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The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on Annual Percentage Yield

Samuel Dominic Chukwuemeka (SamDom For Peace) Formulas Used: Mathematics of Finance
Verify Answers with Calculator: Financial Mathematics Calculators

NOTE: Unless instructed otherwise;
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Round your final answer to $2$ decimal places.
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For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

(1.) Determine the effective rate of interest for a nominal rate of $5\%$ per year compounded:
(a.) Annually
(b.) Semiannually
(c.) Quarterly
(d.) Monthly
(e.) Continuously


$ r = 5\% = \dfrac{5}{100} = 0.05 \\[5ex] (a.) \\[3ex] \underline{Annually} \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.05}{1}\right)^{1} - 1 \\[7ex] = \left(1 + 0.05\right)^{1} - 1 \\[5ex] = (1.05)^1 - 1 \\[3ex] = 1.05 - 1 \\[3ex] = 0.05 \\[3ex] to\:\:percent = 0.05(100) = 5\% \\[3ex] APY = 5\% \\[3ex] (b.) \\[3ex] \underline{Semiannually} \\[3ex] Compounded\:\:semiannually \rightarrow m = 2 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.05}{2}\right)^{2} - 1 \\[7ex] = \left(1 + 0.025\right)^{2} - 1 \\[5ex] = (1.025)^2 - 1 \\[3ex] = 1.050625 - 1 \\[3ex] = 0.050625 \\[3ex] to\:\:percent = 0.050625(100) = 5.0625\% \\[3ex] APY = 5.0625\% \\[3ex] (c.) \\[3ex] \underline{Quarterly} \\[3ex] Compounded\:\:quarterly \rightarrow m = 4 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.05}{4}\right)^{4} - 1 \\[7ex] = \left(1 + 0.0125\right)^{4} - 1 \\[5ex] = (1.0125)^4 - 1 \\[3ex] = 1.05094534 - 1 \\[3ex] = 0.05094534 \\[3ex] to\:\:percent = 0.05094534(100) = 5.094534\% \\[3ex] APY = 5.094534\% \\[3ex] (d.) \\[3ex] \underline{Monthly} \\[3ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.05}{12}\right)^{12} - 1 \\[7ex] = \left(1 + 0.00416666667\right)^{12} - 1 \\[5ex] = (1.00416667)^{12} - 1 \\[3ex] = 1.05116194 - 1 \\[3ex] = 0.05116194 \\[3ex] to\:\:percent = 0.05116194(100) = 5.116194\% \\[3ex] APY = 5.116194\% \\[3ex] (e.) \\[3ex] \underline{Continuously} \\[3ex] APY = e^r - 1 \\[4ex] APY = e^{0.05} - 1 \\[4ex] = 1.0512711 - 1 \\[3ex] = 0.0512711 \\[3ex] to\:\:percent = 0.0512711(100) \\[3ex] APY = 5.12711\% $
(2.) Three banks listed the following money market accounts:
East Bank: $4.36\%$ compounded monthly
South Bank: $4.35\%$ compounded daily
North Bank: $4.31\%$ compounded continuously
(a.) Determine the annual percentage yield.
(b.) Which of the banks has the highest return?


$ (a.) \\[3ex] \underline{East\:\:Bank} \\[3ex] r = 4.36\% = \dfrac{4.36}{100} = 0.0436 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.0436}{12}\right)^{12} - 1 \\[7ex] = \left(1 + 0.00363333333\right)^{12} - 1 \\[5ex] = \left(1.00363333\right)^{12} - 1 \\[5ex] = 1.04448187 - 1 \\[3ex] = 0.04448187 \\[3ex] to\:\:percent = 0.04448187(100) \\[3ex] APY = 4.448187\% \\[3ex] \underline{South\:\:Bank} \\[3ex] r = 4.35\% = \dfrac{4.35}{100} = 0.0435 \\[5ex] Compounded\:\:monthly \rightarrow m = 365 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[7ex] APY = \left(1 + \dfrac{0.0435}{365}\right)^{365} - 1 \\[7ex] = \left(1 + 0.000119178082\right)^{365} - 1 \\[5ex] = \left(1.00011918\right)^{365} - 1 \\[5ex] = 1.04445802 - 1 \\[3ex] = 0.04445802 \\[3ex] to\:\:percent = 0.04445802(100) \\[3ex] APY = 4.445802\% \\[3ex] \underline{North\:\:Bank} \\[3ex] r = 4.31\% = \dfrac{4.31}{100} = 0.0431 \\[5ex] Compounded\:\:continuously \\[3ex] APY = e^r - 1 \\[4ex] APY = e^{0.0431} - 1 \\[4ex] = 1.04404229 - 1 \\[3ex] = 0.04404229 \\[3ex] to\:\:percent = 0.04404229(100) \\[3ex] APY = 4.404229\% \\[3ex] $ (b.) East Bank has the highest return.
(3.) A loan company wants to offer a CD (Certificate of deposit) with a monthly company rate that has an APY of $7.5\%$.
What annual nominal rate compounded monthly should they use?


$ APY = 7.5\% = \dfrac{7.5}{100} = 0.075 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] r = m\left[(APY + 1)^{\dfrac{1}{m}} - 1\right] \\[7ex] r = 12 * \left[(0.075 + 1)^{\dfrac{1}{12}} - 1\right] \\[7ex] = 12 * \left[(1.075)^{\dfrac{1}{12}} - 1\right] \\[7ex] = 12 * \left[(1.075)^{0.0833333333} - 1\right] \\[5ex] = 12 * \left[1.00604492 - 1\right] \\[5ex] = 12 * 0.00604492 \\[3ex] = 0.07253904 \\[3ex] to\:\:percent = 0.07253904(10) = 7.253904 \\[3ex] r \approx 7.25\% $
(4.) Mark bought a thousand shares for twenty eight thousand dollars including commissions.
Two years later, he sold the shares for thirty four thousand and two hundred dollars after deducting commissions.
Calculate the effective rate of return on his investment over the two-year period.


$ r = ? \\[3ex] APY = ? \\[3ex] \underline{Compound\:\:Interest} \\[3ex] m = 1 \\[3ex] A = \$34200 \\[3ex] P = \$28000 \\[3ex] t = 2\:years \\[3ex] r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] \\[7ex] r = 1 * \left[\left(\dfrac{34200}{28000}\right)^{\dfrac{1}{1 * 2}} - 1\right] \\[7ex] r = \left[(1.221428571)^{\dfrac{1}{2}} - 1\right] \\[5ex] r = \left[1.221428571^{0.5} - 1\right] \\[5ex] r = 1.105182596 - 1 \\[3ex] r = 0.105182596 \\[3ex] \underline{APY} \\[3ex] r = 0.105182596 \\[3ex] m = 1 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[5ex] APY = \left(1 + \dfrac{0.105182596}{1}\right)^1 - 1 \\[5ex] APY = (1 + 0.105182596)^1 - 1 \\[3ex] APY = (1.105182596)^1 - 1 \\[3ex] APY = 1.105182596 - 1 \\[3ex] APY = 0.105182596 \\[3ex] to\:\:percent = 0.105182596 * 100 \\[3ex] = 10.5182596\% \\[3ex] APY \approx 10.5183\% $
(5.) Rita invested $\$10,000$ in a mutual fund that pays interest on a daily basis.
The balance in her account at the end of $8$ months ($245$ days) was $\$10,475.25$
Find the effective rate at which Rita's account earned interest over this period.
Assume a $365-day$ year.


$ r = ? \\[3ex] APY = ? \\[3ex] \underline{Compound\:\:Interest} \\[3ex] m = 365 \\[3ex] A = \$10475.25 \\[3ex] P = \$10000 \\[3ex] t = \dfrac{245}{365}\:years \\[5ex] r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] \\[7ex] r = 365 * \left[\left(\dfrac{10475.25}{10000}\right)^{\dfrac{1}{365 * \dfrac{245}{365}}} - 1\right] \\[10ex] r = 365 * \left[(1.047525)^{\dfrac{1}{245}} - 1\right] \\[5ex] r = 365 * \left[1.047525^{0.004081632653} - 1\right] \\[5ex] r = 365 * [1.000189529 - 1] \\[5ex] r = 365 * 0.0001895291375 \\[3ex] r = 0.06917813518 \\[3ex] \underline{APY} \\[3ex] r = 0.06917813518 \\[3ex] m = 365 \\[3ex] APY = \left(1 + \dfrac{r}{m}\right)^m - 1 \\[5ex] APY = \left(1 + \dfrac{0.06917813518}{365}\right)^{365} - 1 \\[5ex] APY = (1 + 0.0001895291375)^{365} - 1 \\[3ex] APY = (1.000189529)^{365} - 1 \\[3ex] APY = 1.071620062 - 1 \\[3ex] APY = 0.07162006234 \\[3ex] to\:\:percent = 0.07162006234 * 100 \\[3ex] = 7.162006234\% \\[3ex] APY \approx 7.1620\% $
(6.)


$ t = 10 \\[3ex] FV = \$1000 \\[3ex] BP = \$514.87 \\[3ex] YTM = ? \\[3ex] YTM = \left(\dfrac{FV}{BP}\right)^{\dfrac{1}{t}} - 1 \\[7ex] YTM = \left(\dfrac{1000}{514.87}\right)^{\dfrac{1}{10}} - 1 \\[7ex] = \left(1.94223785\right)^{0.1} - 1 \\[5ex] = 1.06863708 - 1 \\[3ex] = 0.06863708 \\[3ex] to\:\:percent = 0.06863708(100) \\[3ex] = 6.863708\% \\[3ex] \approx 6.86\% $
(7.)


If the question does not state that the interest is simple interest, then assume compound interest.

First: This is a case of Compound Interest because their retirement savings of $\$25,000$ earns interest compounded annually/yearly at the interest rate of $6\%$.

Second: This is also a case of Annuity Due because the couple would have to deposit $\$1,000$ at the beginning of each year.

$ \underline{Compound\:\:Interest\:\:and\:\:Annuity\:\:Due} \\[3ex] P = \$25000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] PMT = \$1000 \\[3ex] CFV = \$80000 \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] t = ? \\[3ex] t = \dfrac{\log\left[\dfrac{rCFV + PMT(m + r)}{rP + PMT(m + r)}\right]}{m\log\left(1 + \dfrac{r}{m}\right)} \\[10ex] = \dfrac{\log\left[\dfrac{0.06(80000) + 1000(1 + 0.06)}{0.06(25000) + 1000(1 + 0.06)}\right]}{1 * \log\left(1 + \dfrac{0.06}{1}\right)} \\[10ex] = \dfrac{\log\left[\dfrac{4800 + 1000(1.06)}{1500 + 1000(1.06)}\right]}{1 * \log\left(1 + 0.06\right)} \\[10ex] = \dfrac{\log\left[\dfrac{4800 + 1060}{1500 + 1060}\right]}{\log\left(1.06\right)} \\[7ex] = \dfrac{\log\left[\dfrac{5860}{2560}\right]}{\log\left(1.06\right)} \\[7ex] = \dfrac{\log(2.2890625)}{0.0253058653} \\[5ex] = \dfrac{0.359657651}{0.0253058653} \\[5ex] = 14.2124226\:years \\[3ex] Let\:\:us\:\:check\:\:to\:\:confirm \\[3ex] \underline{Check} \\[3ex] \underline{Compound\:\:Interest} \\[5ex] P = \$25000 \\[3ex] r = 0.06 \\[3ex] m = 1 \\[3ex] t = 14.2124226\:years \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] = 25000\left(1 + \dfrac{0.06}{1}\right)^{1(14.2124226)} \\[7ex] = 25000\left(1 + 0.06\right)^{14.2124226} \\[7ex] = 25000(1.06)^{14.2124226} \\[5ex] = 25000(2.2890625) \\[3ex] = \$57226.5625 \\[3ex] \underline{Annuity\:\:Due} \\[3ex] PMT = \$1000 \\[3ex] r = 0.06 \\[3ex] m = 1 \\[3ex] t = 14.2124226\:years \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] * \left(1 + \dfrac{r}{m}\right) \\[10ex] = 1 * 1000 * \left[\dfrac{\left(1 + \dfrac{0.06}{1}\right)^{1(14.2124226)} - 1}{0.06}\right] * \left(1 + \dfrac{0.06}{1}\right) \\[10ex] = 1000 * \left[\dfrac{\left(1 + 0.06\right)^{14.2124226} - 1}{0.06}\right] * \left(1 + 0.06\right) \\[10ex] = 1000 * \left[\dfrac{\left(1.06\right)^{14.2124226} - 1}{0.06}\right] * \left(1.06\right) \\[10ex] = 1000 * 1.06 * \left[\dfrac{2.2890625 - 1}{0.06}\right] \\[7ex] = 1060 * \left[\dfrac{1.2890625}{0.06}\right] \\[7ex] = \dfrac{1060 * 1.2890625}{0.06} \\[5ex] = \dfrac{1366.40625}{0.06} \\[5ex] = \$22773.4375 \\[5ex] \underline{Combined\:\:Amount} \\[3ex] CFV = 57226.5625 + 22773.4375 = \$80000 $
(8.)


$ \underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50 $
(9.)


$ \underline{Amortization} \\[3ex] PV = R1500000 \\[3ex] t = 20\:years \\[3ex] r = 10.5\% = \dfrac{10.5}{100} = 0.105 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] PMT = ? \\[3ex] (9.1) \\[3ex] PMT = \dfrac{PV}{m} * \left[\dfrac{r}{1 - \left(1 + \dfrac{r}{m}\right)^{-mt}}\right] \\[10ex] PMT = \dfrac{1500000}{12} * \left[\dfrac{0.105}{1 - \left(1 + \dfrac{0.105}{12}\right)^{-1 * 12 * 20}}\right] \\[10ex] = 125000 * \left[\dfrac{0.105}{1 - \left(1 + 0.00875\right)^{-240}}\right] \\[7ex] = 125000 * \left[\dfrac{0.105}{1 - \left(1.00875\right)^{-240}}\right] \\[7ex] = 125000 * \left[\dfrac{0.105}{1 - 0.123580101}\right] \\[5ex] = 125000 * \left[\dfrac{0.105}{0.876419899}\right] \\[5ex] = \dfrac{125000 * 0.105}{0.876419899} \\[5ex] = \dfrac{13125}{0.876419899} \\[5ex] = 14975.6983 \\[3ex] \approx R14,975.70 \\[3ex] (9.2) \\[3ex] 12\:months = 1\:year \\[3ex] 144th\:\:payment \rightarrow \dfrac{144}{12} = 12th\:year\:\:payment \\[5ex] Remaining = 20 - 12 = 8\:years \\[3ex] $ Ask your students how they will calculate the outstanding balance after the $144^{th}$ payment.
Note their responses.
Some of the responses may include:
Find the total payment...for 20 years ...which is 240 months
Find the payment for 144 months
Subtract that payment from the total payment
...among other responses.

Remind them that these are Mortgage payments...Fixed Rate Mortgage
Show them how it works...using the example on my website
You can also use the Loan Amortization Calculator to confirm the calculations
Any monthly payment made is used to offset the interest earned during that period...first step
Then, any remaining balance is applied toward the principal.
The same principle applies to car payments and other amortized payments.


We can find the outstanding balance at the end of the $144th$ payemnt (at the end of $12$ years) in at least two ways.
Use any method you prefer.

First Method: Find the present value of the ordinary annuity for the remaining $8$ years.
OR
Second Method:
Compare to the previous response of "probably" one of your students.
But, explain the concept well for your students to understand.


Use the Compound Interest formula to calculate the future value of the loan amount for $12$ years.

Remember that the monthly payment of $14975.70$ was found using the Amortization formula for the entire term of the loan.

So, use the Future Value of an Ordinary Annuity formula and calculate the future value of that payment for $12$ years.
Then, subtract the future value of that payment from the future value of the loan amount.

$ \underline{First\:\:Method} \\[3ex] \underline{Present\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] t = 8\:years \\[3ex] PMT = R14975.6983 \\[3ex] r = 0.105 \\[3ex] m = 12 \\[3ex] PV = m * PMT * \left[\dfrac{1 - \left(1 + \dfrac{r}{m}\right)^{-mt}}{r}\right] \\[10ex] PV = 12 * 14975.6983 * \left[\dfrac{1 - \left(1 + \dfrac{0.105}{12}\right)^{-1 * 12 * 8}}{0.105}\right] \\[10ex] = 179708.38 * \left[\dfrac{1 - \left(1 + 0.00875\right)^{-96}}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{1 - \left(1.00875\right)^{-96}}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{1 - 0.43329075}{0.105}\right] \\[5ex] = 179708.38 * \left[\dfrac{0.56670925}{0.105}\right] \\[5ex] = \dfrac{179708.38 * 0.56670925}{0.105} \\[5ex] = \dfrac{101842.401}{0.105} \\[5ex] = 969927.629 \\[3ex] \approx R969,927.63 \\[3ex] \underline{Second\:\:Method} \\[3ex] \underline{Compound\:\:Interest\:\:and\:\:Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] \underline{Compound\:\:Interest} \\[3ex] P = R1500000 \\[3ex] r = 0.105 \\[3ex] t = 12\:years \\[3ex] m = 12 \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] A = 1500000\left(1 + \dfrac{0.105}{12}\right)^{12 * 12} \\[7ex] = 1500000\left(1 + 0.00875\right)^{144} \\[7ex] = 1500000\left(1.00875\right)^{144} \\[7ex] = 1500000(3.50615308) \\[3ex] = R5259229.62 \\[3ex] \underline{Future\:\:Value\:\:of\:\:Ordinary\:\:Annuity} \\[3ex] m = 12 \\[3ex] PMT = R14975.6983 \\[3ex] r = 0.105 \\[3ex] t = 12\:years \\[3ex] FV = m * PMT * \left[\dfrac{\left(1 + \dfrac{r}{m}\right)^{mt} - 1}{r}\right] \\[10ex] FV = 12 * 14975.6983 * \left[\dfrac{\left(1 + \dfrac{0.105}{12}\right)^{12 * 12} - 1}{0.105}\right] \\[10ex] = 179708.38 * \left[\dfrac{\left(1 + 0.00875\right)^{144} - 1}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{\left(1.00875\right)^{144} - 1}{0.105}\right] \\[7ex] = 179708.38 * \left[\dfrac{3.50615308 - 1}{0.105}\right] \\[5ex] = 179708.38 * \left[\dfrac{2.50615308}{0.105}\right] \\[5ex] = \dfrac{179708.38 * 2.50615308}{0.105} \\[5ex] = \dfrac{450376.71}{0.105} \\[5ex] = R4289302 \\[3ex] \underline{Balance\:\:after\:\:144th\:\:payment} \\[3ex] Balance = 5259229.62 - 4289302 = R969,927.62 \\[3ex] $
Question 9 - Part 1

Question 9 - Part 2
(10.)


$ \underline{First\:\:Semester} \\[3ex] \Sigma CH = 12 \\[3ex] GPA = 3.583333333 \\[3ex] \Sigma GP = 3.583333333(12) = 43 \\[3ex] \underline{Second\:\:Semester} \\[3ex] \Sigma CH = 16 \\[3ex] \Sigma GP = 55 \\[3ex] \underline{First\:\:and\:\:Second\:\:Semesters} \\[3ex] \Sigma CH = 12 + 16 = 28 \\[3ex] \Sigma GP = 43 + 55 = 98 \\[3ex] CGPA = \dfrac{\Sigma GP}{\Sigma CH} \\[5ex] CGPA = \dfrac{98}{28} \\[5ex] CGPA = 3.5 \\[3ex] CGPA = 3.50 $