MATHEMATICS

STUDY GUIDE

Math-Sage

Bring Hope to Our Generation

Grade 12
 Chapter 1 FINANCIAL MATHEMATICS

Financial mathematics is the field of applied mathematics that focuses on the mathematical
modelling of financial markets and instruments. It involves the use of techniques to analyse
and manage financial risk, as well as to price and hedge financial derivatives.

SIMPLE GROWTH AND DECAY: EXPLANATION OF FORMULA

SIMPLE INTERESTS Where:
𝐀 = 𝐏 (𝟏 + 𝒊𝐧) • P – Principal amount (original price)
• A – Final Amount (Accumulated, Book Value)
Simple decay (straight line method)• i – Interest rate (decimals)
• n – Number of time periods
𝐀 = 𝐏 (𝟏 − 𝒊𝐧)
GRAPHICAL REPRESENTATION

interest rate of 20% annually (yearly) interest rate of 20% annually (yearly)

R2000
Amount (R)

R3 600

R400
R2000 4
Time (years)
4
Time (years)
Using formula:
Using formula: A = P (1 – in)
A = P (1 +in) A = R2000(1 – 0.2 × 4)
A = R2000(1 + 0.2× 4) A = R400
A = R3 600
AMOUNT LOST
INTERESTS EARNED L=P–A
I=A–P L = R2000 – R400
I = R3 600 – R2 000 L = R1 600 (over period of 4 years)
I = R1 600 (over period of 4 years)

COMPOUND GROWTH AND DECAY

Compound Decay Compound Interest EXPLANATION:
𝐧 𝐧 We use this formula to
𝐀 = 𝐏(𝟏 − 𝒊) 𝐀 = 𝐏(𝟏 + 𝒊)
calculate accumulated
• Reducing balanced method • Inflation.
• Exponential Growth amount for single principal
• Compound Increases
• Compound decrease amount.
• Compounded

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 GRAPHICAL REPRESENTATION OF COMPOUND GROWTH AND DECAY
COMPOUND GROWTH COMPOUND DECAY
Consider an amount of R2 000 invested at 12% per annum compound interest: decreases rather than increases as with compound interest. Consider, for
Accumulated amount after one year: example, the value of a motor vehicle after 4 years if it depreciates at a rate
of 10% per annum using reducing balance depreciation.

The amount of R2000 is invested using compound The original price of the car is R200 000, which
interest which grows R3524,68 over a period of 5 depreciated to R131 220 over period of 4 years.
years.

DIFFERENT COMPOUNDING PERIODS

COMPOUNDING PERIOD VALUE EXPLANATION

The principal amount gain interest every 4 months

(4 Quarters in a year)
𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
Quarterly 4 Year rate = = 𝑣𝑎𝑙𝑢𝑒 𝑚𝑜𝑛𝑡ℎ𝑙𝑦
4
Year = n × 4 = years (months)

The principal amount gain interest every 6 months

Half-yearly (Semi-annually) 2 (2 6 months per year)
𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
Year rate = = 𝑣𝑎𝑙𝑢𝑒 𝑚𝑜𝑛𝑡ℎ𝑙𝑦
2
Year = n × 2 = years (months)

Annually 1 The principal amount gain interest every year

𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
Year rate = = 𝑣𝑎𝑙𝑢𝑒 𝑚𝑜𝑛𝑡ℎ𝑙𝑦
1
Year = n × 1 = years (months)

Monthly 12 The principal amount gain interest every 12 months

(12 month in a year)
𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
Year rate = = 𝑣𝑎𝑙𝑢𝑒 𝑚𝑜𝑛𝑡ℎ𝑙𝑦
12
Year = n × 12 = years (months)

Daily 365 or 366 The principal amount gain interest every day (there
are 365 or 366 days in 1 year)
𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡
Year rate = = 𝑣𝑎𝑙𝑢𝑒 𝑚𝑜𝑛𝑡ℎ𝑙𝑦
365 𝑜𝑟 366
Year = n × (365 𝑜𝑟 366) = years (months)
RARE CASE

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 RELATIONSHIP BETWEEN THE SIMPLE AND COMPOUND DECAY

ACTIVITY

1. Nelson buys a motor car for R160 000. The car depreciates at a rate of 11% p.a. What
is the car worth after 6 years if depreciation is calculated using:
(a) the straight-line method.
(b) the reducing-balance method.
2. Stephanie invests R5 000 for 6 years. Find the future value of her investment if the
interest she receives is:
(a) 12% p.a. simple interest.
(b) 12% p.a. compound interest.
3. Marc saved an amount of money, and it grew to R15 000 over a period of seven years.
Calculate the amount of money originally invested if the interest received was:
(a) 13% p.a. simple interest.
(b) 13% p.a. compound interest.
4. R90 000 is invested at 10% p.a. simple interest for 4 years. Thereafter, the total
amount is re-invested in a different financial institution at 12% p.a. compound interest
for three more years. What is the future value of the investment at the end of the
seven-year period?
5. Nelson buys a motor car for R160 000. The car depreciates at a rate of 11% p.a. What
is the car worth after 6 years if depreciation is calculated using:
(a) the straight-line method.
(b) the reducing-balance method.

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 TIMELINES

Timelines are useful when dealing with more complicated problems, such as interest rate
changes, or when several deposits are made into a savings account or withdrawals are made
from the account.

𝐓𝟎 𝐓𝟏 𝐓𝟐 𝐓𝟑 𝐓𝐧
Compounding Periods
Number of periods

MORE EXAMPLES ON COMPOUND INTERESTS

1. Calculate the future value of an investment of R15 000 after three years at an interest
rate of 15% per annum compounded:
(a) annually
(b) half-yearly
(c) quarterly
(d) monthly
(e) daily (excluding leap years)
2. R100 000 is invested for five years at an interest rate of 16% per annum compounded
quarterly. Thereafter the accumulated amount is re-invested for a further six years at an
interest rate of 15% per annum compound semi-annually. Calculate the value of the
investment at the end of the eleven-year period.
3. R50 000 is deposited into a savings account. The interest rate for the first three years of
the savings is 12% per annum compounded half-yearly. Thereafter, the interest rate
changes to 18% per annum compounded monthly. The money is left to grow for a
further four years. Calculate the future value of the investment at the end of the seven-
year savings period.
4. Jerome invests a certain sum of money for six years at 16% per annum compounded
quarterly for the first two years and 15% per annum compounded monthly for the
remaining term. The future value of the investment at the end of the 6-year period is
R20 000. How much did he originally invest?
5. Brenda deposits R5500 into a savings account. Five years later, R4 000 is added to the
savings. The interest rate for the first three years is 14% per annum compounded semi-
annually. Thereafter, the interest rate changes to 12% per annum compounded monthly.
Calculate the future value of the savings at the end of the seventh year.
6. Bradley deposits a birthday gift of R14 000 into a savings account in order to save up
for an overseas trip in six years’ time. At the end of the second year, he withdraws R2
000 from the account. How much money will he have saved at the end of the six-year
period, assuming that the interest rate for the whole savings period is 8% per annum
compounded monthly?
7. Brian takes out a loan to finance the purchasing of a new DVD player. He repays the
loan by means of a payment of R5 000 four years after the granting of the loan. Two
years later, he repays a final amount of R6 000. The interest rate during the first four
years of the loan is 16% per annum compounded quarterly. For the remaining two years,
the interest rate changes to 14% per annum compounded half-yearly. How much money
did Brian originally borrow?

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 ACTIVITY

1. Jennifer received a gift of R4 000 from her boyfriend. She deposited the money into
a savings account. Three years later she received another gift of R5 000 from her
boyfriend and deposited the money into the savings account. Calculate how much
money Jennifer will have saved three years after her deposit of R5 000. Assume
that the interest rate for the first two years of the savings period is 18% per annum
compounded monthly and that it changes to 18% per annum compounded half-
yearly for the remaining four years.
2. Mvelo decides to start an emergency savings account for the future. He deposits R6
000 into a savings account. Three years later, he deposits a further R8 000 into the
account. Four years after this, he deposits a further R10 000 into the account. The
interest rate for the first four years is 14% per annum compounded semi-annually.
For the next three years the interest rate increases to 15% per annum compounded
quarterly. Calculate the future value of the savings at the end of the seven-year
period.
3. Mandy has just finished reading a book on the importance of saving for the future.
She immediately opens a savings account and deposits R5 000 into the account.
Two years later, she deposits a further R6 000 into the account. Thirty-six months
later, she withdraws R3 000 to buy a birthday gift for her husband. The interest rate
during the first three years of the investment is 8% per annum compounded
monthly. The interest rate then changes to 9% per annum compounded quarterly.
Calculate the value of Mandy’s investment two years after her withdrawal of R3
000.
4. On the 1st January 2005, Diana deposited R15000 into a savings account. On the 1st
January 2007, she deposited double her first amount into the account. On the 1st
January 2008, she deposited double her second amount into the account. The
interest rate for the first two years (2005-2006) was 9% per annum compounded
monthly. The interest rate then changed to 10% per annum compounded monthly.
Calculate the value of her investment on the 1st January 2009.
5. Michael deposits a gift of R20 000 into a savings account in order to save up for an
overseas trip in six years’ time. The interest rate for the savings period is 8% per
annum compounded monthly for the first two years and 9% per annum compounded
quarterly for the remaining four years.
(a) How much money will he have saved at the end of the six-year period?
(b) Suppose that at the end of the second year, he withdraws R4 000 from the
account. How much money will he have then saved at the end of the six-year period?

(c) Instead of withdrawing money, suppose that at the end of the third year, he adds
R3 000 to the savings. How much money will he have then saved at the end of the
six-year period?

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 NOMINAL AND EFFECTIVE INTEREST RATES

 A nominal rate is an annual rate which financial institutions quote.

 Annual effective interest rates are therefore the equivalent annual rates that yield the same
accumulated amount as rates with different compounding periods.
 Annual effective rates are higher than quoted nominal rates.

FORMULA:
Where: ↷
𝐢𝐧𝐨𝐦 𝐧
𝟏 + 𝐢𝐞𝐟𝐟 = (𝟏 + )  𝐢𝐞𝐟𝐟 – Effective rate
𝐧
Used to Calculate:  𝐢𝒏𝒐𝒎 – Nominal rate
 Effective rate  n – Compounding Periods
 Nominal rate

EXAMPLES:
1.1 Convert a nominal rate of 18% per annum compounded monthly to an annual effective rate.
1.2 Convert an annual effective rate of 13,5% per annum, to a nominal rate per annum
compounded semi-annually.
SOLUTIONS:
1.1 1.2

r = 13,1%

MORE EXAMPLES:
Michael invests R40 000 for five years at 16% per annum compounded monthly.
a) Calculate the future value of the investment using the nominal rate.
b) Convert the nominal rate of 16% per annum compounded monthly to the equivalent
effective rate (annual).
c) Now use the annual effective rate to show that the same accumulated amount will be
obtained as when using the nominal rate.
ANSWERS

a) R88 552,28 b) 0,1722707983 c) R88 552,28

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 ANNUITIES
An annuity is a series of equal investment payments or loan repayments at regular intervals subject to
a rate of interest over a period of time.
TWO TYPES OF ANNUITIES
 Future Value
 Present Value
FUTURE VALUE
In a future value annuity, money is invested at regular intervals in order to save money for
the future. The magic of compound interest makes the investment grow in value, especially if
the interest rate is above the current inflation rate.
FORMULA:
𝒙[(𝟏 + 𝒊)𝐧 − 𝟏] Where:
𝐅=  F – Future value
𝒊  x – equal payments made per period
 Equal payments
 i – interest rate (decimals)
 Regular deposit
 n – number of payments made
 Fixed amount invested
 n = top -bottom + 1 (payments made)
FUTURE VALUE INVESTEMENT EXAMPLES

 Retirement investments (people save money each month to receive pension)

 Education funds (saving money for your education)
 Sinking funds
 And others
EXAMPLE 1:
Suppose that R1 000 is deposited into a bank account. One month later, a further R1 000 is
deposited into the account and then a further R1 000 one month after this. If the interest rate is
6% per annum compounded monthly, how much will have been saved after two months?
SOLUTIONS
There are a few different ways to do this calculation.
Using Compound Interest: Using Future value annuity:
Timeline n = 2 – 0 +1 = 3 payments made
2 months 0.06 3
1000[(1+ ) −1]
12
1 𝐹= 0.06
m 12
o F = R3015.03
0.06 2 0.06 1n NOTICE that in this two-month savings period, a
A = R1000 (1 + ) + 1000((1 + ) t+ 1000 total of three payments were made. This is because
12 12
A = R3015,03 h a payment was made at 𝑇0 .
The last payment did not gain any interest, since the
money is withdrawn after 2 months.

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EXAMPLE 2:
Suppose that R1 500 is invested every month, starting immediately, for a period of 12 months.
Interest is 18% per annum compounded monthly. The future value of the investment after 12
months would be the sum of all payments together with the interest earned.
DATA COLLECTION: Substitution in Correct Formula:
x = R1 500 1500[(1+
0.18 13
) −1]
n = 12 – 0 +1 = 13 payments 𝐹=
12
0.18
0.18
𝑖 = 12 12
∴ 𝐅 = 𝐑𝟐𝟏 𝟑𝟓𝟓, 𝟐𝟒
F =?  There are 13 payments made, and last payment
did not gain any interest.
EXAMPLE 3:
Thomas starts saving money in a Unit Trust fund. He immediately deposits R800 into the
fund. Thereafter, at the end of each month, he deposits R800 into the fund and continues to
do this for ten years. Interest is 8% per annum compounded monthly. Calculate the future
value of his investment at the end of the ten-year period.
SOLUTION: TIMELINE
| | | | | | |
+800 +800 +800

DATA COLLECTION Substitution in Correct Formula:

x = R800 800[(1+
0.08 121
) −1]
n =10× 12 -0+1 = 121 payments 𝐹=
12
0.08
0.08
𝑖 = 12 12
∴ 𝐅 = 𝐑𝟏𝟒𝟖 𝟏𝟑𝟐, 𝟓𝟒
F =?  There are 13 payments made, and last payment did not
gain any interest.

EXAMPLE 4:
Patrick decided to start saving money for a period of eight years starting on 31st December 2009. At
the end of January 2010 (in one month’s time), he deposited an amount of R2 300 into the savings
plan. Thereafter, he continued making deposits of R2 300 at the end of each month for the planned
eight-year period. The interest rate remained fixed at 10% per annum compounded monthly. How
much will he have saved at the end of his eight-year plan which started on the 31st of December 2009?
SOLUTION:

 Number of payments = 96 – 1 + 1 = 96 The payment is made one month later after the account has been
open for the investment (31st Dec 2009 to 31st Jan 2010).
0.1 96
2300[(1+ ) −1]
12
𝐹= 0.1
12
∴ 𝐅 = 𝐑𝟑𝟑𝟔 𝟐𝟏𝟔, 𝟒𝟕

MATHEMATICS HAS A BEAUTY AND ROMMANCE. IT IS NOT A BORING PLACE; MATHEMATICS IS THE WORLD.

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EXAMPLE 5:
Jesse deposits R400 into an account paying 14% per annum compounded half yearly. Six
months later, she deposits R400 into the account. Six months after this, she deposits a further
R400 into the account. She then continues to make half yearly deposits of R400 into the
account for a period of nine years from her first deposit of R400. Calculate the value of her
savings at the end of the savings period.
SOLUTION: TIMELINE

There are a total of 19 payments, and the duration of the investment is 9 years (18 six-
month periods).
0.14 19
400[(1+ ) −1]
12
𝐹= 0.14
12
∴ 𝐅 = 𝐑𝟏𝟒 𝟗𝟓𝟏, 𝟓𝟗

EXAMPLE 6:
Dayna has just turned 20 years old and has a dream of saving R8 000 000 by the time she
reaches the age of 50. She starts to pay equal monthly amounts into a retirement annuity
which pays 18% per annum compounded monthly. Her first payment starts on her 20th
birthday and her last payment is made on her 50th birthday. How much will she pay each
month?

Future Value is given as R8 000 000

How many years between 20 and 50: 50 – 20 = 30 years, hence there 361 x payments that
will be made to obtain the future value of R8 000 000.
0.18 361
𝑥[(1+ ) −1]
12
8 000 000 = 0.18 𝐮𝐬𝐢𝐧𝐠 𝐚𝐥𝐠𝐞𝐛𝐫𝐚𝐢𝐜 𝐦𝐚𝐧𝐢𝐩𝐮𝐥𝐚𝐭𝐢𝐨𝐧:
12
x = R558,41
 There will be 361 payments of R558, 41 to fulfil the dream of having R8 000 000 after
30 years.

SELF CHECK:
1.1 John deposits R500 into a savings account. At the end of each month thereafter, he
deposits R500 into the account and continues to do this for six years. Interest is 7% per
annum compounded monthly. Calculate the future value of his investment at the end of
the six-year period. [R44 580, 47]
1.2 Athaliah wants to save up R400 000 in five years’ time in order to purchase a new car.
She starts making monthly payments into an account paying 12% per annum
compounded monthly, starting immediately. How much will she pay each month?
[R4 791, 20]

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 EXAMPLE 6:
A father decides to start a savings plan for his baby daughter’s future education. On opening
the account, he immediately deposits R2 000 into the account and continues to make monthly
payments at the end of each month thereafter for a period of 16 years. The interest rate
remains fixed at 15% per annum compounded monthly.
1.1 How much money will he have accumulated at the end of the 16th year?
1.2 At the end of the sixteen-year period, he leaves the money in the account for a further
year. How much money will he then have accumulated?
1.3 Calculate the total amount invested.
1.4 What is the total interest earned at the end investment period.

SOLUTION: TIMELINE

There are 192 months in a period of 16 years (16 years×12 months=192 months). The
number of payments made by the father is 193.
0.15 193
2000[(1+ ) −1]
12
𝐹= 0.15
12
∴ 𝐅 = 𝐑𝐑𝟏 𝟓𝟗𝟗 𝟑𝟖𝟔, 𝟎𝟐
1.3 Since there will no longer be any further payments of R2000 into the annuity, all we now
need to do is grow the R1 599 386,02 for 12 months using Compound Interests.
TIMELINE:
Future Value for 193 months
Compound Interests

12 Months
n
A = P(1 + i)
A = 1 599 386,02(1 + 0.0125)12
∴ 𝐀 = R1 856 494,55

Total amount paid = 2000 × 193 = 𝑅386 000 This amount does not include the interest
rate earned (Total amount deposited over a period of 193 months).
TOTAL INTEREST EARNED = FUTURE VALUE – TOTAL PAID

Total interest = R1 856 494,55 – R386 000 = R1 470 494,55

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 ANNUITY PAYMENTS MADE IN ADVANCE:
It is often the case that in annuity investments, payments are made in advance. This means
that the last payment in the annuity is made one month before the investment is paid out.
The next example deals with this type of annuity.

EXAMPLE 1:
In order to supplement his state pension after retirement, a school teacher aged 30 takes out
a retirement annuity. He makes monthly payments of R1 000 into the fund and the payments
start immediately. The payments are made in advance, which means that the last payment of
R1 000 is made one month before the annuity pays out. The interest rate for the annuity is
12% per annum compounded monthly. Calculate the future value of the annuity in twenty-
five years’ time.
SOLUTION:

 Number of payments = 300 – 0 +1 = 301 payments, but one payment is made one
month before the end of investment:
 n = 300 0.12 193 𝐀 = R1 878 846, 626(1 + 0.0125)
1
1000[(1+ ) −1]
 x = 1000 𝐹=
12
0.15 0.12 A = R1 897 635, 09
 𝑖 = 12 12
F = R1 878 846, 626

SELF CHECK:

1.1 Simon starts saving money in a Unit Trust fund. He immediately deposits R900 into the
fund. Thereafter, at the end of each month, he deposits R900 into the fund and continues to
do this for the five years. Interest is 18% per annum compounded monthly.
a) Calculate the future value of his investment at the end of the five-year period.
[R88 792,08]
b) Simon leaves his money in the fund to grow for two years without making any further
payments of R900. The interest rate changes to 10% per annum compounded
quarterly. Calculate the value of his investment after the two-year period.
[R108184,53]

1.3 Irene deposits R1 700 into a savings account. At the end of each month thereafter, she
deposits R1 700 into the account and continues to do this for ten years. Interest is 8% per
annum compounded monthly.
a) Calculate the future value of her investment at the end of the ten-year period.
[R 314 781,65]
b) Irene leaves the accumulated amount in the account for a further six months
without making any further payments of R1 700. If the interest rate changes to
10% per annum compounded half-yearly, calculate the value of the investment
after the further six months. [R 330 520,73]

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 1.4 In order to supplement his pension after retirement, Mpho (aged 20) takes out a
retirement annuity. He makes monthly payments of R2 000 into the fund and the
payments start immediately. The payments are made in advance, which means that the
last payment of R2 000 is made one month before the annuity pays out. The interest rate
for the annuity is 15% per annum compounded monthly. Calculate the future value of
the annuity when he turns 60. [R 62 807 510,92]

DISCUSION OF LOANS:
In Grade 10 and 11, you studied hire purchase loans in which the interest payable is calculated
at the start of the loan using the formula for simple interest. The total interest on the loan must
be paid in full. This means that if you want to pay off the loan earlier than the loan period, you
will still need to pay the full amount of interest owed. The monthly payments are calculated by
dividing the loan amount by the number of months. You are required to pay all of these amounts
and there is no advantage in paying off the loan early.

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