Chapter 7 Review - Maths Past Paper PDF

## Chapter 7 Review - Modelling growth and decay using recursion ### Multiple-choice questions 1. Consider the following recurrence relation: $V_0 = 5$, $V_{n+1} = V_n - 3$ The sequence generated by this recurrence relation is: - A. 5, 15, 45, 135, 405, ... - B. 5, 8, 11, 14, 17, ... - C. 5, 2, -...

## Chapter 7 Review - Modelling growth and decay using recursion ### Multiple-choice questions 1. Consider the following recurrence relation: $V_0 = 5$, $V_{n+1} = V_n - 3$ The sequence generated by this recurrence relation is: - A. 5, 15, 45, 135, 405, ... - B. 5, 8, 11, 14, 17, ... - C. 5, 2, -1, -4, -7, ... - D. 5, 15, 45, 135, 405, ... - E. 5, -15, 45, 135, 405, ... 2. Consider the following recurrence relation: $V_0 = 2$, $V_{n+1} = 2V_n + 8$ The value of the term $V_4$ in the sequence generated by this recurrence relation is: - A. 12 - B. 18 - C. 32 - D. 72 - E. 152 3. Consider the following recurrence relation: $V_0 = 5$, $V_{n+1} = 3V_n - 6$ The value of the term $V_3$ in the sequence of numbers generated by this recurrence relation is: - A. 5 - B. 9 - C. 21 - D. 57 - E. 165 4. Brian has two trees in his backyard. Every month, he will plant three more trees. A recurrence relation for the number of trees, $T_n$, in Brian's backyard after $n$ months is: - A. $T_0 = 2$, $T_{n+1} = 3T_n$ - B. $T_0 = 2$, $T_{n+1} = 3T_n + 3$ - C. $T_0 = 2$, $T_{n+1} = T_n + 3$ - D. $T_0 = 2$, $T_{n+1} = T_n - 3$ - E. $T_0 = 2$, $T_{n+1} = 3T_n - 3$ 5. A graph that shows the value of a simple interest investment of $1000, earning interest of $5 per month is: **(The answer is a graph. This bot cannot generate graphs. The graph will be linear showing the growth of the $1000 investment over time.)** 6. The recurrence relation that generates a sequence of numbers representing the value of a car $n$ years after it was purchased is: $V_0 = 18000$, $V_{n+1} = V_n - 1098$. The car had a purchase price of $18000 and is being depreciated using: - A. flat rate depreciation at 6.1% of its value per annum - B. flat rate depreciation at $6.10 per kilometre travelled - C. flat rate depreciation at $1098 per kilometre travelled - D. unit cost depreciation at $6.10 per kilometre travelled - E. unit cost depreciation at $1098 per kilometre travelled 7. A computer is depreciated using a flat rate depreciation method. It was purchased for $2800 and depreciates at the rate of 8% per annum. The amount of depreciation after 4 years is: - A. $224 - B. $448 - C. $794 - D. $896 - E. $1904 8. Sandra invests $6000 in an account that pays interest at the rate of 4.57% per annum, compounding annually. The number of years it takes for the investment to exceed $8000 is: - A. 5 - B. 6 - C. 7 - D. 8 - E. 9 9. The value of a machine is depreciating by 8% every year. The initial value is 2700. A recurrence relation model for the value of the machine after $n$ years, $P_n$, is: - A. $P_0 = 2700$, $P_{n+1} = 1.8 \times P_n $ - B. $P_0 = 2700$, $P_{n+1} = 1.08 \times P_n $ - C. $P_0 = 2700$, $P_{n+1} = 0.92 \times P_n $ - D. $P_0 = 2700$, $P_{n+1} = 1 + 8 \times P_n $ - E. $P_0 = 2700$, $P_{n+1} = 1.08 + P_n $ 10. An investment of $50 000 is compounding annually over a number of years. The graph that best represents the value of the investment at the end of each year is: **(The answer is a graph. This bot cannot generate graphs. The graph will show a curved line depicting the compounded growth of the investment over time.)** ### Written Response Questions 1. Jack borrows $20000 from a bank and is charged simple interest at the rate of 9.4% per annum. Let $V_n$ be the value of the loan after $n$ years. - a. Write down a recurrence relation for the value of Jack's loan after $n$ years. $V_0 = 20000$, $V_{n+1} = V_n + 1880$ - b. Use the recurrence relation to model how much Jack will need to pay the bank after 5 years. $V_5 = 20000 + 1880\times5 = 29400$ The bank decides to change the loan to a compound interest loan on a yearly basis, with an annual interest rate of 9.4%. Let $W_n$ be the value of the loan after $n$ years. - c. Write a recurrence relation to model the value of Jack's loan. $W_0 = 20000$, $W_{n+1} = 1.094W_n$ - d. Write a rule for $W_n$ in terms of n. $W_n = 20000 \times (1.094)^n$ - e. Use the rule to find the value of the loan after 5 years. Round your answer to the nearest cent. $W_5 = 20000 \times (1.094)^5 = 31341.27$ 2. Ilana uses a personal loan to buy a dress costing $300. Interest is charged at 18% per annum, compounding monthly. If she repays the loan fully after 6 months, how much will she pay? Round your answer to the nearest cent. $300 \times (1 + 0.18/12)^6 = 328.03$ 3. Kelly bought her current car 5 years ago for $22 500. Let $V_n$ be the value of Kelly's car after $n$ years. - a. If Kelly uses a flat rate depreciation of 12% per annum: - i. Write down a recurrence relation for the value of Kelly's car after $n$ years. $V_0 = 22500$, $V_{n+1} = V_n - 2700$ - ii. Use the recurrence relation to find the current value of Kelly's car. $V_5 = 22500 - 2700\times 5 = 7500$ - b. If Kelly uses reducing value depreciation at 16% per annum: - i. Write down a recurrence relation for the value of Kelly's car after $n$ years. $V_0 = 22500$, $V_{n+1} = 0.84V_n$ - ii. Use the recurrence relation to find the current value of Kelly's car using reducing balance depreciation. Round your answer to the nearest cent. $V_5 = 22500 \times (0.84)^5 = 9409.61$ 4. A commercial cleaner bought a new vacuum cleaner for $650. The value of the vacuum cleaner decreases by $10 for every 50 offices that it cleans. - a. How much does the value of the vacuum cleaner depreciate when one office is cleaned? $10/50 = 0.20$ dollars per office - b. Give a recurrence relation for the value of the vacuum cleaner, $V_n$, after $n$ offices have been cleaned. $V_0 = 650$, $V_{n+1} = V_n - 0.2$ - c. The cleaner has a contract to clean 10 offices, 5 nights a week for 40 weeks in a year. What is the value of the vacuum cleaner after 1 year? $10 \times 5 \times 40 = 2000$ offices cleaned $650 - 0.2 \times 2000 = 250$ dollars 5. Meghan has $5000 to invest. - a. Company A offers her an account paying 6.3% per annum simple interest. How much will she have in the account offered by company A at the end of 5 years? $5000 + 0.063 \times 5000 \times 5 = 56575$ - b. Company B offers her an account paying 6.1% per annum compound interest. How much will she have in the account offered by company B at the end of 5 years? Round your answer the nearest cent. $5000 \times (1 + 0.061)^5 = 6722.75$ - c. Find the simple interest rate that company A should offer if the two investments are to have equal values after 5 years. Round your answer to one decimal place. $5000 \times (1+r/100)^5 = 6722.75$ $r = 7.2%$ 6. A sum of $30 000 is borrowed at an interest rate of 9% per annum, compounding monthly. Let $V_n$ be the value of the loan after $n$ months. - a. Write a recurrence relation to model the value of this loan. $V_0 = 30000$, $V_{n+1} = 1.0075V_n$ - b. Use the recurrence relation to find the value of the loan at the end of the first five months. Round your answer to the nearest cent. $V_5 = 30000 \times (1.0075)^5 = 31144.83$ - c. What is the value of the loan after 1 year? Round your answer to the nearest cent. $V_{12} = 30000 \times (1.0075)^{12} = 33418.81$ - d. If the loan is fully repaid after 18 months, how much money is paid? Round your answer to the nearest cent. $V_{18} = 30000 \times (1.0075)^{18} = 36186.55$ 7. On the birth of his granddaughter, a man invests a sum of money at a rate of 11.65% per annum, compounding twice per year. On her 21st birthday he gives all of the money in the account to his granddaughter. If she receives $2529.14, how much did her grandfather initially invest? Round your answer to the nearest cent. $2529.14 = (1 + 0.1165/2)^{21 \times 2} \times Investment$ $Investment = 8000$ 8. Geoff invests $18000 in an investment account. After 2 years the investment account contains $19 300. If the account pays $r%$ interest per annum, compounding quarterly, find the value of $r$, to one decimal place. $19300 = 18000 (1+r/400)^{2\times 4}$ $r = 5.7 %$