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Solved Examples on Continuous Compound Interest

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

NOTE: Unless instructed otherwise;
For all financial calculations, do not round until the final answer.
Do not round intermediate calculations. If it is too long, write it to "at least" $5$ decimal places.
Round your final answer to $2$ decimal places.
Make sure you include your unit.


Solve all questions.
Use at least two methods where applicable.
Show all work.

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.) How long will it take money to double if it is invested at $9\%$ interest rate compounded continuously?
Round to the nearest integer.


$ P = ? \\[3ex] A = 2P...double \\[3ex] r = 9\% = \dfrac{9}{100} = 0.09 \\[5ex] t = ? \\[3ex] t = \dfrac{\ln \left(\dfrac{A}{P}\right)}{r} \\[7ex] t = \dfrac{\ln \left(\dfrac{2P}{P}\right)}{0.09} \\[7ex] = \dfrac{\ln 2}{0.09} \\[5ex] = \dfrac{0.693147181}{0.09} \\[5ex] = 7.70163534 \\[3ex] \approx 8\:years \\[3ex] $ Ceteris paribus, it will take about $8\:years$ for any sum of money invested at $9\%$ to double if the interest is compounded continuously.
(2.) If the population of SamDom For Peace City is growing at $1.5\%$ compounded continuously, how long will it take the population to triple?


$ P = ? \\[3ex] A = 3P...triple \\[3ex] r = 1.5\% = \dfrac{1.5}{100} = 0.015 \\[5ex] t = ? \\[3ex] t = \dfrac{\ln \left(\dfrac{A}{P}\right)}{r} \\[7ex] t = \dfrac{\ln \left(\dfrac{3P}{P}\right)}{0.09} \\[7ex] = \dfrac{\ln 3}{0.015} \\[5ex] = \dfrac{1.09861229}{0.015} \\[5ex] = 73.2408192 \\[3ex] \approx 73.24\:years \\[3ex] $ Ceteris paribus, it will take approximately $73.24\:years$ for the population of SamDom For Peace City to triple in size, if the population is growing at a rate of $1.5\%$ compounded continuously.
(3.) Nahum invested $\$10,000$ in a bank that pays $13.7\%$ compounded continuously.
(a.) How much money will he have after $2$ years?
(b.) If another bank will pay Nahum $14\%$ compounded quarterly, how much would he have after $2$ years?


$ (a.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] A = Pe^{rt} \\[3ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] r = 13.7\% = \dfrac{13.7}{100} = 0.137 \\[5ex] A = 10000 * e^{0.137 * 2} \\[3ex] = 10000 * e^{0.274} \\[3ex] = 10000 * 1.315214802 \\[3ex] = 13152.14802 \\[3ex] A \approx \$13,152.15 \\[3ex] (b.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] m = 4 \\[3ex] r = 14\% = \dfrac{14}{100} = 0.14 \\[5ex] A = 10000 * \left(1 + \dfrac{0.14}{4}\right)^{4 * 2} \\[5ex] = 10000 * (1 + 0.035)^{8} \\[3ex] = 10000 * (1.035)^{8} \\[3ex] = 10000 * 1.316809037 \\[3ex] = 13168.09037 \\[3ex] A \approx \$13,168.09 $
(4.) Matthew has $\$1000$ to invest at $6\%$ per annum compounded quarterly.
(a.) How long will it take before he has $$1450$?
(b.) If the compounding is continuous, how long will it be?


$ (a.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] Compounded\:\:quarterly \rightarrow m = 4 \\[3ex] t = \dfrac{\log \left(\dfrac{A}{P}\right)}{m\log \left(1 + \dfrac{r}{m}\right)} \\[7ex] t = \dfrac{\log \left(\dfrac{1450}{1000}\right)}{4 * \log \left(1 + \dfrac{0.06}{4}\right)} \\[7ex] t = \dfrac{\log (1.45)}{4 * \log (1 + 0.015)} \\[7ex] = \dfrac{\log (1.45)}{4 * \log (1.015)} \\[7ex] = \dfrac{0.161368002}{4 * 0.006466042} \\[7ex] = \dfrac{0.161368002}{0.025864169} \\[5ex] = 6.23905613 \\[3ex] t \approx 6.24\:years \\[3ex] (b.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{1450}{1000}}\right)}{0.06} \\[7ex] = \dfrac{\ln {(1.45)}}{0.06} \\[5ex] = \dfrac{0.371563556}{0.06} \\[5ex] = 6.19272594 \\[3ex] t \approx 6.19\:years $
(5.) Ezra invested $\$2000$ at an interest rate, $k$ compounded continously.
The fund amounted to $\$2504.65$ in $5$ years.
(a.) Calculate the interest rate.
(b.) Calculate the balance after $10$ years.
(c.) After how long will the fund be doubled?
Round to the nearest hundredth as needed.


$ \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] (a.) \\[3ex] P = \$2000 \\[3ex] r = k \\[3ex] A = \$2504.65 \\[3ex] t = 5\:years \\[3ex] r = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{t} \\[7ex] r = \dfrac{\ln \left({\dfrac{2504.65}{2000}}\right)}{5} \\[7ex] = \dfrac{\ln (1.252325)}{5} \\[5ex] = \dfrac{0.225001824}{5} \\[5ex] = 0.045000365 \\[3ex] to\:\:percent = 0.045000365(100) = 4.5000365 \\[3ex] r \approx 4.5\% \\[3ex] (b.) \\[3ex] After\: 10\: years; \\[3ex] t = 10\:years \\[3ex] A = ? \\[3ex] P = \$2000 \\[3ex] r = 0.045000365 \\[3ex] A = Pe^{rt} \\[5ex] A = 2000 * e^{0.045000365 * 10} \\[5ex] = 2000 * e^{0.45000365} \\[5ex] = 2000 * 1.568317907 \\[3ex] = 3136.635813 \\[3ex] A \approx \$3136.64 \\[3ex] (c.) \\[3ex] When\: the\: fund\: is\: doubled \\[3ex] P = \$2000 \: dollars \\[3ex] r = 0.045000365 \\[3ex] A = 2 * 2000 = \$4000 \\[3ex] t = ? \\[3ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{4000}{2000}}\right)}{0.045000365} \\[7ex] = \dfrac{\ln 2}{0.045000365} \\[5ex] = \dfrac{0.693147181}{0.045000365} \\[5ex] = 15.40314574 \\[3ex] t \approx 15.4\: years $
(6.) Malachi invested a sum of $\$800$ into an account that pays interest at the rate of $2.9\%$ per year, compounded continously.
How much money will be in the account after $8$ years?
Calculate the doubling time.
Round to the nearest hundredth as needed.


$ \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] P = \$800 \\[3ex] t = 8\: years \\[3ex] r = 2.9\% = \dfrac{2.9}{100} = 0.029 \\[5ex] A = Pe^{rt} \\[5ex] A = 800 * e^{0.029 * 8} \\[5ex] = 800 * e^{0.232} \\[5ex] = 800 * 1.261119729 \\[2ex] = 1008.895783 \\[3ex] A \approx \$1008.90 \\[3ex] $ Calculating the doubling time means "when" will the fund be doubled?
"How long" will it take for the money to be doubled?

$ When\: the\: fund\: is\: doubled \\[3ex] P = \$800 \\[3ex] r = 0.029 \\[3ex] A = 2 * 800 = \$1600 \\[3ex] t = ? \\[2ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{1600}{800}}\right)}{0.029} \\[7ex] t = \dfrac{\ln 2}{0.029} \\[5ex] t = \dfrac{0.693147181}{0.0029} \\[5ex] t = 23.90162692 \\[3ex] t \approx 23.90\: years $
(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 $