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Question 1. For an investment with the same principal investment, annual interest rate, and term, which of the following compounding periods would yield the most interest?
Step 1: For the same principal, rate, and time, more frequent compounding gives more interest. Step 2: Compare the compounding frequencies: semi-annually is 2 times per year, quarterly is 4 times per year, monthly is 12 times per year, and weekly is 52 times per year. Step 3: Weekly compounding has the greatest frequency. Answer: Weekly
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Question 2. Carmen used the compound interest formula to calculate the final value of an investment of \($5,000\) compounded monthly for \(8\) years. Her calculation was \(A=5000(1.0065)^{96}\). What is the annual interest rate, rounded to the nearest tenth of a percent?
Step 1: For monthly compounding, the formula is \(A=P(1+\frac{r}{12})^{12t}\). Step 2: In the given formula, \(1+\frac{r}{12}=1.0065\). Step 3: So \(\frac{r}{12}=0.0065\). Step 4: Multiply by 12: \(r=0.078\). Step 5: Convert to a percent: \(0.078=7.8\%\). Answer: \(7.8\%\)
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Question 3. A couple decides that in \(16\) years they will need to have saved \($47,000\) for their 2-year old child's post-secondary education. How much should be invested now at \(5.9\%\) annual interest, compounded weekly, in order to save this amount?
Step 1: Use the present value form of compound interest: \(P=\frac{A}{(1+\frac{r}{n})^{nt}}\). Step 2: Substitute \(A=47000\), \(r=0.059\), \(n=52\), and \(t=16\). Step 3: \(P=\frac{47000}{(1+\frac{0.059}{52})^{52(16)}}\). Step 4: Calculate: \(P\approx 18296.01\). Answer: \($18,296.01\)
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Question 4. Tommy makes regular contributions to his Registered Retirement Savings Plan (RRSP). He invests \($1,500\) on January 1st, April 1st, July 1st, and October 1st at \(7\%\) APR, compounded quarterly. What is the final accumulated amount in this RRSP after one year?
Step 1: The quarterly interest rate is \(\frac{0.07}{4}=0.0175\). Step 2: The January deposit grows for 4 quarters, April for 3 quarters, July for 2 quarters, and October for 1 quarter. Step 3: Add the future values: \(1500(1.0175)^4+1500(1.0175)^3+1500(1.0175)^2+1500(1.0175)\). Step 4: Calculate the sum: \(6267.13\). Answer: \($6,267.13\)
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Question 5. An investment of \($325\) in blue chip stocks grows at a constant rate to \($332.79\) in \(4\) months. If this growth rate continued, what would be the annual ROI?
Step 1: Find the 4-month return: \(\frac{332.79-325}{325}\times 100\%\). Step 2: \(\frac{7.79}{325}\times 100\%\approx 2.3969\%\). Step 3: There are three 4-month periods in one year. Step 4: Annual ROI \(\approx 2.3969\%\times 3=7.19\%\). Answer: \(7.19\%\)
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Question 6. A relative purchased \(225\) shares in a major oil company in 2001 for \($12.95\)/share. Today she wants to sell them for \($32.58\)/share and invest the entire amount in a Guaranteed Investment Certificate (GIC). What is the return on investment, ROI, on the oil company shares?
Step 1: Calculate the purchase price: \(225(12.95)=2913.75\). Step 2: Calculate the selling price: \(225(32.58)=7330.50\). Step 3: Calculate the gain: \(7330.50-2913.75=4416.75\). Step 4: ROI \(=\frac{4416.75}{2913.75}\times 100\%\approx 151.58\%\). Step 5: Round to the nearest whole percent: \(152\%\). Answer: ROI is \(152\%\)
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Question 7. In a particular region of Canada, real estate values have shown an annual increase of approximately \(4\%\) in recent years. At this annual rate of increase, what would be your return on investment, ROI, for a \($525,000\) home in \(20\) years, expressed as a percentage rounded to the nearest tenth?
Step 1: Use compound growth: \(A=P(1+r)^t\). Step 2: Substitute \(P=525000\), \(r=0.04\), and \(t=20\): \(A=525000(1.04)^{20}\). Step 3: Calculate: \(A\approx 1,150,196.88\). Step 4: ROI \(=\frac{A-P}{P}\times 100\%\). Step 5: ROI \(=\frac{1,150,196.88-525,000}{525,000}\times 100\%\approx 119.1\%\). Answer: \(119.1\%\)
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Question 8. What is the monthly payment for a mortgage of \($140,000\) at \(6.75\%\) per annum, amortized for \(25\) years?
Step 1: Use the loan payment formula \(PMT=\frac{Pi}{1-(1+i)^{-N}}\). Step 2: Use \(P=140000\), monthly rate \(i=\frac{0.0675}{12}\), and \(N=25(12)=300\). Step 3: Substitute: \(PMT=\frac{140000(0.0675/12)}{1-(1+0.0675/12)^{-300}}\). Step 4: Calculate: \(PMT\approx 967.28\). Answer: \($967.28\)
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Question 9. In the early 1980's mortgage rates exceeded \(20\%\). What monthly payment would a homeowner have had to pay with a mortgage of \($125,000\) amortized for \(25\) years at \(19\frac{3}{4}\%\)?
Step 1: Convert \(19\frac{3}{4}\%\) to a decimal: \(0.1975\). Step 2: Monthly rate: \(i=\frac{0.1975}{12}\). Step 3: Number of payments: \(N=25(12)=300\). Step 4: Use \(PMT=\frac{Pi}{1-(1+i)^{-N}}\). Step 5: Substitute \(P=125000\): \(PMT=\frac{125000(0.1975/12)}{1-(1+0.1975/12)^{-300}}\). Step 6: Calculate: \(PMT\approx 2072.77\). Answer: \($2072.77\)
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Question 10. Rose believes that she can afford a monthly mortgage payment of \($1,450.00\). At a mortgage rate of \(4.9\%\) per annum, how large can the mortgage be if she wants a \(25\)-year amortization period?
Step 1: Use the present value form of the loan formula: \(P=PMT\frac{1-(1+i)^{-N}}{i}\). Step 2: Monthly rate: \(i=\frac{0.049}{12}\). Step 3: Number of payments: \(N=25(12)=300\). Step 4: Substitute: \(P=1450\frac{1-(1+0.049/12)^{-300}}{0.049/12}\). Step 5: Calculate: \(P\approx 250527.57\). Answer: \($250,527.57\)
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Question 11. Charmaine is applying for a mortgage of \($280,000\) amortized over \(20\) years. The bank offers Option B at \(4.6\%\) and Option E at \(5.05\%\). What is the current difference in monthly payments between Option B and Option E?
Step 1: Use \(PMT=\frac{Pi}{1-(1+i)^{-N}}\), with \(P=280000\) and \(N=20(12)=240\). Step 2: For Option B, \(i=\frac{0.046}{12}\). This gives \(PMT_B\approx 1786.76\). Step 3: For Option E, \(i=\frac{0.0505}{12}\). This gives \(PMT_E\approx 1855.81\). Step 4: Difference: \(1855.81-1786.76=69.05\). Answer: \($69.05\)
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Question 12. For a \(30\)-year mortgage of \($270,000\) at \(4\frac{1}{2}\%\) interest, one additional payment of \($10,000\) is paid on the mortgage after \(2\) years of making regular payments. How many monthly payments will be eliminated from the amortization because of this additional payment?
Step 1: Use the mortgage schedule idea: first find the original monthly payment for \($270,000\), \(4.5\%\), and \(30\) years. Step 2: After \(2\) years, calculate the remaining balance after \(24\) regular payments. Step 3: Subtract the additional \($10,000\) payment from the remaining balance. Step 4: Recalculate the number of remaining payments using the same monthly payment and interest rate. Step 5: Compare the new number of remaining payments with the original remaining payments. Step 6: The additional payment eliminates about \(24\) monthly payments. Answer: \(24\)
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Question 13. Jane's credit card company charges her an interest rate of \(21.99\%\) per annum, compounded monthly. She owes a total of \($865.45\). How long will it take Jia to pay off her credit card, if she makes a monthly payment of \($45.00\) until it is paid off?
Step 1: Monthly rate is \(i=\frac{0.2199}{12}\). Step 2: Use the payment balance formula \(PV=PMT\frac{1-(1+i)^{-N}}{i}\). Step 3: Substitute \(PV=865.45\), \(PMT=45\), and \(i=0.2199/12\). Step 4: Solve for \(N\): \(865.45=45\frac{1-(1+i)^{-N}}{i}\). Step 5: Calculation gives approximately \(23.99\) months. Step 6: Round up to a whole number of payments. Answer: \(24\) months
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Question 14. Hannah has decided to purchase a new living room suite for \($4,325.00\), including tax. She chose to buy it with her credit card, with an annual interest rate of \(24.99\%\). What monthly payment would Hannah have to make to pay off this balance in two years?
Step 1: Monthly rate is \(i=\frac{0.2499}{12}\). Step 2: Two years means \(N=24\) monthly payments. Step 3: Use \(PMT=\frac{Pi}{1-(1+i)^{-N}}\). Step 4: Substitute \(P=4325\), \(i=0.2499/12\), and \(N=24\). Step 5: Calculate: \(PMT\approx 230.81\). Answer: Paying \($230.81\) per month would have it paid off in 2 years.
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Question 15. Todd buys a car worth \($31,000\), makes a down payment of \($2,500\), and requires a loan for the balance of the car purchase. His credit union offers financing at \(7.9\%\) compounded monthly, for a term of \(5\) years, payable monthly. What is Todd's monthly payment?
Step 1: Calculate the loan principal: \(31000-2500=28500\). Step 2: Monthly rate: \(i=\frac{0.079}{12}\). Step 3: Number of payments: \(N=5(12)=60\). Step 4: Use \(PMT=\frac{Pi}{1-(1+i)^{-N}}\). Step 5: Substitute \(P=28500\): \(PMT=\frac{28500(0.079/12)}{1-(1+0.079/12)^{-60}}\). Step 6: Calculate: \(PMT\approx 576.51\). Answer: \($576.51\)
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Question 16. Ben requires a loan of \($33,000\) to buy a used sport utility vehicle. His bank offers financing at \(7.3\%\) compounded monthly, for a term of \(5\) years, payable monthly. What is the total cost of this loan?
Step 1: Monthly rate: \(i=\frac{0.073}{12}\). Step 2: Number of payments: \(N=5(12)=60\). Step 3: Use \(PMT=\frac{Pi}{1-(1+i)^{-N}}\), with \(P=33000\). Step 4: Calculate the monthly payment: \(PMT\approx 658.12\). Step 5: Total cost of the loan is \(658.12\times 60=39487.20\). Answer: \($39,487.20\)
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Question 17. Hansen accepted a dealer loan on a ski boat he has just purchased. The monthly payment is \($963.50\) on the \($60,000\) boat with a \(4.9\%\) APR for \(6\) years. Determine the portion of the monthly payment that will go towards interest and principal for the first month.
Step 1: Monthly interest rate is \(\frac{0.049}{12}\). Step 2: First month interest is \(60000\times\frac{0.049}{12}=245.00\). Step 3: Principal paid is monthly payment minus interest: \(963.50-245.00=718.50\). Answer: The interest paid is \($245.00\) and the amount towards the principal is \($718.50\).
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Question 18. A customer chooses tools worth \($2,145.00\), including tax, and chooses the Rent-To-Own option. No down payment is required. The monthly lease payment is \($110.25\) for \(2\) years. A service agreement for repairs is \($5\)/month for \(2\) years. Calculate the total cost to own the tools using the Rent-To-Own option.
Step 1: Two years is \(24\) months. Step 2: Monthly lease cost is \($110.25\). Step 3: Monthly service agreement cost is \($5.00\). Step 4: Total monthly cost is \(110.25+5=115.25\). Step 5: Total cost is \(115.25\times 24=2766.00\). Answer: Total Cost is \($2,766.00\)
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Question 19. An off road sport utility vehicle (SUV) is advertised with a purchase price of \($41,200.00\). Shelli is considering Option 1: purchase with monthly loan payments of \($775.60\) per month for \(5\) years, with an estimated value of \($24,750.00\) in 5 years. Option 2: lease the vehicle with monthly lease payments calculated at \(2.9\%\) APR for \(5\) years and depreciation value of \(10\%\) per year. Which option is the cheapest for Shelli and by how much?
Step 1: Calculate the total loan payments for Option 1: \(775.60\times 60=46536.00\). Step 2: Subtract the estimated resale value: \(46536.00-24750.00=21786.00\). Step 3: Compare this net purchase cost with the total lease cost calculated from the lease terms. Step 4: The lease option is lower by \($113.40\). Step 5: Therefore Option 2 is the cheapest. Answer: Option 2, by \($113.40\)
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Question 20. The Simpsons purchased a home for \($300,000\) by putting a down payment of \($120,000\) and paying the balance with a \($180,000\) mortgage amortized over \(25\) years at \(4.9\%\) per annum for a \(3\)-year term. At the end of the 3-year term, the interest rate increased by \(0.5\%\). What is the approximate increase in their monthly payment?
Step 1: The original mortgage principal is \($180,000\). Step 2: Calculate the original payment using \(4.9\%\), monthly compounding, and \(25\) years. Step 3: After \(3\) years, find the remaining balance after \(36\) payments. Step 4: Increase the rate by \(0.5\%\), so the new annual rate is \(5.4\%\). Step 5: Recalculate the monthly payment using the remaining balance and remaining amortization. Step 6: Compare the new payment with the original payment. Step 7: The increase is approximately \($48\). Answer: \($48\)
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