New Century Maths 12 Mathematics General 2 HSC course
Worked Solutions
Chapter 1: Loans and annuities
SkillCheck
1 a 18 months = 1.5 × 26 fortnights
= 39 fortnights
b 25 years = 25 × 12 months
= 300 months
c 20 years = 20 × 52 weeks
= 1040 weeks
d 6.5 years = 6.5 × 4 quarters
= 26 quarters
2 a 12% p.a. = 12  4% per quarter
= 3% per quarter
b 8% p.a. = 8  12% per month
= 0.6 % per month
3 a 1000 1  0.03  1125.51
4
b
4000
 3693.64
(1  0.008)10
4 P = $5000, r = 0.064, n = 8
I  Prn
 $5000  0.064  8
 $2560
5 a P = $2500, r = 0.05, n = 3
FV  P(1  r ) n
 $2500 1  0.05 
3
 $2894.06
b I = FV – P
 $2894.06  $2500
 $394.06
6 P = $3000, r = 0.0072, n = 12
FV  P(1  r ) n
 $3000 1  0.0072 
12
 $3269.71
I  FV  P
 $3269.71  $3000
 $269.71
Exercise 1-01: Flat rate loans
1 a P = $12 000, r = 0.1, n = 5
I  Prn
 $12 000  0.1  5
 $6000
Total amount to repay  $12 000  $6000
 $18 000
b Monthly repayment = $18 000 ÷ (5 × 12)
= $300
2 P = $3000, r = 0.08, n =
20
12
I  Prn
 $3000  0.08 
20
12
 $400
Total value of investment  $3000  $400
 $3400
The correct answer is B.
3 a P = $13 500, r = 0.15, n = 3
I  Prn
 $13 500  0.15  3
 $6075
Total cost  $13 500  $6075
 $19 575
b Fortnightly repayment = $19 575 ÷ (3 × 26)
 $250.96153...
 $250.97
c P = $13 500, r = 0.14, n = 3
I  Prn
 $13 500  0.14  3
 $5670
Amount saved = $6075  $5670
= $405
4 a P = $2900, r = 0.112, n = 2
I  Prn
 $2900  0.112  2
 $649.60
b Total cost = $2900 + $649.60
= $3549.60
Monthly repayment  $3549.60  (2  12)
 $147.90
c P = $2900, r = 0.107, n = 2
I  Prn
 $2900  0.107  2
 $620.60
Total amount saved = $649.60  $620.60
= $29
Amount saved per month = $29  24
= $1.21
5 P = $2300, r = 0.146, n = 1.5
I  Prn
 $2300  0.146 1.5
 $503.70
Total cost  $2300  $503.70
 $2803.70
Monthly repayment  $2803.70  39
 $71.89
The correct answer is C.
6 a Total amount borrowed = $2850 + $29 + $125 + $45
= $3049
b P = $3049, r = 0.1855, n = 1.5
I  Prn
 $3049  0.1855 1.5
 $848.38
Total to repay  $3049  $848.38
 $3897.38
Weekly repayment = $3897.38  (1.5 × 52)
= $49.97
c P = $2850, r = 0.1855, n = 1.5
I  Prn
 $2850  0.1855 1.5
 $793.01
Amount saved  $848.38  $793.01
 $55.37
7 Total amount borrowed = $14 500 + $37.50 + $115 × 3
= $14 882.50
P = $14 882.50, r = 0.135, n = 3
I  Prn
 $14 882.50  0.135  3
 $6027.41
Total to repay  $14 882.50  $6027.41
 $20 909.91
Monthly repayment = $20 909.91  (3 × 12)
= $580.83083...
≈ $580.84
8 Ali:
a P = $22 500, r = 0.094, n = 5
I  Prn
 $22 500  0.094  5
 $10 575
b Total to repay = $22 500 + $10 575
= $33 075
Monthly repayment = $33075  (5× 12)
= $551.25
Bryce: a P = $9000, r = 0.1075, n =
20
12
I  Prn
 $9000  0.1075 
20
12
 $1612.50
b Total to repay = $900 + $1612.50
= $10 612.50
Monthly repayment = $10 612.50  20
= $530.63
Chloe: a P = $24 000, r = 0.1049, n = 4.5
I  Prn
 $24 000  0.1049  4.5
 $11 329.20
b Total to repay = $24 000 + $11 329.20
= $35 329.20
Monthly repayment = $35 329.20  (4.5 × 12)
= $654.244 44...
≈ $654.25
9 a P = $28 400, r = 0.0925, n = 3
I  Prn
 $28 400  0.0925  3
 $7881
Total to repay  $28 400  $7881
 $36 281
Quarterly repayment = $36 281  (3 × 4)
= $3023.42
b P = $32 000, r = 0.085, n = 4
I  Prn
 $32 000  0.085  4
 $10 880
Total to repay  $32 000  $10 880
 $42 880
Quarterly repayment = $42 880  (4 × 26)
= $412.31
c P = $2000, r = 0.0725, n = 1.5
I  Prn
 $2000  0.0725  1.5
 $217.50
Total to repay  $2000  $217.50
 $2217.50
Quarterly repayment = $2217.50  18
= $123.194 44...
≈ $123.20
Exercise 1-02: Term payments
1 Total repaid =
10
 $11 000  $360  4  12
100
= $18 380
Interest paid  $18 380  $11 000
 $7380
The correct answer is C.
2 P = $7460, r = 0.072, n = 1.5
I  Prn
 $7460  0.072  1.5
 $805.68
Total amount paid = $7560 + $805.68
= $8365.68
The correct answer is A.
3 Plan A:
20
 $19 500  36  $595
100
 $25 320
Amount repaid 
Interest  $25 320  $19 500
 $5820
Monthly instalment = $595
Plan B: P = $19 500, r = 0.089, n = 4
I  Prn
 $19 500  0.089  4
 $6942
Total to repay = $19 500 + $6942
= $26 442
Monthly instalment = $26 442  (4 × 12)
= $550.88
a Plan B charges more interest.
b Plan B has the lower monthly instalment.
4 a Deposit =
10
 $26 990
100
= $2699
Amount outstanding  $26 990  $2699
 $24 291
P  $24 291, r  0.119, n  3
I  Prn
 $24 291  0.119  3
 $8671.89
Total to repay  $8671.89  $24 291
 $32 962.89
Cost of car  $2699  $32 962.89
 $35 661.89
b Cash price  $29 990
P  $29 990, r  0.087, n  2
I  Prn
 $29 990  0.087  2
 $5218.26
The amount saved by paying cash is the interest of $5218.26.
c Amount repaid = $500 + $92 × (5 × 52)
= $24 420
Interest  Amount repaid  Cost price
 $24 420  $15 790
 $8630
Principal borrowed = Price after deposit
= $15 790 - $500
= $15 290
P  $15 290, r  ?, n  5
I  Prn
$8630  $15 290  r  5
$8630  $76 450r
$8630
r
$76 450
r  0.1128...
The annual flat interest rate is 11.3%.
d Popularity of car; excess stock is cheaper; price reductions on old stock due to new
models being released.
5 a Total cost = $47 500 + $475 + $1203 – $5400
= $43 778
b Cash price = $47 500
Deposit = $6500
Trade in = $5400
Amount owing  $47 500  $6500  $5400
 $35 600
P  $35 600, r  0.158, n  5
I  Prn
 $35 600  0.158  5
 $28124
Total to repay  $35 600  $28124
 $63 724
Monthly repayment  $63 724   5  12 
 $1062.07
6 a i Deposit =
10
 $1600
100
= $160
Amount borrowed  $1600  $160
 $1440
P  $1440, r  0.002, n  365
I  Prn
 $1440  0.002  365
 $1051.20
Amount owing  $1440  $1051.20
 $2491.20
ii P = $1440, r = 0.002, n = 396
I  Prn
 $1440  0.002  396
 $1140.48
Amount owing  $1440  $1140.48
 $2580.48
b i P = $1350, r = 0.002, n = 90
I  Prn
 $1350  0.002  90
 $243
Amount owing  $1350  $243
 $1593
ii P = $1350, r = 0.002, n = 101
I  Prn
 $1350  0.002  101
 $272.70
Amount owing  $1350  $272.70
 $1622.70
c Local Loans:
Positive: 12 months to save the balance
Negative: interest is charged per day on the amount owing
Feelgood Finance:
Positive: 100 days to save the price of the computer
Negative: more interest is paid because no deposit is paid
7 a Cost = $150 + 24 × $75
= $1950
b Cost = $1600 + 24 × $3.95
= $1694.80
c Cost = $800 + 30 × $3.95 + $800 + $800 × 2% × 30 (2 years 6 months = 30 months)
= $2198.50
8 Deposit =
15
 $1360
100
= $204
Amount owing  $1360  $204
 $1156
P  $1156, r  0.168, n  2
I  Prn
 $1156  0.168  2
 $388.42
The amount saved by paying cash is the interest of $388.42.
9 Amount paid = $900 + 24 × $130.18
= $4024.32
Interest  $4024.32  $3299
 $725.32
P  $2399, r  ?, n  2
I  Prn
$725.32  $2399  r  2
$725.32  $4798r
$725.32
r
$4798
r  0.15117
The interest rate is 15.117%  15%
The correct answer is B.
10 a No interest, but you must have the money to pay.
b No cash is required upfront, but interest may be charged.
c There is time for saving, but high interest is charged from the day of purchase if not
repaid on time.
Exercise 1-03: Reducing balance loans
1 Amount paid off the principal after 4 months  $395 000  $391 902.55
= $3097.45
The correct answer is C.
2
Month
(n)
Principal
(P)
Interest
(I)
5
6
$391 902.55
$391 115.24
$2612.69
$391 115.24 ×
0.08
12
Amount owing before
repayment
(P + I)
$394 515.24
$391 115.24 + $2607.43
= $393 722.67
Balance
(P + I – R)
$391 115.24
$393 722.67 – $3400
= $390 322.67
$390 322.67 + $2602.15
= $392 924.82
$392 924.82 – $3400
= $389 524.82
$389 524.82 + $2596.83
= $392 121.65
$392 121.65 – $3400
= $388 721.65
$388 721.65 + $2591.48
= $391 313.13
$391 313.13 – $3400
= $387 913.13
= $2607.43
7
$390 322.67
$390 322.67 ×
0.08
12
= $2602.15
8
$389 524.82
$389 524.82 ×
0.08
12
= $2596.83
9
$388 721.65
$388 721.65 ×
0.08
12
= $2591.48
3 The correct answer is B.
Month
(n)
Principal
(P)
1
$370 000
Interest
(I)
$370 000 ×
0.064
12
= $1973.33
4 a Total to repay = $932.75 × (12 × 26)
= $291 018
b Interest payable = $291 018 – $180 000
= $111 018
c P = $180 000, r = ?, n = 12
Amount owing before
repayment
(P + I)
$370 000 + $1973.33
= $371 973.33
Balance
(P + I – R)
$371 973.33 – $2100
= $369 873.33
I  Prn
$111 018  $180 000  r  12
r
$111 018
$180 000  12
r  0.051
The equivalent flat rate of interest is 5.1% p.a.
d P = $180 000, r = 0.09, n = 12
I  Prn
 $180 000  0.09  12
 $194 400
5 a
Fortnights
(n)
Principal (P)
Interest
(I)
1
2
$450 000
$449 770.77
$1730.77
$449 770.77 ×
0.10
26
Amount owing before
repayment
(P + I)
$451 730.77
$449 770.77 + $1729.89
= $451 500.66
Balance
(P + I – R)
$449 770.77
$451 500.66 – $1960
= $449 540.66
$449 540.66 + $1729.00
= $451 269.66
$451 269.66 – $1960
= $449 309.66
$449 309.66 + $1728.11
= $451 037.77
$451 037.77 – $1960
= $449 077.77
$449 077.77 + $1727.22
= $450 804.99
$450 804.99 – $1960
= $448 844.99
$448 844.99 + $1726.33
= $450 571.32
$450 571.32 – $1960
= $448 611.32
= $1729.89
3
$449 540.66
$449 540.66 ×
0.10
26
= $1729.00
4
$449 309.66
$449 309.66 ×
0.10
26
= $1728.11
5
$449 077.77
$449 077.77 ×
0.10
26
= $1727.22
6
$448 844.99
$448 844.99 ×
0.10
26
= $1726.33
b Amount paid off the principal after 6 fortnights  $450 000  $448 611.32
= $1388.68
c Interest paid in the first 12 weeks  6  $1960  $1388.68
= $10 371.32
6 a Monthly interest rate = 0.139 ÷ 12
= 0.011 583
b I  $15 749.33  0.011 583
= $182.42
c
Month
1
2
P
$16 000
$15 749.33
I
$185.33
$182.42
3
$15 495.75
$15 495.75 × 0.011 583
= $179.49
P+I
$16 185.33
$15 749.33 + $182.42
= $15 931.75
$15 495.75 + $179.49
= $15 675.24
P+I –R
$15 749.33
$15 931.75 – $436
= $15 495.75
$15 675.24 – $436
= $15 239.24
d After 3 months, Goran is charged a loan administration fee of $30.
e Interest = $185.33 + $182.42 + $179.49
= $547.24
7 a Fortnightly interest rate = 0.125 ÷ 26
= 0.004 81
b
Fortnight
1
2
P
$17 800
$17 738.62
I
$17 800 × 0.004 81
= $85.62
$17 738.62 × 0.004 81
= $85.32
c Interest = $85.62 + $85.32
= $170.94
d Amount owing after 2 fortnights  $17 676.94
Exercise 1-04: Compound interest
1 PV = $27 000, r = 0.07, n = 16
FV  PV (1  r )n
 $27 000 1  0.07 
16
 $79 708.42
The correct answer is D.
2 PV = $86 200, r = 0.03, n = 28
FV  PV (1  r )n
 $86 200 1  0.03
28
The correct answer is A.
P+I
$17 800 + $85.62
= $17 885.62
$17 738.62 + $85.32
= $17 823.94
P+I–R
$17 885.62 – $147
= $17 738.62
$17 823.94 – $147
= $17 676.94
3 a i PV = $134 000, r = 0.065, n = 12
FV  PV (1  r )n
 $134 000 1  0.065 
12
 $285 298.90
ii I = FV – PV
 $285 298.90  $134 000
 $151 298.90
PV  $5000, r 
b i
0.09
, n  5  12  60
12
FV  PV (1  r ) n
 0.09 
 $5000 1 

12 

 $7828.41
60
ii I = FV – PV
 $7828.41  $5000
 $2828.41
PV  $14 000, r 
c i
0.08
 0.02, n  8  4  32
4
FV  PV (1  r )n
 $14 000 1  0.02 
32
 $26 383.57
ii I = FV – PV
 $26 383.57  $14 000
 $12 383.57
PV  $18 200, r 
d i
0.083
 0.0415, n  12  2  24
2
FV  PV (1  r )n
 $18 200 1  0.0415 
 $48 294.09
ii I = FV – PV
24
 $48 294.09  $18 200
 $30 094.09
PV  $34 300, r 
e i
0.078
, n  20  52  1040
52
FV  PV (1  r ) n
1040
0.078 

 $34 300 1 

52 

 $163 036.89
ii I = FV – PV
 $163 036.89  $34 300
 $128 736.89
4 a FV = $50 000, r = 0.075, n = 4
FV  PV (1  r ) n
$50 000  PV 1  0.075 
PV 
4
$50 000
1.075
4
 $37 440.03
b FV  $50 000, r 
0.092
, n  8  12  96
12
FV  PV (1  r ) n
0.092 

$50 000  PV 1 

12 

$50 000
PV 
96
0.092 

1 

12 

96
 $24 018.64
c FV  $50 000, r 
0.10
, n  2 4 8
4
FV  PV (1  r ) n
$50 000  PV 1  0.025 
8
PV 
$50 000
8
1.025
 $41 037.33
5 Option A:
PV  $5000, r  0.06, n  4
FV  PV (1  r ) n
 $5000 1  0.06 
4
 $6312.38
Option B:
P  $5000, r  0.062, n  4
I  Prn
 $5000  0.062  4
 $1240
Total value of investment  $5000  $1240
 $6240
Option C:
PV  $5000, r 
0.058
 0.0145, n  4  4  16
4
FV  PV (1  r ) n
 $5000 1  0.0145 
16
 $6295.10
Option D:
PV  $5000, r 
0.052
, n  4  12  48
12
FV  PV (1  r ) n
0.052 

 $5000 1 

12 

 $6153.30
48
The best interest rate is A.
PV  $250 000, r 
6
0.10
, n  13
12
FV  PV (1  r ) n
13
0.10 

 $250 000 1 

12 

 $278 479.75
7 FV = $31 737, r = 0.08, n = 6
FV  PV (1  r ) n
$31 737.48  PV 1  0.08 
PV 
6
$31 737.48
1.08
6
 $20 000
PV  $1000, r 
8
0.05
 0.025, n  20  2  40
2
FV  PV (1  r ) n
 $1000 1  0.025
40
 $2685.06
Yes, Michaela’s money had doubled (in fact more than doubled) in the 20 years.
Exercise 1-05: Credit cards
1 P = $1200, r = 0.0004, n = 30
A  P 1  r 
n
 $1200  1  0.0004 
30
 $1214.483 8327
I = $1214.483 8327 – $1200
= $214.483 8327
≈ $214.48
The correct answer is B.
2 Interest is paid from 21 December to 30 December.
30 – 21 = 9 days
P  $235, r 
0.225
,n9
365
Interest  Prn
 $235 
0.225
9
365
 $1.30
Total paid  $235  $1.30
 $236.30
The correct answer is C.
3
Item
Purchase
amount
$45.95
Swim trunks
$88.00
30 – 18 + 1 = 13
$275.00
30 – 25 + 1 = 6
Swim goggles
Wetsuit
Number of days’
interest
30 – 4 + 1 = 27
Interest
$45.95 × 0.000 453 × 27
= $0.562 01…
$88 × 0.000 453 × 13
= $0.518 232
$275 × 0.000 453 × 6
= $0.747 45
Total interest = $0.562 01… + $0.518 232 + $0.747 45
= $1.827 692…
= $1.83
Total to repay = $45.95 + $88.00 + $275.00 + $1.83
= $410.78
4 a Number of days of interest charged on the football tickets = 31  20 + 1 = 12 days
b
Item
Purchase
amount
$465.95
Coffee maker
Number of days’
interest
31 – 15 + 1 = 17
Interest
$465.95 ×
0.17
365
× 17
= $3.6893
Football tickets
$139.37
31 – 20 + 1 = 12
$139.37 ×
0.17
365
× 12
= $0.7789
Car tyres
$446.79
31 – 28 + 1 = 4
Totalinterest  $3.6893  $0.7789  $0.8324
 $5.30
$446.79 ×
= $0.8324
0.17
365
×4
Totaldue  $465.95  $139.37  $446.79  $5.30
 $1057.41
5 a Closing balance = $3256.43 – $3000.00 + $2195.48 + $3.84
= $2455.75
b $3000 payment on 15 June 2015
c The interest-free period starts on 1 June 2016 (the start of the statement period) and
ends on 25 July 2016 (the due date).
d i No interest is paid if the account is paid in full on 20 July, since this is in the
interest-free period.
ii 30 July is 5 days outside of the interest-free period.
Item
Purchase
Amount
$1346.58
Bike repairs
Number of days’
interest
30  18  30  42
Interest
A  $1346.58 1  0.0005
42
 $1375.1499...
I  $1375.1499...  $1346.58
 $28.5699...
Cycling outfit
$665.50
30  27  30  33
A  $665.50 1  0.0005 
33
 $676.5690...
I  $676.5690...  $665.50
 $11.0690...
Bicycle helmet
$183.40
30  27  30  33
A  $183.40 1  0.0005 
33
 $186.4504...
I  $186.4504...  $183.40
 $3.0504...
Total interest  $28.5699...  $11.0690...  $3.0504...
 $42.6893...
 $42.69
6 Interest of $886.50 is charged from 25 July to 28 August inclusive.
This is 7 + 28 = 35 days.
Interest  $886.50  0.000 575  35
 $17.84
Closing balance  $886.50  $17.84
 $904.34
5
 $904.34
100
 $45.22
5% of closing balance 
The minimum payment due is $50 since this is more than 5% of the closing balance.
The correct answer is D.
7 Number of days’ interest = 21 + 19
= 40
Interest  $2100 
0.2035
 40
365
 $46.83
Total paid  $2100  $46.83
 $2146.83
The correct answer is D.
Exercise 1-06: Future value of an annuity
1 a I  $400, P  $5000, r  ?, n  1
I  Prn
$400  $5000  r  1
$400  $5000  r
$400
r
$5000
r  0.08
The annual interest rate is 8%.
b P  $16 232, r  0.08, n  1
I  Prn
X  $16232  0.08  1
 $1298.56
c
Year
5
Balance at
start of year
(P)
$22 530.56
Interest
(I)
Contribution
(a)
Balance at end of year
(P + I + a)
$22 530.56 × 0.08
= $1802.44
$5000
$22 530.56 + $1802.44 + $5000
= $29 333.00
2 a
Interest rate per year = 10%
Number of years = 5
Future value of $1 per year for 5 years  $6.1051
Future value of $8000 per year for 5 years  $6.1051  8000
 $48 840.80
b
8%
 2%
4
9
Number of quarters   3
3
Future value of $1 per quarter for 3 quarters  $3.0604
Interest rate per quarter 
Future value of $600 per quarter for 3 quarters  $3.0604  600
 $1836.24
c
Interest rate per half year  6%  2  3%
Number of half years  5  2  10
Future value of $1 per year for 10 half years  $11.4639
Future value of $5400 per year for 10 half years  $11.4639  5400
 $61 905.06
d
Interest rate per month  12% 12  1%
Number of months  10
Future value of $1 per month for 10 months  $10.4622
Future value of $900 per month for 10 months  $10.4622  900
 $9415.98
3
Interest rate per half year  8%  2  4%
Number of half years  5  2  10
Future value of $1 per year for 10 half years  $12.0061
Future value of $5000 per year for 10 half years  $12.0061  5000
 $60 030.50
4
Interest rate per month  12% 12  1%
Number of months  12
Future value of $1 per month for 12 months  $12.6825
Future value of $660 per month for 12 months  $12.6825  660
 $8370.45
Tom does not have enough to buy the boat.
He is short of the purchase price by $9300  $8370.45  $929.55.
5 a FV  $10 000, Interest rate  8% p.a., Number of years  3
Future value of $1 per year for 3 years = $3.2464
$10000  $3.2464  a
$10000
$3.2464
a  3080.34
a
The contribution will be $3080.34 per year.
b FV  $10 000, Interest rate  8%  4  2% per quarter, Number of quarters  3  4  12
Future value of $1 per quarter for 12 quarters = $13.4121
$10000  $13.4121 a
$10000
$13.4121
a  745.60
a
The contribution will be $745.60 per quarter.
c FV  $26 500, Interest rate  10%  2  5% per half year,
Number of half years  6  2  12
Future value of $1 per half year for 12 half years = $15.9171
$26 500  $15.9171  a
$26 500
$15.9171
a  1664.88
a
The contribution will be $1664.88 per half year.
d FV  $9800, Interest rate  1% per month, Number of months  3
Future value of $1 per month for 3 months = $3.0301
$9800  $3.0301 a
$9800
$3.0301
a  3234.22
a
The contribution will be $3234.22 per month.
6
FV  $40 000, Interest rate  6%  2  3% per half year, Number of half years  4.5  2  9
Future value of $1 per half year for 9 half years = $10.1591
$40 000  $10.1591  a
$40 000
$10.1591
a  3937.36
a
Her contribution will be $3937.36 per half year.
7 FV  $400 000, Interest rate = 8% per year, Number of years  11
Future value of $1 per year for 11 years = $16.6455
$400 000  $16.6455  a
$400 000
$16.6455
a  24 030.52
a
Her contribution will be $24 030.52 per year.
8 FV  $50 000, Interest rate = 10% per year, Number of years  10
Future value of $1 per year for 10 years = $15.9374
$50 000  $15.9374  a
$50 000
$15.9374
a  3137.27
a
They should deposit $3137.27 into the account each birthday.
9 a PV  $7600, r  0.08  4  0.02, n  3  4  12
FV  PV (1  r ) n
 $7600 1  0.02 
12
 $9638.64
The final amount of her investment is $9639.
b
8%
 2%
4
Number of quarters  3  4  12
Interest rate per quarter 
Future value of $1 per quarter for 12 quarters  $13.4121
Future value of $760 per quarter for 12 quarters  $13.4121  760
 $10 193.20
She would have achieved a better financial result with the annuity.
10 a FV  $50 000, Interest rate  8%  4  2% per quarter, Number of quarters  2  4  8
Future value of $1 per quarter for 8 quarters = $8.5830
$50 000  $8.5830  a
$50 000
$8.5830
a  5825.47
a
The amount to be deposited will be $5825.47 per quarter.
b Amount deposited = 8 × $5825.47
= $46 603.76
Interest  $50 000  $46 606.76
 $3396.24
8%
 2%
4
Number of quarters  2  4  8
Interest rate per quarter 
c
Future value of $1 per quarter for 8 quarters  $8.5830
Future value of $4000 per quarter for 8 quarters  $8.5830  4000
 $34 332
11
FV  $6000, Interest rate  4%  4  1% per quarter, Number of quarters  1  4  4
Future value of $1 per quarter for 4 quarters = $4.0604
$6000  $4.0604  a
$6000
$4.0604
a  1477.69
a
The amount to be deposited will be $1477.69 per quarter.
12
16%
 4%
4
Number of quarters  3  4  12
Interest rate per quarter 
Future value of $1 per quarter for 8 quarters  $15.0258
Future value of $4000 per quarter for 8 quarters  $15.0258  4000
 $60 103.20
The deposit required is $60 000, so they will have saved just over this amount in 4 years.
Exercise 1-07: Present value of an annuity
1 a
Interest rate per year  4%
Number of years  10
Present value of $1 per year at 4% for 10 years  $8.1109
Present value of $760 per year at 4% for 10 years  $8.1109  760
 $6164.28
Interest rate per month  24%  12  2%
b
Number of months  8
Present value of $1 per month at 2% for 8 months  $7.3255
Present value of $400 per month at 2% for 8 months  $7.3255  400
 $2930.20
Interest rate per month  16%  2  8%
c
Number of half years  5.5  2  11
Present value of $1 per half year at 8% for 11 half years  $7.1390
Present value of $1600 per half year at 8% for 11 half years  $7.1390  1600
 $11 422.40
Interest rate per year  6%
d
Number of years  4
Present value of $1 per year at 6% for 4 years  $3.4651
Present value of $1140 per year at 6% for 4 years  $3.4651  1140
 $3950.21
Interest rate per half year  6%  2  3%
2
Number of half years  2  2  4
Present value of $1 per half year at 3% for 4 half years  $3.7171
Present value of $800 per half year at 3% for 4 half years  $3.7171  800
 $2973.68
$2973.68 invested now at the same rate will achieve the same result.
Interest rate per year  8%
3
Number of years  5
Present value of $1 per year at 8% for 5 years  $3.9927
Present value of $1000 per year at 8% for 5 years  $3.9927  1000
 $3992.70
 $3993
The correct answer is B.
4
Interest rate per quarter  16%  4  4%
Number of quarters  2  4  8
Present value of $1 per quarter at 4% for 8 quarters  $6.7327
Present value of $200 per quarter at 4% for 8 quarters  $6.7327  200
 $1346.54
$1346.54 invested now at the same rate will achieve the same result.
Interest rate per half year  6%  2  3%
5 a
Number of half years  5  2  10
Future value of $1 per half year at 3% for 10 half years  $11.4639
Future value of $500 per half year at 3% for 10 half years  $11.4639  500
 $5731.95
Lulu will have $5731.95 at the end of 5 years.
b FV  $5731.95, PV  ?, r 
0.06
 0.03, n  5  2  10
2
FV  PV (1  r ) n
$5731.95  PV 1  0.03
10

$5731.95
1  0.03
10
 $4265.11
Lulu could invest $4265.11 now to grow to the same amount after 5 years.
6 a PV  $7400, Interest rate  2% per month, Number of periods  9
Present value of $1 per month at 2% for 9 months  $8.1622
Let $a be the repayment that will give a present value of $7400.
$7400  $8.1622  a
$7400
$8.1622
 906.62
a
The monthly repayment will be $906.62
b PV  $38 000, Interest rate  4% per quarter, Number of periods  2  4  8
Present value of $1 per quarter at 4% for 8 quarters  $6.7327
Let $a be the repayment that will give a present value of $38 000.
$38 000  $6.7327  a
$38 000
$6.7327
 5644.10
a
The quarterly repayment will be $5644.10
c PV  $29 500, Interest rate 
16%
 8% per half year, Number of periods  3.5  2  7
2
Present value of $1 per half year at 8% for 7 half years  $5.2064
Let $a be the repayment that will give a present value of $38 000.
$29 500  $5.2064  a
$29 500
$5.2064
 5666.10
a
The half-yearly repayment will be $5666.10
7 a PV  $20 000, Interest rate  2% per quarter, Number of periods  3  4  12
Present value of $1 per quarter at 2% for 12 quarters  $10.5753
Let $a be the repayment that will give a present value of $20 000.
$20 000  $10.5753  a
$20 000
$10.5753
 1891.20
a
The quarterly repayment will be $1891.20
b Total of repayments = $1891.20 × 12
= $22 694.40
8 a PV  $16 000, Interest rate  3% per quarter, Number of periods  3  4  12
Present value of $1 per quarter at 3% for 12 quarters  $9.9540
Let $a be the repayment that will give a present value of $16 000.
$16 000  $9.9540  a
$16 000
$9.9540
 1607.39
a
The quarterly repayment will be $1607.39.
b Total of repayments = $1607.39 × 12
= $19 288.68
Interest  $19 288.68  $16 000
 $3288.68
9 a FV  $12 000, Interest rate  12%  4  3% per quarter, Number of quarters  3  4  12
Future value of $1 per quarter for 12 quarters = $14.1920
$12 000  $14.1920  a
$12 000
$14.1920
a  845.55
a
The amount to be deposited will be $845.55 per quarter.
b FV  $12 000, PV  ?, r 
0.12
 0.03, n  3  4  12
4
FV  PV (1  r ) n
$12 000  PV 1  0.03 
12

$12000
1  0.03
12
 $8416.5585...
 $8416.56
10 a PV  $25 000, Interest rate 
20%
 10% per half year, Number of periods  5  2  10
2
Present value of $1 per half year at 10% for 10 half years  $6.1446
Let $a be the repayment that will give a present value of $25 000.
$25 000  $6.1446  a
$25 000
$6.1446
 4068.61
a
The half-yearly repayment will be $4068.61
b Amount repaid = $4068.61 × 2 × 5
= $40 686.10
11
Interest rate per year = 10%
Number of years = 9
Present value of $1 per year at 10% for 9 years = $5.7590
Present value of $10 000 per year at 10% for 9 years  $5.7590  10 000
 $57 590
Peter’s lotto win was $57 590.
12 a Amount repaid = $225 × 10
= $2250
b
Interest rate per month = 12% ÷ 12 = 1%
Number of months = 10
Present value of $1 per month at 1% for 10 months = $9.4713
Present value of $225 per month at 1% for 9 months  $9.4713  225
 $2131.0425
 $2131.04
The amount borrowed was $2131.04
Exercise 1-08: Loan repayment tables
1 Term  15 years, Interest rate  9% p.a.
Monthly repayment for $1000 = $10.14
Monthly repayment for $450 000  $10.14  450
 $4563
The correct answer is B.
2 Term  30 years, Interest rate  6% p.a.
Monthly repayment for $1000 = $6.00
Monthly repayment for $500 000  $6.00  500
 $3000
Total amount repaid  $3000  30  12
 $1 080 000
Interest paid  $1 080 000  $500 000
 $580 000
The correct answer is A.
3 a
i Term  15 years, Interest rate  7% p.a.
Monthly repayment for $1000 = $8.99
Monthly repayment for $380 000  $8.99  380
 $3416.20
ii Total amount repaid = $3416.20 × 15 × 12
= $614 916
iii Interest paid = $614 916 – $380 000
= $234 916
b
i Term  25 years, Interest rate  9% p.a.
Monthly repayment for $1000 = $8.39
Monthly repayment for $925 000  $8.39  925
 $7760.75
ii Total amount repaid = $7760.75 × 25 × 12
= $2 328 225
iii Interest paid = $2 328 225 – $925 000
= $1 403 225
c
i Term  20 years, Interest rate  6% p.a.
Monthly repayment for $1000 = $7.16
Monthly repayment for $1 230 000  $7.16  1230
 $8806.80
ii Total amount repaid = $8806.80 × 20 × 12
= $2 113 632
iii Interest paid = $2 113 632 – $1 230 000
= $883 632
d
i Term  10 years, Interest rate  10% p.a.
Monthly repayment for $1000 = $13.22
Monthly repayment for $697 000  $13.22  697
 $9214.34
ii Total amount repaid = $9214.34 × 10 × 12
= $1 105 720.80
iii Interest paid = $1 105 720.80 – $697 000
= $408 720.80
4 a $20 000 at 8% p.a.
Monthly repayment = $242.66
Total amount repaid  $242.66  10  12
 $29 119.20
Interest paid  $29119.20  $20 000
 $9119.20
b $100 000 at 6% p.a.
Monthly repayment = $1110.21
Total amount repaid  $1110.21  10  12
 $133 225.20
Interest paid  $133 225.20  $100 000
 $33 225.20
c $5000 at 7% p.a.
Monthly repayment = $58.06
Total amount repaid  $58.06  10  12
 $6967.20
Interest paid  $6967.20  $5000
 $1967.20
d $50 000 at 10% p.a.
Monthly repayment = $660.76
Total amount repaid  $660.76  10  12
 $79 291.20
Interest paid  $79 291.20  $50 000
 $29 291.20
5 a $5000 at 6% p.a.
Monthly repayment = $55.52
Reduction in monthly payment = $58.06  $55.52 = $2.54
b $20 000 at 10% p.a.
Monthly repayment = $262.31
Increase in monthly payment = $262.31  $242.66 = $19.65
Increase in interest paid  $19.65  10  12
 $2358
6 a i $60 000 for 19 years
Monthly repayment = $531.27
Amount repaid per year  $531.27  12
 $6375.24
ii $80 000 for 17 years
Monthly repayment = $742.64
Amount repaid per year  $742.64  12
 $8911.68
b To find the monthly repayment for a loan of $180 000 for 16 years, either add the
monthly payments for $100 000 and $80 000 for 16 years or double the monthly
payment for $90 000 for 16 years.
c $300 000 for 15 years:
Monthly repayment = 3 × $984.74 = $2954.22
Total amount repaid  $2954.22  15  12
 $531 759.60
$300 000 for 20 years:
Monthly repayment = 3 × $867.83 = $2603.49
Total amount repaid  $2603.49  20  12
 $624 837.60
Interest saved  $624 837.60  $531 759.60
 $93 078
7 a $30 000 over 2 years with monthly repayments:
Monthly repayment = $1402
Total to repay  $1402  2  12
 $33 648
b $30 000 over 2 years with fortnightly repayments
Fortnightly repayment = $646
Total to repay  $646  2  26
 $33 592
Amount saved = $33 648  $33 592 = $56
c Jamal: $40 000 over 1 year with monthly repayments
Monthly repayment = $3540
Total to repay  $3540  12
 $42 480
Nadine: $40 000 over 2 years with monthly repayments
Monthly repayment = $2425
Total to repay  $2425  24
 $58 200
Difference  $58 200  $42 480
 $15 720
Exercise 1-09: Repaying a home loan
1 a $9000 borrowed for a second-hand car requires a car loan with an interest rate of
10.25% p.a.
b The establishment fee for a car loan is $295.
c The early termination fee is $750, since she borrowed less than $12 000.
2 a A $500 000 home loan has an interest rate of 6.8% p.a.
b The upfront charges are the ‘Registration of mortgage’, ‘Registration of transfer’ and
‘Loan establishment fee’.
Upfront charges  $102  $204  $600
 $906
c Late payment and service charges = 2 × $20 + 2 × $10
= $60
3 a $4700 for furniture requires a personal loan with an interest rate of 14.50% p.a.
b The only upfront charge is the loan approval fee of $150.
c She has to pay four loan service fees of $30 and one late payment fee of $20.
Total in charges  4  $30  $20
 $140
4 a The loan is paid off when the loan amount is $0. The time taken to pay off the loan is
read from where the graph touches the ‘Months’ axis. It takes approximately
288 months or 24 years.
b Draw a horizontal line at $250 000 on the vertical axis. Extend this line to intersect
with the graph and then draw a vertical line down to the horizontal axis. It takes
approximately 135 months, or 11 years and 3 months, to reduce the loan to $250 000.
c Each unit on the ‘Months’ axis represents 5 months. Twenty years is
20  12  240 months . Locate 240 on the horizontal axis. Draw a vertical line that
extends up to the graph. Then draw a horizontal line across to the vertical axis. The
amount owing after 20 years is $100 000.
d Total repaid = $2490 × 288
= $717 120
e Interest = $717 120 – $350 000
= $367 120
f
The loan is paid off when the loan amount is $0. The time taken to pay off the loan is
read from where the graph touches the ‘Weeks’ axis. It takes approximately
1200 weeks = 1200 ÷ 52.18 ≈ 23 years and 3 months.
g Draw a horizontal line at $300 000 on the vertical axis. Extend this line to intersect
with the graph and then draw a vertical line down to the horizontal axis. It takes
approximately 76 months, or 6 years and 4 months, to reduce the loan to $300 000.
h Each unit on the ‘Weeks’ axis represents 20 weeks. Fifteen years is
15  52  780 weeks . Locate 780 on the horizontal axis. Draw a vertical line that
extends up to the graph. Then draw a horizontal line across to the vertical axis. The
amount owing after 15 years is $190 000.
i Total repaid = $580 × 1200
= $696 000
j Interest = $696 000 – $350 000
= $346 000
k Time saving is 9 months. The loan is repaid in 23 years 3 months instead of 24 years.
l Interest saving = $367 120 – $346 000
= $21 120
5 a i Locate 5 on the horizontal axis. Draw a vertical line that extends up to the graph.
Then draw a horizontal line across to the vertical axis. The amount owing after
5 years is $370 000.
ii Locate 15 on the horizontal axis. Draw a vertical line that extends up to the graph.
Then draw a horizontal line across to the vertical axis. The amount owing after
15 years is $250 000.
b i Draw a horizontal line at $300 000 on the vertical axis. Extend this line to intersect
with the graph and then draw a vertical line down to the horizontal axis. In 12 years
the amount owing is $300 000.
ii Half of $400 000 is $200 000. Draw a horizontal line at $200 000 on the vertical
axis. Extend this line to intersect with the graph and then draw a vertical line down
to the horizontal axis. In 18 years the amount owing is $200 000.
6 a It takes 30 years to pay off the loan with monthly payments but only 24 years to repay
the loan with fortnightly payments. The loan can be repaid 6 years sooner by making
fortnightly repayments.
b i Draw a horizontal line at $100 000 on the vertical axis. Extend this line to intersect
with the blue graph and then draw a vertical line down to the horizontal axis. In
22.5 years the amount owing is $100 000.
ii Draw a horizontal line at $100 000 on the vertical axis. Extend this line to intersect
with the red graph and then draw a vertical line down to the horizontal axis. In
17 years the amount owing is $100 000.
c i Locate 10 on the horizontal axis. Draw a vertical line that extends up to the red
graph. Then draw a horizontal line across to the vertical axis. The amount owing
after 10 years with fortnightly repayments is $160 000.
ii Locate 10 on the horizontal axis. Draw a vertical line that extends up to the red
graph. Then draw a horizontal line across to the vertical axis. The amount owing
after 20 years with monthly repayments is $112 000.
7 a The vertical axis intercept is $250 000, so the Hayes family borrowed $250 000.
b Locate 150 on the horizontal axis. Draw a vertical line that extends up to the graph.
Then draw a horizontal line across to the vertical axis. The amount owing after
150 months is $195 000.
c The loan is paid off when the loan amount is $0. The time taken to pay off the loan is
read from where the blue graph touches the ‘No. of repayments’ axis. It takes
approximately 352 months, or 29 years 4 months.
d Total repaid = $1497 × 352
= $526 944
e Interest = $526 944 – $250 000
= $276 994
f The loan is paid off when the loan amount is $0. The time taken to pay off the loan with
the extra repayment is read from where the red graph touches the ‘No. of repayments’
axis. It takes approximately 345 months, or 28 years 9 months.
g Total repaid = $1497 × 345 + $10 000
= $526 465
h Interest = $526 465 – $250 000
= $266 465
i The time saved is the difference between the original 29 years 4 months and the
28 years 9 months, which is 7 months.
j Interest = $276 994 – $266 465
= $10 479
Sample HSC problem
a $320 000 at 5.5% p.a. for 10 years.
Monthly repayment for $1000 = $10.85
Monthly repayment for $320000 = $10.85 × 320 = $3472
b Amount repaid = $3472 × 10 × 12
= $416 640
c Interest = $416 640 – $320 000
= $96 640
d P  $320 000, r  ?, n  10, I  $96 640
I  Prn
$96 640  $320 000  r  10
$96 640  $3 200 000r
r
$96 640
$3 200 000
r  0.0302
The equivalent flat rate of interest is 3.02% p.a.
Revision
1 a Deposit 
10
 $15 000
100
= $1500
Remainder  $15 000  $1500
 $13 500
Total amount borrowed  $13 500  $100  $35  $250  4  $80
 $14 205
b P  $14 205, r  0.12, n  4
I  Prn
 $14 205  0.12  4
 $6818.40
c Total to repay = $14 205 + $6818.40
= $21 023.40
Monthly instalment 
$21 023.40
4  12
 $437.99
2 P  $1900, r  0.127, n  2
I  Prn
 $1900  0.127  2
 $482.60
The correct answer is C.
3 a Deposit 
10
 $8000
100
= $800
Total paid  $800  $238  $1.5  26
 $10 082
b Interest = $10 082 – $8000
= $2082
c P  $7200, r  ?, n  1.5, I  $2082
I  Prn
$2082  $7200  r  1.5
$2082  $10 800r
r
$2082
$10 800
r  0.1927
The equivalent flat rate of interest is 19.3%.
4 $10 000 with 10% deposit and 14.6% p.a. on the balance over 4 years.
10
 $10 000
100
 $1000
Deposit 
Amount owing  $10 000  $1000
 $9000
P  $9000, r  0.146, n  4
I  Prn
 $9000  0.146  4
 $5256
Total to repay  $9000  $5256
 $14 256
Monthly repayment 
$14 256
4  12
 $297
5 a
n
P
I
P+I
P+I–R
1
$18 000
$18 000 ×
0.09
12
$18 000 + $135
= $18 135
$18 135 – $420
= $17 715
$17 715 + $132.86
= $17 847.86
$17 847.86 – $420
= $17 427.86
$17 427.86 – $130.71
= $17 558.57
$17 558.57 – $420
= $17 138.57
$17 138.57 – $128.54
= $17 267.11
$17 267.11 – $420
= $16 847.11
= $135
2
$17 715
$17 715 ×
0.09
12
= $132.86
3
$17 427.86
$17 427.86 ×
0.09
12
= $130.71
4
$17 138.57
$17 138.57 ×
0.09
12
= $128.54
b Amount paid off principal = $18 000 – $16 847.11
= $1152.89
c Interest = 4 × $420 – $1152.89
= $527.11
d P  $18 000, r 
0.09
,n4
12
I  Prn
 $18 000 
0.09
4
12
 $540
Interest saved  $540  $527.11
 $12.89
6 PV  $9500, r 
0.0475
,n6
12
FV  PV (1  r ) n
0.0475 

 $9500 1 

12 

 $9727.87
6
Interest  $9727.87  $9500
 $227.87
The correct answer is B.
7 FV  $10 000, PV  ?, r 
0.08
, n  8  12  96
12
FV  PV (1  r ) n
0.08 

$10 000  PV 1 

12 

$10 000
PV 
96
0.08 

1 

12 

 $5284.14
96
8
Item
Golf lesson
Golf shoes
Massage
Purchase
amount
$325.50
$295.00
$137.00
Number of days’
interest
30 – 15 + 5 + 1 = 21
30 – 20 + 5 + 1 = 16
30 – 28 + 5 + 1 = 8
Interest
$325.50 × 0.000 449 × 21 = $3.07
$295 × 0.000 449 × 16 = $2.12
$137 × 0.000 449 × 8 = $0.49
Total interest  $3.07  $2.12  $0.49
 $5.68
9 a
Interest rate per year = 8%
Number of years  12
Future value of $1 per year for 12 years  $18.9771
Future value of $5000 per year for 12 years  $18.9771  5000
 $94 885.50
b Interest = $94 885.50 – 12 × $5000
= $34 885.50
10
12%
 3%
4
Number of quarters  1  4  4
Interest rate per quarter 
Future value of $1 per quarter for 4 quarters  $4.1836
Future value of 16 000 per quarter for 4 quarters  $4.1836  16 000
 $66 937.60
11
Interest rate per month  12%  12  1%
Number of months  9
Present value of $1 per month at 1% for 9 months  $8.5660
Present value of $2000 per month at 2% for 8 months  $8.5660  2000
 $17 132
She could invest $17 132 now to produce the same financial result.
12 a PV  $13 500, Interest rate  2% per month
Number of months = 6
Present value of $1 per month for 6 months = $5.6014
$13 500  $5.6014  a
$13 500
$5.6014
 2410.11
a
The monthly repayment is $2410.11
b PV  $45 000, Interest rate  6% per quarter
Number of quarters = 2.5 × 4 = 10
Present value of $1 per quarter for 10 quarters = $7.3601
$45 000  $7.3601  a
$45 000
$7.3601
 6114.05
a
The quarterly repayment is $6114.05
13 a 4 years = 4 × 12 = 48 months
For $20 000 the monthly repayment is $547.
b Total repaid = $547 × 48
= $26 256
c Interest paid = $26 256 – $20 000
= $6256
14 a The vertical axis intercept is $400 000, so Ante borrowed $400 000.
b 10 years = 10 × 12 months
= 120 months
Locate 120 on the horizontal axis. Draw a vertical line that extends up to the graph,
then draw a horizontal line across to the vertical axis. The amount owing after 10 years
is $290 000.
c The loan is paid off when the loan amount is $0. The time taken to pay off the loan is
read from where the graph touches the horizontal axis. It takes approximately
260 months, or 21 years 8 months to repay the loan.
d Half of the amount borrowed = $400 000 ÷ 2
= $200 000
Draw a horizontal line at $200 000 on the vertical axis. Extend this line to intersect
with the graph and then draw a vertical line down to the horizontal axis. It takes
approximately 174 months for the loan to be half paid.
© Cengage Learning Australia Pty Ltd 2013 MAT12FMWS10021
Financial Mathematics: Credit and
borrowing; Annuities and loan repayments www.nelsonnet.com.au