated by $5 per annum. It also has a scrap value of $00. The other costs $3600 and is to be depreciated by $30 per annum. This one has a scrap value of $500. a Which machine would need to be replaced first? b How much later would the other machine need to be replaced? 9 multiple choice A car valued at $0 000 was bought 5 years ago for $ The straight-line depreciation model is represented by: A V = 0 000T B V = 5000T C V = T D V = T E V = T

42 686 Further Mathematics Reducing balance depreciation If an item depreciates by the reducing balance method then its value decreases by a fixed rate each unit time interval, generally each year. This rate is a percentage of the previous book value of the item. Reducing balance depreciation is also known as diminishing value depreciation. WORKED Example 0 Suppose the new $5 000 printing press considered in worked example 8 was depreciated by the reducing balance method at a rate of 0% p.a. of the previous book value. a Draw a depreciation schedule for the first years of work for the press. b What is the book value after years? c Draw a graph of book value against time. WRITE a Find the depreciation for the first year. a d = 0% of = $ Find the book value after the first year. Book value = cost price depreciation. Calculate year depreciation and book value after years. BV = = $ 000 d = 0% of $ 000 = $00 BV = = $9600 Repeat the process for the next years. d 3 = 0% of 9600 = $90 BV 3 = = $7680 d = 0% of 7680 = $536 BV = BV = $6 Draw the depreciation schedule. Time, T (years) Depreciation, d ($) Book value, V ($)

43 Chapter Growth and decay 687 b State the book value from the depreciation schedule. c Draw a graph of the book value against time. WRITE b The book value of the press after years will be $6. c V T Time (years) Book value ($ '000) It is clear from the graph and the schedule that the reducing balance depreciation results in greater depreciation during the early stages of the asset s life (the book value drops more quickly at the start since the annual depreciation falls from $3000 in year to $536 in year ). Now, the Australian Taxation Office allows depreciation of an asset as a tax deduction. This means that the annual depreciation reduces the amount of tax paid by a business in that year. The higher the depreciation, the greater the tax benefit. Therefore, depreciating an asset by the reducing balance method allows a greater tax benefit for a business in the beginning of an asset s life rather than towards the end. In contrast, flat rate depreciation remains constant throughout the asset s life. People have a choice as to whether they depreciate an item by the flat rate or reducing balance methods, but once a method is applied to an article it cannot be changed for the life of that article. The percentage depreciation rates, which are set by the Australian Taxation Office, vary from one item to another but for each item the rate applied for the reducing balance method is greater than that for the flat rate method. Let us compare depreciation for both methods. A transport business has bought a new bus for $ The business has the choice of depreciating the bus by a flat rate of 0% of the cost price each year or by 30% of the previous book value each year. a Draw depreciation schedules using both methods for a life of 5 years. b c WORKED Example Draw graphs of the book value against time for both methods on the same set of axes. After how many years does the reducing balance book value become greater than the flat rate book value? a Calculate the flat rate depreciation per year. WRITE a d = 0% of $ = $ 000 per year Continued over page

44 688 Further Mathematics Draw and complete a flat rate depreciation schedule for 0 5 years. WRITE Time, T (years) Depreciation, d ($) Book value, V ($) Draw and complete a reducing balance depreciation schedule. Annual depreciation is 30% of the previous book value. Subtract this from the previous book value to ascertain the present book value. Continue to calculate the book value for a period of 5 years. Time, T (years) Depreciation, d ($) Book value, V ($) % of = $ % of 000 = $ % of 9 00 = $ % of = $ % of 06 = $ b Draw graphs using values for V and T from the schedules. b Book value ($ '000) V T Time, (years) c Look at the graph to see when the reducing balance curve lies above the flat rate line. State the first whole year after this point of intersection. c The book value for the reducing balance method is greater than that of the flat rate method after years. The graph of book value against time in worked example 0 suggests that reducing balance depreciation may be an example of exponential decay which was discussed earlier. Let us check this by analysing the schedule.

45 Chapter Growth and decay Growth or compounding factor = = Now = 9600 and = 7680 and = 6. So reducing balance depreciation is an example of exponential decay and in this case the growth or compounding factor is 0.8. As we saw in the first section of this chapter an exponential decay situation could be represented by the equation: y = ka x where k, a are constants, a <. In this situation the growth or compounding factor, a, represents 00% of the previous book value less the rate of depreciation per unit time interval. In worked example 0 the annual depreciation is 0% of the previous book value. Hence, growth or compounding factor = 00% 0% = 0. = 0.8 We can write a general formula for reducing balance depreciation which is similar to the compound interest formula (and can be derived in the same way) except that the rate is subtracted rather than added to. The reducing balance depreciation formula is: BV T = P r T BV T = book value after time, T 00 r = rate of depreciation P = cost price T = time since purchase That is, given the cost price and depreciation rate we can find the book value (including scrap value) of an article at any time after purchase. Let us now see how we can use this formula. The printing press from worked example 0 was depreciated by the reducing balance method at 0% p.a. What will be the book value and total depreciation of the press after years if it cost $5 000 new? WRITE State P, r and T. P = 5 000, r = 0, T = Substitute into the depreciation formula and simplify. 3 5 WORKED Example BV T = P r = 5 000( ) = 5 000(0.8) Evaluate. BV = $6 Total depreciation is: Total depreciation = P BV cost price current book value. = = $8856 Write a summary statement. The book value of the press after years will be $6 and total depreciation will be $8856. T 00

46 690 Further Mathematics Effective life The situation may arise where the scrap value is known and we want to know how long it will be before an item reaches this value, that is, its useful or effective life. So, in the reducing balance formula BV = P r , T is needed. A similar 00 situation arose in the section Finding time in compound interest. On that occasion, two different methods were used: trial and error, and logarithms. We will use both methods here to find T. A photocopier purchased for $8000 depreciates by 5% p.a. by the reducing balance method. If the photocopier has a scrap value of $00, how long will it be before this value is reached? Method : Trial and error WRITE State BV, P and r. BV = 00, r = 5, P = 8000 Substitute into the depreciation formula BV = P r T 00 and simplify. 00 = 8000(0.75) T Substitute values for T and compare with 0.5. Since 0.75 <, increasing T value decreases the value of 0.75 T. Let T = T 00 = T = = > 0.5. Try T = = 0.78 Try T = = 0.33 In whole years, the time needed is 7. T = 7 years Write a summary statement. The time needed to reach scrap value is 7 years. Method : Logarithms State BV, P and r. BV = 00, r = 5, P = 8000 Substitute into the depreciation formula BV = P r T 00 and simplify. 00 = 8000(0.75) T 0.75 T 00 = T = Take the log of both sides. log 0.75 T = log 0.5 Apply the log rule and isolate T. T log 0.75 = log 0.5 (log x n = n log x) log 0.5 T = log Evaluate T. T = State T in whole years. T = 7 years Write a summary statement. The time needed to reach scrap value is 7 years. 7 WORKED Example 3 T

47 remember remember Chapter Growth and decay 69. To make calculations for reducing balance depreciation, use the formula: BV T = P r T. 00. To find time in reducing balance depreciation use either trial and error or logarithms. 3. Reducing balance depreciation is also known as diminishing value depreciation. F Reducing balance depreciation WORKED Example 0 WORKED Example A farming company chose to depreciate its new $ bulldozer by the reducing balance method at a rate of 0% p.a. of the previous book value. a Draw a depreciation schedule for the first years of the bulldozer s life. Reducing b What is its book value after years? balance depreciation c Draw a graph of book value against time. A retail store chose to depreciate its new $000 computer by the reducing balance method at a rate of 0% p.a. of the previous book value. a Draw a depreciation schedule for the first years of the computer s life. b What is its book value after years? c Draw a graph of book value against time. 3 A carpenter chose to depreciate her new set of electric hand tools, valued at $500, by the diminishing value method at a rate of 0% p.a. of the previous book value. a Draw a depreciation schedule for the first years of the tool set s life. b What is its book value after years? c Draw a graph of book value against time. An accounting firm chose to depreciate a set of new electronic calculators, valued at $000 in total, by the diminishing value method at a rate of 5% p.a. of the previous book value. a Draw a depreciation schedule for the first years of the calculators life. b What is the book value after years? 5 A café buys a cash register for $550. The owner has the choice of depreciating the register by the flat rate method (at 0% of the cost price each year) or the reducing balance method (at 30% of the previous book value each year). a b c Draw depreciation schedules for both methods for a life of 5 years. Draw graphs of book value against time for both methods on the same set of axes. After how many years does the reducing balance book value become greater than the flat rate book value? 6 Speedy Cabs taxi service has bought a new taxi for $ The company has the choice of depreciating the taxi by the flat rate method (at % of the cost price each 3 year) or the diminishing value method (at 50% of the previous book value each year). a Draw depreciation schedules for both methods for 3 years. b Draw graphs of book value against time for both methods on the same set of axes. c After how many years does the reducing balance book value become greater than the flat rate book value? Comparing depreciations Mathcad Mathcad

48 69 Further Mathematics Mathcad WORKED Example Reducing balance depreciation formula 7 Using the reducing balance formula, find BV (to decimal places) given: a P = 0 000, r = 0, T = b P = , r = 5, T = c P = 5 000, r = 5, T = 6 d P = 000, r = 30, T = 5 e P = 500, r = 0, T = 6 f P = 980, r = 7, T = 5 g P = 675, r =.5, T = 5 h P = 8650, r = 3.5, T = A refrigerator costing $00 new is depreciated by the reducing balance method at 0% a year. After years its book value will be: A $0 B $9.5 C $960 D $05 E $88.3 multiple choice A hot water system, purchased for $600, was depreciated by the reducing balance method at 0% p.a. Its book value after 6 years would be closest to: A $500 B $000 C $000 D $700 E $500 0 The items below are depreciated by the reducing balance method at 5% p.a. What will be the book value and total depreciation of: a a TV after 8 years, if it cost $50 new? b a photocopier after years, if it cost $370 new? c carpets after 6 years, if they cost $730 new? d an electric heater after 5 years, if it cost $975 new? The items below are depreciated at 30% p.a. by the diminishing value method. What will be the book value and total depreciation of: a a lawn mower after 5 years, if it cost $685 new? b a truck after years, if it cost $3 500 new? c a washing machine after 3 years, if it cost $075 new? d a bus after 6 years, if it cost $6 000 new? The items below are depreciated at.5% p.a. by the diminishing value method. What will be the book value and total depreciation of: a a mini bus after 5 years, if it cost $38 00 new? b a small truck after years, if it cost $6 80 new? c a car after 6 years, if it cost $ 770 new? d a machine after 3 years, if it cost $ 300 new? 3 multiple choice After 7 years, a new $3000 photocopier, which devalues by 5% of its book value each year, will have depreciated by: A $00.5 B $750 C $50 D $ E $750. multiple choice New office furniture valued at $7 500 is subjected to reducing balance depreciation of 0% p.a. and will reach its scrap value in 5 years. The scrap value will be: A less than $300 B between $300 and $00 C between $00 and $500 D between $500 and $600 E between $600 and $ multiple choice multiple choice A new chainsaw bought for $50 has a useful life of only 3 years. If it depreciates annually at 60% diminishing value rate, its scrap value will be: A zero B $60 C $80 D $50 E $70.

49 Chapter Growth and decay 693 WORKED Example 3 6 Use the reducing balance formula to find T (to decimal places), given: a BV = $3000, P = $0 000, r = 0 b BV = $500, P = $3000, r = 30 c BV = $900, P = $500, r = 5 d BV = $500, P = $7600, r = 5 e BV = $50, P = $600, r = 0 f BV = $00, P = $0 500, r = 5 g BV = $650, P = $985, r = 7 h BV = $0, P = $699, r =.5 7 Use the reducing balance formula to find the time, T, (to the appropriate integral value) needed to reach scrap value for an item which has: a a purchase price $3 000, depreciation of 0% p.a. and a scrap value of $800 b a purchase price $6 000, depreciation of 30% p.a. and a scrap value of $3000 c a purchase price $39 700, depreciation of 5% p.a. and a scrap value of $5000 d a purchase price $90, depreciation of 5% p.a. and a scrap value of $50 e a purchase price $70, depreciation of % p.a. and a scrap value of $50 f a purchase price $60, depreciation of 7% p.a. and a scrap value of $350 g a purchase price $7 300, depreciation of 33% p.a. and a scrap value of $500 h a purchase price $5 900, depreciation of 30% p.a. and a scrap value of $ A tractor was bought for $ 000 and was depreciated by the diminishing value method at % p.a. Find: a the book value after 5 years b the total depreciation after 5 years c how many years it takes to reach the scrap value of $ A car bought for $ was depreciated by the diminishing value method at 5% p.a. Find: a the book value after years b the total depreciation after years c how many years it takes to reach the scrap value of $ A boat was bought for $5 670 and depreciated by the reducing balance method at 30% p.a. Find: a the book value after years b the total depreciation after years c how many years it takes to reach the scrap value of $3500. The owner of a bus, bought for $8 900, has the choice of depreciating it by the flat rate method at 0% p.a. of cost price or the reducing balance method at 30% p.a. Using the depreciation formulas, find how many years it will take before the reducing balance book value becomes greater than the flat rate book value. A diving instructor buys new equipment for $60. She may depreciate the gear at 7% p.a. of cost price or 5% p.a. of the previous book value. Using the depreciation formulas, find how many years it will take for the diminishing value book value to exceed the corresponding prime cost value. 3 A business accountant decided to depreciate a $9 000 company vehicle by the reducing balance method at the rate of 0% p.a. until the car reached its scrap value of $5000. When the vehicle was bought the business invested money in an account which paid interest at 0% p.a., compounded monthly. a How long would it take the vehicle to reach its scrap value? b How much was invested (to the nearest $00) if the company was able to buy a new $9 000 vehicle by the time the old one was scrapped?

50 69 Further Mathematics Unit cost depreciation The flat rate and reducing balance depreciations of an item are based on the age of the item. With the unit cost method, the depreciation is based on the possible maximum output (units) of the item. For instance, the useful life of a truck could be expressed in terms of the distance travelled rather than a fixed number of years for example, kilometres rather than 6 years. The actual depreciation of the truck for the financial year would be a measure of the number of kilometres travelled. WORKED Example A taxi is bought for $3 000 and it depreciates by an average of 8. cents per kilometre driven. In one year the car is driven 5 6 km. Find: a the annual depreciation for this particular year b its useful life if its scrap value is $ 000. a b Depreciation amount = distance travelled rate Write a summary statement. Total depreciation = cost price scrap value total depreciation Distance travelled = rate of depreciation where rate of depreciation = 8. cents/km where rate of depreciation = $0.8 per km State your answer. WRITE a d = cents d = $3.38 Annual depreciation for the year was $3.38. b Total depreciation = = $ Distance travelled = = km The taxi has a useful life of km.

51 Chapter Growth and decay 695 WORKED Example 5 A photocopier purchased for $0 800 depreciates at a rate of 0 cents for every 00 copies made. In its first year of use copies were made and in its second year, Find: a the depreciation each year b the book value at the end of the second year. WRITE a To find the depreciation, identify the rate and number of copies made. Express the rate of 0 cents per 00 copies in a simpler form of dollars per 00 copies, 0.0 that is $0.0 per 00 copies or copies b Book value = cost price total depreciation a Depreciation = copies made rate 0.0 d = copies = $000 Depreciation in the first year is $ d = copies = $00 Depreciation in the second year is $00. b Total depreciation after years = = $00 Book value = = $8700 WORKED Example 6 The initial cost of a vehicle was $7 850 and its scrap value is $5050. If the vehicle needs to be replaced after travelling km (useful life): a Find the depreciation rate (depreciation ($) per km). b Find the amount of depreciation in a year when 6 97 km were travelled. c Find the book value after it has been used for a total of km. d Set up schedule table listing book value for every km. WRITE a To find the depreciation rate, first find the total depreciation. Total amount of depreciation a Total amount of depreciation = = $ 800 = Cost price Scrap value Find the rate of depreciation. Depreciation rate total depreciation = total distance travelled 800 = It is common to express rates in cents per use if less than a dollar = $0.85 per km = 8.5 cents per km Continued over page

52 696 Further Mathematics b Find the amount of depreciation using the rate calculated. c 3 Amount of depreciation is always expressed in dollars. To find the book value first calculate the amount of depreciation for a use of km. Calculate the book value. In this case the km has been travelled from when the car was new so the previous book value is $ Write your answer. d Calculate the book value for every km of use and summarise in a table. WRITE b Amount of depreciation = amount of use rate of depreciation = = cents = $70.65 c Amount of depreciation = = cents = $ Book value = previous value amount of depreciation = = $0 750 Book value after the car has been used for km is $ d For every km, amount of depreciation = = $5700 Use (km) Book value ($) 0 $ $ $ $ $ Alternatively, using a graphics calculator, enter the equation in Y =. The depreciation equation is BV = km. Set up TABLE in TABLE SET and display the table. Start table at 0 km and increment at every km.

53 remember remember Chapter Growth and decay 697. Unit cost depreciation is based on how much an item is used.. Current book value ($) = previous book value ($) amount of depreciation ($) 3. Amount of depreciation ($) = amount of use rate of depreciation ($ per use) amount of depreciation ($). Rate of depreciation ($ per use) = amount of use G Unit cost depreciation WORKED Example WORKED Example 5 Below are depreciation details for 6 vehicles. In each case find: i the annual depreciation ii the useful life (km). Purchase price ($) Scrap value ($) Average rate of depreciation (c/km) a b c d e f In each situation in questions, find: a the annual depreciation b the item s useful life. Unit cost depreciation Distance travelled in first year (km) A company buys a $3 000 car which depreciates at a rate of 3 cents per km driven. It covers 5 30 km in the first year and has a scrap value of $ A new taxi is worth $9 500 and it depreciates at 7. cents per km travelled. In its first year of use it travelled 8 6 km. Its scrap value was $800. A government car was purchased for $36 99 and depreciated at a rate of.5 cents per km travelled. It travelled 9560 km in its first year of use. It has a scrap value of $ In each situation in questions 5 8, find: a the depreciation for each year b the book value at the end of the second year. 5 A photocopier is bought for $8600 and it depreciates at a rate of cents for every 00 copies made. In its first year of use copies are made and in its second year, copies are made. 6 A photocopier purchased for $700 depreciates at a rate of $.50 per 000 copies made. In its first year of use copies were made and in its second year, were made. Mathcad

54 698 Further Mathematics 7 A metal stamping machine purchased for $5 800 depreciates at a rate of.5 cents for every 00 completed stamps. In its first year of use million stamps were completed and in its second year. million were made. 8 A printing machine was purchased for $ and depreciated at a rate of $.50 per million pages printed. In its first year 385 million pages were printed and 96 million in its second year. 9 A photocopier bought for $ 300 depreciates at a rate of.5 cents for every 0 copies made. Copy and complete the table below. Time (years) Copies made per year Annual depreciation ($) Book value at end of year ($) 0 A corking machine bought for $ 750 depreciates at a rate of $.50 for every 00 bottles corked. Copy and complete the table below. Time (years) Bottles corked per year Depreciation ($) Book value at end of year ($) multiple choice A vehicle is bought for $5 900 and it depreciates at a rate of.6 cents per km driven. After its first year of use, in which it travelled km, the book value of the vehicle will be closest to: A $000 B $3000 C $0 000 D $3 000 E $5 000 In each situation in questions and 3, find: a the depreciation for each year b the book value at the end of the second year. A company van is purchased for $3 600 and it depreciates at a rate of.8 cents per km driven. In its first year of use the van travelled 5 60 km and it travelled 6 05 km in its second year. 3 A taxi is bought for $ and it depreciates at a rate of 9. cents per km driven. It travelled 6 km in its first year of use and km in its second year.

55 Chapter Growth and decay 699 A car bought for $8 395 depreciates at a rate of 3.6 cents for every km travelled. Copy and complete the table below. Time (years) Distance travelled (km) Depreciation ($) Book value at end of year ($) 5 A farm vehicle bought for $33 99 depreciates at a rate of 3.5 cents per km driven. Copy and complete the table below. Time (years) Distance travelled (km) Depreciation ($) Book value at end of year ($) WORKED Example 6a 6 The following information provides depreciation details about various vehicles. In each case find the depreciation rate (depreciation per km). a Initial cost = $5 50, scrap value = $5000, useful life = km b Initial cost = $3 00, scrap value = $6000, useful life = km c Initial cost = $36 000, scrap value = $6000, useful life = km d Cost price = $ 005, scrap value = $000, useful life = km e Cost price = $9 960, scrap value = $000, useful life = km f Cost price = $9 80, scrap value = $5000, useful life = km 7 The following information provides depreciation details about various photocopiers. In each case find the depreciation rate (depreciation per 00 copies). a Initial cost = $8000, scrap value = $000, useful life = million copies b Initial cost = $7800, scrap value = $000, useful life = million copies c Initial cost = $5900, scrap value = $500, useful life = million copies d Cost price = $0 00, scrap value = $000, useful life = 3 million copies e Cost price = $6800, scrap value = $500, useful life = million copies f Cost price = $7500, scrap value = $500, useful life = million copies 8 multiple choice A machine costing $8500 is estimated to have scrap value of $500 and believed to have a maximum output of units. The depreciation rate (charged per unit) is: A 0 cents per unit B dollars per unit C cents per unit D.5 cents per unit E.5 dollars per unit

56 700 Further Mathematics WORKED Example 6 9 multiple choice An $8500 machine depreciates by cents/unit. By the time the machine had depreciated by $5000, it would have produced: A units B units C units D units E units 0 multiple choice A machine which was bought for $8500 was depreciated at the rate of cents per unit produced. By the time the book value had decreased to $000, the number of units produced would be: A B C D E A delivery service purchases a van for $ and it is expected that the van will be written off after travelling km. It is estimated that the van will travel 600 km each week. a Find the depreciation rate (charge per km). b Find how long it will take for the van to be written off. c Find the distance travelled for the van to depreciate by $ d Find its book value after it has travelled kilometres. e Set up a schedule table for the value of the van for every kilometres. A car is bought for $ and a scrap value of $0 000 is set for it. The following three options for depreciating the car are available: i flat rate of 0% of the purchase price each year ii 0% p.a. of the reducing balance iii 5 cents per km driven (the car travels an average of km per year). a b c Which method will enable the car to reach its scrap value sooner? If the car is used in a business the annual depreciation can be claimed as a tax deduction. What would the tax deduction be in the first year of use for each of the depreciation methods? How would your answers to part b vary for the 5th year of use?

57 Chapter Growth and decay 70 Inflation The term inflation is often used when talking about prices. It is a measure of the average increase in the price of goods and services from one year to the next. The effect of inflation (price increases) is that money loses its value and therefore its purchasing power (how much we can buy with it). Inflation is often expressed as a rate per year and so indicates the annual increase in the price of a fixed set of goods and services. An average of the annual inflation rates over several years can be used to make price calculations of items. Since the prices increase by a fixed rate each year, inflation is another example of exponential growth. For example, consider an average inflation rate of 8% p.a. and a current price of $.65 for a carton of milk. What will be the price in years time? Now, growth factor = 00% + 8% = =.08 Price in year s time = = $.78 Price in years time = = $.9 So, inflation rate works in much the same way as compound interest. In fact we can use the compound interest formula as follows: A = PR T where A = price after time, T P = original price ($) T = time in years r R = where r is the inflation rate 00 WORKED Example 7 Assuming an average inflation rate of 5% p.a. find the price for a $7.50 theatre ticket in 6 years time. WRITE List P, r and T. P = 7.5, r = 5, T = 6 Substitute into the formula and evaluate. A = P r T 00 = 7.5(.05) 6 = $0.05 The price of a ticket after 6 years is $0.05.

58 70 Further Mathematics remember remember. Inflation rate is a measure of the average percentage growth in the cost of goods and services over a period of years.. It is another example of exponential growth and we can use the compound interest formula as follows: A = PR T where A = price after time, T P = original price ($) T = time in years r R = where r is the inflation rate 00 H Inflation Mathcad Inflation WORKED Example 7 In each case below find the price of the item in: a 5 years time if it now costs $5 and the average inflation is % p.a. b 6 years time if it now costs $90 and the average inflation is 5% p.a. c years time if it now costs $ 000 and the average inflation is.3% p.a. d years time if it now costs $6870 and the average inflation is 3.8% p.a. e 7 years time if it now costs $8.95 and the average inflation is.% p.a. f 0 years time if it now costs $.5 and the average inflation is.7% p.a. g 5 years time if it now costs $5 600 and the average inflation is.9% p.a. h years time if it now costs $05.50 and the average inflation is 5.% p.a. What could you have expected to pay for an item: a 5 years ago if it now costs $5 and the average inflation has been % p.a.? b 6 years ago if it now costs $90 and the average inflation has been 5% p.a.? c years ago if it now costs $ 000 and the average inflation has been.3% p.a.? d years ago if it now costs $6870 and the average inflation has been 3.8% p.a.? e 7 years ago if it now costs $8.95 and the average inflation has been.% p.a.? f 0 years ago if it now costs $.5 and the average inflation has been.7% p.a.? g 5 years ago if it now costs $5 600 and the average inflation has been.9% p.a.? h years ago if it now costs $05.50 and the average inflation has been 5.% p.a.? 3 How many years (to the nearest year) would it take for the price of an item to go from: a $5.00 to 3. if the average inflation during the time was % p.a.? b $.30 to $.7 if the average inflation during the time was 5% p.a.? c $ to $63.09 if the average inflation during the time was.5% p.a.? d $ 000 to $7 796 if the average inflation during the time was 8.% p.a.? e 80 cents to $.08 if the average inflation during the time was.5% p.a.? f $695 to $370 if the average inflation during the time was.8% p.a.? g $95 to $089 if the average inflation during the time was 3.6% p.a.? h $89.00 to $.87 if the average inflation during the time was 5.% p.a.? The price of a standard cricket bat changes from $0 to $90 over a 6-year period. Find the average annual inflation rate during this time.

59 Chapter Growth and decay The price of 00W light bulbs changes from 85 cents to $.03 over a 5-year period. Find the average annual inflation rate during this time (to decimal places). 6 Over a -month period the price of a certain cereal changes from $.85 to $5.5. Find the annual inflation rate during the year (to decimal places). 7 Calculate the average inflation rate for the 9-year period of the items in the table (to decimal place): 8 At the start of 99 the price of a tennis racquet was $0.50. Using an average inflation rate of % per annum, find the price of this racquet at the start of In 985 the cost of a house was $ Find the 995 cost of the house (to the nearest $00) if the inflation rate has been 5.% per annum. 0 Item 989 Price ($) 998 Price ($) a Video b Pen c Can of coke.67.0 d Council rates e Cough medicine f Freddo frog g Pair of socks h Toaster multiple choice In a year when the inflation rate was.% and the cost of a book at the start of the year was $8.95, the cost at the end of the year would be closest to: A $6 B $8 C $0 D $ E $3 multiple choice In a year when the price of a cubic metre of firewood was $35, the inflation rate over the past three years when its cost rose from $5 would be closest to: A 9% B 0% C % D % E 3%

60 70 Further Mathematics summary Growth and decay functions Growth and decay can be linear or exponential. Linear growth and decay can be represented by the equation y = a + bx where y is the dependent variable x is the independent variable (usually time) a is the initial or starting value of y and b is the rate of growth or decay. Exponential growth and decay means an initial value multiplied by a growth or compounding factor for each unit time interval. Exponential growth and decay can be represented by the equation y = ka x where a = growth or compounding factor a > for growth, a < for decay and k = the initial or starting value of y. Compound interest formula Compound interest is an example of exponential growth and is calculated using the formula: A = PR n where A = final amount ($) P = principal ($) R = growth or compounding factor = r = interest rate per period n = number of interest bearing periods For compound interest, I = A P Finding time in compound interest using trial and error Finding time by trial and error involves educated guessing for the value of n. Time = n number of times interest is compounded each year. Finding time in compound interest using logarithms When using logs to find time, use the rule: log x n = n log x. Change the n value to time in more meaningful terms. Depreciation There are 3 methods by which depreciation can be calculated:. flat rate depreciation. reducing balance depreciation 3. unit cost depreciation. An item is written off when its book value becomes zero. Scrap value is an item s book value when it is no longer used. r

61 Chapter Growth and decay 705 Flat rate depreciation To make calculations in flat rate depreciation, use the formula BV T = P dt where BV T = book value after T years ($) P = purchase or cost price ($) T d = time (years of depreciation) = rate of depreciation ($ per year) = fixed amount per year or = percentage of P per year Flat rate depreciation is also known as prime cost depreciation. Total depreciation = cost price current value Rate of depreciation = total depreciation number of years Reducing balance depreciation Reducing balance depreciation is an example of exponential decay and is calculated using the formula: BV T = P r T where BV T is the book value after time, T 00 P is the cost price r is the rate of depreciation T is the time since purchase Unit cost depreciation Unit cost depreciation is based on how much an item is used Current book value ($) = previous book value ($) amount of depreciation ($) Amount of depreciation ($) = amount of use rate of depreciation ($ per use) amount of depreciation ($) Rate of depreciation ($ per use) = amount of use Inflation Inflation rate is a measure of the average percentage growth in the cost of goods and services over a period of years. It is another example of exponential growth and we can use the compound interest formula as follows: A = PR T where A = price after time, T P = original price ($) T = time in years r R = where r is the inflation rate 00

62 706 Further Mathematics CHAPTER review A A Multiple choice What type of growth or decay does the graph at right display? A Linear growth B Linear decay C Exponential growth D Exponential decay E Steady/no change Which of the following graphs displays an exponential decay? A B C D E FM 7 The following information refers to questions 3 and. An investment of $500 earns compound interest at a rate of 6.% p.a. and is made for 5 years. B B B C,D 3 The balance in the account at the end of the investment period, if interest is compounded quarterly, is: A $68.0 B $87.7 C $ D $ E $859.0 The principal plus interest accrued during the investment, if interest is credited weekly, is: A $ B $ C $ D $ E $ After -- years $00 has grown to $750 in an account where interest is compounded monthly. The annual interest rate is: A 7.0% B 0.7% C 8.% D 3.% E 38% 6 A sum of $850 is invested at 8% p.a. compound interest, credited fortnightly. For the balance to grow to $00 the investment should be left for a minimum of: A years B 3 years C years 8 fortnights D years 9 fortnights E 5 years

63 Chapter Growth and decay In an account which pays compound interest at % p.a., credited daily, an investment of $3 000 will accrue $000 interest in: A years 86 days B years 87 days C 3585 days D 9 years 30 days E 0 years 8 A machine bought for $8500 is depreciated by the flat rate method. If its useful life is 6 years and its scrap value is $000 then the annual depreciation will be: A $7 B $583 C $7500 D $79 E $50 D E The following information refers to questions 9. A refrigerator valued at $50 is depreciated at 5% p.a. of the previous book value. 9 The book value of the refrigerator after years will be: A $395.5 B zero C $50 D $.88 E $5. F 0 The total depreciation after years will be: A $50 B $5. C $85.9 D $.88 E $00 F If the refrigerator has a scrap value of $00 then its useful life will be closest to: A 5 years B 6 years C 7 years D years E 3 years F The following information refers to questions and 3. A car was purchased for $3 600 and depreciated at a rate of 5.6 cents per km driven. The depreciation in a year when the car travelled km would be: A $ B $ C $500 D $ E $66. 3 If the car travelled an average of km each year then it would be written off after: A years B 3 years C years D 8 years E 9 years The inflation rate for the past 6 years has averaged 3.75% per annum. If the salary of a teacher 6 years ago was $ then the teacher, to maintain the standard of earnings at the present time, should be receiving close to: A $ 000 B $5 000 C $6 000 D $7 000 E $8 000 G G H

64 708 Further Mathematics A B B B C,D E Short answer If the average inflation rate during a 5-year period had been 3.5% p.a., what would be the cost of a jar of peanut butter at the end of the period if it cost $.60 at the start? If $500 is to be invested for 5 years, which of the options below would be the most productive to use? a % p.a. simple interest b compound interest at.8% p.a., credited quarterly c compound interest at.7% p.a., credited monthly. 3 What amount must be invested at 9.5% p.a., interest compounded 6-monthly, if it is to grow to $5000 over years? How much interest would $950 earn if it was invested for 3 years at % p.a., interest credited daily? 5 How long would it take for $000 to amount to $350 by earning interest at 6.8% p.a., compounded monthly? 6 A computer is depreciated by the prime cost method at 5% p.a. If it was bought for $900 how many years elapsed before it was written off? F 7 Furniture is bought for $5 000 and depreciated at 8% p.a. by the reducing balance method. What would its value be in 6 years time? G G H 8 A taxi depreciates at 9.5 cents per km driven. If it was bought for $9 600, how far would it have travelled for it to be valued at $0 000? 9 A taxi was purchased for $ and it depreciates at an average rate of 30 cents per km driven. During its first year the taxi travelled 650 km and during its second it travelled km. Find: a the depreciation in each of the first years b how far the car had travelled if its total depreciation was $0 000 c how far the car had travelled when it reached its scrap value of $ A stamp collection was purchased years ago for $3350. It has increased in value at the same rate as inflation. It is currently valued at $5680. a Find the inflation rate. b How long will it take for the stamp collection to double in value? c How long ago did it have a value of $980? d Graph its value over the first 0 years on a graphics calculator or spreadsheet.

65 Chapter Growth and decay 709 Analysis A small number of rabbits have migrated to Mr Smith s farm. Mr Smith has brought in experts to make some investigations. a The small number of rabbits is estimated to be 80. From experience, the experts know that the rabbit population increases by 0% on each previous month s population. i Copy and complete the table of the expected population of the rabbits for the first three months. Month 0 3 Rabbit population 80 b ii A population of 000 or more is considered to be harmful to the ecosystem. How soon (in months) will this occur on Mr Smith s farm? The area of land used (and devastated) by a rabbit population is given in the table below. Number of rabbits Area of land used (hectares) c d i State whether it is an exponential or straight line growth. ii Give evidence to justify your response in part i. The cost (C in dollars) to the farmer for every hectare of land (H) lost is given by the equation C = 000. H. i What is the cost to the farmer if there are 00 rabbits on the farm? ii Mr Smith can only sustain about $000 loss. How much land is this equivalent to (to the nearest hectare)? At the completion of the investigation, the experts suggest Mr Smith budget $ 00 for a rabbit eradication program. Mr Smith needs to borrow this amount and he has loan options: 0% p.a. simple interest with full payment in years 7% p.a. compound interest (adjusted monthly) with full payment in 3 years time. i Calculate the total payment to be made at the end of the term for each of the loan options. ii Which is the best option? Explain why. Another option is to use the cash set aside for a tractor and get the tractor on hire-purchase. iii On hire-purchase he can get the $3 000 tractor with a deposit of $600 and monthly instalments at 6% p.a. over 3 years. Find the monthly instalments.