3.0 FINANCE

3. FINANCE

TABLE OF CONTENTS

3.1 Senior Mathematics Assessment and the Graphic Calculator ..................................................................................... 3

3.2 Finance Approach 1 - A Real Life Problem ................................................................................................................... 7

3.3 Borrowing money – Application of a Geometric Progression ...................................................................................... 8

3.31 Save Save Save ................................................................................................................................................................ 11

Example ................................................................................................................................................................................... 11

3.4 Programming graphic calculators for finance calculations ........................................................................................ 12

3.5 Finance with the EL-9600 .............................................................................................................................................. 16

Example 1 ................................................................................................................................................................................ 16Example 2 ................................................................................................................................................................................ 17

3. FINANCE

3.1 Senior Mathematics Assessment and the Graphic Calculator

Jim Mc Cartan

Senior school teachers have long recognised that there can be more than one way to enhancelearning and to assess their students’ capacity to demonstrate what they know. It would, however,be safe to suggest that mathematics teachers have tended to be overly conservative in theirreliance on chalk and talk and paper and pencil assessment tasks.

In defence of maths teachers, especially senior teachers, not a great deal of resource material hasbeen available to assist them. The advent of computers and graphing calculators, has greatlyimproved this situation.

Today teachers are becoming increasingly aware of the need, not only to change their teaching /assessment style, but also to assist the learner to better understand how he / she learns best andhow best they may demonstrate their knowledge, skills and attitudes. Mathematics teachers areincreasingly recognising the need to involve their students in written and oral communication andthe use of technology to investigate and convey ideas. Across Australia syllabus developers areencouraging the move towards rich assessment tasks that are open ended, assessing the full rangeof outcomes. WA, Victoria and Queensland have all moved in this direction. The NSW juniorsyllabus has introduced this type of assessment and this, I believe, argues well for future changesin senior courses.

Experience, reading, discussion and listing to teachers and students has led me to recognise thatthe teaching and learning of mathematics may be enhanced by:

• creating a sense of challenge.• encouraging a sense of personal control in problem solving.• employing well planned co operative group work strategies.• ensuring the necessary cognitive as well as the physical tools for solving the problem are

available to the learner.• giving feedback and encouraging self and, where appropriate, group evaluation. If a

piece of work is to be submitted for assessment those involved ought to be encouraged toprovide a critical reflection of the work.

• students should be trained and encouraged to become involved in the assessmentprocess.

• relating the problem to real life.• displaying work, even that which is incomplete.• asking for more than one type of solution.• asking open ended questions and encouraging diversity in the response.• rewarding an appropriate approach, even though the final answer may be incorrect.• getting out of the classroom to train the learner to observe, to record and to make use of

real data.• encouraging different approaches eg drawing, model building, experimentation,

physical activity and the use of technology.• having the learner become the teacher.

3. FINANCE

One teaching resource that accommodates many of the above ideas is the graphic calculator.Spend just a few minutes investigating this tool and you will realise its power. The real value ofthe graphic calculator is that it can give the learner power over the learning environment. Not onlycan the calculator be used to perform tedious operations, or indeed, add a graphical dimension tothe learning environment, it may and ought to be used as a powerful tool for investigation.

One creative application for the graphic calculator is to use it as an assessment tool. In thisapplication the student is asked to use the graphic calculator to present a lesson to their peers, to inservice the maths or other staff or to teach a junior class.

For the past year I have been involved in the organisation of mathematics conferences wherestudents from year 9 to 11 have trained their peers and teachers in the application of graphiccalculators to mathematics. Testimony as to the success of this approach was very evident in thewritten feedback received.

Most importantly the success from the students’ perspective was that they felt they:

• Had significant control over the content of the material they presented.

• where using modern technology to help get a message across.

• had to ‘ REALLY ‘ know the content.

• had to think out an effective way to convey their message.

• were challenged to create a learning environment that stimulated others to come to anunderstanding of something new.

• were able to express their knowledge in a domain usually reserved for teachers.

• acknowledged the difficulties of trying to stimulate others to grasp new ideas.

• had a sense of personal achievement.

Listed below is a small number of assessment ideas associated with the graphic calculator. Thereare many more for you to try. Clearly each assessment task would need to be further developed.Notice some of these activities provide the opportunity for the student to apply their knowledgeand skills outside the immediate maths environment. The tasks may be individual or groupfocused, the tasks may be presented to a group, in the form of a written report, instructional notes,posters or a lesson plan.

A Sample of Assessment Tasks:

1 Write a program to simulate the roll of a die so as to help 28 Year 8 students grasp the ideaof probability. Other programs may simulate the reading of a tax table, home loans, a watermeter or electricity billing table. There are obviously many more.

2 Develop a lesson to introduce the notion of testing for maximum and minimum turningpoints.

3. FINANCE

3 Investigate the graphing of trig functions y = a sin ( bx + π ).

4 Develop a year nine lesson to investigate y = nx + b, as n and b vary.

5 Introduce the concept of a limiting sum to a year 11 class.

6 Demonstrate for the science department the use of the CBL to collect temperature readingsand model the data obtained.

7 Using real data demonstrate to a year 10 class how to draw a box plot to describe theinformation.

8 Use the regression models to demonstrate to a year 11 class how town planners may planfuture developments based on information gained from a survey.

9 Using the zoom feature develop a process to demonstrate the notion of instantaneous rateof change of a function.

10 Investigate the function f ( x ) = ax2 + bx + c and present a report to your class.

These tasks represent only a small sample of useful ways to assess students understanding of aconcept. They also allow the student to develop their self confidence, communication andpresentation skill as well as demonstrating the application of their knowledge in differing contexts.

The following is an example that illustrates how a graphic calculator may be used toassess year 11 students’ knowledge of a specific mathematical concept.

Task

Use your graphic calculator to present a lesson on one of the following topics:

arithmetic progressionsgeometric progressionssketching curvestrigonometric functionsline of best fitregionsarea and volumeproblem solving

You may use any other resource that you consider may assist you to complete your task.

Assumptions

You may assume that your class has a good working knowledge of the graphic calculators and thatthey have been introduced to sequences and series.

Equipment

Graphic calculator, OH projection panel, computer and computer link.

3. FINANCE

Guidelines

You may choose to work individually or work in pairs. Note, each member must justify the extentof his / her involvement in the task.

Your presentation must be in the form of :

* Written report that will include - a Max of 1000 words.- a statement of the problem.- a statement describing the major mathematical outcomes.- a list of any formula to be used.- a summary of your approach- an outline of calculator steps.- one real life application.

* Presentation - you must convey a problem situation and present a solution.- you may choose to involve the class or adopt a lecture style.- you must prepare a handout for your students

3. FINANCE

3.2 Finance Approach 1 - A Real Life Problem

Assume that your mortgage of $150000 is set at a fixed rate of 8.5% p.a., payable monthly over 25years.

(a) Explain how you could use your knowledge of geometric progressions to calculate yourmonthly repayments.

(b) Describe how you would use the finance function of your graphic calculator to calculateyour monthly repayments. Calculate the payment per month.

(c) Today it is possible to package your loan differently, ie you may wish to pay quarterly,monthly or weekly. Choosing one method over another may well save you a lot of money.Examine the following:

(i) If the total monthly repayment is to remain as above, but payments are to be madefortnightly, calculate how long it will take to pay off the loan.

(ii) Repeat the exercise using, quarterly, weekly payments.

(d) Show that the formula for the balance (y) owed at any given time (T) is given by:

Y = P ( 1 + R ) NT - Q ( ( 1 + R ) NT – 1 )

R

P = principal $150000Q = monthly payment $ 1207.85N = number of payments per year 12R = annual interest divided by N 0.085/12

(i) Use your graphic calculator to illustrate graphs for each of the above paymentoptions.

(ii) Use the table function to compare the various packages.

(iii) Comment on your observations made from the graphs and tables.

(e) Based on the mathematical knowledge and any other information you have gathered,design an information pamphlet, poster or other medium, to inform prospective loanseekers as to how best they should purchase a loan.

3. FINANCE

3.3 Borrowing money – Application of a Geometric Progression

It costs to borrow moneyEasy loans

12 %

High Finance ------ Mega Bucks Borrowing

Lets say I borrow $10 at an interest rate of 12% for a period of 3 years.

Loan plus interest = $14.05

3 equal annual payments of $p

How much is each payment worth?

Step 1 Calculate the total own after the first year 10 + 10 x 12 x 1 = 10 ( 1 + 0.12 ) 100

Step 2 Make the 1st payment 10 ( 1.12 ) - p

Step 3 Calculate the total own after the second year ( 10 ( 1.12 ) – p ) x 1.12

Step 4 Make the 2nd payment ( 10 x 1.122 – 1.12p ) – p

Step 5 Calculate the total ( 10 x 1.122 – 1.12p – p ) x 1.12

Step 6 Make the final payment ( 10 x 1.123 – 1.122p – 1.12p ) – p

Now the debt is paid ie the balance is 0

Step 7 Rewrite the step 6 10 x 1.123 – ( 1.122 + 1.12 + 1 ) p = 0

Solve for p p = 10 1.123

( 1.122 + 1.12 + 1 )

is a geometric progression Sn = 1 ( 1.123 – 1 ) a = 1 , r = 1.12 , n = 3

1.12 – 1

= 3.37

Therefore p = 4.17

This problem could easily be solved without the application of a geometric progression.The calculation would be much more tedious if n was large, the amount borrowed more significantand the interest more complex.In such applications the application of the sum of a gp is of great use.

$10

PAY 1

12%

PAY 2 PAY 3

1.122 + 1.12 + 1

3. FINANCE

Now let’s look at a different problem.

This time let’s borrow the same amount only this time we will pay it off monthly.

Amount $10Interest 12% / 12 = 1% per monthNo. payments 3x 12 = 36Payments $p per month

After year 1 the amount owing is 10 x 1.01 – p After year 2 the amount owing is (10 x 1.01– p) 1.01 – pAfter year 3 the amount owing is ( ( 10 x 1.01 – p) 1.01 – p) 1.01 – pAfter year 4 the amount owing is ( ( (10 x 1.01 – p) 1.01 – p) 1.01 – p) 1.01 – p

Now lets expand the last set out

10x 1.014 – p 1.013 - p 1.012 - p 1.011 - p= 10x 1.014 – p (1.013 + 1.012 + 1.01 + 1)

When all 36 payments have been made the pattern would be

10x 1.0136 – p (1.0135 + 1.0134 + 1.0133 +1.0132 + 1.0131 + 1.0130 + …….. + 1)

Using the sum of a gp this becomes 1( 1.0136 – 1 ) = 100 x ( 1.0136 – 1 )

0.01

When all payments are made then the loan is paid off and the balance is 0.

∴ p = 10x 1.0136

(1.0135 + 1.0134 + 1.0133 +1.0132 + 1.0131 + 1.0130 + …….. + 1)

= 10x 1.0136

100 (1.0136 - 1)

= 33 cents

Notice in the first case the total payback is 3 x 4.17 = $ 12.51In the second case the total payback is 36 x 33 = $ 11.88

When you borrow money you pay for it in the form of interest.

This is a geometric progression a = 1 , r = 1.1 , n = 36

3. FINANCE

The interest is usually compounded over the period of the loan and is recalculated after each payment.

It is possible to calculate the period payments PMT if you know

Present Value of the loan PV $100000Interest rate charged I 12% per annumThe number of years n 20

NB The PMT may be paid annuallyQuarterlyMonthlyFortnightlyWeeklyDaily

Most domestic loans are paid monthly.

It is possible to use any graphic calculator with a finance feature to solve the problem of calculating themonthly payments for the figures above.

This solution uses the Sharp EL9600 graphic calculator.

Step 1 PRESS 2ndF and ÷ to set the calculator to finance mode.

Step 2 Edit to option C and choose 1 (Pmtend), press enter. This will allow for payments to be made atthe end of each month ie., payments will be made in arrears.

Step 3 Press 2nd F and ÷, edit to option A and press enter.

Step 4 At n type 240 and press enter. This is the number of payment periods.

Step 5 At I type 12 and press enter. This is the yearly interest.

Step 6 At PV type 100000 and press enter. This is the present value of the loan ie the amountbeing borrowed.

Step 7 Miss PMT

Step 8 At FV type 0 and press enter.

Step 9 At P/Y type 12 this is the number of payments in a year.

Step 10 At C/Y type 12 this represents the cumulative interest per year and has the same numericalvalue as P/Y.

3. FINANCE

3.31 Save Save Save

Pay your loan off quickly and cheaply.

Yes it’s possible.

If instead of paying your loan off monthly you pay an equivalent amount fortnightly you willSave! Save! Save! and you will pay your loan off faster.

The graphs below illustrate the opportunity to save without altering the total amount you paymonthly.

Example

Borrow PV = $100000Rate of interest I = 6.5% per monthMonthly periods n = 20 x 12 = 240Payments PMV = ?

Using your graphical calculator the PMV = $

Now simply halve your monthly payments ie PMV = $

Using the graphic calculator you will pay your loan in years and save $

The equation below represents the graph of the time taken to pay off a fixed loan given a fixedinterest rate and the number of payment periods. The graphs are based upon the fact that the totalmonthly payment remains fixed. The variable is the number of payments to be made.

Graph 1 illustrates a loan paid off at $ P / month.Graph 2 illustrates a loan paid off fortnightly at $P / 2 per fortnight.

Clearly it is better to pay the loan fortnightly than monthly.

A LITTLE MATHEMATICS CAN GO A LONG WAY

3. FINANCE

3.4 Programming graphic calculators for finance calculations

Most calculators use a form of basic for programming that makes them relatively easy to set up.An advantage with programming over the in-built modules is the transparency of the process - andthus the possibility that students will be able to design and modify their own programs and choiceof calculation/formula.

If the calculator can be linked easily to a desktop or laptop computer, then programs can beconveniently written and downloaded (usually as text files) to the students’ calculators. Providedprograms are not too complex, direct programming is an option but be prepared to spend sometime helping students correct and debug their programs.

The following program has been written specifically for the Sharp EL-9600.

A basic contributory savings scheme calculator

This eight line program will calculate the future value of an investment scheme, assuming aconstant rate of return, a regular deposit, and no fees or charges. Payment is made at thebeginning of each period.

The program

• requires the user to enter the investment parameters• calculates the effective interest rate per compounding• stores this in memory (R)• calculates the total number of compounding periods and stores this in memory (N)

(note: these two actions are not strictly necessary but may be useful in other situations)• calculates the future value• prints or outputs the future value

The data required are:

A the value, or amount, of the regular depositI the nominal annual interest (in 5%)Y the number of years the investment is to be retainedQ the number of times interest is calculated and deposits made and

credited to the account each year

Here is the program. It’s called Annuity1 (or any other name you prefer)

Input AInput IInput YInput Q1+I/100/Q RYQ NAR(1-R^N)/(1-R) PPrint P

3. FINANCE

Entering the program directly

The program is entered using the calculators pre-defined menus and keypad entry of thecalculators normal functions.

1. 2ndF : MATRIX (second function PRGM) takes you to the program menu. Choose option 2(NEW) and enter a suitable title (eg. Annuity1) and ENTER.

2. Pressing 2ndF : MATRIX a second time takes you to the programming instruction page. Inthis case select option A (PRGM) and sub-option 3 Input from the list and ENTER.

3. At the cursor enter A -do this as ALPHA : sin.4. Use 2ndF : MATRIX, option A (PRGM), sub-option 3 Input to enter the next three lines; your

program should now look like:

ANNUITY1Input AInput IInput YInput Q

To check the program so far, use 2ndF: CL to first save your work. Then 2ndF : MATRIX (secondfunction PRGM), Option A (EXEC), program name Annuity1 and [ENTER]. The program shouldrun, and the screen show:

ANNUITY1A=?

Enter any value for A, and work through the other variables until the screen shows DONE at thebottom right (don’t forget to press ENTER after each variable).

Now go back to the program edit mode 2ndF : CL, 2ndF : MATRIX, Option B (EDIT), program nameAnnuity1 and ENTER. Move the cursor down to the next line with the ‘down’ cursor control key,and ..

5. using a combination of the numberpad and [ALPHA]pad type in:

1+I/100/Q R note: to enter R, use STO : ALPHA : Rdon’t forget ENTER at the end of each line.

6. YQ N don’t forget ENTER at the end of each line. 7. AR(1-R^N)/(1-R) P the ^ symbol is entered by pressing the ab key after R don’t forget ENTER at the end of each line. 8. Print P ‘Print’ is accessed through pressing

2ndF : MATRIX a second time, choosing option A,sub-option 1 Printdon’t forget ENTER after the last line.

Provided you have used the [ENTER] key after each line, and after the last line, your program isnow saved. To exit edit mode, press 2ndF : CL.

3. FINANCE

To run the program

To run the program select 2ndF : MATRIX, Option A (EXEC), program name Annuity1 and ENTER.

Enter the variable values. After the last variable has been entered the program calculates and printsthe result:

eg. ANNUITY1A=? value of each payment1000I =? nominal interest rate7.5Y=? how many years10Q=? how many payments each year12

179042.4066

If you wish to do further calculations on any of the variables used in the program, these are nowavailable in the calculation mode.

This means that the output from the program, stored in P could be used as an input in anotherprogram, or a subsequent calculation. The outputs from programs can also be stored in the LISTSdisplay in STAT mode. This enables graphs to be drawn and data to be analysed.

Using programs to store data in lists for graphing

As an example, the program annuity1 has been modified to provide a range of outputs (FutureValues) where the interest rate rises incrementally by 0.5% from an initial value (I%) up to(I+5)%. The outputs, Future Value and interest rate, are stored in lists L1 and L2.

The program, annuity2, now

• inputs the annuity variables as before, but this time the interest rate is the lower value ofthe desired range

• incrementally increase the interest by 0.5% ten times and calculates the future value of thescheme fro this rate

• prints the future values on the screen• stores the future value and its corresponding rate in the LISTS L1 and L2

To view the LISTS and to use then to draw graphs, use [STAT], option A, edit list.

It would be long-winded and possibly confusing to try to explain all the steps in this (inelegant-to-be-clear) program. It’s form can be seen from the text:

1 LInput AInput IInput YInput QLabel LOOPI+(L-1)*0.5 I

3. FINANCE

1+I/100/Q RYQ NAR(1-R^N)/(1-R) L1(L)Print L1(L)I L2(L)If L=11 Goto END1+L=LGoto LOOPLabel END

Note: any program command shown in this format in the above program is a pre-definedcommand, and is accessed through the 2ndF : MATRIX, Options A (PRGM) and B(BRNCH) sub-menus. Entering these commands through the ALPHA facility is not an option. The List names (L1and L2) are accessed through 2ndF : 1 and 2ndF : 2.

These programs are available for download if you want to run them on the EL9600 through theCE-LK1 link Kit (see annuity1.xxx and annuity2.xxx)

3. FINANCE

3.5 Finance with the EL-9600

We must first establish how the SHARP EL-9600 interprets the various aspects of finance.

Cash Flow think of cash flow as the amount of money paid to you (positive) versusthe money that you have to pay (negative).

No. Payments (N) is the number of payments to be madeInterest % is the interest rate (usually per annum)Present Value (PV) is the amount that you have nowPayments (PMT) come in two forms• payments coming in eg. a money gift (positive PMT) or• payments going out eg. paying a bill (negative PMT as money coming

out of your personal account)• Payments can be made at the beginning of each payment period or at the end of each paymentperiod.

(To set PMT period …2nd Finance, C Period then 1 PmtEnd or 2 PmtBegin, ENTER,ENTER)

Future Value (FV) is the amount that you have after a period of time in today’s dollar valuePMT’s /Year (P/Y) the number of payments to be made in one yearCumul/Year (C/Y) the number of times that interest is accumulated/calculated per year

Let’s begin by looking at a simple example.

Example 1

You invest $100 a month (at the end of the month) for 6 months at 6%pa (0.5% per month)compounded monthly.

The tedious way using compound interest (A=P(1+i)n) and a table.

TimePeriod

OpeningBalance

+ Interest on OpeningBalance + Payment

= ClosingBalance

Month 1 $0.00 0.00(1.005)1= $0.00 $100 $100.00Month 2 $100.00 100.00(1.005)1= $100.50 $100 $200.50Month 3 $200.50 200.50(1.005)1= $201.50 $100 $301.50Month 4 $301.50 301.50(1.005)1= $303.01 $100 $403.01Month 5 $403.01 403.01(1.005)1= $405.03 $100 $505.03Month 6 $505.03 505.03(1.005)1= $507.56 $100 $607.56

Now let’s let the SHARP EL-9600 do the work for us.

• Set the Payment Period to end of each period as outlined previously• 2nd Finance, A Solver, ENTER.• Enter appropriate information into each of the fields followed by ENTER but skip FV. ie:

• N=6• I%=6• PV=0• PMT=-100 (as you are paying not receiving)

3. FINANCE

• FV= don’t enter, just down arrow• P/Y=12• C/Y=12

• Place the cursor on FV and press 2nd EXE• Result…….607.5501879 or $607.55 when rounded to 2 DP.

* Note the slight discrepancy due to rounding in the table.

Example 2

You borrow $100,000 from your local bank for your dream house. The bank is currently charginginterest at a rate of 6% pa (compounding fortnightly). Find how much your repayments will beover a 25 year loan where your repayments are made at the end of each fortnight.

This problem could be solved using a time line diagram as follows:At the end of the fortnight scenario:Interest rate per F/N is r= 1+ 6/26/100=1.002307692 (the common ratio)Although 1.0023 is written below the actual value for r above will be used.

x(1.0023)0

x(1.0023)1

x (1.0023)648

x (1.0023)649

The sum of the terms above will be Sn for 650 terms of a Geometric Progression

FV of $100000 = PV(1+i)n

= 100,000(1.0023)650

= $447,395.0902

For the GP, Sn =1

)1(−

−rra n

S650 =10023.1

)10023.1( 650

−−x

447,395.0902 = 1,505.378724xx = $297.1976972

= $297.20 (DP)

* This technique relies on the fact that the $100,000 left as an investment will accumulate to thesame as a fortnightly investment of $x over the same period.

Now let’s let the SHARP EL-9600 do the work for us.

• Set the Payment Period to the end of each period as outlined previously• 2nd Finance, A Solver, ENTER

$100000 650th F/N

$x $x $x $x

1st F/N 2nd F/N 649th F/N

3. FINANCE

• Enter appropriate information into each of the fields followed by ENTER but skip FV. ie:• N=25*26• I%=6• PV=100,000• PMT= don’t enter, just down arrow• FV=0• P/Y=26• C/Y=26

• Place the cursor on PMT and press 2nd EXE• Result…….-$297.1976972 or -$297.20 when rounded to 2 DP (-ve indicates a payment!)

Part B – The Finance CalculatorLet us investigate further using the other features of the EL-9600’s Finance CALCulator.NOTE: All of the calculations will be based on the settings in the TVM Solver Screen.

B CALC Entry Format User Notes 01slv_pmt used as constants from the calculator screen 02slv_I% used as constants from the calculator screen 03slv_PV% used as constants from the calculator screen 04slv_N% used as constants from the calculator screen 05slv_FV% used as constants from the calculator screen 06Npv( Not applicable to this type of Q 07Irr( Not applicable to this type of Q 08Bal( Bal(after ? payments, to ? DP’s) DP’s – Decimal Places 09∑Prn( ∑Prn(from ? payment, to ? payment,

DP’s)Sum of Payments and Interest during thespecified period-’ve answer will indicate money out+’ve answer will indicate money in

10∑Int( ∑Int(from ? payment, to ? payment,DP’s)

Total interest paid during specified period-’ve answer will indicate money out+’ve answer will indicate money in

11→Apr( →Apr(effective interest rate, numberof payments)

Converts an Annual Simple InterestRate(one payment only) to the equivalentAnnual Compound Interest Rate calculatingover the given number of periods→Apr(12,12)=11.38655152ie: Investing 1 payment of $20 at 12% pa(Simple Interest) will amount to the same asinvesting 1 payment of $20 at11.38655152% pa (compounded monthly).

12→Eff( →Eff(nominal interest rate, numberof payments)

Converts an Annual Compound InterestRate to the equivalent Annual SimpleInterest Rate using one payment only→Eff(12,12)=12.68250301ie: Investing 1 payment of $20 at 12% pa(compounded monthly) will amount to thesame as investing 1 payment of $20 at12.68250301% pa (Simple Interest).

13days( days(initial month.day year, lastmonth.day year)

eg: days between 1st Aug 1998 and 2nd Dec1999 is entered as…days(8.0198,12.0299)=488 days

3. FINANCE

Part C - Bringing it all together, the Questions

1. You borrow $200,000 from your local bank. The bank is currently charging interest at arate of 6% pa (compounded monthly). Find how much your repayments will be over a 25year loan where you pay at the end of each month?

2. You have just landed a 5 year plumbing contract with the Government and as a result youwant to buy a new car, that you have had your eye on, worth $28,700. You can afford topay $300 a fortnight at the beginning of each fortnight for the 5 years. Can you afford topurchase the car if the bank is currently charging a 6% pa interest rate that is compoundedmonthly?

3. If you receive a bill dated the 1st Feb 1999 and it is due to be paid on the 15th Aug 1999,how many days do you have to pay?

4. You invest $300 at the beginning of each fortnight at 5%pa (compounding fortnightly).How long will it take your investment to reach $20,000?

5. You are currently paying off a loan of $100,000 that you borrowed exactly one year agofrom a financing company where the interest rate was 7.5% pa (compounded monthly).Your loan repayments are currently $427.13(entered as 427.1342138) a fortnight (paid atthe end each fortnight).

a. What was the initial term of your loan?b. What was today’s loan balance?c. (i)* Interest rates dropped today and as a result you will pay out the loan 2 years

earlier. Assuming that your repayments remain the same, what is the newinterest rate?

(ii)** If you choose to increase your loan repayment by $50 a fortnight to takeadvantage of this new interest rate, how much time will you cut off theremaining term of your loan?

*6. You wish to set up a 3 year investment fund into which you are prepared to pay $800 amonth (at the start of each month). After shopping around, you find that you can get 4%pa (compounded monthly) from a bank and an effective interest rate of 4.1% pa fromanother financial institution. Which of the 2 will you choose if they both accept monthlydeposits? Justify your choice.

*7. You borrow $150,000 from your local bank. The bank is currently charging interest at arate of 6% pa (compounded monthly). If the loan term is 30 years,

a. calculate the loan repayments if you pay them at the end of each fortnight.b. investigate the interest paid for each 5 year block during the course of the loan.c. how much do you therefore end up paying back over the 30 year term?e. investigate the actual amount paid off the original loan amount in each 5 year block

using the ∑Prn( calculator function.f. Verify that the sum of your answers in d) is $150000

*8. How much money will you save by paying a loan of $152000 (taken over 25 years) in just10 years if the bank is currently charging 6.7% pa (compounded monthly) and you arepaying at the end of each fortnight?

3. FINANCE

Part D - Solutions1. Finding Repayments

Payment : ENDN=25*12=300I%=6PV=200000

*PMT=-1288.602803FV=0P/Y=12C/Y=12

∴ repayments will be $1288.60 a month.

NB: -‘ve indicates money out ∴ payment

2. Can you purchase the car?

Payment : BEGINN=26*5=130I%=6

*PV=28769.61427PMT=-300FV=0P/Y=26C/Y=12

∴ yes, you can afford repayments as PV exceeds $28700.

NB: +‘ve answer indicates money in ie. the bank will giveyou $28769.61

3. How many days?

Using the function days(2.0199,8.1599)=195 days.

4. How long will it take?

Payment : BEGIN*N=62.67330868

I%=5PV=0PMT=-300FV=20000P/Y=26C/Y=26

∴ It will take 62.67330868 payments.

=26

67330868.62 =2.410511827 years

≈2 years 4 months and 28 days

NB: FV is positive as you will receive the money at the end of theinvestment. Interest earned along the way would also be –‘ve(payment) as you are paying the interest earned straight back to theinvestment.

5. One year down the tracka) Initial term of loan

Payment : END*N=390

I%=7.5PV=100000PMT=-427.1342138FV=0P/Y=26C/Y=12

∴ Initial term was 390 (slv_N) payments.

=26

360 = 15 years

3. FINANCE

b) Today’s Loan Balance

Using the function bal(26)=96248.6393, today’s balance will be $96248.64

c) (i) Required interest rate

Payment : ENDN=390-26-(26*2)=312

*I%=5.74618608PV=96248.6393PMT=-427.1342138FV=0P/Y=26C/Y=12

∴ An interest rate of ≈5.75% would reduce the loan termby 2 Yrs from this point onwards

NB: 1 year had passed and 2 years were cut thereforereducing N to 312 from today. The PV was also changedto reflect today’s balance as calculated in part b).

(ii) How much time will it cut off the loan term?

Payment : END*N=267.2505578

I%=5.74618608PV=96248.6393PMT=-477.1342138 (use ↑key)FV=0P/Y=26C/Y=12

∴ Remaining term now 267.2505578 (slv_N) payments.

=26

2505578.267 =10.27886761 years ≈ 10 years 3 months

∴ Reducing loan term by 12 years -(10 years 3 months)= 1 year 9 months

NB: 1 year had passed therefore reducing the PV to today’sloan balance as calculated in part b). Note that the loan termwas reduced to 13 years from part c)(i) leaving only 12 yearsfrom today.

6. Which will you choose?

Method 1: 4 % pa (compounded monthly) converted to an effective interest rate is…→Eff(4,12)=4.074154292%.∴ as 4.074154292<4.1, choose the 2nd institution.

Method 2: 4.1 % pa (effective) converted to an annual compounded monthly interest rate is…→Apr(4.1,12)=4.024913879%.∴ as 4.1>4.024913879, choose the 2nd institution.

7. $150000 Loana) Loan repayments

Payment : END*N=30*26=780

I%=6PV=150000PMT=-414.5161159FV=0P/Y=26C/Y=12

∴ repayments will be $414.52 a fortnight.

3. FINANCE

b) Investigating interest in 5 Yr Blocks

5 Yr Block Payment No’s Calculation Interest Paid1 1-130 ∑Int(1,130) -43468.630312 131-260 ∑Int(131,260) -39834.147283 261-390 ∑Int(261,390) -34931.77434 391-520 ∑Int(391,520) -28319.207765 521-650 ∑Int(521,650) -19399.846376 651-780 ∑Int(651,780) -7368.964407

∑Int(1,780) -173322.5704

c) Total Paid

Total repayments will be slv_PMT*780= -323322.5704∴ Total payed during loan term is $323322.57

d) Investigating actual amounts paid in 5 Yr Blocks

5 Yr Block Payment No’s Calculation Interest Paid1 1-130 ∑Prn(1,130) -10418.464772 131-260 ∑Prn(131,260) -14052.947793 261-390 ∑Prn(261,390) -18955.320774 391-520 ∑Prn(391,520) -25567.887315 521-650 ∑Prn(521,650) -34487.24876 651-780 ∑Prn(651,780) -46518.13066

∑Prn(1,780) -150000.00

8. How much will you save?

Payment : ENDN=26*25=650I%=6.7PV=152000

*PMT=-481.7651165FV=0P/Y=26C/Y=12

For the 25 year case…

∑Int(1,650)=-1611477.3257

NB. you will need to enter all of this information prior tothe interest calculation from the calculator screen. Notethat the –ve answer is because you have to pay interestfrom your own account

Payment : ENDN=26*5=130I%=6.7PV=152000

*PMT=-1377.15476FV=0P/Y=26C/Y=12

For the 5 year case…

∑Int(1,130)=-27030.11882

NB. you will need to enter all of this information prior tothe interest calculation from the calculator screen. Notethat the –ve answer is because you have to pay interestfrom your own account