Year 10 General Maths Unit 2

Transcription

1 Year 10 General Mathematics Unit 2 - Financial Arithmetic II Topic 2 Linear Growth and Decay In this area of study students cover mental, by- hand and technology assisted computation with rational numbers, practical arithmetic and financial arithmetic, including estimation, order of magnitude and accuracy. This topic includes: applications of simple interest and compound interest cash flow in common savings and credit accounts including interest calculation compound interest investments and loans Key knowledge concepts of simple and compound interest and their application compound interest investments and debts. Key skills apply simple interest to analyse cash flow in common savings and credit accounts apply compound interest to solve problems involving compound interest investments and loans compare the costs of a range of purchase options such as cash, credit and debit cards, personal loans, and time payments (hire purchase). Chapter Sections Questions to be completed 1. Simple Interest Tables In the notes (4 questions) 2. Simple Interest Formula (3.5 Textbook) 1bd, 2bd, 3bd, 4bd, 12, 13, Simple interest - Loans (3.5 Textbook) 5bd, 8, 10, 14, 17, 20, Simple Interest Recurrence relation In the Notes (3 questions) 5. Simple interest General recurrence relation In the Notes (5 questions) 6. Flat rate depreciation In the Notes (5 questions) 7. Unit- cost depreciation In the Notes (3 questions) More resources available at Page 1 of 20

2 Table of Contents 1. SIMPLE INTEREST TABLES... 3 SIMPLE INTEREST... 3 EXAMPLE EXAMPLE EXERCISE SIMPLE INTEREST FORMULA... 5 EXAMPLE FINDING V O, R AND N... 5 EXAMPLE EXAMPLE WORKED EXAMPLE 9: SIMPLE INTEREST - LOANS... 7 WORKED EXAMPLE CASH FLOW... 7 NON- ANNUAL INTEREST CALCULATIONS... 7 WORKED EXAMPLE MINIMUM BALANCE CALCULATIONS... 8 WORKED EXAMPLE SIMPLE INTEREST - RECURSION RELATION... 9 EXAMPLE 4.1 RECURRENCE RELATION... 9 SIMPLE INTEREST USING CAS CALCULATOR EXAMPLE EXAMPLE EXERCISE 4. MODELLING SIMPLE INTEREST WITH RECURRENCE RELATIONS GENERAL RECURRENCE RELATION FOR THE NTH TERM IN LINEAR GROWTH EXAMPLE 5.1 USING A RULE TO DETERMINE THE VALUE OF A SIMPLE INTEREST INVESTMENT EXAMPLE EXERCISE FLAT RATE DEPRECIATION FLAT RATE (STRAIGHT LINE DEPRECIATION) EXAMPLE EXERCISE UNIT- COST DEPRECIATION UNIT COST DEPRECIATION RECURRENCE RELATION EXAMPLE EXERCISE Page 2 of 20

3 1. Simple Interest Tables When you borrow money from a bank or financial institution, taking out a loan or mortgage (a home loan), you must repay the original amount borrowed plus an extra amount you are charged for using the banks money. This extra amount is called interest. Similarly, if you put money into a bank (i.e. a savings account), you are paid by the bank for letting them use your money. The money you are paid is the interest. (The money you put in the bank is used by the bank to loan out to other people and other investments.) There are two common ways that interests are calculated, they are known as Simple Interest and Compound Interest. In this section, we will only be concerned with Simple Interest. We will investigate Compound Interest later. Simple Interest If you invest or borrow money from the bank and you are reward or charged a fixed amount of interest at regular time periods it is called a Simple Interest. Simple Interest is an example of linear growth where the interest is usually calculated as a percentage of the amount invested or borrowed, the starting value (V 0 ) or Principal. This constant amount is added at each payment over the duration of the loan. Simple interest involves a calculation based on the original amount borrowed or invested; as a result, the amount of simple interest for the duration of loan is constant. For this reason, simple interest is often called flat rate interest. Example 1.1 John invests $5 000 into a bank to save for his first car. The bank gives him 4 % interest per year. Using simple interest, calculate how much money will he have after 5 years if interest was to be calculated yearly? Principal = Amount invested = Interest rate = Time/Period invested = Time Principal Interest Interest Accumulation Total interest earnt after 5 years = Total amount earnt = Page 3 of 20

4 Example 1.2 Kayla borrows $3 000 from a bank to put towards her savings for her oversea trip. The bank lends her for a period of 18 months with an interest rate of 7.8% pa. Using simple interest, how much does she owe after 18 months if the interest is calculated quarterly? Interest rate = Time/Period borrowed = Time Principal Interest Interest Accumulation Total interest owed = Total amount of money Kayla owed = Exercise For the following questions use a table to calculate: i the amount of simple interest, I, earned ii the total amount, A, at the end of the term. (a) $1200 for 5 years at 10.5% p.a. calculated yearly. (b) $8320 for 3 years at 6.45% p.a. calculated quarterly. (c) $960 for 1 ½ years at 9.20% p.a. calculated monthly. 2. Calculate the simple interest that has to be paid if $4 650 is invested on term deposit for 7 years at 5.75% p.a. 3. Jenny wishes to invest $3 500 for 2 years at 4.95% p.a. but is not quite sure which option will give her the ultimate reward. Calculate the return for each of the investment and advice Jenny which option is best for her. (a) annually (b) half- yearly (c) monthly 4. For the following questions use CAS calculator to calculate the total amount, A, owed at the end of the term when borrowing: (a) $ for 10 years at 6.89% p.a. calculated yearly. (b) $2460 for 4 years at 5.75% p.a. calculated quarterly. (c) $3730 for 4 # $ years at 7.39% p.a. calculated monthly. (d) $ for 3 years at 8.35% p.a. calculated daily. Page 4 of 20

5 2. Simple Interest Formula By obse0rving how simple interest is calculated from the above table, we can see that under the Interest Accumulation column it follows a general rule: Simple Interest = I = V ( * n = -.*/ +(( +(( Where I = the amount of interest earned, or owed ($) fortnights 2 ) V 0 = Principal the starting amount of money invested or borrowed ($) r = Rate of the interest per year or per annum (p.a.) (%) n= Term - borrowed or investment period (could be years, quarters 1, months, Example 2.1 Calculate the amount of simple interest earned on an investment of $4450 that returns 6.5% per annum for 3 years. Finding V o, r and n We can rearrange the simple interest formula to find the initial amount or principal (V ( ), the interest rate (r) or the term of the loan or investment (n) Example 2.2 A bank offers 4.5% p.a. simple interest on an investment. At the end of 4 years the total interest earned was $90. How much was invested? Example 2.2 on CAS calculator On a calculator use the nsolve function, Enter the equation I = - 1 * /, V +(( ( and set the values of I, r and n using, I=$90, r=4.5 and n=4 1 A quarter of a year is ¼ x 12 months = 3 months 2 A fortnight is 2 weeks, there are 26 fortnights in a year (52 weeks in a year/2 = 26) Page 5 of 20

6 Example 2.3 What simple interest rate is required to earn $86.70 interest on an original amount of $255 in 4 years? Example 2.3 on CAS calculator On a calculator use the nsolve function, Enter the equation I = - 1 * / +((, and set the values of V 0, I and r using, I=$86.70, V 0 =255 and n=4. Worked Example 9: How long will it take an investment of $2500 to earn $1100 with a simple interest rate of 5.5% p.a.? Page 6 of 20

7 3. Simple Interest - Loans At the end of the borrowing or investing period, the total amount of money repaid or awarded to you is calculated as followed. Amount = Principal + Interest Where A = Total Amount repaid or reward ($) P = Principal ($) or starting amount I = Simple Interest ($) Worked Example 3.1 Calculate the monthly payments for a $ loan that is charged simple interest at a rate of 8.45% p.a. for 4 years. Cash flow Non- annual interest calculations Although interest rates on investments and loans are frequently quoted in terms of an annual rate, calculations on interest rates are made more frequently throughout a year. Quarterly, monthly, weekly and even daily calculations are not uncommon. Worked Example 3.2 How much interest is paid on a monthly balance of $665 with a simple interest rate of 7.2% pa? Page 7 of 20

8 Minimum Balance Calculations Banks and financial institutions need to make decisions about when to apply interest rate calculations on accounts of their customers. For investment accounts, it is common practice to use the minimum balance in the account over a set period of time. Worked Example 3.3 Interest on a savings account is earned at a simple rate of 7.5% p.a. and is calculated on the minimum monthly balance. How much interest is earned for the month of June if the opening balance is $1200 and the following transactions are made? Give your answer correct to the nearest cent. Date Details Amount June 2 Deposit $500 June 4 Withdraw $150 June 12 Withdraw $620 June 18 Deposit $220 June 22 Withdraw $500 June 29 Deposit $120 Date Details Withdraw Deposit Balance June 1 Opening Balance June 2 Deposit $500 June 4 Withdraw $150 June 12 Withdraw $620 June 18 Deposit $220 June 22 Withdraw $500 June 29 Deposit $120 Page 8 of 20

9 4. Simple Interest - Recursion Relation Simple interest can be represented by a first- order linear recurrence relation. You can also calculate the total amount of a simple interest loan or investment by using: where: In the case of simple interest, the total value of investment increases by the same amount per period. Therefore, if the values of the investment at the end of each time period are plotted, a straight- line graph is formed. Example 4.1 Recurrence relation Example 2 can be approached using recurrence relations, where V n is the amount owed at the end of the n th period and V n+1 is the amount owed in the next period. Kayla borrows $3 000 from a bank to put towards her savings for her oversea trip. The bank lends her for a period of 18 months with an interest rate of 7.8% pa. Using simple interest, how much does she owe after 18 months if the interest is calculated quarterly? n + 1 V n Balance at the end of n + 1 time (V n+1 ) 1 V 0 = V 1 = 2 V 1 = V 2 = 3 V 2 = V 3 = 4 V 3 = V 4 = 5 V 4 = V 5 = 6 V 5 = V 6 = Page 9 of 20

10 Simple Interest using CAS calculator Example 4.2 Jan invests $210 with building society in a fixed deposit account that paid 8% p.a. simple interest for 18 months. a) How much did she receive after the 18 months? Step 1. Make sure r and n are in the same units Example 4.2 on CAS Calculator (calculator page) Start with a blank calculator page Press Type 210 press enter Initial Value V 0 Next type +1^40 Press enter Note: when you press enter, the CAS converts ANS to the value of the previous answer (in this case 210) Pressing repeatedly applies the rule to the last calculated value, in the process generating the amount of the investment at the end of each year as shown. 1 st year 2 nd year 3 rd year 4 th year 5 th year After 18 months of investment, Jane will receive Page 10 of 20

11 b) Represent the account balance for each of the 18 months graphically. Example on CAS Calculator (Lists and Spreadsheet Page) Label column A month Enter 0 in cell A1 In cell A2 enter: =a1+1 Fill down until the 18 th month Label column B total Enter $210 in cell b1 In the next cell (B2) enter the equation = x a2 Now fill down this equation to the cells below. Press Menu b, data 3, fill 3 Add a data and statistics page /~ Put the month on the x axis and total on the y axis Example 4.3 $1000 is invested in a simple interest account for 5 years at 5.5% p.a. a) Set up a recurrence relation to find the value of the investment after n years. b) Use the recurrence relation from part (a) to find the value of the investment at the end of each of the first 5 years. n V n ($) V n+1 ($) 0 V 0 = Page 11 of 20

12 Exercise 4. Modelling simple interest with recurrence relations 1. The following recurrence relation can be used to model a simple interest investment of $2000 paying interest at the rate of 3.8% per annum. V 0 =2000, V n+1 =V n +76 In the recurrence relation, V n is the value of the investment after n years. a) Use the recurrence relation to find the value of the investment after 1, 2 and 3 years. b) Use your calculator to determine how many weeks it takes for the value of the investment to be more than $3000. c) Write down a recurrence relation model if $1500 was invested at the rate of 6.0% per annum. 2. The following recurrence relation can be used to model a simple interest loan of $7000 charged interest at the rate of 7.4% per annum. V 0 =7000,V n+1 =V n +518 In the recurrence relation, V n is the value of the loan after n years. a) Use the recurrence relation to find the value of the loan after 1, 2 and 3 years. b) Use your calculator to determine how many weeks it takes for the value of the loan to be more than $ c) Write down a recurrence relation model if $12000 was borrowed at the rate of 8.2% per annum. 3. The following recurrence relation can be used to model a simple interest investment. In the recurrence relation, C n is the value of the loan after n years. C 0 =15000, C n+1 =C n +525 a) i What is the principal of this investment? ii How much interest is paid every year? iii What is the annual interest rate of this investment? b) Use your calculator to determine how many weeks it takes for the value of the investment to be more than double the principal. Page 12 of 20

13 5. General recurrence relation for the nth term in linear growth. While we can generate as many terms as we like in a sequence using a recurrence relation for linear growth and decay, it is possible to derive a rule for calculating any term in the sequence directly. This is most easily seen by working with a specific example. For instance, if you invest $2000 in a simple interest investment paying 5% interest per annum, your investment will increase by the same amount, $100, each year. If we let V n be the value of the investment after n years, we can use the following recurrence relation to model this investment: V 0 =2000, V n+1 =V n +100 Using this recurrence relation, we can write out the sequence of terms generated as follows: V 0 = 2000 = V (no interest paid yet) V 1 = V = V (after 1 years interest paid) V 2 =V =(V ) +100 = V (after 2 years interest paid) V 3 =V =(V ) +100 = V (after 3 years interest paid) V 4 =V =(V ) +100 = V (after 4 years interest paid) and so on. Following this pattern, after n years interest has been added, we can write: V n =2000+n 100 With this rule, we can now predict the value of the nth term in the sequence without having to generate all the other terms first. This rule can be readily generalised to apply to any linear growth or decay situation. For a recurrence rule of the form: V 0 = initial value, V n+1 = V n + d (d is constant) the value of the nth term generated by this recurrence relation is: V / = V ( + n d, where V 0 is the initial value of the investment or loan, r is the rate of depreciation per year Note that the starting Value V 0 must always be stated also. where d = V (r 100 Page 13 of 20

14 Example 5.1 Using a rule to determine the value of a simple interest investment The following recurrence relation can be used to model a simple interest investment: V 0 =3000, V n+1 =V n +260 where V n is the value of the investment after n years. a) What is the principal of the investment? How much interest is added to the investment each year? b) Write down the rule for the value of the investment after n years. c) Use a rule to find the value of the investment after 15 years. d) Use a rule to find when the value of the investment first exceeds $ Example 5.2 Amie invests $4000 in a simple interest investment of paying interest at the rate of 6.5% per year. Use a rule to find the value of the investment after 10 years. Page 14 of 20

15 Exercise 5 1. Write down a rule for the value of the nth term of the sequence generated by each of the following recurrence relations. In each case calculate A 20. a) A 0 = 4, A n+1 = A n + 2 b) A 0 =10, A n+1 =A n +3 c) A 0 = 5, A n+1 = A n + 8 d) A 0 =300, A n+1 =A n The value of a simple interest loan after n years, F n, can be calculated from the rule: a) What is the principal of this loan? F n = n b) How much interest is charged every year in dollars? c) Use the rule to find: i) the value of the loan after 5 years ii) how long it takes for the initial investment to double. d) The annual interest rate for this loan is 6.4%. Use this information to find the value of the loan after 15 years. 3. The value of a simple interest investment after n years, V n, can be calculated from the rule: Vn = n. a) How much interest is awarded every year in dollars? b) Use the rule to find: i) the value of the investment after 6 years ii) how long it takes for the initial investment to double. c) The annual interest rate for this loan is 3.5%. Use this information to find the value of the loan after 10 years. 4. Webster borrows $5000 from a bank at an annual simple interest rate of 5.4%. a) Determine how much interest is charged each year in dollars. b) Write down: i a recurrence relation to model the value of the loan, V n, from year to year ii a rule for the value of the loan, V n, after n years. c) Use your rule in part b) ii) to find how much Webster will owe the bank after 9years. 5 Anthony borrows $ from a bank at an annual simple interest rate of 7.2%. a) Determine how much interest is charged each year in dollars. b) Write down: i a recurrence relation to model the value of the loan, B n, from year to year ii a rule for the value of the loan, B n, after n years. c) Use your rule in part b) ii) to find how much Anthony will owe the bank after 11 years. Page 15 of 20

16 6. Flat rate depreciation Some items such as antiques, jewellery and real estate increase in value (appreciate or increase in capital gain. Computers, mobile phones, cars or machinery decrease in value (depreciate) with time due to wear and tear, advances in technology or lack of demand. Depreciation is the estimated loss in value of assets. The estimated value of an item at a point in time is called its future value (book value). When the value becomes zero, the item is written off. At the end of an item s useful life its future value is called its scrap value. There are 3 methods in which to calculate depreciation: flat rate depreciation (studied here); reducing balance depreciation (studied later) and unit cost depreciation (studied next) Flat rate (straight line depreciation) If an item depreciated by the flat rate method, then the value decreases by a fixed amount each time interval. It may be expressed in dollars or as a percentage of the original cost price. As the depreciation value is the same for each interval, it is an example of straight line decay. This relationship can be expressed in the following recurrence relation: The future value can also be calculated after n periods of depreciation. We can use the above relationships or a depreciation schedule (table) to analyse flat rate depreciation. Example 6.1 Fast Word Printing Company bought a new printing press for $ and chose to depreciate it by the flat rate method. The depreciation was 15% of the prime cost each year and its useful life was 5 years. a) Find the annual depreciation. b) Set up a recurrence relation to represent the depreciation Page 16 of 20

17 c) Complete a depreciation schedule for the useful life of the press and use it to draw a graph of book value against time. Time n (yrs) Depreciation d ($) Future value V n ($) d) Generate the relationship between the book value and time and use it to find the scrap value. Example 6.1(c) and (d) on CAS calculator (c) On a lists & spreadsheet page Label column A n and column B V n Enter 0 to 5 in the n column and the starting value (V 0 ) in cell b1. In cell b2: Enter the equation =b This equation is just V n+1 =V n found in part (b) Note: the 2250 is the annual depreciation found in part (a) Press enter, then fill down (b33) until n=5 V n =3750 when n=5. So, this is the scrap value Add a Data & Statistics page Label the x- axis n and the y- axis V n In this worked example the depreciation schedule gives the scrap value, when n=5 V n =$3750. This can also be seen in the 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. Page 17 of 20

18 Exercise 6 1. The following recurrence relation can be used to model the flat rate of deprecation of a set of office furniture: V 0 =12000, V n+1 =V n +1200, where V n is the value of the furniture after n years. a) What is the initial value of the furniture? By how much does the furniture decrease in value each year? b) Write down the rule for the value of the furniture after n years. c) Use a rule to find the value of the investment after 6 years. d) How long does it take for the furniture s value to decrease to zero? e) A photocopier in the office costs $6000 when new. Its value depreciates at the flat rate of 17.5% of its new value each year. What is its value after 4 years? 2. The value of a sewing machine after n years, V n, can be calculated from the rule: V n = n. a) By how much is the value of the sewing machine depreciated each year in dollars? b) Usetheruletofindthevalueofthesewingmachineafter4years. c) The sewing machine depreciates by 12.5% of its purchase price each year. Use this information to determine its value after 8 years. 3. The value of a harvester after n years, V n, can be calculated from the rule: V n = n. a) What is the percentage depreciation for the harvester? b) Use the rule to find the value of the harvester after 7 years. c) How long does it take the harvester to reach a value of $29250? 4. A computer is purchased for $5600 and is depreciated at a flat rate of 22.5% per year. a) Determine the annual depreciation in dollars. b) Write down a recurrence relation to model the value of the computer, V n, from year to year c) Write down a rule for the value of the computer, V n, after n years. d) Use your rule in part c) to find: i the value of the computer after 3 years ii the number of years it takes for the computer to be worth nothing. 5. A machine costs $7000 new and depreciates at a flat rate of 17.5% per annum. The machine will be written off when its value is $875. a) Write down a recurrence relation to model the value of the machine, V n, from year to year b) Write down a rule for the value of the machine, V n, after n years. c) Use your rule in part b) to find: i. the value of the machine after 2 years ii the number of years the machine will be used. Page 18 of 20

19 7. Unit- cost depreciation The unit cost method is based upon the maximum output (units) of the item. For example, the useful life of a truck could be expressed in terms of the distance travelled rather than number of years. The actual depreciation per year would be a measure of the number of kilometres travelled. Unit cost depreciation recurrence relation The future value over time using unit cost depreciation can be expressed by the recurrence relation: A future value after n outputs using unit cost depreciation can be expressed as: Example 7.1 A hairdryer in a salon was purchased for $850. The value of the hairdryer depreciates by 25 cents for every hour it is in use. Let V n be the value of the hairdryer after n hours of use. a) Write down a rule to find the value of the hairdryer after n hours of use. b) What is the value of the hairdryer after 50 hours of use? c) On average, the salon will use the hairdryer for 17 hours each week. How many weeks will it take the value of the hairdryer to halve? Page 19 of 20

20 Exercise 7 1. The value of a taxi after n kilometres, V n, can be calculated from the rule: V n = n. a) What is the purchase price of the taxi? b) By how much is the value of the taxi depreciated per kilometre of travel? c) What is the value of the taxi after 20,000 kilometres of travel? d) Find how many kilometres have been travelled if the taxi is valued at $ A car is valued at $ at the start of the year, and at $ at the end of that year. During that year, the car travelled kilometres. a) Find the total depreciation of the car in that year in dollars. b) Find the depreciation per kilometre for this car. c) Using V 0 = , write down a rule for the value of the car, V n, after n kilometres. d) How many kilometres are travelled when the value of the car is $6688? e) Including the first year of travel, how many kilometres in total is this car expected to drive before it has a value of zero? 3. A printing machine costing $ has a scrap value of $2500 after it has printed 4 million pages. a) Find: i. the unit cost of using the machine ii. the value of the machine after printing 1.5 million pages iii. the annual depreciation of the machine if it prints pages per year. b) Find the value of the machine after 5 years if it prints, on average, 750,000 pages per year. c) How many pages has the machine printed by the time the value of the machine is $70 000? Page 20 of 20