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- 1. Finance Formulae
- 2. Finance Formulae Simple Interest I PRT I simple interest R interest rate as a decimal (or fraction) P principal T time periods
- 3. Finance Formulae Simple Interest I PRT I simple interest R interest rate as a decimal (or fraction) P principal T time periods e.g. If $3000 is invested for seven years at 6% p.a. simple interest, how much will it be worth after seven years?
- 4. Finance Formulae Simple Interest I PRT I simple interest R interest rate as a decimal (or fraction) P principal T time periods e.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. Finance Formulae Simple Interest I PRT I simple interest R interest rate as a decimal (or fraction) P principal T time periods e.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. Compound Interest An PR n P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time
- 7. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time
- 8. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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 7 A7 3000 1.06 A7 4510.89
- 10. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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 7 A7 3000 1.06 A7 4510.89 Investment is worth $4510.89 after 7 years
- 11. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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 7 A7 3000 1.06 A7 4510.89 Investment is worth $4510.89 after 7 years
- 12. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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 7 84 A7 3000 1.06 A84 3000 1.005 A7 4510.89 A84 4561.11 Investment is worth $4510.89 after 7 years
- 13. Compound Interest An PR n Note: general term of a geometric series P principal An amount after n time periods R 1 interest rate as a decimal(or fraction) n time periods Note: interest rate and time periods must match the compounding time e.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 7 84 A7 3000 1.06 A84 3000 1.005 A7 4510.89 A84 4561.11 Investment is worth Investment is worth $4510.89 after 7 years $4561.11 after 7 years
- 14. Depreciation An PR n P principal An amount after n time periods R 1 depreciation rate as a decimal(or fraction) n time periods Note: depreciation rate and time periods must match the depreciation time
- 15. Depreciation An PR n P principal An amount after n time periods R 1 depreciation rate as a decimal(or fraction) n time periods Note: 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. Depreciation An PR n P principal An amount after n time periods R 1 depreciation rate as a decimal(or fraction) n time periods Note: 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 A9 4509.87 9
- 17. Depreciation An PR n P principal An amount after n time periods R 1 depreciation rate as a decimal(or fraction) n time periods Note: 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 A9 4509.87 9 Machine is worth $4509.87 after 9 years
- 18. b) During which year will the value drop below 10% of the original cost?
- 19. b) During which year will the value drop below 10% of the original cost? An PR n n 15000 0.875 1500
- 20. b) During which year will the value drop below 10% of the original cost? An PR n n 15000 0.875 1500 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. b) During which year will the value drop below 10% of the original cost? An PR n n 15000 0.875 1500 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. Investing Money by Regular Instalments
- 23. Investing Money by Regular Instalments 2002 HSC Question 9b) (4) A superannuation fund pays an interest rate of 8.75% p.a. which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003. What will be the value of Stephanie’s superannuation when she retires on 31 December 2023?
- 24. Investing Money by Regular Instalments 2002 HSC Question 9b) (4) A superannuation fund pays an interest rate of 8.75% p.a. which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003. What will be the value of Stephanie’s superannuation when she retires on 31 December 2023? A21 5000 1.0875 21 amount invested for 21 years
- 25. Investing Money by Regular Instalments 2002 HSC Question 9b) (4) A superannuation fund pays an interest rate of 8.75% p.a. which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003. What will be the value of Stephanie’s superannuation when she retires on 31 December 2023? A21 5000 1.0875 21 A20 5000 1.0875 20 amount invested for 21 years amount invested for 20 years
- 26. Investing Money by Regular Instalments 2002 HSC Question 9b) (4) A superannuation fund pays an interest rate of 8.75% p.a. which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003. What will be the value of Stephanie’s superannuation when she retires on 31 December 2023? A21 5000 1.0875 21 A20 5000 1.0875 20 A19 5000 1.0875 19 amount invested for 21 years amount invested for 20 years amount invested for 19 years
- 27. Investing Money by Regular Instalments 2002 HSC Question 9b) (4) A superannuation fund pays an interest rate of 8.75% p.a. which compounds annually. Stephanie decides to invest $5000 in the fund at the beginning of each year, commencing on 1 January 2003. What will be the value of Stephanie’s superannuation when she retires on 31 December 2023? A21 5000 1.0875 21 A20 5000 1.0875 20 A19 5000 1.0875 19 1 A1 5000 1.0875 amount invested for 21 years amount invested for 20 years amount invested for 19 years amount invested for 1 year
- 28. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875
- 29. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875 a 5000 1.0875 , r 1.0875, n 21 S21
- 30. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875 a 5000 1.0875 , r 1.0875, n 21 S21 5000 1.0875 1.087521 1 0.0875 $299604.86
- 31. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875 a 5000 1.0875 , r 1.0875, n 21 S21 5000 1.0875 1.087521 1 0.0875 $299604.86 c*) Find the year when the fund first exceeds $200000.
- 32. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875 a 5000 1.0875 , r 1.0875, n 21 S21 5000 1.0875 1.087521 1 0.0875 $299604.86 c*) Find the year when the fund first exceeds $200000. Amount 5000 1.0875 5000 1.0875 2 5000 1.0875 n
- 33. Amount 5000 1.0875 5000 1.0875 21 20 5000 1.0875 a 5000 1.0875 , r 1.0875, n 21 S21 5000 1.0875 1.087521 1 0.0875 $299604.86 c*) Find the year when the fund first exceeds $200000. Amount 5000 1.0875 5000 1.0875 2 Sn i.e Sn 200000 5000 1.0875 n
- 34. 5000 1.0875 1.0875n 1 0.0875 200000
- 35. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87
- 36. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87 367 n 1.0875 87
- 37. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87 367 n 1.0875 87 367 n log 1.0875 log 87
- 38. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87 367 n 1.0875 87 367 n log 1.0875 log 87 367 n log 1.0875 log 87 367 log 87 n log 1.0875
- 39. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87 367 n 1.0875 87 367 n log 1.0875 log 87 367 n log 1.0875 log 87 367 log 87 n log 1.0875 n 17.16056585 n 18
- 40. 5000 1.0875 1.0875n 1 0.0875 200000 1.0875n 1 280 87 367 n 1.0875 87 367 n log 1.0875 log 87 367 n log 1.0875 log 87 367 log 87 n log 1.0875 n 17.16056585 n 18 Thus 2021 is the first year when the fund exceeds $200000
- 41. d*) What annual instalment would have produced $1000000 by 31st December 2020?
- 42. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 18 17 P 1.0875
- 43. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 P 1.0875 a P 1.0875 , r 1.0875, n 18 18 i.e. S18 1000000 17
- 44. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 P 1.0875 a P 1.0875 , r 1.0875, n 18 18 17 i.e. S18 1000000 P 1.0875 1.087518 1 0.0875 1000000
- 45. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 P 1.0875 a P 1.0875 , r 1.0875, n 18 18 17 i.e. S18 1000000 P 1.0875 1.087518 1 0.0875 1000000 1000000 0.0875 P 1.0875 1.087518 1
- 46. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 P 1.0875 a P 1.0875 , r 1.0875, n 18 18 17 i.e. S18 1000000 P 1.0875 1.087518 1 0.0875 1000000 1000000 0.0875 P 1.0875 1.087518 1 22818.16829
- 47. d*) What annual instalment would have produced $1000000 by 31st December 2020? Amount P 1.0875 P 1.0875 P 1.0875 a P 1.0875 , r 1.0875, n 18 18 17 i.e. S18 1000000 P 1.0875 1.087518 1 0.0875 1000000 1000000 0.0875 P 1.0875 1.087518 1 22818.16829 An annual instalment of $22818.17 will produce $1000000
- 48. Loan Repayments
- 49. Loan Repayments The amount still owing after n time periods is; An principal plus interest instalments plus interest
- 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. 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. 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. 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 71 1st repayment invested for 71 months 300 1.01 72 Repayments are an investment in your loan
- 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 71 1st repayment invested for 71 months 300 1.01 2nd repayment invested for 70 months 300 1.01 70 72 Repayments are an investment in your loan
- 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 71 1st repayment invested for 71 months 300 1.01 2nd repayment invested for 70 months 300 1.01 72 70 2nd last repayment invested for 1 month 300 1.01 300 last repayment invested for 0 months 1 Repayments are an investment in your loan
- 56. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 72 300 1.01 300 1.01 70 71
- 57. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 a r n 1 72 20000 1.01 r 1 70 71
- 58. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 a r n 1 72 20000 1.01 r 1 300 1.0172 1 72 20000 1.01 0.01 $9529.01 70 71
- 59. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 a r n 1 72 20000 1.01 r 1 300 1.0172 1 72 20000 1.01 0.01 $9529.01 b) How much interest will they have paid in six years? 70 71
- 60. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 a r n 1 72 20000 1.01 r 1 300 1.0172 1 72 20000 1.01 0.01 $9529.01 b) How much interest will they have paid in six years? Total repayments = 300 72 $21600 70 71
- 61. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 a r n 1 72 20000 1.01 r 1 300 1.0172 1 72 20000 1.01 0.01 $9529.01 b) How much interest will they have paid in six years? Total repayments = 300 72 $21600 Loan reduction = 20000 9529.01 $10470.99 70 71
- 62. An principal plus interest instalments plus interest A72 20000 1.01 300 300 1.01 300 1.01 300 1.01 a 300, r 1.01, n 72 72 70 71 a r n 1 72 20000 1.01 r 1 300 1.0172 1 72 20000 1.01 0.01 $9529.01 b) 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. (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. (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. (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 60 Initial loan is borrowed for 60 months 30000 1.0075
- 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 60 Initial loan is borrowed for 60 months 30000 1.0075 1st repayment invested for 59 months M 1.0075 59
- 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 60 Initial loan is borrowed for 60 months 30000 1.0075 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 last repayment invested for 0 months M 1
- 68. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 59
- 69. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 a r n 1 60 30000 1.0075 r 1 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 59
- 70. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 59
- 71. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 But A60 0 59
- 72. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 But A60 0 M 1.007560 1 60 30000 1.0075 0 0.0075 59
- 73. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 But A60 0 M 1.007560 1 60 30000 1.0075 0 0.0075 1.007560 1 60 M 30000 1.0075 0.0075 59
- 74. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 But A60 0 M 1.007560 1 60 30000 1.0075 0 0.0075 1.007560 1 60 M 30000 1.0075 0.0075 60 30000 1.0075 0.0075 M 1.007560 1 59
- 75. An principal plus interest instalments plus interest A60 30000 1.0075 M M 1.0075 60 M 1.0075 M 1.0075 58 a M , r 1.0075, n 60 a r n 1 60 30000 1.0075 r 1 M 1.007560 1 60 30000 1.0075 0.0075 But A60 0 M 1.007560 1 60 30000 1.0075 0 0.0075 1.007560 1 60 M 30000 1.0075 0.0075 60 30000 1.0075 0.0075 M 1.007560 1 M $622.75 59
- 76. (iii) Finding the length of the loan 2005 HSC Question 8c) Weelabarrabak Shire Council borrowed $3000000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. 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. (iii) Finding the length of the loan 2005 HSC Question 8c) Weelabarrabak Shire Council borrowed $3000000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. 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 Initial loan is borrowed for 2 years 3000000 1.12 2
- 78. (iii) Finding the length of the loan 2005 HSC Question 8c) Weelabarrabak Shire Council borrowed $3000000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. 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 Initial loan is borrowed for 2 years 3000000 1.12 1st repayment invested for 1 year 480000 1.12 480000 2nd repayment invested for 0 years 2 1
- 79. (iii) Finding the length of the loan 2005 HSC Question 8c) Weelabarrabak Shire Council borrowed $3000000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. 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 Initial loan is borrowed for 2 years 3000000 1.12 1st repayment invested for 1 year 480000 1.12 480000 2nd repayment invested for 0 years 2 1 An principal plus interest instalments plus interest A2 3000000 1.12 480000 480000 1.12 2
- 80. (iii) Finding the length of the loan 2005 HSC Question 8c) Weelabarrabak Shire Council borrowed $3000000 at the beginning of 2005. The annual interest rate is 12%. Each year, interest is calculated on the balance at the beginning of the year and added to the balance owing. 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 Initial loan is borrowed for 2 years 3000000 1.12 1st repayment invested for 1 year 480000 1.12 480000 2nd repayment invested for 0 years 2 1 An principal plus interest instalments plus interest A2 3000000 1.12 480000 480000 1.12 2 A2 3 10 6 1.12 4.8 10 1 1.12 2 5
- 81. (ii) Show that An 106 4 1.12 n
- 82. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 n
- 83. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 1st repayment invested for n – 1 years 480000 1.12 n n1
- 84. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 1st repayment invested for n – 1 years 480000 1.12 2nd repayment invested for n – 2 years 480000 1.12 n n1 n2
- 85. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 1st repayment invested for n – 1 years 480000 1.12 2nd repayment invested for n – 2 years 480000 1.12 2nd last repayment invested for 1 year 480000 1.12 last repayment invested for 0 years 480000 1 n n1 n2
- 86. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 1st repayment invested for n – 1 years 480000 1.12 2nd repayment invested for n – 2 years 480000 1.12 n n1 n2 2nd last repayment invested for 1 year 480000 1.12 last repayment invested for 0 years 480000 1 An principal plus interest instalments plus interest An 3000000 1.12 480000 1 1.12 n 1.12 n2 1.12 n1
- 87. (ii) Show that An 106 4 1.12 n Initial loan is borrowed for n years 3000000 1.12 1st repayment invested for n – 1 years 480000 1.12 2nd repayment invested for n – 2 years 480000 1.12 n n1 n2 2nd last repayment invested for 1 year 480000 1.12 last repayment invested for 0 years 480000 1 An principal plus interest instalments plus interest An 3000000 1.12 480000 1 1.12 1.12 1.12 a 480000, r 1.12, n n a r n 1 n 3000000 1.12 r 1 n n2 n1
- 88. 480000 1.12n 1 n An 3000000 1.12 0.12
- 89. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n
- 90. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n n
- 91. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 n n
- 92. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n n
- 93. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n n (iii) In which year will Weelabarrabak Shire Council make the final repayment?
- 94. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An 0
- 95. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n n (iii) In which year will Weelabarrabak Shire Council make the final repayment? An 0 10 4 1.12 6 n 0
- 96. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n 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. 480000 1.12n 1 n An 3000000 1.12 0.12 3000000 1.12 4000000 1.12n 1 n 3000000 1.12 4000000 1.12 4000000 n 4000000 1000000 1.12 106 4 1.12 n n 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. log 1.12 log 4 n
- 99. log 1.12 log 4 n log1.12 log 4 n
- 100. log 1.12 log 4 n log1.12 log 4 log 4 n log1.12 n n 12.2325075
- 101. log 1.12 log 4 n log1.12 log 4 log 4 n log1.12 n n 12.2325075 The thirteenth repayment is the final repayment which will occur at the end of 2017
- 102. log 1.12 log 4 n log1.12 log 4 log 4 n log1.12 n n 12.2325075 The thirteenth repayment is the final repayment which will occur at the end of 2017 Exercise 7B; 4, 6, 8, 10 Exercise 7C; 1, 4, 7, 9 Exercise 7D; 3, 4, 9, 11

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