Samuel Dominic Chukwuemeka (SamDom For Peace)

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


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] $

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.

Solve all the word problems.
Use at least two methods as applicable.
Check your solutions 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} \\[3ex] P = 750 \: dollars \\[2ex] t = 16 \: years \\[2ex] m = 1 \\[2ex] r = 3\% = 0.03 \\[2ex] A = 750 * \left(1 + \dfrac{0.03}{1}\right)^{1 * 16} \\[3ex] = 750 * (1 + 0.03)^{16} \\[2ex] = 750 * (1.03)^{16} \\[2ex] = 750 * 1.604706439 \\[2ex] = 1203.53 $
$A$ = $$1203.53$
(2.) Nahum invested $$10,000$ in a bank that pays $13.7\%$ compounded continuously.
How much money will he have after $2$ years?
If another bank will pay Nahum $14\%$ compounded quarterly, how much would he have after $2$ years?


Continuous Compound Interest
$ A = Pe^{rt} \\[2ex] P = 10000 \: dollars \\[2ex] t = 2 \: years \\[2ex] r = 13.7\% = 0.137 \\[2ex] A = 10000 * e^{0.137 * 2} \\[3ex] = 10000 * e^{0.274} \\[3ex] = 10000 * 1.315214802 \\[2ex] = 13152.15 $
$A$ = $$13152.15$

Compound Interest
$ A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[3ex] P = 10000 \: dollars \\[2ex] t = 2 \: years \\[2ex] m = 4 \\[2ex] r = 14\% = 0.14 \\[2ex] A = 10000 * \left(1 + \dfrac{0.14}{4}\right)^{4 * 2} \\[3ex] = 10000 * (1 + 0.035)^{8} \\[2ex] = 10000 * (1.035)^{8} \\[2ex] = 10000 * 1.316809037 \\[2ex] = 13168.09 $
$A$ = $$13168.09$
(3.) Matthew has $$1000$ to invest at $6\%$ per annum compounded quarterly.
How long will it take before he has $$1450$?
If the compounding is continuous, how long will it be?


Compound Interest
$ t = \dfrac{\log \left(\dfrac{A}{P}\right)}{m\log \left(1 + \dfrac{r}{m}\right)} \\[7ex] P = 1000 \: dollars \\[2ex] r = 6\% = 0.06 \\[2ex] A = 1450 \: dollars \\[2ex] m = 4 \\[2ex] 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] t = \dfrac{\log (1.45)}{4 * \log (1.015)} \\[7ex] t = \dfrac{0.161368002}{4 * 0.006466042} \\[7ex] t = \dfrac{0.161368002}{0.025864169} \\[5ex] t = 6.24 $
$t$ = $6.24$ years

Continuous Compound Interest
$ t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[5ex] P = 1000 \: dollars \\[2ex] r = 6\% = 0.06 \\[2ex] A = 1450 \: dollars \\[2ex] t = \dfrac{\ln \left({\dfrac{1450}{1000}}\right)}{0.06} \\[5ex] t = \dfrac{\ln {(1.45)}}{0.06} \\[5ex] t = \dfrac{0.371563556}{0.06} \\[5ex] t = 6.19 $
$t$ = $6.19$ 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?


$ r = m\left[(APY + 1)^{\dfrac{1}{m}} - 1\right] \\[3ex] APY = 7.5\% = 0.075 \\[2ex] m = 12 \\[2ex] r = 12 * \left[(0.075 + 1)^{\dfrac{1}{12}} - 1\right] \\[3ex] r = 12 * \left[(1.075)^{\dfrac{1}{12}} - 1\right] \\[3ex] r = 12 * \left[\sqrt[12]{1.075} - 1\right] \\[3ex] r = 12 * \left[1.006044919 - 1\right] \\[3ex] r = 12 * 0.006044919 \\[3ex] r = 0.072539028 $
$r$ = $7.25\%$
(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.


Continuous Compound Interest
$ r = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{t} \\[5ex] P = 2000 \: dollars \\[2ex] r = k \\[2ex] A = 2504.65 \: dollars \\[2ex] t = 5 \\[2ex] r = \dfrac{\ln \left({\dfrac{2504.65}{2000}}\right)}{5} \\[5ex] r = \dfrac{\ln (1.252325)}{5} \\[5ex] r = \dfrac{0.225001824}{5} \\[5ex] r = 0.045000365 \\[2ex] r = 4.5\% \\[7ex] After\: 10\: years; \\[2ex] t = 10 \\[2ex] A = ? \\[2ex] P = 2000 \: dollars \\[2ex] r = 0.045000365 \\[2ex] A = Pe^{rt} \\[5ex] A = 2000 * e^{0.045000365 * 10} \\[3ex] A = 2000 * e^{0.45000365} \\[3ex] A = 2000 * 1.568317907 \\[2ex] A = 3136.635813 \\[2ex] $ A = $3136.64


$ When\: the\: fund\: is\: doubled \\[2ex] P = 2000 \: dollars \\[2ex] r = 0.045000365 \\[2ex] A = 2 * 2000 = 4000 \: dollars \\[2ex] t = ? \\[2ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[5ex] t = \dfrac{\ln \left({\dfrac{4000}{2000}}\right)}{0.045000365} \\[5ex] t = \dfrac{\ln 2}{0.045000365} \\[5ex] t = \dfrac{0.693147181}{0.045000365} \\[5ex] t = 15.40314574 \\[2ex] t = 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.


Continuous Compound Interest
$ A = Pe^{rt} \\[2ex] P = 800 \: dollars \\[2ex] t = 8\: years \\[2ex] r = 2.9\% = 0.029 \\[2ex] A = 800 * e^{0.029 * 8} \\[3ex] = 800 * e^{0.232} \\[3ex] = 800 * 1.261119729 \\[2ex] = 1008.895783 $
$A$ = $$1008.90$

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 \\[2ex] P = 800 \: dollars \\[2ex] r = 0.029 \\[2ex] A = 2 * 800 = 1600 \: dollars \\[2ex] t = ? \\[2ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[5ex] t = \dfrac{\ln \left({\dfrac{1600}{800}}\right)}{0.029} \\[5ex] t = \dfrac{\ln 2}{0.029} \\[5ex] t = \dfrac{0.693147181}{0.0029} \\[5ex] t = 23.90162692 \\[2ex] t = 23.90\: years $