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Financial Mathematics Notes - Part 2

Page 1 of 27

HSC Mathematics General
Financial Mathematics Notes - Part 2
CREDIT AND BORROWING
Review of Simple Interest
You should remember from your preliminary year
that simple interest is the cost for borrowing an
amount, P, at a given percentage rate, r, for a
certain number of time periods, n
...
A loan that charges simple interest is
sometimes called a “flat-rate loan”
...
This
addition of interest to the principal is called
compounding
...
5% p
...
compound
interest over a 5 year term, what is
the future value of his investment?
Answer:
PV = $25 000, r = 7
...
075, n = 5
Using,
𝐹𝑉 = 𝑃𝑉(1 + 𝑟) 𝑛
𝐹𝑉 = $25 000(1 + 0
...
4356 … )
𝐹𝑉 = $𝟑𝟓 𝟖𝟗𝟎
...
a
...
06
4

= 0
...
015)20
𝑃𝑉 = 14 000
...

𝑷𝑽 = $𝟏𝟒 𝟎𝟎𝟎 (to the nearest dollar)

Credit Cards








Credit cards allow you to borrow money up
to a certain limit as long as you make regular
minimum repayments
...

They can be used to pay for goods and
services and for cash advances (i
...
withdraw
cash from an ATM or bank branch)
The difference from a debit card is that a
debit card uses money that is in your account
whereas a credit card borrows money that
you will need to pay back later
Many credit cards have an interest free
period (e
...
55 days)
...


© Eoin O'Malley 2015

Page 2 of 27
Note: An outstanding amount is an
amount of money that was due for
payment previously but has not been paid
 Usually interest-free periods apply only to
purchases whereas cash advances usually
attract interest straight away
...
Therefore,
the annual rate of interest must be divided
by 365 to get the daily rate
...

 Note: Even if simple interest is charged,
you will be charged interest on interest in
the following month if the outstanding
amount isn’t paid in full
...
The interest rate is 16
...
a
...
His monthly bill shows a
total of $3580 owing
...
What will his bill be the
following month (31 days later) if
he doesn’t use his credit card
during that time?

Financial Mathematics Notes - Part 2

Page 3 of 27

Example 3 (contd
...
(as no interest is charged if the
bill is paid by the due date)

b) Assuming simple interest,
𝐼 = 𝑃𝑟𝑛
P = $3580; r = 16
...
169;
n = 1 day
0
...
𝟔𝟔 𝒑𝒆𝒓 𝒅𝒂𝒚 (to nearest cent)
𝐼 = $3580 𝑥

c) Min
...
02 𝑥 $3580
= $71
...
amount payable is $72, i
...

rounded up to the nearest dollar)
𝑅𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑏𝑎𝑙𝑎𝑛𝑐𝑒
= $3580 − $72
= $3508
Interest for next 31 days:
𝐼 = 𝑃𝑟𝑛
P = $3508; r = 0
...
169
𝑥 31
365

= $50
...
27
= $𝟑𝟓𝟓𝟖
...
00
$88
...
60

Total
$199
...
40
$364
...
9% p
...
, charged daily
from the day of purchase to the day of
payment inclusive
...
00;
21
...
9% p
...
=
= 0
...
0006
n = 34 days (from 4 Aug – 7 Sept)
𝐼 = 199 𝑥 0
...
0596
 Amount of interest accrued on
Zara’s JB Hi-fi purchase = $𝟒
...
40; r = 0
...
)
𝐼 = 88
...
0006 𝑥 27
𝐼 = 1
...
𝟒𝟑 (to
nearest cent)

Ticketmaster purchase:
P = $76
...
0006
n = 19 days (from 19 Aug – 7 Sept)
𝐼 = 76
...
0006 𝑥 19
𝐼 = 0
...
𝟖𝟕 (to nearest cent)
 Total interest she pays =
$4
...
43 + $0
...
𝟑𝟔
c) Total amount she pays back = total
principle amount + total interest
= $364
...
36 = $𝟑𝟕𝟎
...
The credit card
company charges a fee of $2
...
5% of the amount, whichever is
greater for cash advances
...
9% p
...
compound interest
...
50?
c) If he pays off the debt after 28
days, how much does he pay?
© Eoin O'Malley 2015

Page 4 of 27

Example 5 (contd
...
50 or 2
...

2
...
025 𝑥 $500
= $12
...
50,  the fee to be paid is $12
...
50, the
maximum amount would be:
𝐹𝑒𝑒
$2
...
5%
$2
...
025
c) To pay off the debt, he must pay
back the principle borrowed ($500)
+ the fee ($12
...

For compund interest,
 𝐴 = 𝑃(1 + 𝑟) 𝑛
P = $500
...
9%
r = 17
...
a
...
049041
...
00049041…
n = 28 days
 𝐴 = 500(1 + 0
...
00049041)28
𝐴 = $506
...
91 + $12
...
𝟒𝟏

Financial Mathematics Notes - Part 2

Page 5 of 27

Reducing-balance loans

Example 6 (contd
...
With a reducingbalance loan (or reducible–interest loan), the
rate of interest stays the same but the actual
amount of interest reduces between time periods
as the balance of the loan reduces due to
payments being made
...

Tables of loan repayments can be used to see
the amount owing over the term of the loan
...
The interest is calculated on that
principal
...
These tables clearly
show how the interest reduces over the term of
the loan as the balance reduces, hence why the
loan can also be called a reducible-interest loan
...
They handover a
table of repayments, partially
reproduced below
...
00
$4814
...
41

$50
...
15
$46
...
00
$4862
...
69

$4814
...
41
$4438
...
e
...
01

= 𝟏% 𝒑𝒆𝒓 𝒎𝒐𝒏𝒕𝒉
(We could have gotten the same answer
by taking the respective amounts for
months 2 and 3 also)

b) The monthly repayments (R) can be
worked out by taking the Principle
+ Interest (P+I) amounts away from
the final amount in a particular
month, i
...
:
Repayments, R:
R = $5050
...
63 =
= $𝟐𝟑𝟓
...
)

Example 6 (contd
...
37 = $𝟕𝟎𝟔
...
37, therefore:

d) Over the first three months, the
loan reduces by:
$5000 − $4438
...
𝟔𝟖

The interest, I, is 1% of the
Principle, P, so:

𝑃 + 𝐼 = $235
...
01𝑃
𝑃 + 0
...
37

This difference between this and
the total repayments made is the
amount which has been paid as
interest
...
01𝑃 = $235
...
37
1
...
04
𝑃=

e) The Principle for month 4 is the
final amount after the repayment
in month 3, i
...
P = $4438
...

The interest for the month is 1% of
this Principle, i
...


∴ 𝐼 = 0
...
04
𝐼 = $2
...
01 x $4438
...
38

24

$233
...
33

$235
...
00

∴ 𝑃 + 𝐼 = $4482
...
70 − $235
...
33
Therefore, the next line in the table
is:
Month

Principle

Interest

P+I

P+I-R

4

$4438
...
38

$4482
...
33

f) To work out the last line in the
table, we start at the end, i
...
the
fact that there will be zero amount
owing at the end of the loan term:
𝑃 + 𝐼 − 𝑅 = $0

© Eoin O'Malley 2015

Alex takes out a loan from a bank for
$10 000 at a reducible interest rate of
9
...
a
...
36 per year, by drawing up a
table of installments, find out how
many years it will take him to repay
the loan
...
Note

Financial Mathematics Notes - Part 2

Page 7 of 27

Example 7 (contd
...
e
...
5%
...

The interest rate is 11
...
a
...
The interest for each
month is then charged on the first day
of the following month
...
00
$8345
...
11
$4550
...
41

$950
...
84
$620
...
30
$225
...
00
$9138
...
85
$4982
...
36

$8345
...
11
$4550
...
41
$0
...


Daily Interest
Some loans may calculate the interest daily
...

For example, if we have a balance of $10 000 on a
loan with an interest rate of 10% p
...
, then the
following shows how the number of days in the
month affects the interest to be charged:


31 days:
(



30 days:
(



31
) x 0
...
93
365

30
) x 0
...
19
365

28 days:
28
(
) x 0
...
71
365

Date
Details
Amount
Balance
Balance brought forward
$16 914
...
21 $17 079
...
79 $16 641
...
30 $16 799
...
79 $16,361
...
80 $16 521
...
79 $16 083
...

b) What amount of interest will be
charged on 1 July?
Answer:
a) The interest in June is higher as it is
calculated for May which has 31
days whereas the interest in May is
calculated for April which only has
30 days
...

Therefore the interest is:
30
(
) x 0
...
46
365
= $152
...
There are many factors to consider,
which could make a difference of thousands of
dollars over the term of a loan
...
)
d) The interest rate
e) Whether the interest rate is fixed or variable
(or a combination)
f) Whether the interest is flat or reducible
g) Any fees that may be charged
h) The flexibility of the loan (e
...
whether
additional payments are allowed, whether
you can redraw against additional
repayments, take repayment holidays etc
...
He goes to a
broker who presents him with the
following three options:
1
...
2% p
...
compound
interest over 30 years
2
...
a
...
Loan C: Monthly repayments of
$5079
...
)
Answer:
To answer these questions, we need
to find the total amount, A, to be
repaid for each of the loans
...

We are told that the interest is
compound interest, therefore we can
use the compound interest formula,
𝑨 = 𝑷(𝟏 + 𝒓) 𝒏
to find the total amount to be repaid
...
042/12 = 0
...
0035)360
𝐴 = $𝟏 𝟓𝟖𝟐 𝟗𝟓𝟑
...
We are told that the loan is
repaid in equal monthly installments,
therefore the monthly installment is:
=

$1 582 953
...
𝟎𝟗
360

Loan B:
P = $450 000; r = 0
...
005;
n = 20 x 12 = 240
𝐴 = 450 000 𝑥 (1 + 0
...
𝟎𝟏 (to nearest cent)
Therefore the monthly installment is:
=

$1 489 592
...
𝟔𝟑
12

Financial Mathematics Notes - Part 2

Page 9 of 27

Example 9 (contd
...
53

Brooke wants to borrow $20 000 to
buy a new car
...
94 over 5 years
...
53 = $𝟏 𝟓𝟐𝟑 𝟖𝟓𝟗
...
09
...
01, which is less than
the other two loan options
...
55 − 450 000
= $1 132 953
...
01 − 450 000
= $1 039 592
...
55 − $1 039 592
...
𝟓𝟑
Alternatively, since the Principle is the
same for both loans ($450 000), we
can calculate the difference in interest
by working out the difference in total
amounts to be repaid:
= $1 582 953
...
01
= $𝟗𝟑 𝟑𝟔𝟏
...
4%
p
...

a) How much interest will she pay if
she goes with the hire purchase
option?
b) How much interest will she pay if
she chooses the loan from her
bank?
c) Which option will cost her the least
over the full term and how much
will she save over the alternative
option?
d) Find the monthly repayment for
the bank loan option and suggest a
reason why she might go with this
option?
Answer:
a) Total amount to be repaid through
the hire purchase option:
= $424
...
40
Therefore, total interest that would
be paid:
= $25 494
...
𝟒𝟎

Financial Mathematics Notes - Part 2

Example 10 (contd
...
084/12 = 0
...
085 𝑥 84
∴ 𝐼 = $𝟏𝟏 𝟗𝟎𝟎
c) The hire purchase option will cost
her the least over the full term
...
40 = $𝟔𝟒𝟎𝟓
...
𝟕𝟔 (to nearest cent)
As this monthly repayment is
cheaper than the hire purchase
option, she may decide to choose
this option in order to reduce her
monthly repayment
...

A variable interest rate is when the interest rate
can go up or down over the term of a loan
...
It is important for borrowers to
consider the effect of possible interest rate rises
on their loans to ensure that they will still be able
to make the increased repayments
...
This is common for home loans where the
fixed period is often for the first 1 to 5 years of
the loan term
...


Financial Mathematics Notes - Part 2

Page 11 of 27

ANNUITIES AND LOAN REPAYMENTS
Annuities
Generally speaking, there are two types of
annuities - immediate annuities and deferred
annuities
...

Immediate annuities convert a cash pool into a
lifelong income stream, providing you with a
guaranteed monthly allowance as you grow old
...
In this course, you
will deal only with deferred annuities
...

It is a form of insurance or investment
which entitles the holder to a guaranteed
income stream for a certain number of
years or even the rest of their life
...

The future value of an annuity is the total
value of the investment at the end of the last
contribution period
...
These usually give the
amount for an investment of $1 (see below
for an example table)
...

The present value of an annuity is the single
sum of money that can be invested now
under the same terms (interest rate and

© Eoin O'Malley 2015





length of time) as an annuity with regular
contributions, M, and will produce the same
financial outcome
...
The one with the greater
present value will produce the greater
financial outcome over time
...
These tables
usually give the present values of an annuity
of $1 for different interest rates and periods
of time
...
0604 4
...
2465
6
6
...
3081 6
...
2857 8
...
2142
12
12
...
412 15
...
258 18
...
825
24
26
...
422 39
...
223 79
...
26

6%
4
...
9753
9
...
870
25
...
816
256
...
9901 0
...
9434
3
2
...
8286 2
...
7955 5
...
9173
9
8
...
7861 6
...
255 9
...
3838
18
16
...
754 10
...
108 21
...
621

9%
0
...
5313
4
...
9952
7
...
7556
10
...
)
Interest rate, r = 6%; No
...
673 for an annuity contribution of
$1
...
673
= $𝟐𝟏𝟖 𝟐𝟐𝟎
...
a
...
a
...
a
...
)
FV = $15 305
...
75 − (48 𝑥 $250)
= $15 305
...
𝟕𝟓
Option B:
Interest rate = 8% p
...

= 2% per quarter
No
...
639
= $20 502
...
90 − (16 𝑥 $1100)
= $20 502
...
𝟗𝟎
Option C:
Interest rate = 8% p
...

= 4% per half year
No
...
a
...
of periods = 4 years x 12 months
= 48 periods
Future Value, FV = $250 x 61
...
2142
= $16 585
...
56 − (8 𝑥 $1800)
= $16 585
...
𝟓𝟔
 Option A would earn Clarissa the
most interest over the three years
...
)

Liam wants to save for a deposit to
buy an apartment in two years time
...
If he can invest
into an account which pays out
compound interest at 2% per month,
how much will he need to invest per
month (rounded up to the nearest
dollar) in order to reach his goal?

Interest rate = 3% per month

Answer:
Future Value, FV = $45 000
Interest rate = 2% per month
No
...
422
$𝑀 =

𝐹𝑉
30
...
422

𝑀 = $1479
...
of periods = 3 years x 12 months
= 36 periods
Present Value, PV = $250 x 21
...
Sally has worked out
that if she invests $2000 every 6
months over that time into an account
paying interest of 12% p
...

compounded half-yearly, that she will
have enough for the trip
...
a
...
of periods = 3 x 2 half-years
= 6 periods
𝑃𝑉 = $2000 𝑥 4
...
𝟔𝟎

Financial Mathematics Notes - Part 2

Example 15 (contd
...
60(1 + 0
...
4%
0
...
65%
60
53
...
726 49
...
156 90
...
144
180
128
...
50 105
...
09 139
...
35
300
174
...
21 131
...
60 166
...
91

0
...
173
78
...
593
111
...
16
124
...
60 𝑥 1
...
568 …
𝐹𝑉 = $𝟏𝟑 𝟗𝟓𝟏 (to nearest dollar)
OR
By using the future values table:
𝐹𝑉 = $𝑀 𝑥 6
...
9753
𝐹𝑉 = $13 950
...

His bank will charge compound
interest each month of 4
...
a
...
g
...
Interest compounds on the
balance owing and then a set repayment is made,
thus reducing the balance owing and the interest
the following period
...

The total amount repaid on the loan can then be
calculated by multiplying the amount of each
repayment, M, by the number of repayments to
be made, n
...

The following table is used for Examples 16, 17
and 19 that follow:

© Eoin O'Malley 2015

a) Interest rate = 4
...
4% per month
No
...
60

= $𝟑𝟒𝟏𝟎
...
38 𝑥 360
= $1 227 736
...
80 − $650 000
= $𝟓𝟕𝟕 𝟕𝟑𝟔
...
Some may offer a low introductory
interest rate for the first months or years but
then increases after this period
...

Another consideration is whether the interest
rate is fixed or variable or perhaps even a
combination of both
...

 An interest-only period means that only
interest is paid on the loan rather than
interest + principal
...

 Downside: the repayments will be higher for
the rest of the term of the loan and ultimately
the loan will cost more
...

 A fixed interest rate means that the interest
rate will stay the same for the term of the
loan (or the amount of time that it is fixed)
...

 It is common for lending institutions to offer
mortgages which are fixed for the first few
years and then turn into a variable rate loan
for the remainder of the term
...

 However, fixed rate loans will often have a
higher interest rate than a variable rate loan
...
)
at an interest rate of 7
...
a
...
Her bank is
giving her the option of paying
interest only for the first 5 years
...

f) Suggest a reason why Tara should
not take the interest only option
...
8% per annum
= 0
...
0065 𝑥 $ 480 000
= $𝟑𝟏𝟐𝟎
b) PV = $480 000 (no principle was paid
off during the interest only period)

Interest rate = 0
...
of periods = 25 years x 12 months
= 300 periods

Financial Mathematics Notes - Part 2

Example 17 (contd
...
82

= $3641
...
33 − $3120
= $𝟓𝟐𝟏
...
of periods = 30 years x 12 months
= 360 periods
∴ 𝑀=

$480 000
138
...
𝟒𝟕 (to nearest cent)
d) Cost of loan with interest only
option:
= (5 𝑥 12 𝑥 $3120)
+ (25 𝑥 12 𝑥 $3641
...
47
= $1 243 969
...
20
= $𝟑𝟓 𝟔𝟐𝟗
...
)
opposed to $3455
...
The money
saved during this period may be
useful for one-off purchases such
as furniture or home
improvements
...
33 per month as opposed to
$3455
...
In addition, she would also
pay $35 629
...

Example 18
Brendan wants to borrow $30 000
over 5 years
...
a
...
a
...

Use the following table to answer the
questions below
...
25%
2
...
75%
8
7
...
1701 7
...
415 10
...
104
16
13
...
055 12
...
964 15
...
227

3
...
0197
9
...
561
14
...
)

Example 18 (contd
...
a
...
of periods as above for
part a

Period
1
2
3
4

P
$30 000

I

R

Interest rate = 9% p
...

= 2
...
23 (to nearest cent)

 Total amount he’d repay with the
variable rate option at 9% p
...
for the
full term:
= $1879
...
60

P+I-R

d) How much would he end up
repaying if he took the variable
rate option and the interest rate
increased to 12% p
...
after 1 year?

∴ 𝑆𝑎𝑣𝑖𝑛𝑔 = $38 488
...
60 = $𝟗𝟎𝟒
c) From part b:
Interest rate = 2
...
43

Answer:
b) PV = $30 000
Interest rate = 10% p
...

= 2
...
of periods = 5 years x 4 quarters
= 20 periods
∴ 𝑀=

$30 000
15
...
43 (to nearest cent)
 Total amount he’d repay with the
fixed rate option:
= $1924
...
𝟔𝟎

© Eoin O'Malley 2015

$30 000
15
...
00
$28 825
...
78
$26 387
...
00
$720
...
54
$659
...
43
$1 924
...
43
$1 924
...
57
$27 621
...
89
$25 123
...
a
...

PV = $25 123
...
a
...
of periods = 4 years x 4 quarters
= 16 periods

Financial Mathematics Notes - Part 2

Example 18 (contd
...
16
∴ 𝑀=
12
...
09 (to nearest cent)
Total amount he’d end up repaying:
= (4 𝑥 $1924
...
09)
= $7697
...
44
= $𝟑𝟗 𝟔𝟗𝟗
...
These
include:
 Loan establishment fees
 Loan service fees
 Stamp duty
 Penalties for late or missed payments

Example 19
Julie is looking to buy a home that is
valued at $375 000
...
The term of the loan is
20 years and will be repaid monthly
...
50 for
every $100 that the value of the home
exceeds $300 000
...
In addition, Julie’s bank also
charges loan establishment fees of
$660 and an ongoing service fee of $8
per month
...
)
b) How much in fees will her bank
charge over the full term?
c) If she pays an interest rate of 6%
p
...
, compounded monthly, find
her monthly repayment amount
including service fees
...

d) If after 5 years, Julie wins the
lottery and decides to pay off the
remaining balance on the loan,
how much of an early settlement
fee would she pay if her bank
charges 1
...

Answer:
a) 𝑆𝑡𝑎𝑚𝑝 𝑑𝑢𝑡𝑦 = $8990 +
($375 000 − $300 000)
[$4
...
50 𝑥 750)
= $12 365
 Total government fees:
= $12 365 + $107 + $214
= $𝟏𝟐 𝟔𝟖𝟔

b) Loan term = 20 years
= 20 x 12 months
= 240 months
 Total bank fees:
= $660 + ($8 𝑥 240) = $660 + $1920
= $𝟐𝟓𝟖𝟎

Financial Mathematics Notes - Part 2

Page 19 of 27

Example 19 (contd
...


Total loan amount:
= $375 000 + $12 686 + $660 − $75 000
= $313 286

Interest rate = 6% per annum
= 0
...
of periods = 240 months
∴ 𝑀=

$313 286
139
...
49 (to nearest cent)
Total amount she pays per month:

The tables usually show how the repayment
differs depending on the term of the loan and
may also show how it differs depending on the
interest rate also
...
We must
then multiply this amount by the number of
thousands in the loan to get the actual
repayment required, e
...
:
 If the table shows that the repayment is
$19
...
5 x $19
...
90

Example 20

= $2244
...
𝟒𝟗

c) Total that would’ve been repaid
over 20 years not including service
fee:
= 240 𝑥 $2244
...
60
Total she had already repaid over 5
years not including service fee:
= 60 𝑥 $2244
...
40
Early settlement fee she’ll pay:
= 1
...
60

− $134 669
...
019 𝑥 $404 008
...
𝟏𝟔 (to nearest cent)

© Eoin O'Malley 2015

A bank supplies the following table of
monthly repayments per $1000
borrowed for a range of different loan
terms and interest rates
...
5%
5
...
50%
5
19
...
57
19
...
10
11
...
61
15
8
...
711
8
...
164
7
...
753

6
...
04
11
...
270
8
...
a
...
a
...
)

Example 20 (contd
...
5% p
...

interest?
d) Alex wants to take out a loan of
$350 000
...
5%
...
5% p
...


Answer:
a) From the table supplied, the
monthly repayment per $1000
borrowed for a 15 year term and
6% p
...
interest rate is $9
...

 The monthly repayment for a loan
of $40 000 is:
= 40 𝑥 $9
...
𝟖𝟎
b) From the table supplied, the
monthly repayment for a loan of
$33 000 borrowed for a 10 year
term and 5% p
...
interest rate is:
= 33 𝑥 $11
...
50
 Total amount repaid is:
= no
...
50
= $𝟒𝟎 𝟖𝟔𝟎
...
753
= $2093
...
of months x monthly repayment
= 20 𝑥 12 𝑥 $2093
...
40
 Total amount of interest paid:
= total repaid – total borrowed*
*(assuming no fees or taxes etc
...
40 − $270 000
= $𝟐𝟑𝟐 𝟑𝟗𝟒
...
a
...
711
= $3048
...
85
= $548 793
...
5% p
...

 Monthly repayment is:
= 350 𝑥 $7
...
40
 Total amount repaid is:
= 20 𝑥 12 𝑥 $2507
...
00
 Option 1 would cost the Alex the
least over the full term
...
00 − $548 793
...
𝟎𝟎

© Eoin O'Malley 2015

Financial Mathematics Notes - Part 2

Page 21 of 27

Graphs comparing repayment options

Example 21 (contd
...

Sometimes the graphs also show how the total
amount to be repaid (principal + interest) varies
over time
...

These graphs can also sometimes show the
effects of changes in repayments, such as making
more frequent repayments (e
...
fortnightly vs
monthly), increasing the repayment amounts or
making a lump sum repayment
...

d) How many years it will take to
reduce the balance owing by a half
...


b)

450
400
300

d)

200
100
0
0

10

15

20

23 25

Years

Example 21
600

c)

30

a) Balance owing at the start is
$500 000
...


300

200
100
0

0

5

10

15

20

25

30

c) It will take 30 years to pay off the
full amount
...

b) The balance owing after 10 years
to the nearest $10 000
...
This occurs after about
23 years
...
)

Paul and Liz each take out a home
loan of $800 000 and pay the same
monthly repayments
...

800
700

800
700
600

Loan Balance ($000's)

Example 22

500

b)

450
400
300

200

Loan Balance ($000's)

600

100
500

0

0

400

5

10

15

20

Years

a)

25

30

300

Paul's loan

Liz's loan

200
100

Example 23

0
5

10

15

20

25

30

Years
Paul's loan

Liz's loan

Using the graph above, find:
a) How many years earlier Liz will
have paid off her loan compared to
Paul
...

Answer:
From the graph, we get:

Sophie takes out a home loan
...

2 000
1 800
1 600
1 400
Loan Balance ($000's)

0

1 200
1 000
800

600

a) Liz’s loan will be paid off 10 years
earlier than Paul’s
...

© Eoin O'Malley 2015

400
200
0

0

5

10

Principle

15
Years

20

Total owing

25

30

Financial Mathematics Notes - Part 2

Page 23 of 27

Example 23 (contd
...
)

Using the graph above, find:
a) The amount she borrows
...

c) Comment on why one line is
straight while the other is curved
...
e
...

b) The amount of interest she pays is
the difference between the total
owing and the Principle at the start
of the loan
...
The
“Principle” line is curved as
towards the start of the loan she
pays off more interest than
principle
...
Therefore the rate
at which the principle is paid off
increases over time
...

Therefore the total interest that
she pays is:

d) By using a ruler (or counting the
number of squares) on the graph,
we see that the amount of
principle owed is roughly equal to
the amount of interest owed after
8 years
...

 A flat-rate loan is a loan that charges
simple interest
...
This addition of interest to
the principal is called compounding
...

 It is a means of borrowing money to pay
for goods and services
...
e
...

 Even if simple interest is charged, you
will be charged interest on interest in
the following month if the outstanding
amount isn’t paid in full
...

 Sometimes called reducible–interest
loans
 With a reducing-balance loan the rate of
interest stays the same but the actual
amount of interest reduces between time
periods as the balance of the loan reduces
due to payments being made
...


Page 25 of 27

Comparing Loans
 There are many factors to consider,
including:
a)
b)
c)
d)
e)

The term (length) of the loan
The repayment amount
The frequency of the repayments
The interest rate
Whether the interest rate is fixed or
variable (or a combination)
f) Whether the interest is flat or
reducible
g) Any fees that may be charged
h) The flexibility of the loan (e
...

additional repayments, redraw
facility, repayment holidays etc
...

the loan
...
The interest is calculated
the term of a loan
...



𝑪𝒍𝒐𝒔𝒊𝒏𝒈 𝒃𝒂𝒍𝒂𝒏𝒄𝒆:
= 𝑃𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 + 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 − 𝑅𝑒𝑝𝑎𝑦𝑚𝑒𝑛𝑡
= 𝑃+ 𝐼− 𝑅

 The closing balance from one time period
then becomes the opening principal at
the beginning of the following time
period
...

© Eoin O'Malley 2015

 A fixed period is the length of time
when a fixed interest rate is charged on
a loan before changing to a variable
interest rate
...
It is a form of insurance or
investment which entitles the holder to
a guaranteed income stream for a
certain number of years or even the
rest of their life
...


Page 26 of 27

Loan Repayments
 A loan with regular equal instalments
works a lot like an annuity

 The future value of an annuity is the total  Interest compounds on the balance
value of the investment at the end of the
owing and then a set repayment is
last contribution period
...

 There are tables that we can use to help
us find the future values
...

Present Value, then we can use a table
This amount should then be multiplied by
of present values to find the regular
the number of dollars invested to give the
instalment, M, required to pay off the
required value
...

 The present value of an annuity is the
 Total amount to be repaid on a loan:
single sum of money that can be invested
= 𝑀x 𝑛
now under the same terms (interest rate
where,
and length of time) as an annuity with
M is the instalment amount and;
regular contributions, M, and will produce
n is the number of repayments
the same financial outcome
...
The one
with the greater present value will
produce the greater financial outcome
over time
...

These tables usually give the present
values of an annuity of $1 for different
interest rates and periods of time
...

 Benefits:
o repayments will be lower during
this period
 Downside:
o the repayments will be higher for
the rest of the term of the loan
and
o the loan will cost more overall
...

 A variable interest rate loan means that
the interest rate may go up or down (or
both) over the term of the loan
...

 A fixed rate loan offers peace of mind to
the borrower as they know exactly what
the repayments will be and don’t have to
worry about interest rate rises
...

Costs involved with Loans
 Costs and fees associated with include:
o Loan establishment fees
o Loan service fees
o Stamp duty
o Penalties for late or missed
payments
Tables of instalments
 Tables of instalments give the
repayments required to repay reducingbalance loans of certain amounts
...
We must then multiply this
amount by the number of thousands in
the loan to get the actual repayment
required
Graphs comparing repayment options
 Graphs can show the balance owing on
a loan over its term
...

 Sometimes the graphs also show how
the total amount to be repaid
(principal + interest) varies over time
...

 These graphs can also sometimes show
the effects of changes in repayments,
such as making more frequent
repayments (e
...
fortnightly vs
monthly), increasing the repayment
amounts or making a lump sum
repayment

---