Chapter 15 FM Page 711 Monday, November 13, 2000 3:36 PM
Reducing
balance
loans
15
VCE coverage
Area of study
Units 3 & 4 • Business
related
mathematics
In this chapter
15A
15B
15C
15D
Loan schedules
The annuities formula
Number of repayments
Effects of changing the
repayment
15E Frequency of
repayments
15F Changing the rate
15G Reducing balance and
flat rate loans
Chapter 15 FM Page 712 Monday, November 13, 2000 3:36 PM
712
Further Mathematics
Introduction
Amount owing
When we invest money with a financial institution the institution pays us interest
because it is using our money to lend to others. Conversely, when we borrow
money from an institution we are using the institution’s money and so it charges
us interest.
In reducing balance loans, interest is usually charged every month by the financial institution and repayments are made by the borrower also on a regular basis.
These repayments nearly always amount to more than the interest for the same
period of time and so the amount still owing is reduced. Since the amount still
owing is continually decreasing and interest is calculated on a daily balance but
debited monthly, the amount of interest charged decreases as well throughout the
life of the loan.
This means that less of the amount borrowed is paid off
p
in the early stages of the loan compared to the end.
If we graphed the amount owing against time for a loan
it would look like the graph at right. That is, the rate at
which the loan is paid off increases as the loan progresses.
Time
The terms below are often used when talking about reducing balance loans:
Principal, P = amount borrowed ($)
Balance, A = amount still owing ($)
Term = life of the loan (years)
To discharge a loan = to pay off a loan (that is A = $0)
It is possible to have an ‘interest only’ loan account whereby the repayments
equal the interest added and so the balance doesn’t reduce. This option is available
to a borrower who wants to make the smallest repayment possible.
Though the focus of this chapter is reducing balance loans, note that the theory
behind reducing balance loans can also be applied to other situations such as
superannuation payouts, for people during retirement, and bursaries. In each of
these situations a lump sum is realised at the start of a period of time and regular
payments are made during that time. Regular payments are called annuities. So
these situations are often called annuities in arrears because the annuity follows
the realisation of the lump sum.
Chapter 15 FM Page 713 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
713
Loan schedules
The first amount of interest is added to the balance of a loan account one month after
the funds are provided to the customer. The first repayment is usually made on the
same day.
Consider a loan of $800 that is repaid in 5 monthly instalments of $165.81 at an
interest rate of 1.2% per month, interest debited each month. A loan schedule can be
drawn for this information, showing all interest debits and repayments.
From the schedule the amount owing after each month is shown and the total interest
charged can be calculated.
For any period of the loan:
Total repayments = Interest paid + Principal repaid
Month
Balance at
start of
month
($)
Interest
(1.2% of
monthly
starting
balance)
($)
Total owing
at end of
month
($)
Repayment
($)
Balance
after
repayment
($)
1
800.00
9.60
809.60
165.81
643.79
2
643.79
7.73
651.52
165.81
485.71
3
485.71
5.83
491.54
165.81
325.73
4
325.73
3.91
329.64
165.81
163.83
5
163.83
1.97
165.80
165.80
0.00
Each month interest of 1.2% of the monthly starting balance is added to that balance
and the repayment value is subtracted, leaving the starting balance for the next month.
This process continues until the loan is paid off after the 5 months.
Note that the amount of interest charged falls each month and so the amount of
principal paid each month increases as outlined earlier.
Another method can be used to analyse this account, but it doesn’t display interest amounts.
Since the interest rate is 1.2% per month the balance increases by this rate each
month. Recalling the work covered in the previous chapter about the growth factor, we
can write:
r
Growth factor, R = 1 + --------where 1 represents the original amount and
100
=1+
1.2
--------100
r represents the increase per period
= 1.012
So:
Balance at start of second month = balance at start of first month × R − repayment
A2 = Α1 × R – Q
where Q is the regular repayment.
Chapter 15 FM Page 714 Monday, November 13, 2000 3:36 PM
714
Further Mathematics
WORKED Example 1
An $800 loan is repaid in 5 monthly instalments of $165.81 at an interest rate of 1.2% per
month, interest debited each month. Calculate:
a the amount still owing after the 4th month
b the total interest charged during the 5 months.
THINK
WRITE
a
a
r
R = 1 + --------100
1.2
= 1 + --------100
= 1.012
1
Calculate the growth factor.
r
R = 1 + --------100
2
Find the balance, A1, at the start of the
2nd month.
A0 = starting principal
A0 = $800
A1 = A0 × R − Q
= 800(1.012) − 165.81
A1 = $643.79
3
Find the balance, A2, at start of the 3rd
month.
A2 = A1 × R − Q
= 643.79(1.012) − 165.81
A2 = $485.71
4
Continue this process to find A3, A4
and A5.
A3 = A2 × R − Q
= 485.71(1.012) − 165.81
A3 = $325.73
A4 = A3 × R − Q
= 325.73(1.012) − 165.81
A4 = $163.83
5
The amount still owing at the end of the
4th month is A4.
The amount still owing at the end of the
4th month is $163.83
b Total interest = Total repayments −
Principal repaid
b Total interest = 165.81 × 5 − 800
= 829.05 − 800
= $29.05
As mentioned earlier, institutions usually debit a loan account with interest each
month. In this chapter we also consider situations in which interest is debited fortnightly
and quarterly. The frequency with which a customer can make repayments may be
weekly, fortnightly or monthly, and we also consider quarterly repayments.
In all cases in this chapter the frequency of debiting interest will be the same as the
frequency of making repayments, although this is not necessary in practice. It simply
makes calculations easier.
The calculations outlined for monthly repayments would follow exactly the same
pattern for other repayment frequencies.
In worked example 1, the loan was paid off with only a few repayments. In practice,
the repayment of most loans takes considerably longer than this. The process outlined
in the example continues throughout any part of the term of the loan.
Chapter 15 FM Page 715 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
715
WORKED Example 2
A loan of $16 000 is repaid by monthly instalments of $430.83 over 4 years at an interest
rate of 1.1% per month, interest debited monthly. Calculate:
a the amount still owing after the 5th repayment
b the decrease in the principal during the first 5 repayments
c the interest charged during this time.
THINK
WRITE
a
a
1
Calculate the growth factor, R.
r
R = 1 + --------100
1.1
= 1 + --------100
= 1.011
b
c
2
Find the balance, A1, at the end of the
1st month (or the start of the 2nd
month).
A0 = 16 000, Q = 430.83.
A1 = A0 × R − Q
= 16 000(1.011) − 430.83
A1 = $15 745.17
3
(a) Find A2 from A1.
(b) Repeat until A5 is found.
(A5 is the balance at the end of the
5th month.)
A2 = A1 × R − Q
= 15 745.17(1.011) − 430.83
A2 = $15 487.54
A3 = 15 487.54(1.011) − 430.83
= $15 227.07
A4 = 15 227.07(1.011) − 430.83
= $14 963.74
A5 = 14 963.74(1.011) − 430.83
= $14 697.51
4
Write a statement.
The amount owing after 5 months is
$14 697.51.
1
The decrease in the principal is the
difference between the amount owing
initially, A0, and after the 5th month, A5.
2
Write a statement.
1
2
Interest charged = Total repayments −
Principal repaid
Write a statement.
b Decrease in principal
= A0 − A5
= 16 000 − 14 697.51
= $1302.49
The principal has decreased by
$1302.49 in the first 5 months of
the loan.
c Interest charged = 430.83 × 5 − 1302.49
= $851.66
The interest charged during the first
5 months is $851.66.
Chapter 15 FM Page 716 Monday, November 13, 2000 3:36 PM
716
Further Mathematics
More often than not a financial institution provides the nominal interest rate per year
rather than the interest rate per period. As outlined in the previous chapter in the compound interest formula section, the rate per period can be obtained from the nominal
annual rate as follows:
Nominal interest rate per annum
Interest rate per period, r = ------------------------------------------------------------------------------------Number of interest periods per year
It is important to note that while a loan can be drawn at a certain interest rate, that
rate will generally not remain the same for the life of the loan. This means that when
we consider borrowing we should be aware that the amount of the repayments may
increase (due to an increase in the interest rate) during the term of the loan and we
should be confident that repayments can be met even if the rate rises.
It has been said that if a potential borrower can maintain repayments for a rate of
11% p.a. over the term of the loan then the borrower can withstand rate changes that
may range from perhaps 5% p.a. to 17% p.a.
Let us now look at how quickly the principal decreases at the end of a loan compared
with the earlier stages.
WORKED Example 3
a A family take out a loan of $40 000 to extend their home. The loan is made at a rate of
interest of 10% p.a. (debited monthly) and is repaid over 10 years by monthly instalments of $528.60. For the 3rd repayment find:
i the amount of principal repaid
ii the amount of interest paid.
b After 8 years the amount still owing is $11 455.71. Assuming the same conditions apply
as in part a, for the 97th repayment find:
i the principal repaid
ii the interest paid.
THINK
a i
1
WRITE
Calculate the monthly interest rate, r.
10
a i r = -----12
= 0.833 33% per month
2
3
(a) Calculate the monthly growth
factor, R.
(b) Store in your calculator memory
if it is recurring.
Calculate the amount owing after
each of the first 3 months — A1, A2
and A3.
r
R = 1 + --------100
0.833 33
= 1 + -------------------100
= 1.008 333 3
A1 = A0 × R − Q
= 40 000(1.008 333 3) − 528.60
A1 = $39 804.73
A2 = 39 804.73(1.008 333 3) − 528.60
= $39 607.84
A3 = 39 607.84(1.008 333 3) − 528.60
= $39 409.31
Chapter 15 FM Page 717 Monday, November 13, 2000 3:36 PM
Chapter 15
THINK
4
1
2
717
WRITE
Principal repaid = A2 − A3
(3rd repayment)
ii Interest paid = Total repayments −
Principal repaid
b i
Reducing balance loans
Monthly repayment =
8 years × 12 payments/year = 96.
So, A96 = $11 455.71. Find A97.
Principal repaid = A96 − A97
(97th repayment)
ii Interest paid = Repayments
− Principal repaid
Principal repaid = 39 607.84 − 39 409.31
= $198.53
ii
Interest = 528.60 × 1 − 198.53
= $330.07
b i A97 = A96 × R − Q
= 11 455.71(1.008 333 3) − 528.60
= $11 022.57
Principal repaid = 11 455.71 − 11 022.57
= $433.14
ii
Interest = 528.60 − 433.14
= $95.46
As mentioned in the introduction, a greater percentage of each repayment made in
the early part of a loan is interest, compared with the repayments toward the end. This
is confirmed by the calculations made in the last example.
In summary, with each of 120 repayments being $528.60;
for the 3rd repayment: interest = $330.07, principal repaid = $198.53
for the 97th repayment: interest = $95.46, principal repaid = $433.14.
That is, the principal decreases faster towards the end of the loan.
remember
remember
1. In a loan schedule:
(a) the interest charged each period increases the amount owed
(b) the repayment each period decreases the amount owed.
r
2. Growth factor, R = 1 + --------- where 1 represents the original amount and
100
r represents the increase per period in %.
3. Balance at the end of the month = balance at start of the month × R − Q
An + 1 = An × R − Q where Q = repayment
4. Total repayments = Interest paid + Principal repaid
Nominal interest rate per annum
5. Interest rate per period, r = ----------------------------------------------------------------------------------Number of interst periods per year
Chapter 15 FM Page 718 Monday, November 13, 2000 3:36 PM
718
Further Mathematics
15A
WORKED
Example
1
d
hca
Mat
EXCE
et
reads
L Sp he
Reducing balance loans
Loan schedules
1 A loan of $1000 is repaid in five monthly instalments of $206.04 at a rate of
1% per month, interest debited monthly. Calculate:
a the amount still owing after the 4th repayment
b the total interest charged during the 5 months.
2 Dimitri takes out a loan of $1500 and repays it in five monthly instalments of
$309.97 at a rate of 1.1% per month, interest debited monthly. Calculate:
a the amount still owing after the 4th repayment
b the total interest charged during the 5 months.
3 A loan of $2000 is repaid in four quarterly instalments of $525.25 at a rate of
2% per quarter, interest debited quarterly. Calculate:
a the amount still owing after the 3rd repayment
b the total interest charged during the 4 quarters.
4 Gaetana borrows $900 which she repays in five quarterly instalments of $193.72 at
a rate of 2.5% per quarter, interest debited quarterly. Calculate:
a the amount still owing after the 4th repayment
b the total interest charged during the
5 quarters.
5 Josh’s loan of $3000 is repaid in four halfyearly instalments of $807.08 at a rate of
3% per half-year, interest debited half-yearly.
Calculate:
a the amount still owing after the
3rd repayment
b total interest charged during the
4 repayments.
6 Rebecca takes out a loan of
$2500 to purchase a
new computer. The
loan is repaid
in four
6-monthly
instalments of
$696.86 at a rate of
4.5% per 6-months, interest
debited 6-monthly. Calculate:
a the amount still owing after the
3rd repayment
b the total interest charged during the
4 repayments.
Chapter 15 FM Page 719 Monday, November 13, 2000 3:36 PM
Chapter 15
WORKED
Example
2
Reducing balance loans
719
7 a A loan of $20 000 is repaid by monthly instalments of $444.89 over 5 years at an
interest rate of 1% per month, interest debited monthly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
b A loan of $20 000 is repaid by quarterly instalments of $1344.31 over 5 years at
an interest rate of 3% per quarter, interest debited quarterly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
8 a
Jose borrows $30 000 which he repays in fortnightly instalments of $206.45 over
10 years at an interest rate of 0.5% per fortnight, interest debited fortnightly.
Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
b A loan of $30 000 is repaid by quarterly instalments of $1350.84 over 10 years at
an interest rate of 3.25% per quarter, interest debited quarterly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
9 a Angela takes out a loan of $20 000 to set up a catering business. The loan is repaid
by monthly instalments of $664.29 over 3 years at an interest rate of 1% per
month, interest debited monthly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
b Emad borrows $20 000 to establish a pet-minding business. The loan is repaid by
monthly instalments of $325.06 over 8 years at an interest rate of 1% per month,
interest debited monthly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
c Hank takes out a loan of $20 000 which he repays in monthly instalments of
$286.94 over 10 years at an interest rate of 1% per month, interest debited
monthly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
d In parts a–c above the three loan accounts are the same except for the term. As the
term of the loan increases how does this affect:
i the repayment?
ii the amount still owing after the 5th repayment?
iii the amount of interest paid during the 5 repayments?
Chapter 15 FM Page 720 Monday, November 13, 2000 3:36 PM
720
Further Mathematics
10 a Jaques borrows $30 000 which he repays in quarterly instalments of $1373.05
over 8 years at an interest rate of 2.5% per quarter, interest debited quarterly.
Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
b Isabel borrows $30 000 and repays it by quarterly instalments of $1195.09 over 10
years at an interest rate of 2.5% per quarter, interest debited quarterly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
c George takes out a loan of $30 000 which he repays in quarterly instalments of
$1080.18 over 12 years at an interest rate of 2.5% per quarter, interest debited
quarterly. Calculate:
i the amount still owing after the 5th repayment
ii the decrease in the principal during the first 5 repayments
iii the interest charged during this time.
d In parts a–c above the 3 loan accounts are the same except for the term. As the
term of the loan increases how does this affect:
i the repayment?
ii the amount still owing after the 5th repayment?
iii the amount of interest paid during the 5 repayments?
In questions 11–13 find:
i the amount still owing after the 4th repayment
ii the decrease in the principal during the first 4 repayments
iii the total interest paid during this time.
11 A loan of $50 000 is to be paid by monthly instalments of:
a $525.13 over 15 years at 0.8% per month (interest debited monthly)
b $487.13 over 18 years at 0.8% per month (interest debited monthly)
c $440.33 over 25 years at 0.8% per month (debited monthly)
d $639.22 over 15 years at 1.1% per month (debited monthly)
e $607.15 over 18 years at 1.1% per month (debited monthly)
f $571.46 over 25 years at 1.1% per month (debited monthly).
12 A loan of $60 000 is to be repaid by monthly instalments of:
a $429.86 over 20 years at 0.5% per month (interest debited monthly)
b $472.41 over 20 years at 0.6% per month (interest debited monthly)
c $516.90 over 20 years at 0.7% per month (interest debited monthly)
d $563.20 over 20 years at 0.8% per month (interest debited monthly)
e $635.73 over 20 years at 0.95% per month (interest debited monthly)
f $685.92 over 20 years at 1.05% per month (interest debited monthly).
13 A loan of $60 000 is to be repaid by quarterly instalments of:
a $1292.90 over 20 years at 1.5% per quarter (debited quarterly)
b $1421.02 over 20 years at 1.8% per quarter (debited quarterly)
c $1554.87 over 20 years at 2.1% per quarter (debited quarterly)
d $1694.06 over 20 years at 2.4% per quarter (debited quarterly)
e $1911.89 over 20 years at 2.85% per quarter (debited quarterly)
f $2062.53 over 20 years at 3.15% per quarter (debited quarterly).
Chapter 15 FM Page 721 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
721
14 The loan accounts outlined in question 12 are the same except for the interest rate.
The same applies to question 13. In these cases, as the interest rate increases, what
happens to:
a the repayment?
b the amount still owing after the 4th repayment?
c the amount of interest paid during the 4 repayments?
15 a Madako’s loan of $50 000 has interest charged at a rate of 9% p.a. (debited
monthly) and it is repaid over 10 years by monthly instalments of $633.38. For the
3
3rd repayment find:
i the principal repaid
ii the interest paid.
b After 8 years the amount still owing is $13 863.96. Assuming the same conditions
apply as in part a, for the 97th repayment find:
i the principal repaid
ii the interest paid.
WORKED
Example
16 a Pina’s loan of $60 000 has interest charged at a rate of 8% p.a. (debited monthly)
and it is repaid over 20 years by monthly instalments of $501.86. For the
3rd repayment find:
i the principal repaid
ii the interest paid.
b After 18 years the amount still owing is $11 098.43. Assuming the same conditions
apply as in part a, for the 217th repayment find:
i the principal repaid
ii the interest paid.
17 a Katharine’s loan of $80 000 has interest charged at a rate of 12% p.a. (debited
quarterly) and it is repaid over 20 years by quarterly instalments of $2648.94. For
the 3rd repayment find:
i the principal repaid
ii the interest paid.
b After 18 years the amount still owing is $18 594.66. Assuming the same conditions
apply as in part a, for the 73rd repayment find:
i the principal repaid
ii the interest paid.
18 a Tony and Marietta take out a loan
of $90 000 as part payment on
their new house. The loan is to be
repaid over 25 years at 13% p.a.
(debited fortnightly) and with fortnightly instalments of $468.31. For
the 3rd repayment find:
i the principal repaid
ii the interest paid.
b If the principal is reduced to
$80 268.49 after 10 years (use the
same conditions as in part a), for
the 261st repayment find:
i the principal repaid
ii the interest paid.
c If the principal is reduced to
$44 676.17 after 20 years (use the
same conditions as in part a), for
the 521st repayment find:
i the principal repaid
ii the interest paid.
Chapter 15 FM Page 722 Monday, November 13, 2000 3:36 PM
722
Further Mathematics
19 multiple choice
If the quarterly instalments for a $15 000 loan, which is to be repaid over 4 years, are
$1148.98 and interest is debited quarterly at 2.5% per quarter, the decrease in the
principal in the first year would be (to the nearest dollar):
A $11 786
B $3214
C $1382
D $774
E $375
20 multiple choice
John’s $23 000 loan has interest charged at 9% p.a., debited fortnightly, and is repaid
over 8 years by fortnightly instalments of $155.30. For the 3rd repayment the amount
of interest paid is:
A $13.98
B $75.95
C $76.21
D $79.09
E $155.30
21 multiple choice
The term of a loan is 120 monthly instalments. Which of the following repayments
will reduce the principal by the greatest amount?
A 10th
B 20th
C 30th
D 100th
E 110th
22 multiple choice
Which of the following loan terms would have the greatest amount of interest debited?
(Assume other conditions are the same.)
A 20 years
B 22 years
C 14 years
D 12 years
E 10 years
23 Voula’s loan of $55 000 starts with quarterly repayments of $1396.64 and is due to
run for 15 years at 6% p.a., interest debited quarterly. However, after 1 year the
interest rate rises to 7% p.a. and consequently the quarterly repayments rise to
$1482.84 to maintain the 15 year term.
a What amount is still owing after 2 years?
b What amount would have still been owing after 2 years if the rate had remained at
6% p.a.?
c What would be the difference in interest charged between the two scenarios?
24 Cynthia takes out a loan of $85 000 to set up an outdoors adventure business. She starts
with quarterly repayments of $2300.42 and the loan is due to run for 20 years at 9%
p.a., interest debited quarterly. However, after 1 year the interest rate falls to 8% p.a.
and consequently the quarterly repayments fall to $2143.88 to maintain the 20 year term.
a What amount is still owing after 2 years?
b What amount would have still been owing after 2 years if the rate had remained at
9% p.a.?
c What would be
the difference in
interest charged
between the two
scenarios?
Chapter 15 FM Page 723 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
723
The annuities formula
In the previous section step-by-step calculations were made to determine the amount
still owing. The process was restrictive in that the previous balance was needed to
calculate subsequent balances. A method is needed to enable calculation of the amount
still owing at any point in time during the term of the loan.
An annuities formula can be used to enable such calculations to be made. An annuity
is a regular payment. When a consumer borrows money from a financial institution that
person contracts to make regular payments or annuities in order to repay the sum
borrowed over time.
Let us now use, in general terms, the process adopted in the previous section to
develop this annuities formula.
Let P = amount borrowed (principal)
R = growth factor for amount borrowed
r
= 1 + --------(r = interest rate period)
100
n = number of repayments
Q = amount of regular repayments made per period
An = amount owing after n repayments
Assuming interest is debited to the account before a repayment is credited, then:
A0 = P
A1 = A0R − Q
A2 = A1R − Q
= PR − Q
= (PR − Q)R − Q
= PR2 − QR − Q
A3 = A2R − Q
= PR2 − Q(R + 1)
= (PR2 − QR − Q)R − Q
= PR3 − QR2 − QR − Q
A4 = A3R − Q
= PR3 − Q(R2 + R + 1)
= (PR3 − QR2 − QR − Q)R − Q
= PR4 − QR3 − QR2 − QR − Q
= PR4 − Q(R3 + R2 + R + 1)
In general,
An = PRn − Q(Rn − 1 + . . . + R2 + R + 1)
The term in the bracket (Rn − 1 + . . . + R2 + R + 1) is the sum of n terms of a geometric
progression (GP) (refer to chapter 6: Arithmetic and geometric sequences).
First term, a = 1
Common ratio, r = R
Now, the sum of n terms of a geometric progression is:
a(rn – 1)
Sn = ---------------------r–1
Hence, in this case,
1( Rn – 1 )
1 + R + R2 + . . . + Rn − 1 = ----------------------R–1
Q( Rn – 1 )
An = PR n – -----------------------R–1
Chapter 15 FM Page 724 Monday, November 13, 2000 3:36 PM
724
Further Mathematics
So, in general, the amount owing in a loan account for n repayments is given by the
annuities formula:
Number of repayments made
Amount still owing
Repayment value
Interest rate per period
Q( Rn – 1 )
A = PR n – -----------------------R–1
Amount borrowed
r
where R = 1 + --------100
Growth factor
WORKED Example 4
A loan of $50 000 is taken out over 20 years at a rate of 6% p.a. (interest debited monthly)
and is to be repaid with monthly instalments of $358.22. Find the amount still owing after
10 years.
THINK
WRITE
1
State the loan amount, P, and regular
repayment, Q.
P = 50 000
Q = 358.22
2
Find the number of payments, n,
interest rate per month, r, and
growth factor, R.
n = 10 × 12
= 120
6
r = -----12
= 0.5
r
R = 1 + --------100
0.5
= 1 + --------100
= 1.005
3
Substitute into the annuities formula.
4
Evaluate A.
5
Write a statement.
Q( Rn – 1 )
A = PRn − -----------------------R–1
358.22 ( 1.005 120 – 1 )
= 50 000(1.005)120 − --------------------------------------------------1.005 – 1
A = $32 264.98
The amount still owing after 10 years will be
$32 264.98.
Note: If R is a recurring decimal, place the value in the calculator memory and bracket R if
needed when evaluating A.
Chapter 15 FM Page 725 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
725
Note that, even though 10 years is the halfway point of the term of the loan, more
than half of the original $50 000 is still owing.
When we consider borrowing money we usually know how much is needed and we
choose a term which requires a repayment that we can afford.
To find the repayment value, Q, the annuities formula is used where A is zero, that is,
the loan is fully repaid. Q is then isolated.
Q( Rn – 1 )
A = PRn − -----------------------R–1
When A = 0,
Q( Rn – 1 )
0 = PRn − -----------------------R–1
Q( Rn – 1 )
------------------------ = PRn
R–1
PR n ( R – 1 )
Q = --------------------------Rn – 1
WORKED Example 5
Rob wants to borrow $2800 for a new hi-fi system from a building society at 7.5% p.a.,
interest adjusted monthly.
a What would be Rob’s monthly repayment if the loan is fully repaid in 1 1--2- years?
b What would be the total interest charged?
THINK
WRITE
a
a P = 2800
n = 18
7.5
r = ------12
= 0.625
0.625
R = 1 + ------------100
= 1.00625
1
(a) Find P, n, r and R.
(b) Store in your calculator memory the
growth factor, R.
2
Substitute into the annuities formula to
find the regular monthly repayment, Q.
3
Evaluate Q.
Write a statement.
4
b
1
Total interest = Total repayments −
Amount borrowed
2
Write a statement.
PR n ( R – 1 )
Q = --------------------------Rn – 1
2800 ( 1.00625 ) 18 ( 1.00625 – 1 )
= -------------------------------------------------------------------------1.00625 18 – 1
Q = $164.95
The monthly regular payment is $164.93
over 18 months.
b Total interest = 164.95 × 18 − 2800
= 2969.10 − 2800
= $169.10
The total interest on a $2800 loan over
18 months is $169.10.
Chapter 15 FM Page 726 Monday, November 13, 2000 3:36 PM
726
Further Mathematics
Alternative method using a graphics calculator
The Texas Instrument graphics calculators TI–83 and TI–86 have a FINANCE
function: TVM Solver. This allows quick analysis of reducing balance loans using the
annuities formula.
To use the TVM Solver, press 2nd [FINANCE] and select 1:TVM Solver.
From this screen we define the following:
where N
= the number of repayments
I% = the nominal interest rate (must
enter as % per annum)
PV = the amount borrowed or the
current amount owed (enter as
a positive number as cash is
flowing to you from the bank; a positive cashflow)
PMT = regular payment amount (enter as a negative number as the cash is
flowing from you to the bank; a negative cashflow)
FV = the final amount owing (enter as ‘0’ if the loan is fully repaid or enter
the amount still owing as a negative number)
P/Y = number of payments per year, for example quarterly; P/Y = 4.
C/Y = number of compounds per year, for example monthly adjusted C/Y = 12
(Note: In this chapter, P/Y and C/Y are to be of the same frequency.)
PMT:END BEGIN — Leave END highlighted as normally interest is charged at
the end of the month.
WORKED Example 6
Josh borrows $12 000 for some home office equipment. He agrees to repay the loan over
4 years with monthly instalments at 7.8% p.a. (adjusted monthly). Find:
a the instalment value
b the principal repaid and interest paid during the:
i 10th repayment
ii 40th repayment.
THINK
WRITE
a
a P = 12 000
n = 4 × 12
= 48
7.8
r = ------12
= 0.65
0.65
R = 1 + ---------100
1
(a) Find P, n, r and R.
(b) Store R in your calculator memory.
2
Substitute into the annuities formula to
find the monthly repayment, Q.
3
Evaluate Q.
= 1.0065
PR n ( R – 1 )
Q = --------------------------Rn – 1
12 000 ( 1.0065 ) 48 ( 1.0065 – 1 )
= ------------------------------------------------------------------------1.0065 48 – 1
Q = $291.83
Chapter 15 FM Page 727 Monday, November 13, 2000 3:36 PM
Chapter 15
THINK
Reducing balance loans
727
WRITE/DISPLAY
If using the TVM Solver on
the TI–83, enter the
appropriate values. Identify A,
P, r and R.
N = 48
r (= I%) = 7.8
P(= PV) = 1200
Q(= PMT) = unknown
A(= FV) = 0
P/Y = 12
C/Y = 12
Place cursor on PMT =.
Press ALPHA [SOLVE] to solve.
4
b i
Write a statement.
1
Find the amount owing after
9 months.
(a) State P, n, R.
(b) Substitute into the annuities
formula.
2
Evaluate A9.
The monthly repayment over a 4-year
period is $291.83.
b i
P = 12 000, n = 9, R = 1.0065
Q( Rn – 1 )
A = PRn − -----------------------R–1
291.83 ( 1.0065 9 – 1 )
= 12 000(1.0065)9 − ------------------------------------------------1.0065 – 1
A9 = $10 024.73
If using the TVM Solver on
the TI–83, enter the
appropriate values. Place
cursor on FV =. Press ALPHA
[SOLVE] to solve.
3
Find the amount owing after 10
months. Substitute (change n = 9 to
n = 10) and evaluate.
A10 = 12 000(1.0065)10 −
291.83 ( 1.0065 10 – 1 )
--------------------------------------------------A10 =
1.0065 – 1
A10 = $9798.06
Continued over page
Chapter 15 FM Page 728 Monday, November 13, 2000 3:36 PM
728
Further Mathematics
THINK
WRITE/DISPLAY
If using the TVM Solver on
the TI–83, enter the
appropriate values. Place
cursor on FV =. Press ALPHA
[SOLVE] to solve.
4
6
Interest paid = Total repayments −
Principal repaid
Write a statement.
1
Repeat steps 1–6 for A39 and A40.
2
Write a statement.
5
b ii
Principal repaid = A9 − A10
Principal repaid = 10 024.73 − 9798.06
= $226.67
Total interest = $291.83 − 226.67
= $65.16
In the 10th repayment $226.67 principal is
repaid and $65.16 interest is paid.
b ii A39 = 12 000(1.0065)39 −
291.83 ( 1.0065 39 – 1 )
--------------------------------------------------A39 =
1.0065 – 1
A39 = $2543.10
A40 = $2267.80
Principal repaid = A39 − A40
= 2543.10 − 2267.80
= $275.30
Interest = 291.83 − 275.30
= $16.53
In the 40th repayment $275.30
principal is repaid and $16.53 interest
is paid.
remember
remember
1. To calculate the amount in a loan account use the formula:
Q( Rn – 1 )
A = PR n – -----------------------R–1
2. To calculate the repayment value use the formula:
PR n ( R – 1 )
Q = --------------------------Rn – 1
where P = amount borrowed (principal) ($)
R = growth factor for amount borrowed
r
= 1 + --------(r = interest rate per period)
100
n = number of repayments
Q = amount of regular repayments made per period ($)
An = amount owing after n repayments ($)
Chapter 15 FM Page 729 Monday, November 13, 2000 3:36 PM
Chapter 15
The annuities formula
Example
4
d
hca
Mat
reads
L Sp he
et
1 Use the annuities formula to find A, given:
a P = $50 000, n = 100, Q = $550, r = 1
b P = $50 000, n = 200, Q = $550, r = 1
c P = $60 000, n = 100, Q = $650, r = 1
d P = $60 000, n = 200, Q = $650, r = 1
e P = $20 000, n = 50, Q = $300, r = 0.5
f P = $40 000, n = 100, Q = $400, r = 0.8
g P = $80 000, n = 150, Q = $700, r = 0.75
h P = $100 000, n = 200, Q = $720, r = 0.65.
WORKED
729
EXCE
15B
Reducing balance loans
Reducing balance loans
2 A loan of $65 000 is taken out over 20 years at a rate of 12% p.a. (interest debited
monthly) and is to be repaid with monthly instalments of $715.71. Find the amount
still owing after:
a 5 years
b 10 years
c 15 years
d 18 years.
3 Matthew takes out a reducing balance loan of $75 000 over 25 years at a rate of 10%
p.a. (interest debited quarterly) and is to be repaid with quarterly instalments of
$2048.39. Find the amount still owing after:
a 5 years
b 10 years
c 15 years
d 20 years.
4 A loan of $52 000 is taken out over 15 years at a rate of 13% p.a. (interest debited
fortnightly) and is to be repaid with fortnightly instalments of $303.37. Find the
amount still owing after:
a 4 years
b 8 years
c 12 years
d 14 years.
5 Link borrows $48 000, taken out over 10 years and to be repaid in monthly instalments. (Note: As the interest rate increases, the monthly repayment increases if the
loan period is to remain the same.) Find the amount still owing after 5 years if interest
is debited monthly at a rate of:
a 6% p.a. and the repayment is $532.90
b 9% p.a. and the repayment is $608.04
c 12% p.a. and the repayment is $688.66
d 15% p.a. and the repayment is $774.41.
6 A loan of $20 000 has interest charged monthly at a rate of 9% p.a. What will be the
amount still owing after 3 years if the term of the loan is:
a 4 years and monthly repayments of $497.70 are made?
b 5 years and monthly repayments of $415.17 are made?
c 6 years and monthly repayments of $360.51 are made?
d 7 years and monthly repayments of $321.78 are made?
e 8 years and monthly repayments of $293 are made?
7 Pablo’s loan of 30 000 has interest charged quarterly at a rate of 10% p.a. What will
be the amount still owing after 5 years if the term of the loan is:
a 6 years and quarterly repayments of $1677.38 are made?
b 7 years and quarterly repayments of $1502.64 are made?
c 8 years and quarterly repayments of $1373.05 are made?
d 9 years and quarterly repayments of $1273.55 are made?
e 10 years and quarterly repayments of $1195.09 are made?
Chapter 15 FM Page 730 Monday, November 13, 2000 3:36 PM
730
Further Mathematics
8 multiple choice
Peter wants to borrow $8000 for a second-hand car and his bank offers him a personal
loan for that amount at an interest rate of 13% p.a., interest debited fortnightly, with
fortnightly repayments of $124.11 over 3 years. After 2 years he wants to calculate
how much he still owes by using the annuities formula.
a Which of the following equations should he use?
124.11 ( 1.005 78 – 1 )
A A = 8000 ( 1.005 ) 78 – -----------------------------------------------1.005 – 1
124.11 ( 1.05 52 – 1 )
B A = 8000 ( 1.05 ) 52 – --------------------------------------------1.05 – 1
124.11 ( 1.005 52 – 1 )
C A = 8000 ( 1.005 ) 52 – -----------------------------------------------1.005 – 1
124.11 ( 1.05 78 – 1 )
D A = 8000 ( 1.05 ) 78 – --------------------------------------------1.05 – 1
124.11 ( 0.005 52 – 1 )
E A = 8000 ( 0.005 ) 52 – -----------------------------------------------0.005 – 1
b The actual amount that Peter still owes after 2 years is closest to:
A $2500
B $3000
C $3500
D $4000
E $4500
Chapter 15 FM Page 731 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
731
9 multiple choice
Gwendoline has borrowed $14 000 for renovations to her house. The terms of this
loan are monthly instalments of $297.46 over 5 years with interest debited monthly at
10% p.a. of the outstanding balance.
a The amount still owing after 3 years is given by:
297.46 ( 1.008 333 3 36 – 1 )
A A = 14 000 ( 1.008 333 3 ) 36 – --------------------------------------------------------------1.008 333 3 – 1
297.46 ( 1.008 333 3 60 – 1 )
B A = 14 000 ( 1.008 333 3 ) 60 – --------------------------------------------------------------1.008 333 3 – 1
297.46 ( 1.1 60 – 1 )
C A = 14 000 ( 1.1 ) 60 – -----------------------------------------1.1 – 1
297.46 ( 0.083 333 36 – 1 )
D A = 14 000 ( 1.083 333 ) 36 – ----------------------------------------------------------0.083 333 – 1
297.46 ( 1.083 333 36 – 1 )
E A = 14 000 ( 0.083 333 ) 36 – ----------------------------------------------------------1.083 333 – 1
b The actual amount that Gwendoline still owes after 3 years is closest to:
A $5000
B $5500
C $6000
D $6500
E $7000
10 multiple choice
Ben took out a loan for $20 000 to buy a new car. The contract required that he repay
the loan over 5 years with monthly instalments of $421.02. After 2 1--2- years Ben used
the annuities formula to obtain the expression below to calculate the amount he still
owed.
421.02 ( 1.008 30 – 1 )
A = 20 000 ( 1.008 ) 30 – -----------------------------------------------0.008
The interest rate per annum charged by the bank for this reducing balance loan is:
A 1.008%
B 0.008%
C 0.096%
D 9.6%
E 12.096%
11 Use the annuities formula to find the repayment value, Q, given:
a P = $5000, r = 1, n = 12
5
b P = $3000, r = 2, n = 8
c P = $1500, r = 3, n = 4
d P = $9000, r = 0.5, n = 30
e P = $14 000, r = 0.8, n = 24
f P = $120 000, r = 0.6, n = 240
g P = $95 000, r = 2.5, n = 100
h P = $64 000, r = 0.5, n = 520.
WORKED
Example
12 Sergio’s reducing balance loan of $12 000 has interest charged at 9% p.a., interest
adjusted monthly. Find:
5
i the monthly repayment
ii the total interest charged
if the loan is fully repaid in:
a 2 years
b 3 years
c 4 years
d 4 1--2- years
e 5 1--2- years.
WORKED
Example
13 Conchita’s loan of $85 000 is charged interest at 7% p.a., interest adjusted monthly.
Find:
i the monthly repayment
ii the total interest charged
if the loan is fully repaid in:
a 10 years b 12 years c 15 years d 18 years e 20 years f 25 years.
Chapter 15 FM Page 732 Monday, November 13, 2000 3:36 PM
732
Further Mathematics
14 In each of questions 12 and 13 the only quantity which varied was the term of the
loan. As the term of the loan increases, what happens to:
a the repayment value?
b the amount of interest paid?
15 Declan borrows $32 000 and contracts to repay the loan over 10 years. Find:
i the repayment value
ii the total interest charged
if the loan is repaid quarterly at:
a 6% p.a., interest charged quarterly
b 8% p.a., interest charged quarterly
c 10% p.a., interest charged quarterly
d 10.5% p.a., interest charged quarterly
e 11% p.a., interest debited quarterly
f 12.5% p.a., interest debited quarterly.
16 Felice borrows $46 500 and contracts to repay the loan over 15 years. Find:
i the repayment value
ii the total interest charged
if the loan is repaid fortnightly, with interest adjusted fortnightly at:
a 6% p.a.
b 8% p.a.
c 10% p.a.
d 10.5% p.a.
e 11% p.a.
f 12.5% p.a.
17 A loan of $94 000 is to be repaid over 20 years. Find:
i the repayment value
ii the total interest charged
if the loan is repaid:
a weekly at 13% p.a., interest adjusted weekly
b fortnightly at 13% p.a., interest adjusted fortnightly
c monthly at 13% p.a., interest adjusted monthly
d quarterly at 13% p.a., interest adjusted quarterly
e weekly at 6.5% p.a., interest adjusted weekly
f fortnightly at 6.5% p.a., interest adjusted fortnightly.
18 Based on your answers to question 17 a–d when the frequency of repayments (and
interest charged) decreases, how does this affect:
a the repayment value?
b the total interest paid?
19 multiple choice
Which of the following would decrease the total amount of interest paid during the
life of a loan? (There may be more than one answer.)
A A fall in the interest rate
B A decrease in the frequency of repayment (repay less often)
C A greater amount borrowed
D A decrease in the term of the loan
E A rise in the interest rate
20 multiple choice
Which of the equations below would enable the quarterly repayment value, Q, to be
determined for a loan of $16 000 to be repaid over 5 years at 7.8% p.a., interest
debited quarterly?
Q ( 0.0195 20 – 1 )
A 0 = 16 000 ( 0.0195 ) 20 – --------------------------------------0.0195 – 1
Chapter 15 FM Page 733 Monday, November 13, 2000 3:36 PM
Chapter 15
Reducing balance loans
733
Q ( 1.078 5 – 1 )
B 0 = 16 000 ( 1.078 ) 5 – ---------------------------------1.078 – 1
Q
( 1.0195 5 – 1 )
C 0 = 16 000 ( 1.0195 ) 5 – ------------------------------------1.0195 – 1
Q
( 1.0195 20 – 1 )
D 0 = 16 000 ( 1.0195 ) 20 – --------------------------------------1.0195 – 1
20 – 1 )
Q
(
1.078
E 0 = 16 000 ( 1.078 ) 20 – -----------------------------------1.078 – 1
21 Grace has borrowed $18 000 to buy a car. She agrees to repay the reducing balance
loan over 5 years with monthly instalments at 8.1% p.a. (adjusted monthly). Find:
6
a the instalment value
b the principal repaid and the interest paid during:
i the 10th repayment
ii the 50th repayment.
WORKED
Example
22 Tim has borrowed $45 000 to buy a house. He agrees to repay the reducing balance
loan over 15 years with monthly instalments at 9.3% p.a. (adjusted monthly). Find:
a the instalment value
b the principal repaid and the interest paid during:
i the 20th repayment
ii the 150th repayment.
23 Gail has agreed to repay a $74 000 reducing balance loan with fortnightly instalments
over 20 years at 9.75% p.a. (adjusted fortnightly). Find:
a the instalment value
b the principal repaid and the interest paid during:
i the 1st repayment
ii the 500th repayment.
24 Terry is repaying a $52 000 loan over 15 years with quarterly instalments at 6.25% p.a.
(adjusted quarterly). Currently, 5 1--2- years have passed since the loan was drawn down
(money borrowed). How much does Terry still owe?
25 Stefanie borrowed $18 000 exactly 3 1--2- years ago. The reducing balance loan was for a
term of 5 years and was to be repaid in monthly instalments of 10.2% p.a. (adjusted
monthly). How much does Stefanie still owe?
Questions 26 and 27 refer to the following information. The interest charged to a housing
loan account during a financial year (1 July–30 June) is a tax deduction against income if
the house is rented to tenants.
26 Bruce borrowed $80 000 to finance the purchase of a rental property and he is repaying
the loan over 20 years by quarterly instalments at 8.6% p.a. (adjusted quarterly). By
1 July last year he had made 24 repayments. Find:
a the amount Bruce owed after 24 repayments
b the amount he owes by 30 June this year
c the total interest that Bruce can claim as a tax deduction for this particular
financial year.
ET
SHE
Work
27 Lyn is repaying a $55 000 housing loan over 15 years by monthly instalments at
9.9% p.a. (adjusted monthly). By 1 July last year she had made 42 repayments. If Lyn
rented the house to tenants, find:
a the amount she owed after 42 repayments
b the amount she owes by 30 June this year
c the total interest that Lyn can claim as a tax deduction for this particular financial
year.
15.1