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] $

Samuel Dominic Chukwuemeka (SamDom For Peace) You may verify your answers as applicable with the: Mathematics of Finance Calculators
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(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 $