MATHEMATICS - WordPress.com€¦ · Revision Guide for Grade 12 ... Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required

MATHEMATICS

Learner’s Study and

Revision Guide for

Grade 12

FINANCIAL MATHS

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents

Unit 8

Revision notes 3

Exercise 8 lets you revise everything you need to know 5

Answers 7

Examination questions with solution hints and answers 8

More questions from past examination papers 12

Answers 16

How to use this revision and study guide

1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Financial Mathematics

3

REVISION UNIT 8: FINANCIAL MATHEMATICS

What you must make sure you understand very clearly when working with compound growth or compound depreciation questions is that you must choose the correct formula to use. This means reading the question very carefully in order to choose the appropriate formula.

In all the formulae below:

• i is the interest for the time period; and • n is the number of time periods.

Example: Interest rates are usually stipulated as say r % yearly or per annum but compounded six-monthly or quarterly or monthly or daily over a time period of say t years. Then the following calculations must be made to substitute for i and n in the formulae:

Monthly Quarterly Bi-annually Daily

i =

12100×r

n =

12 t×

Note: r % must be changed to a decimal. That is divide r by 100.

Here are the formulae:

Simple Interest: )1( inPA ±= for growth and decay respectively – also called the straight line method.

Compound Interest: also called the reducing balance mthod.

1. niPA )1( += which is used if a fixed sum of money P grows in value to become .A

2. niPA )1( −= which is used if a fixed sum P decreases in value in value to become .A

3. iixF

n ]1)1[( −+= which is used if regular payments equal to x are made (such as in

savings)and it is F the future value of the amount that needs to be known. Or, the formula is also used when x the amount of the regular payments is known and what needs to be known is F the amount of the future value. (Note: F for Future value).


4. i

ixPn ])1(1[ −+−

= which is used if regular payments equal to x are made (such as in

bond repayments) and it is P the present value of the amount (such as the amount borrowed) that needs to be known. Or, the formula is also used when the amount of the loan P is known and what needs to be known is x the amount of the regular payments. (Note: P for Present value).

You must be able to distinguish between problems involving future and present value so as to choose the correct formula. Future value, as the word “future” implies, usually refers to a future benefit such as the growth in savings. Present value usually refers to the money in hand now such as a loan received to purchase a car or a house.

Sinking fund

A sinking fund is a savings plan.

For sinking fund questions such a savings to be made in order to pay for the replacement of a present asset at some time in the future, it will be required to work out the compound decrease on the present asset and also work out the future value of an intended purchase. The difference between the two answers will be the value of the sinking fund.

Amortisation

Loans like bonds are usually amortised. This means that each repayment that is made is part capital and part interest.

Effective interest rate

The nominal rate is the annual rate that is usually quoted and does not take into account the effect of the compounding periods which may be daily, weekly, monthly, quarterly or semi-annually.

Take as an example: 12% per annum compounded weekly. This is a nominal rate. The actual interest earned in a year will be higher than 12% because it will be an accumulation of interest obtained from applying the compound interest formula using the interest period as one week so making n =52 (and, of course, changing the interest rate by dividing it by 52). The actual interest earned is called the effective rate and is calculated using the formula:

( ) 11 min −+= nalnoeffective ii


5

Exercise 8

Simple Interest

1. Lehlohonolo opens a savings account that bears 15% simple interest per annum. If he deposits R 5 700 how much money will he have after 11 years and 7months?

2. How long would it take for R8 400 to grow to R15 000 at a simple interest rate of 7.5% p.a.? Give your answer correct to the nearest month.

3. How much should be invested now at 8% simple interest per annum to accumulate to R12 000 in 6 years?

4. A car worth R145 000 depreciates at the rate of 16% annually according to the straight line method. What will the value of the car be after 4.5 years?

5. How long would it take for the car in the previous question to depreciate in value to R100 000? Give your answer correct to the nearest month.

6. If Sipho buys a camera for R899 at the beginning of the year 2010 and intends to sell it for R350 at the end of 2014, what would the rate of depreciation be assuming simple interest decay?

Compound Interest

7. Kwena opts for an investment with an interest rate of 8% per annum compounded quarterly. How much will he have after 6 years if he invests R2000?

8. Juju invests R18 000 into a savings account with an interest rate of 9% per annum compounded monthly. If he invests a further R10 000 into the account after 4 years, how much will he have after 6 years and 8 months?

9. How much should be invested now at 8% interest per annum compounded annually to accumulate to R12 000 in 6 years?

10. What is the annual interest rate if R8 700 is compounded semi‐annually and grows to R12 500 after 4 years?

11. A car worth R145 000 depreciates at a compound rate of 16% per annum. Calculate the book value of the car after 4,5 years.

12. A car dealer uses the straight‐line method to calculate a car’s value. The car was originally bought for R150 000. The car dealer calculates the car to be worth R85 667.89 as it is now 5 years old.

a) What rate of depreciation has the dealer used? b) What rate of depreciation would have to be used to give the same value using

the reducing balance method?


Converting nominal to effective interest rate

13. Calculate the effective interest rate, if the nominal interest rate is 12.5% compounded monthly.

14. Calculate the effective interest rate, if the nominal interest rate is 12.5% compounded quarterly.

15. Calculate the effective interest rate, if the nominal interest rate is 12.5% compounded biannually.

16. Given an effective rate of 11.65%, what would the annual nominal rate compounded monthly be?

Comparing two investments

17. You are offered two savings schemes. Bank A will pay you 12% compounded monthly, and Bank B 12.5% compounded quarterly. Which is the better investment? (Hint: compare their effective rates.)

Future value of an ordinary annuity

18. What will the value of an annuity be after 30 years if R500 is deposited each year into an investment earning 6.75% per annum compounded monthly?

Sinking fund

19. You reckon that in 4 years’ time you will need to replace your car at a cost of R300 000. What fixed monthly savings must you make into an account paying 6% p.a. compounded monthly?

Present value of an annuity

20. What is the present value of an annuity that offers to pay R1000 per month for 10 years if the interest is 7% per annum compounded monthly?

Ammortisation

21. You take out a bond worth R500 000 in order to buy a house. The terms of the bond are that you pay equal monthly instalments over a period of 20 years. The interest on the bond is 9.5% per annum compounded monthly. What will your monthly repayments be?


7

ANSWERS

EXERCISE 8

1. R15 603,75 2. 11years 8 months 3. R8 108,11 4. R40 600 5. 1 year 11 months 6. 15,13% 7. R3 216,87 8. R45 425,90 9. R7 562,04 10. 9,27% 11. R66 164,46 12. (a) 8,58% (b) 10,60% 13. 13,24% 14. 13,10% 15. 12,89% 16. 11,07% 17. Bank A: 12,68% Bank B: 13,10% Better investment with Bank B 18. R580 732,93 19. R5 545,51 20. R86 126,35 21. R4 660,66


PAPER 1 QUESTION 9 DoE/ADDITIONAL EXEMPLAR 2008

PAPER 1 QUESTION 7 DoE/NOVEMBER 2008


9

PAPER 1 QUESTION 9 DoE/ADDITIONAL EXEMPLAR 2008

Number Hints and answers Work out the solutions in the boxes below 9.1 You must convert the nominal rate of

interest which is given into an effective rate of interest by dividing it by the number of quarters in a year. The time period nmust be multiplied by the number of quarters in a year The problem is now all yours to do using the compound interest formula found in the formula sheet. Answer: n = 20,51 years

9.2.1 Read the question carefully and ask yourself: is this a Present Value problem or a Future Value problem? The formulae are on the formula sheet: F for the Future Value formula and P for the Present Value formula. Having worked out which formula to use, convert the nominal interest rate given in the problem by dividing by the number of months in a year and multiply the n years by the number of months in a year. The problem is all yours now using the right formula. Answer: x = R4 270,93

9.2.2 Balance outstanding = Present value of the remaining 15 payments to be made.

= ( )i

ix n ]11[ −+−

Now you do the working out. Answer: R59 216,95


PAPER 1 QUESTION 7 DoE/NOVEMBER 2008

Number Hints and answers Work out the solutions in the boxes below 7.1 Problem tells you that that it is about

compound interest. Substitute into the appropriate formula and work out n . You will need to use logs. Answer: 23,7 years

7.2 Always a good a practice to start off by listing all the relevant information you are given in the statement of the problem. P = what? n = what? Trade‐in value = what? Replacement value increases annually by what rate? That is r = what?

7.2.1 Write down the formula : A = ? Work out A . What he will need to pay = A minus replacement value. Answer: R975 462,46 or R975 462,50

7.2.2 Is this a present or future value problem? That is the first question you must ask yourself. Write down the formula for P or F value according to which is the correct one for this question. Calculate i . Determine n . Substitute for i , n and (is it P or F?) to work out the value of x . Answer: x = R11 944,00


11

Number Hints and answers Work out the solutions in the boxes below 7.2.3 You can argue this problem in several

ways. One way is to assume that you are saving R5 000 at the end of each year with interest at 12% compounded monthly. In this way you will be calculating the growth of each R5000 deposited at the end of each year. Because the interest is compounded monthly the years must be converted into months and the annual rate into a monthly rate of interest. Adding the growth of each R5000 leads to a geometric series. Sum up the series. This will be the cost of the maintenance over the 5 years. Maintenance cost = R 32 197,77 The new monthly deposit x can now be worked out by reasoning that: What he will need to pay in 5 years’ time (which is what you will have worked out in 7.2.1 above) = {Future value} – {Maintenance cost} Replace Future Value by its formula and work out x . An easier method is to work out the amount by which the monthly payments must be increased. This can be done by treating R5 000 as a future value after 12 months of monthly deposits of x rands Work the x out which must then be added to the answer in 7.2.2 above. Answer: x = R12 338,24


MORE QUESTIONS FROM PAST EXAMINATION PAPERS

Exemplar 2008

Preparatory Examination 2008


13

Feb – March 2009

November 2009 (Unused)


November 2009 (1)


15

Feb – March 2010


ANSWERS

Exemplar 2008 2.1 =n 5,88 years 2.2.1 =r 20,00% 2.2.2 =vF R1 607 287,50

2.2.3 increase = 74,90% 2.2.4 =x R27 942,76 Additional Exemplar 2008 9.1 =n 20,51 years 9.2.1 =x R4 270,93 9.2.2 Outstanding balance = R59 216,95 Preparatory Examination 2008 2.1 =n 5,66 years 2.2.1 R650 000,00 2.2.2 =x R10 467,74 2.2.3 Balance = R597 114,88 November 2008 7.1 =n 23,7 years 7.2.1 =A R975 462,46 or R975 462,50 7.2.2 =x R11 944,00 7.2.3 =x R12 338,24 Feb/March 2009 9.1 =i 24,08% 9.2 =vF R1 244,99 so that Farouk will

not be able to buy the bike in January 2009.

November 2009 (Unused papers) 8.1 Depreciation value = R3 037,50 8.2.1 =n 154,65 or 155 8.2.2 Outstanding balance = R3 230,50 8.2.3 Last payment = R3 278,96 8.2.4 Total repaid = R773 278,96 November 2009(1) 9.1 =n 2,37 years 9.2.1 =A R133 929,25 9.2.2 (a) =x R3 636,36 9.2.2 (b) Total = R196 363,66 9.2.3 Total payments = R190 602,72 9.2.4 R5 760,96 Feb/March 2010 4.1 30 000 + 27 000 +24 000 + . . . . + 0 4.2 After 11 years 4.3 =x 13 250

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MATHEMATICS - WordPress.com€¦ · Revision Guide for Grade 12 ... Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required