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PARTICIPANT HAND-OUT – FET PHASE 1
Module 4: Financial Mathematics
CAPS extraction indicating progression from Grades 10-
Grade 10 Grade 11 Grade 12
Use simple and compound
growth formulae
A P 1 in and
n A P 1 i^ to solve
problems (including interest,
hire purchase, inflation,
population growth and other
real life problems).
Use simple and compound
growth formulae
A P 1 in and
n A P 1 i^ to solve
problems (including straight
line depreciation and
depreciation on a reducing
balance)
The effect of different periods
of compounding growth and
decay (including effective
and nominal interest rates).
Calculate the value of n in
the formula
n A P 1 i and
n A P 1 i
Apply knowledge of
geometric series to solve
annuity and bond repayment
problems.
Critically analyse different
loan options.
Introduction
The study of Financial Mathematics is centred on the concepts of simple and compound
growth. The learner must be made to understand the difference in the two concepts at Grade
10 level. This may then be successfully built upon in Grade 11, eventually culminating in the
concepts of Present and Future Value Annuities in Grade 12.
One of the most common misconceptions found in the Grade 12 examinations is the lack of
understanding that learners have from the previous grades (Grades 10 and 11) and the lack
of ability to manipulate the formulae. In addition to this, many learners do not know when to
use which formulae, or which value should be allocated to which variable. Mathematics is
becoming a subject of rote learning that is dominated by past year papers and
memorandums which deviate the learner away from understanding the basic concepts,
which make application thereof simple.
Let us begin by finding ways in which we can effectively communicate to learners the
concept of simple and compound growth.
PARTICIPANT HANDOUT – FET PHASE 2
Simple and Compound Growth
What is our understanding of simple and compound growth?
How do we, as educators, effectively transfer our understanding of these concepts to our
learners?
What do the learners need to know before we can begin to explain the difference in
simple and compound growth?
The first aspect that learners need is to understand the terminology that is going to be used.
Activity 1: Terminology for Financial Maths
Group organisation: Time: Resources: Appendix:
Groups of 6 30 min Flipchart
Permanent markers
None
In your groups you will:
- Select a scribe and a spokesperson for this activity only – should rotate from activity
to activity.
- Use the flipchart and permanent markers to write down definitions/explanations that
you will use in your classroom to explain to your learners the meaning of the
following terms:
Interest
Principal amount
Accrued amount
Interest rate
Term of investment
Per annum
- Every group will have an opportunity to provide feedback.
A star educator always takes into account the dynamics of his/her classroom
PARTICIPANT HANDOUT – FET PHASE 4
Cindy will have an ACCRUED AMOUNT of R700. Her PRINCIPAL AMOUNT was R500.
Compound Growth Plan:
The compound growth plan has interest that is recalculated every year based on the money
that is in the account. The interest WILL CHANGE every year of her investment.
Year 1:
50
100
10 500
R
Interest
Therefore, at the end of the 1st year Cindy will have R500 + R50 = R
Year 2:
55
100
10 550
R
Interest
Therefore, at the end of the 2nd year Cindy will have R550 + R55 = R
Year 3:
- 50
100
10 605
R
Interest
Therefore, at the end of the 3rd year Cindy will have R605 + R60.50 = R665.
Year 4:
- 55
100
10
- 50
R
Interest
Therefore, at the end of the 4th year Cindy will have R665.50 + R66.55 = R732.
Cindy will have an ACCRUED AMOUNT of R732.05. Her PRINCIPAL AMOUNT was R500.
N Notice that the interest is recalculated based on the amount present in the account.
N Notice that the interest is recalculated based on the amount present in the account.
N Notice that the interest is recalculated based on the amount present in the account.
PARTICIPANT HAND-OUT – FET PHASE 5
Now that we understand the difference between simple and compound growth it is evident
that if we are required to perform a simple or compound growth calculation, it would be
tiresome to conduct that calculation in the above manner. We will use the following formulae
to help us simplify our calculations.
In both formulae: A = Accrued amount P = Principal amount i = Interest rate n = Number of times interest is calculated
Common Errors:
In applying these formulae, some of the most common errors found are as follows:
- The accrued amount is the amount that will be received at the end of the investment
period. This is NOT the same as the interest earned. Many times a question will ask
what was the interest earned and the learner will provide the accrued amount as the
answer. The accrued amount is actually the principal amount plus the interest:
( A P I ).
- The interest rate is always divided by 100 in all calculations, since it is given as a
percentage. We should perhaps modify the equation to be
n
i A P 100
1 and
n i A P
100
1 respectively so that the learners do not forget to divide the interest
rate.
- Learners need to understand that interest can work for and against an individual. It
works to an individual’s benefit when they invest a sum of money and works against
them when they borrow a sum of money. Ensure that the learner understands that
when money is borrowed the ACCRUED AMOUNT is the amount that has to be paid
back and the PRINCIPAL AMOUNT is the initial amount that was borrowed.
SIMPLE GROWTH: A^ ^ P ^1 in COMPOUND GROWTH:
n
A P 1 i
PARTICIPANT HAND-OUT – FET PHASE 7
Solution:
Option 1:
6500
5000 ( 1. 3 )
50001 0. 15 2
2 100
15 50001
100
1
R
n
i A P
- 83 24
6500 R
R
If Thabo chose OPTION 1 he would pay back R270.83 per month and a total amount of
R6500.
Option 2:
50
05
100
5 50001
100
1
24
24
R
i A P
n
90 24
50 R
R
From the calculations it is evident that Thabo made the wrong choice. With option 2, he will
pay a total of R9625.90 more and a monthly amount of R401.07 more than if he had chosen
Option 1.
Activity 2 : Calculations on Misconceptions and
Errors
Group organisation: Time: Resources: Appendix:
Individual 30 min (^) Participant hand-out
Pens/pencils
Calculators
None
PARTICIPANT HANDOUT – FET PHASE 8
This activity is to be completed individually. You are already aware of the common errors
and misconceptions that we have covered. Use your calculator to answer the questions
below. Note the types of misconceptions that could occur and how you would deal with
these. Selected participants will share their responses with the entire group.
Answer the following questions. Show all calculations.
I. Pavan wants to invest money at an institute that offers him 3.75% interest compounded quarterly. He invests R8 000 for a period of 4 years. What is the interest that he will earn in this time period?
Misconception:_______________________________________________________
Addressing the misconception:
II. A certain amount invested at 4.2% interest compounded semi-annually yields a return of R12 400 after five years. How much was initially invested?
Misconception:_______________________________________________________
Addressing the misconception:
PARTICIPANT HANDOUT – FET PHASE 10
V. Alex wants to save money to go for a holiday in three years’ time. He needs R8 000 for his holiday. He has three options of saving his money:
Option A: At 10% per annum simple interest Option B: At 3.25% compounded quarterly Option C: At 7.5% per annum compound interest
Which option will allow Alex to save the least amount of money presently so he can still enjoy his planned holiday in three years’ time?
Misconception:_______________________________________________________
Addressing the misconception:
Nominal and Effective Interest Rates
As we have seen from the work covered thus far, very often interest can be compounded
more than once a year. Notice that in the examples we looked at where the interest was
calculated more than once a year, the interest rate was simply stated as compounded
quarterly etcetera, and the words ‘per annum’ were omitted.
Why do you think this was done?
If the interest is compounded quarterly at 12%, this means that interest was calculated every
quarter at 12%. The difference is now if the interest is calculated quarterly at 12% per
annum. This implies that for the year the interest is 12%, therefore, it should be compounded
at 12 4 = 3% at every quarter.
Let us investigate these concepts which we call nominal and effective interest rates to gain
an understanding of them.
PARTICIPANT HAND-OUT – FET PHASE 11
Worked Example 3: Nominal and Effective Interest Rates
( 1 0 min)
The facilitator will now explain this example to you.
Remember to make notes as the facilitator is talking and ask as many questions as possible to clarify any misconceptions that may occur.
Example 3:
Suppose that R10 000 is invested at 12% per annum compounded quarterly. The growth of the investment can be tabulated as follows:
Since the interest rate is 12% per annum, the rate at which interest will be calculated per quarter will be 12 4 = 3%.
QUARTER PASSED
VALUE OF INVESTMENT 0 10 000 1
10300
100
3 100001
100
1
1
R
i A P
n
2
10609
100
3 103001
100
1
1
R
i A P
n
3
- 27
100
3 106091
100
1
1
R
i A P
n
4
- 09
100
3
- 271
100
1
1
R
i A P
n
If we look at the final amount of R11 255.09 we can determine that the investment actually
grew by R1 255.09. This equates to a percentage increase of 100 12. 5509 % 10000
- 09 .
Therefore, it can be seen from this example that the quoted interest rate of 12% is the
nominal interest rate, and the actual interest rate with which the investment grew in the
one-year period was 12.5509%, which is the effective interest rate.
PARTICIPANT HAND-OUT – FET PHASE 13
Example 5:
Determine the nominal interest rate, compounded monthly, which results in an effective interest rate of 12.4%.
Solution:
- 75 %
( 1. 009788745 1 ) 1200
1200
- 124 1
1200
- 124 1
100 12
1 0. 124 1
100
1 1
12
12
12
m
m
m
m
m
m m
eff
i
i
i
i
i
m
i i
Now that we clearly see the difference between nominal and effective interest rates, it is
necessary to modify our previous equations for compound growth. This is done so that
learners do not forget that nominal and effective rates must apply in examples wherein the
investment period and the principle investment amount are of relevance.
Using the principles we have just investigated, we can modify the compound growth formula
to the following:
m t m
m
i
A P
Where:
m i Nominal interest rate
m Number of times interest is calculated per annum
t Time period for the investment
This formula takes into account all of the aspects which the learner is required to know. If the
learner uses this formula, he/she will always be prompted by the equation itself and aspects
will not be left out, which would cause him/her to lose marks.
PARTICIPANT HANDOUT – FET PHASE 14
Activity 3 : Nominal and Effective Interest
Rates
Group organisation: Time: Resources: Appendix:
Groups of 6 20 min (^) Flipchart
Permanent markers
None
In your groups you will:
- Select a scribe and a spokesperson for this activity only – should rotate from activity
to activity.
- Use the flipchart and permanent markers to answer the questions that follow. Be sure
to explain to the groups during your report back how you would teach these
concepts.
- Every group will have an opportunity to provide feedback.
I. Shristi inherited a sum of money which she invested at 11.5% per annum calculated
monthly. She kept her money in a fixed investment for a period of 10 years. At the
end of the 10th year she had R22 000 in the bank. How much money did Shristi
invest initially?
II. What is the interest on R12 000 in three years at 6% per annum compounded bi-
annually?
Common Errors:
Many learners find manipulation of the formulae to be a barrier. A simple way is to use
variables up until the second-last step and then simply solve for the unknown. Too often
learners spend a lot of time trying to manipulate the formulae incorrectly, putting all the
correct values into an incorrect formula.
Learners often forget to take into consideration the nominal and effective interest rates
as it sounds too complicated. In fact, these are simple concepts to understand. In
addition to this, learners often forget to divide the interest rate by 100. They also confuse
PARTICIPANT HANDOUT – FET PHASE 16
II. Up to the end of the second year:
83477
025
100 4
10 120001
100
1
8
24
R
m
i A P
m t^ m
At the end of the second year Ramil deposited a further R5 000 into the account.
R14 620.83477 + R5 000 = R19 620.
73034
- 025
100 4
10
- 834771
100 4
1
:
4
14
14
2 3
R
i A P
T toT
m
91927
- 054
100 2
- 8
- 730441
:
2
12
3 4
R
A
T toT
76
- 2544
- 12
100 1
12
- 919271
:
2
21
4 6
R
A
T toT
At the end of the 6th year Ramil will have R30 180.76 in his account.
NB: The above example has shown us a few very important things that are often
found to be a stumbling block for many learners.
PARTICIPANT HAND-OUT – FET PHASE 17
When using time lines, always start with T 0. T 0 is the start of the first year and then T 1 is
the end of the first year. T 1 is also the start of the second year, and so forth.
There was one equation that was used for all calculations (the modified compound
growth formula). When there are too many formulas being used, learners tend to
become confused and very often use the incorrect formula.
Intermediate answers are never rounded off as this reduces the accuracy at the end.
Only the final answer is rounded off.
Notice how the accrued amount at the end of a particular time period becomes the
principle amount for the next period. This requires understanding of the question and the
terminology.
The easiest way to successfully complete an example of this nature is to break down the
scenario into a number of different transactions.
Activity 4 : Calculations
Group organisation: Time: Resources: Appendix:
Groups of 6 45 min (^) Flipchart
Permanent markers
None
In your groups you will:
- Select a scribe and a spokesperson for this activity only – should rotate from activity to
activity.
- Use the flipchart and permanent markers to answer the questions that follow. Be sure to
explain to the groups during your report back how you would teach these concepts and
what the possible misconceptions are.
- Every group will have an opportunity to provide feedback.
PARTICIPANT HAND-OUT – FET PHASE 19
IV. Nelly deposited R2 500 into a savings scheme at an interest rate of 8.8% per annum
compounded bi-annually for three years. At the end of the 2nd year she deposited
R5 000 into the savings scheme. At the end of the 3rd year Nelly withdrew R3 000
and transferred her balance to another scheme that offered her an interest rate of
10.75% per annum compound interest for two years. She moved her investment
again and earned a further 7% per annum compounded quarterly for a period of
three years.
a) Show the following investment on a timeline.
b) Calculate the value of the investment after six-and-a-half years.
Simple and Compound Decay
Thus far we have looked at aspects which showed compound or simple growth of an
investment. Not all investments will grow in time. Certain investments lose value over time.
Examples of these types of investments are the purchase of assets such as motor vehicles
and office equipment. This brings us to the concept of depreciation.
PARTICIPANT HANDOUT – FET PHASE 20
Again the terminology that is used in this section is important. It will allow learners to
understand what the questions are asking. As an educator, please ensure that your learners
always understand the terminology before every section. This allows the subsequent lessons
to be more beneficial as they will understand the terminology that is being used.
The two types of decay that are relevant are as follows:
- Simple decay: also known as straight line depreciation
- Compound decay: also known as reducing balance depreciation
It is essential that learners understand the difference between these two types of decay.
Let us investigate some simple ways to explain effectively the difference between the two
concepts.
Simple Decay: Straight Line Depreciation
Simple decay or straight line depreciation are the terms given to an investment type that
loses value over time. The loss of value is calculated as a percentage of the original value.
The value per year, for example, will decrease by the same amount. It is theoretically valid
that in this type of depreciation there will come a point when the value of a certain item will
reach zero. This, in practise, is not very likely, as vehicles or machinery will always have
some value.
Let us investigate this type of depreciation through the following example.
Worked Example 7: Simple and Compound Decay
(15 min)
Pay careful attention, make notes and ask questions.
Example 7:
Ruvi bought a new car for R300 000. The car depreciates at 12.5% p.a. on a straight line
depreciation method. What will the value of Ruvi’s car be in five years’ time?
Solution:
This can be easily calculated using a simple table:
The car depreciates at 12.5% per annum. Therefore, 300000 37500 100
- 5 R. This implies
that the car will lose R37 500 in value every year.
