If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

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

# Solved Examples on the Mathematics of Finance

• ### Symbols and Meanings

• You can verify all answers with the
• To solve for a specified variable for each formula, please review
• $A$ = amount or future value (in dollars, naira, any currency)
• $P$ = principal or present value (in dollars, naira, any currency)
• $r$ = annual interest rate (in percent)
• $m$ = number of compounding periods per year
• $t$ = time (in years)
• $i$ = annual interest rate per period
• $n$ = total number of compounding periods

### Formulas

If Compounded: $m$ =
Annually $1$ ($1$ time per year)
Semiannually $2$ ($2$ times per year)
Quarterly $4$ ($4$ times per year)
Monthly $12$ ($12$ times per year)
Weekly $52$ ($52$ times per year)
Daily (Ordinary) $360$ ($360$ times per year)
Daily (Exact) $365$ ($365$ times per year)

#### Compound Interest

(1.) $A = P\left(1 + \dfrac{r}{m}\right)^{mt}$

(2.) $P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}}$

(3.) $r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1 \right]$

(4.) $r = m\left[10^{\dfrac{\log \left(\dfrac{A}{P}\right)}{mt}} - 1 \right]$

(5.) $t = \dfrac{\log \left(\dfrac{A}{P}\right)}{m\log \left(1 + \dfrac{r}{m}\right)}$

(6.) $A = P(1 + i)^n$

(7.) $i = \dfrac{r}{m}$

(8.) $n = mt$

#### Continuous Compound Interest

(9.) $A = Pe^{rt}$

(10.) $P = \dfrac{A}{{e^{rt}}}$

(11.) $t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r}$

(12.) $r = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{t}$

#### Annual Percentage Yield (APY) for Compound Interest

(13.) $APY = \left(1 + \dfrac{r}{m}\right)^{m} - 1$

(14.) $r = m\left[(APY + 1)^{\dfrac{1}{m}} - 1\right]$

#### Annual Percentage Yield (APY) for Continuous Compound Interest

(15.) $APY = e^r - 1$

(16.) $r = \ln[APY + 1]$

You may verify your answers as applicable with the: Mathematics of Finance Calculators
For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB, NZQA, and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

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.

Solve all the word problems.
Use at least two methods as applicable.
Show all Work.
Round all answers to two decimal places.

(1.) Esther's parents deposited a sum of $\$750$in a prepaid college account. How much is the value of this money after a period of sixteen years if it is invested at$3\%$compounded annually?$ A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] P = 750 \: dollars \\[3ex] t = 16 \: years \\[3ex] Compounded\:\:annually\rightarrow m = 1 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] A = 750 * \left(1 + \dfrac{0.03}{1}\right)^{1 * 16} \\[7ex] = 750 * (1 + 0.03)^{16} \\[3ex] = 750 * (1.03)^{16} \\[3ex] = 750 * 1.60470644 \\[3ex] = 1203.52983 \\[3ex] \approx \$1,203.53 \\[3ex]$ Ceteris paribus, the amount in the prepaid account at the end of sixteen years will be $\$1203.53$(2.) 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?

$(a.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] A = Pe^{rt} \\[3ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] r = 13.7\% = \dfrac{13.7}{100} = 0.137 \\[5ex] A = 10000 * e^{0.137 * 2} \\[3ex] = 10000 * e^{0.274} \\[3ex] = 10000 * 1.315214802 \\[3ex] = 13152.14802 \\[3ex] A \approx \$13,152.15 \\[3ex] (b.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] m = 4 \\[3ex] r = 14\% = \dfrac{14}{100} = 0.14 \\[5ex] A = 10000 * \left(1 + \dfrac{0.14}{4}\right)^{4 * 2} \\[5ex] = 10000 * (1 + 0.035)^{8} \\[3ex] = 10000 * (1.035)^{8} \\[3ex] = 10000 * 1.316809037 \\[3ex] = 13168.09037 \\[3ex] A \approx \$13,168.09$
(3.) 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?

$(a.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] Compounded\:\:quarterly \rightarrow m = 4 \\[3ex] t = \dfrac{\log \left(\dfrac{A}{P}\right)}{m\log \left(1 + \dfrac{r}{m}\right)} \\[7ex] t = \dfrac{\log \left(\dfrac{1450}{1000}\right)}{4 * \log \left(1 + \dfrac{0.06}{4}\right)} \\[7ex] t = \dfrac{\log (1.45)}{4 * \log (1 + 0.015)} \\[7ex] = \dfrac{\log (1.45)}{4 * \log (1.015)} \\[7ex] = \dfrac{0.161368002}{4 * 0.006466042} \\[7ex] = \dfrac{0.161368002}{0.025864169} \\[5ex] = 6.23905613 \\[3ex] t \approx 6.24\:years \\[3ex] (b.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{1450}{1000}}\right)}{0.06} \\[7ex] = \dfrac{\ln {(1.45)}}{0.06} \\[5ex] = \dfrac{0.371563556}{0.06} \\[5ex] = 6.19272594 \\[3ex] t \approx 6.19\:years$
(4.) 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?

$APY = 7.5\% = \dfrac{7.5}{100} = 0.075 \\[5ex] Compounded\:\:monthly \rightarrow m = 12 \\[3ex] r = m\left[(APY + 1)^{\dfrac{1}{m}} - 1\right] \\[7ex] r = 12 * \left[(0.075 + 1)^{\dfrac{1}{12}} - 1\right] \\[7ex] = 12 * \left[(1.075)^{\dfrac{1}{12}} - 1\right] \\[7ex] = 12 * \left[(1.075)^{0.0833333333} - 1\right] \\[5ex] = 12 * \left[1.00604492 - 1\right] \\[5ex] = 12 * 0.00604492 \\[3ex] = 0.07253904 \\[3ex] to\:\:percent = 0.07253904(10) = 7.253904 \\[3ex] r \approx 7.25\%$
(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 \$