Upcoming SlideShare
×

# 11 x1 t14 13 finance formulas (2012)

889 views

Published on

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
889
On SlideShare
0
From Embeds
0
Number of Embeds
507
Actions
Shares
0
15
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 11 x1 t14 13 finance formulas (2012)

1. 1. Finance Formulae
2. 2. Finance FormulaeSimple Interest I  PRT I  simple interest R  interest rate as a decimal (or fraction) P  principal T  time periods
3. 3. Finance FormulaeSimple Interest I  PRT I  simple interest R  interest rate as a decimal (or fraction) P  principal T  time periodse.g. If \$3000 is invested for seven years at 6% p.a. simple interest, how much will it be worth after seven years?
4. 4. Finance FormulaeSimple Interest I  PRT I  simple interest R  interest rate as a decimal (or fraction) P  principal T  time periodse.g. If \$3000 is invested for seven years at 6% p.a. simple interest, how much will it be worth after seven years? I  PRT I   3000  0.06  7   1260
5. 5. Finance FormulaeSimple Interest I  PRT I  simple interest R  interest rate as a decimal (or fraction) P  principal T  time periodse.g. If \$3000 is invested for seven years at 6% p.a. simple interest, how much will it be worth after seven years? I  PRT I   3000  0.06  7   1260  Investment is worth \$4260 after 7 years
6. 6. Compound Interest An  PR n An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periodsNote: interest rate and time periods must match the compounding time
7. 7. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periodsNote: interest rate and time periods must match the compounding time
8. 8. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually?
9. 9. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually? An  PR n A7  3000 1.06  7 A7  4510.89
10. 10. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually? An  PR n A7  3000 1.06  7 A7  4510.89  Investment is worth \$4510.89 after 7 years
11. 11. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually? b) compounded monthly? An  PR n A7  3000 1.06  7 A7  4510.89  Investment is worth \$4510.89 after 7 years
12. 12. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually? b) compounded monthly? An  PR n An  PR n A7  3000 1.06  A84  3000 1.005 7 84 A7  4510.89 A84  4561.11  Investment is worth \$4510.89 after 7 years
13. 13. Compound Interest Note: general term of a An  PR n geometric series An  amount after n time periods P  principal R  1  interest rate as a decimal(or fraction) n  time periods Note: interest rate and time periods must match the compounding timee.g. If \$3000 is invested for seven years at 6% p.a, how much will it be worth after seven years if;a) compounded annually? b) compounded monthly? An  PR n An  PR n A7  3000 1.06  A84  3000 1.005 7 84 A7  4510.89 A84  4561.11  Investment is worth  Investment is worth \$4510.89 after 7 years \$4561.11 after 7 years
14. 14. Depreciation An  PR n An  amount after n time periods P  principal R  1  depreciation rate as a decimal(or fraction) n  time periodsNote: depreciation rate and time periods must match the depreciation time
15. 15. Depreciation An  PR n An  amount after n time periods P  principal R  1  depreciation rate as a decimal(or fraction) n  time periodsNote: depreciation rate and time periods must match the depreciation time e.g. An espresso machine bought for \$15000 on 1st January 2001 depreciates at a rate of 12.5%p.a. a) What will its value be on 1st January 2010?
16. 16. Depreciation An  PR n An  amount after n time periods P  principal R  1  depreciation rate as a decimal(or fraction) n  time periodsNote: depreciation rate and time periods must match the depreciation time e.g. An espresso machine bought for \$15000 on 1st January 2001 depreciates at a rate of 12.5%p.a. a) What will its value be on 1st January 2010? An  PR n A9  15000  0.875 9 A9  4509.87
17. 17. Depreciation An  PR n An  amount after n time periods P  principal R  1  depreciation rate as a decimal(or fraction) n  time periodsNote: depreciation rate and time periods must match the depreciation time e.g. An espresso machine bought for \$15000 on 1st January 2001 depreciates at a rate of 12.5%p.a. a) What will its value be on 1st January 2010? An  PR n A9  15000  0.875 9 A9  4509.87  Machine is worth \$4509.87 after 9 years
18. 18. b) During which year will the value drop below 10% of the original cost?
19. 19. b) During which year will the value drop below 10% of the original cost? An  PR n 15000  0.875  1500 n
20. 20. b) During which year will the value drop below 10% of the original cost? An  PR n 15000  0.875  1500 n  0.875  0.1 n log  0.875  log 0.1 n n log 0.875  log 0.1 log 0.1 n log 0.875 n  17.24377353
21. 21. b) During which year will the value drop below 10% of the original cost? An  PR n 15000  0.875  1500 n  0.875  0.1 n log  0.875  log 0.1 n n log 0.875  log 0.1 log 0.1 n log 0.875 n  17.24377353  during the 18th year (i.e. 2018) its value will drop to 10% the original cost
22. 22. Investing Money by Regular Instalments
23. 23. Investing Money by Regular2002 HSC Question 9b) InstalmentsA superannuation fund pays an interest rate of 8.75% p.a. which (4)compounds annually. Stephanie decides to invest \$5000 in the fund at thebeginning of each year, commencing on 1 January 2003.What will be the value of Stephanie’s superannuation when she retires on31 December 2023?
24. 24. Investing Money by Regular2002 HSC Question 9b) InstalmentsA superannuation fund pays an interest rate of 8.75% p.a. which (4)compounds annually. Stephanie decides to invest \$5000 in the fund at thebeginning of each year, commencing on 1 January 2003.What will be the value of Stephanie’s superannuation when she retires on31 December 2023? A21  5000 1.0875 21 amount invested for 21 years
25. 25. Investing Money by Regular2002 HSC Question 9b) InstalmentsA superannuation fund pays an interest rate of 8.75% p.a. which (4)compounds annually. Stephanie decides to invest \$5000 in the fund at thebeginning of each year, commencing on 1 January 2003.What will be the value of Stephanie’s superannuation when she retires on31 December 2023? A21  5000 1.0875 21 amount invested for 21 years A20  5000 1.0875 20 amount invested for 20 years
26. 26. Investing Money by Regular2002 HSC Question 9b) InstalmentsA superannuation fund pays an interest rate of 8.75% p.a. which (4)compounds annually. Stephanie decides to invest \$5000 in the fund at thebeginning of each year, commencing on 1 January 2003.What will be the value of Stephanie’s superannuation when she retires on31 December 2023? A21  5000 1.0875 21 amount invested for 21 years A20  5000 1.0875 20 amount invested for 20 years A19  5000 1.0875 19 amount invested for 19 years
27. 27. Investing Money by Regular2002 HSC Question 9b) InstalmentsA superannuation fund pays an interest rate of 8.75% p.a. which (4)compounds annually. Stephanie decides to invest \$5000 in the fund at thebeginning of each year, commencing on 1 January 2003.What will be the value of Stephanie’s superannuation when she retires on31 December 2023? A21  5000 1.0875 21 amount invested for 21 years A20  5000 1.0875 20 amount invested for 20 years A19  5000 1.0875 19 amount invested for 19 years   A1  5000 1.0875 1 amount invested for 1 year
28. 28. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20
29. 29. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20 a  5000 1.0875 , r  1.0875, n  21  S21
30. 30. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20 a  5000 1.0875 , r  1.0875, n  21  S21 5000 1.0875  1.087521  1  0.0875  \$299604.86
31. 31. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20 a  5000 1.0875 , r  1.0875, n  21  S21 5000 1.0875  1.087521  1  0.0875  \$299604.86c*) Find the year when the fund first exceeds \$200000.
32. 32. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20 a  5000 1.0875 , r  1.0875, n  21  S21 5000 1.0875  1.087521  1  0.0875  \$299604.86c*) Find the year when the fund first exceeds \$200000. Amount  5000 1.0875  5000 1.0875   5000 1.0875 2 n
33. 33. Amount  5000 1.0875  5000 1.0875  5000 1.0875 21 20 a  5000 1.0875 , r  1.0875, n  21  S21 5000 1.0875  1.087521  1  0.0875  \$299604.86c*) Find the year when the fund first exceeds \$200000. Amount  5000 1.0875  5000 1.0875   5000 1.0875 2 n  Sn i.e Sn  200000
34. 34. 5000 1.0875  1.0875n  1  200000 0.0875
35. 35. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87
36. 36. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87 367 1.0875  n 87
37. 37. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87 367 1.0875  n 87  367  log 1.0875   log  n   87 
38. 38. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87 367 1.0875  n 87  367  log 1.0875   log  n   87   367  n log 1.0875  log    87  log  367    n  87  log 1.0875 
39. 39. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87 367 1.0875  n 87  367  log 1.0875   log  n   87   367  n log 1.0875  log    87  log  367    n  87  log 1.0875  n  17.16056585  n  18
40. 40. 5000 1.0875  1.0875n  1  200000 0.0875 1.0875n  1  280 87 367 1.0875  n 87  367  log 1.0875   log  n   87   367  n log 1.0875  log    87  log  367    n  87  log 1.0875  n  17.16056585  n  18Thus 2021 is the first year when the fund exceeds \$200000
41. 41. d*) What annual instalment would have produced \$1000000 by 31st December 2020?
42. 42. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17
43. 43. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17 a  P 1.0875 , r  1.0875, n  18 i.e. S18  1000000
44. 44. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17 a  P 1.0875 , r  1.0875, n  18 i.e. S18  1000000 P 1.0875  1.087518  1  1000000 0.0875
45. 45. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17 a  P 1.0875 , r  1.0875, n  18 i.e. S18  1000000 P 1.0875  1.087518  1  1000000 0.0875 P 1000000  0.0875 1.0875 1.087518  1
46. 46. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17 a  P 1.0875 , r  1.0875, n  18 i.e. S18  1000000 P 1.0875  1.087518  1  1000000 0.0875 P 1000000  0.0875 1.0875 1.087518  1  22818.16829
47. 47. d*) What annual instalment would have produced \$1000000 by 31st December 2020? Amount  P 1.0875  P 1.0875   P 1.0875 18 17 a  P 1.0875 , r  1.0875, n  18 i.e. S18  1000000 P 1.0875  1.087518  1  1000000 0.0875 P 1000000  0.0875 1.0875 1.087518  1  22818.16829 An annual instalment of \$22818.17 will produce \$1000000
48. 48. Loan Repayments
49. 49. Loan Repayments The amount still owing after n time periods is;An   principal plus interest    instalments plus interest 
50. 50. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years?
51. 51. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years? Initial loan is borrowed for 72 months  20000 1.01 72
52. 52. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years? Initial loan is borrowed for 72 months  20000 1.01 72 Repayments are an investment in your loan
53. 53. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years? Initial loan is borrowed for 72 months  20000 1.01 72 Repayments 1st repayment invested for 71 months  300 1.01 71 are an investment in your loan
54. 54. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years? Initial loan is borrowed for 72 months  20000 1.01 72 Repayments 1st repayment invested for 71 months  300 1.01 71 are an 2nd repayment invested for 70 months  300 1.01 70 investment in your loan
55. 55. Loan Repayments The amount still owing after n time periods is; An   principal plus interest    instalments plus interest e.g. (i) Richard and Kathy borrow \$20000 from the bank to go on an overseas holiday. Interest is charged at 12% p.a., compounded monthly. They start repaying the loan one month after taking it out, and their monthly instalments are \$300. a) How much will they still owe the bank at the end of six years? Initial loan is borrowed for 72 months  20000 1.01 72 Repayments 1st repayment invested for 71 months  300 1.01 71 are an 2nd repayment invested for 70 months  300 1.01 70 investment in your loan 2nd last repayment invested for 1 month  300 1.01 1 last repayment invested for 0 months  300
56. 56. An   principal plus interest    instalments plus interest  72 A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71 
57. 57. An   principal plus interest    instalments plus interest  72 A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1   
58. 58. An   principal plus interest    instalments plus interest  72 A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1     300 1.0172  1     20000 1.01   72    0.01    \$9529.01
59. 59. An   principal plus interest    instalments plus interest  72  A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1     300 1.0172  1     20000 1.01   72    0.01    \$9529.01b) How much interest will they have paid in six years?
60. 60. An   principal plus interest    instalments plus interest  72  A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1     300 1.0172  1     20000 1.01   72    0.01    \$9529.01b) How much interest will they have paid in six years? Total repayments = 300  72  \$21600
61. 61. An   principal plus interest    instalments plus interest  72  A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1     300 1.0172  1     20000 1.01   72    0.01    \$9529.01b) How much interest will they have paid in six years? Total repayments = 300  72  \$21600 Loan reduction = 20000  9529.01  \$10470.99
62. 62. An   principal plus interest    instalments plus interest  72  A72  20000 1.01  300  300 1.01  300 1.01  300 1.01 70 71  a  300, r  1.01, n  72  a  r n  1     20000 1.01   72   r 1     300 1.0172  1     20000 1.01   72    0.01    \$9529.01b) How much interest will they have paid in six years? Total repayments = 300  72  \$21600 Loan reduction = 20000  9529.01  Interest = 21600  10470.99  \$10470.99  \$11129.01
63. 63. (ii) Finding the amount of each instalment Yog borrows \$30000 to buy a car. He will repay the loan in five years, paying 60 equal monthly instalments, beginning one month after he takes out the loan. Interest is charged at 9% p.a. compounded monthly. Find how much the monthly instalment shold be.
64. 64. (ii) Finding the amount of each instalment Yog borrows \$30000 to buy a car. He will repay the loan in five years, paying 60 equal monthly instalments, beginning one month after he takes out the loan. Interest is charged at 9% p.a. compounded monthly. Find how much the monthly instalment shold be. Let the monthly instalment be \$M
65. 65. (ii) Finding the amount of each instalment Yog borrows \$30000 to buy a car. He will repay the loan in five years, paying 60 equal monthly instalments, beginning one month after he takes out the loan. Interest is charged at 9% p.a. compounded monthly. Find how much the monthly instalment shold be. Let the monthly instalment be \$M Initial loan is borrowed for 60 months  30000 1.0075 60
66. 66. (ii) Finding the amount of each instalment Yog borrows \$30000 to buy a car. He will repay the loan in five years, paying 60 equal monthly instalments, beginning one month after he takes out the loan. Interest is charged at 9% p.a. compounded monthly. Find how much the monthly instalment shold be. Let the monthly instalment be \$M Initial loan is borrowed for 60 months  30000 1.0075 60 1st repayment invested for 59 months  M 1.0075 59
67. 67. (ii) Finding the amount of each instalment Yog borrows \$30000 to buy a car. He will repay the loan in five years, paying 60 equal monthly instalments, beginning one month after he takes out the loan. Interest is charged at 9% p.a. compounded monthly. Find how much the monthly instalment shold be. Let the monthly instalment be \$M Initial loan is borrowed for 60 months  30000 1.0075 60 1st repayment invested for 59 months  M 1.0075 59 2nd repayment invested for 58 months  M 1.0075 58 2nd last repayment invested for 1 month  M 1.0075 1 last repayment invested for 0 months M
68. 68. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59 
69. 69. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1  
70. 70. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075  
71. 71. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075   But A60  0
72. 72. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075   But A60  0  M 1.007560  1     30000 1.0075    0 60   0.0075  
73. 73. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075   But A60  0  M 1.007560  1     30000 1.0075    0 60   0.0075    1.007560  1     30000 1.0075  60 M   0.0075   
74. 74. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075   But A60  0  M 1.007560  1     30000 1.0075    0 60   0.0075    1.007560  1     30000 1.0075  60 M   0.0075    30000 1.0075   0.0075  60 M 1.007560  1
75. 75. An   principal plus interest    instalments plus interest  60 A60  30000 1.0075  M  M 1.0075    M 1.0075   M 1.0075  58 59  a  M , r  1.0075, n  60  a  r n  1     30000 1.0075    60    r 1    M 1.007560  1     30000 1.0075    60    0.0075   But A60  0  M 1.007560  1     30000 1.0075    0 60   0.0075    1.007560  1     30000 1.0075  60 M   0.0075    30000 1.0075   0.0075  60 M  M  \$622.75 1.007560  1
76. 76. (iii) Finding the length of the loan2005 HSC Question 8c)Weelabarrabak Shire Council borrowed \$3000000 at the beginning of2005. The annual interest rate is 12%. Each year, interest is calculatedon the balance at the beginning of the year and added to the balanceowing. The debt is to be repaid by equal annual repayments of \$480000,with the first repayment being made at the end of 2005. Let An be the balance owing after the nth repayment. (i) Show that A2   3 106  1.12    4.8 105  1  1.12  2
77. 77. (iii) Finding the length of the loan2005 HSC Question 8c)Weelabarrabak Shire Council borrowed \$3000000 at the beginning of2005. The annual interest rate is 12%. Each year, interest is calculatedon the balance at the beginning of the year and added to the balanceowing. The debt is to be repaid by equal annual repayments of \$480000,with the first repayment being made at the end of 2005. Let An be the balance owing after the nth repayment. (i) Show that A2   3 106  1.12    4.8 105  1  1.12  2  3000000 1.12  2 Initial loan is borrowed for 2 years
78. 78. (iii) Finding the length of the loan2005 HSC Question 8c)Weelabarrabak Shire Council borrowed \$3000000 at the beginning of2005. The annual interest rate is 12%. Each year, interest is calculatedon the balance at the beginning of the year and added to the balanceowing. The debt is to be repaid by equal annual repayments of \$480000,with the first repayment being made at the end of 2005. Let An be the balance owing after the nth repayment. (i) Show that A2   3 106  1.12    4.8 105  1  1.12  2  3000000 1.12  2 Initial loan is borrowed for 2 years  480000 1.12  1 1st repayment invested for 1 year 2nd repayment invested for 0 years  480000
79. 79. (iii) Finding the length of the loan2005 HSC Question 8c)Weelabarrabak Shire Council borrowed \$3000000 at the beginning of2005. The annual interest rate is 12%. Each year, interest is calculatedon the balance at the beginning of the year and added to the balanceowing. The debt is to be repaid by equal annual repayments of \$480000,with the first repayment being made at the end of 2005. Let An be the balance owing after the nth repayment. (i) Show that A2   3 106  1.12    4.8 105  1  1.12  2  3000000 1.12  2 Initial loan is borrowed for 2 years  480000 1.12  1 1st repayment invested for 1 year 2nd repayment invested for 0 years  480000 An   principal plus interest    instalments plus interest  A2   3000000 1.12   480000  480000 1.12  2
80. 80. (iii) Finding the length of the loan2005 HSC Question 8c)Weelabarrabak Shire Council borrowed \$3000000 at the beginning of2005. The annual interest rate is 12%. Each year, interest is calculatedon the balance at the beginning of the year and added to the balanceowing. The debt is to be repaid by equal annual repayments of \$480000,with the first repayment being made at the end of 2005. Let An be the balance owing after the nth repayment. (i) Show that A2   3 106  1.12    4.8 105  1  1.12  2  3000000 1.12  2 Initial loan is borrowed for 2 years  480000 1.12  1 1st repayment invested for 1 year 2nd repayment invested for 0 years  480000 An   principal plus interest    instalments plus interest  A2   3000000 1.12   480000  480000 1.12  2 A2   3 10  1.12   4.8 10  1  1.12  6 2 5
81. 81. (ii) Show that An  106 4  1.12  n 
82. 82. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years
83. 83. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years  480000 1.12  n1 1st repayment invested for n – 1 years
84. 84. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years  480000 1.12  n1 1st repayment invested for n – 1 years  480000 1.12  n2 2nd repayment invested for n – 2 years
85. 85. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years  480000 1.12  n1 1st repayment invested for n – 1 years  480000 1.12  n2 2nd repayment invested for n – 2 years 2nd last repayment invested for 1 year  480000 1.12  1 last repayment invested for 0 years  480000
86. 86. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years  480000 1.12  n1 1st repayment invested for n – 1 years  480000 1.12  n2 2nd repayment invested for n – 2 years 2nd last repayment invested for 1 year  480000 1.12  1 last repayment invested for 0 years  480000 An   principal plus interest    instalments plus interest  n  An   3000000 1.12    480000  1  1.12   1.12  n2  1.12  n1 
87. 87. (ii) Show that An  106 4  1.12  n   3000000 1.12  n Initial loan is borrowed for n years  480000 1.12  n1 1st repayment invested for n – 1 years  480000 1.12  n2 2nd repayment invested for n – 2 years 2nd last repayment invested for 1 year  480000 1.12  1 last repayment invested for 0 years  480000 An   principal plus interest    instalments plus interest  n  An   3000000 1.12    480000  1  1.12   1.12   1.12  n2 n1  a  480000, r  1.12, n  n  a  r n  1     3000000 1.12    n    r 1  
88. 88.  480000 1.12n  1   An  3000000 1.12    n    0.12  
89. 89.  480000 1.12n  1   An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n
90. 90.  480000 1.12n  1   An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n
91. 91.  480000 1.12n  1   An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n
92. 92.  480000 1.12n  1   An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n 
93. 93.  480000 1.12n  1    An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n (iii) In which year will Weelabarrabak Shire Council make the final repayment?
94. 94.  480000 1.12n  1    An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An  0
95. 95.  480000 1.12n  1    An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An  0  10 4  1.12  6 n 0
96. 96.  480000 1.12n  1    An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An  0  10 4  1.12  6 n 0 4  1.12   0 n
97. 97.  480000 1.12n  1    An  3000000 1.12    n    0.12    3000000 1.12   4000000 1.12n  1 n  3000000 1.12   4000000 1.12   4000000 n n  4000000  1000000 1.12  n   106 4  1.12  n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An  0  10 4  1.12  6 n 0 4  1.12   0 n 1.12   4 n
98. 98. log 1.12   log 4 n
99. 99. log 1.12   log 4 n n log1.12  log 4
100. 100. log 1.12   log 4 n n log1.12  log 4 log 4 n log1.12 n  12.2325075
101. 101. log 1.12   log 4 n n log1.12  log 4 log 4 n log1.12 n  12.2325075The thirteenth repayment is the final repayment which will occur atthe end of 2017
102. 102. log 1.12   log 4 n n log1.12  log 4 log 4 n log1.12 n  12.2325075The thirteenth repayment is the final repayment which will occur atthe end of 2017 Exercise 7B; 4, 6, 8, 10 Exercise 7C; 1, 4, 7, 9 Exercise 7D; 3, 4, 9, 11