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

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.) ACT When a bank pays $I\%$ interest, compounded annually, a deposit of $\$P$ increases to $\$P\left(1 + \dfrac{I}{100}\right)^y$ at the end of $y$ years, where $y$ is a whole number.
Lou initially deposits $\$600$ in an account that pays $4.75\%$ interest compounded annually.
Lou does not make any further deposits or withdrawals.
How much money, in dollars, is in Lou's account after $8$ years?

$ A.\:\: 6 * (1 + 4.75)^8 \\[3ex] B.\:\: 6 * (1 + 0.0475)^8 \\[3ex] C.\:\: 600 * (1 + 4.75)^8 \\[3ex] D.\:\: 600 * (1 + 0.0475)^8 \\[3ex] E.\:\: 600 * (1 + 0.0475) * 8 \\[3ex] $

$ P = 600 \\[3ex] I = 4.75\% \\[3ex] \dfrac{I}{100} = \dfrac{4.75}{100} = 0.0475 \\[5ex] y = 8 \\[3ex] Amount = P\left(1 + \dfrac{I}{100}\right)^y \\[5ex] \therefore Amount = 600 * (1 + 0.0475)^8 $
(2.) An investment of $\$520.00$ was made for $63$ months in a financial institution that gives an interest rate of $3.5\%$.
Calculate the amounts and the dividends of the investment if the interest rate is compounded:
(I.) Annually
(II.) Semiannually
(III.) Quarterly
(IV.) Monthly
(V.) Weekly
(VI.) Ordinary daily
(VII.) Exact daily


$ P = \$520.00 \\[3ex] t = 63\:months = \dfrac{63}{12} = 5.25\:years \\[5ex] r = 3.5\% = \dfrac{3.5}{100} = 0.035 \\[5ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] (I.) \\[3ex] Annually \rightarrow m = 1 \\[3ex] A = 520\left(1 + \dfrac{0.035}{1}\right)^{1(5.25)} \\[5ex] A = 520(1 + 0.035)^{(5.25)} \\[4ex] A = 520(1.035)^{(5.25)} \\[4ex] A = 520(1.197944882) \\[3ex] A = 622.9313386 \\[3ex] A \approx \$622.93 \\[5ex] (II.) \\[3ex] Semiannually \rightarrow m = 2 \\[3ex] A = 520\left(1 + \dfrac{0.035}{2}\right)^{2(5.25)} \\[5ex] A = 520(1 + 0.0175)^{(10.5)} \\[4ex] A = 520(1.0175)^{(10.5)} \\[4ex] A = 520(1.19980699) \\[3ex] A = 623.899635 \\[3ex] A \approx \$623.90 \\[5ex] (III.) \\[3ex] Quarterly \rightarrow m = 4 \\[3ex] A = 520\left(1 + \dfrac{0.035}{4}\right)^{4(5.25)} \\[5ex] A = 520(1 + 0.00875)^{(21)} \\[4ex] A = 520(1.000875)^{(21)} \\[4ex] A = 520(1.200755273) \\[3ex] A = 624.3927418 \\[3ex] A \approx \$624.39 \\[5ex] (IV.) \\[3ex] Monthly \rightarrow m = 12 \\[3ex] A = 520\left(1 + \dfrac{0.035}{12}\right)^{12(5.25)} \\[5ex] A = 520(1 + 0.002916666667)^{(63)} \\[4ex] A = 520(1.002916667)^{(63)} \\[4ex] A = 520(1.201394002) \\[3ex] A = 624.7248812 \\[3ex] A \approx \$624.72 \\[5ex] (V.) \\[3ex] Weekly \rightarrow m = 52 \\[3ex] A = 520\left(1 + \dfrac{0.035}{52}\right)^{52(5.25)} \\[5ex] A = 520(1 + 0.0006730769231)^{(273)} \\[4ex] A = 520(1.000673077)^{(273)} \\[4ex] A = 520(1.20164108) \\[3ex] A = 624.8533613 \\[3ex] A \approx \$624.85 \\[5ex] (VI.) \\[3ex] Ordinary\:Daily \rightarrow m = 360 \\[3ex] A = 520\left(1 + \dfrac{0.035}{360}\right)^{360(5.25)} \\[5ex] A = 520(1 + 0.00009722222222)^{(1890)} \\[4ex] A = 520(1.000097222)^{(1890)} \\[4ex] A = 520(1.201704623) \\[3ex] A = 624.8864042 \\[3ex] A \approx \$624.89 \\[5ex] (VII.) \\[3ex] Exact\:Daily \rightarrow m = 365 \\[3ex] A = 520\left(1 + \dfrac{0.035}{365}\right)^{365(5.25)} \\[5ex] A = 520(1 + 0.00009589041096)^{(1916.25)} \\[4ex] A = 520(1.00009589)^{(1916.25)} \\[4ex] A = 520(1.20170477) \\[3ex] A = 624.8864806 \\[3ex] A \approx \$624.89 $
(3.) Habakkuk wants to have $\$16000$ in the future.
He intends to deposit some money in a company that pays $9\%$ interest rate compounded semiannually.
(a.) How much should he deposit $5$ years from now?
(b.) How much should he deposit $10$ years from now?


The question is asking us to calculate the principal.

$ (a.) \\[3ex] A = 16000 \\[3ex] r = 9\% = \dfrac{9}{100} = 0.09 \\[5ex] m = 2 \:\:(compounded\:\: semiannually) \\[3ex] t = 5 \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{16000}{\left(1 + \dfrac{0.09}{2}\right)^{2 * 5}} \\[7ex] P = \dfrac{16000}{(1 + 0.045)^{10}} \\[5ex] P = \dfrac{16000}{(1.045)^{10}} \\[5ex] P = \dfrac{16000}{1.552969422} \\[5ex] P = 10302.84291 \\[3ex] P = \$10302.84 \\[3ex] $ He should deposit $\$10302.84$ now in order to earn $\$16000.00$ in $5$ years ceteris paribus


$ (b.) \\[3ex] A = 16000 \\[3ex] r = 9\% = \dfrac{9}{100} = 0.09 \\[5ex] m = 2 \:\:(compounded\:\: semiannually) \\[3ex] t = 10 \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{16000}{\left(1 + \dfrac{0.09}{2}\right)^{2 * 10}} \\[7ex] P = \dfrac{16000}{(1 + 0.045)^{20}} \\[5ex] P = \dfrac{16000}{(1.045)^{20}} \\[5ex] P = \dfrac{16000}{2.411714025} \\[5ex] P = 6634.285755 \\[3ex] P = \$6634.29 \\[3ex] $ He should deposit $\$6634.29$ now in order to earn $\$16000.00$ in $10$ years ceteris paribus
(4.) WASSCE A principal of GH¢$5,600.00$ was deposited for $3$ years at compound interest.
If the interest earned was GH¢$1,200.00$, find, correct to $3$ significant figures, the interest rate per annum.


$ P = 5600 \\[3ex] t = 3 \\[3ex] CI = 1200 \\[3ex] r = ? \\[3ex] CI = A - P \\[3ex] A - P = CI \\[3ex] A = CI + P \\[3ex] A = 1200 + 5600 = 6800 \\[3ex] per\:\: annum\:\: means\:\: compounded\:\: annually \\[3ex] m = 1 \\[3ex] r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] \\[7ex] r = 1 * \left[\left(\dfrac{6800}{5600}\right)^{\dfrac{1}{1 * 3}} - 1\right] \\[7ex] r = \left(\dfrac{6800}{5600}\right)^{\dfrac{1}{3}} - 1 \\[5ex] r = (1.214285714)^{\dfrac{1}{3}} - 1 \\[5ex] r = \sqrt[3]{1.214285714} - 1 \\[3ex] r = 1.066858844 - 1 \\[3ex] r = 0.0668588443 \\[3ex] $ $r = 0.0669$ to $3$ s.f OR $r = 6.69\%$
(5.) CSEC Mr Williams bought a plot of land for $\$40,000$.
The value of the land appreciated by $7\%$ each year.
Calculate the value of the land after a period of two years.


We can solve this question in two ways - by using Quantitative Reasoning and the Compound Interest Formula

$ \underline{First\:\:Method - Quantitative\:\:Reasoning} \\[3ex] Present\:\:value = \$40000 \\[3ex] Appreciation\:\:rate = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] 1st\:\:year\:\:appreciation = 0.07(40000) = 2800 \\[3ex] New\:\:value = 40000 + 2800 = 42800 \\[3ex] 2nd\:\:year\:\:appreciation = 0.07(42800) = 2996 \\[3ex] Newest\:\:value = 42800 + 2996 = 45796 \\[3ex] $ The value of the land after two years is $\$45796.00$

This question can also be solved by the Compound Interest formula.
You can view it as $7\%$ interest compounded yearly/annually

$ \underline{Second\:\:Method - Compound\:\:Interest\:\:Formula} \\[3ex] P = 40000 \\[3ex] r = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] t = 2\:years \\[3ex] m = 1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 40000\left(1 + \dfrac{0.07}{1}\right)^{1(2)} \\[5ex] A = 40000(1 + 0.07)^2 \\[4ex] A = 40000(1.07)^2 \\[4ex] A = 40000(1.1449) \\[3ex] A = 45796 \\[3ex] $ The value of the land after two years is $\$45796.00$

Ask students their preferred method.
They should give reasons for their choices.
(6.) Five and a half years ago, Peter invested $\$20,000$ in a retirement fund that grew at the rate of $12.45\%$ per year compounded quarterly.
How much does he have in his account today?


$ t = 5\dfrac{1}{2}\: years = 5.5\: years \\[5ex] P = \$20000 \\[3ex] r = 12.45\% = \dfrac{12.45}{100} = 0.1245 \\[5ex] Compounded\:\:Quarterly \implies m = 4 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 20000 * \left(1 + \dfrac{0.1245}{4}\right)^{4(5.5)} \\[5ex] A = 20000(1 + 0.031125)^{22} \\[4ex] A = 20000(1.031125)^{22} \\[4ex] A = 20000(1.9626776) \\[3ex] A = 39253.552 \\[3ex] $ Peter has $\$39253.55$ in his account today.
(7.) How long will it take an investment of $\$10,000$ to double if the investment earns interest at the rate of $10\%$ per year compounded monthly?


$ P = \$10000 \\[3ex] A = double\:\:P = 2(10000) = \$20000 \\[3ex] r = 10\% = \dfrac{10}{100} = 0.1 \\[5ex] Compounded\:\:Monthly \implies m = 12 \\[3ex] t = ? \\[3ex] t = \dfrac{\log\left(\dfrac{A}{P}\right)}{m\log\left(1 + \dfrac{r}{m}\right)} \\[7ex] \dfrac{A}{P} = \dfrac{20000}{10000} = 2 \\[5ex] \dfrac{r}{m} = \dfrac{0.1}{12} = 0.0083333 \\[5ex] t = \dfrac{\log 2}{12 * \log(1 + 0.0083333)} \\[5ex] t = \dfrac{0.301029995}{12 * \log(1.0083333)} \\[5ex] t = \dfrac{0.301029995}{12 * 0.003604124} \\[5ex] t = \dfrac{0.301029995}{0.043249491} \\[5ex] t = 6.96031301 \\[3ex] t \approx 6.96 \\[3ex] $ It will take about $6.96\:years$ for the investment of $\$10,000$ to double at the rate of $10\%$ per year compounded monthly
(8.) CSEC A credit union pays $8\%$ per annum compound interest on all fixed deposits.
A customer deposited $\$24,000$ in an account.
Calculate the TOTAL amount of money in the account at the end of two years.


$ r = 8\% = \dfrac{8}{100} = 0.08 \\[5ex] P = \$24000 \\[3ex] A = ? \\[3ex] t = 2\:years \\[3ex] Compounded\:\:per\:\:annum \implies m = 1 \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = P\left(1 + \dfrac{0.08}{1}\right)^{1(2)} \\[5ex] A = 24000(1 + 0.08)^2 \\[4ex] A = 24000(1.08)^2 \\[4ex] A = 24000(1.1664) \\[3ex] A = 27993.6 \\[3ex] $ The total amount of money in the account at the end of two years is $\$27,993.60$
(9.) Joseph launched his construction business in $2008$
The revenue of his firm for that year was $\$250,000$
The revenue grew by $25\%$ in $2009$ and by $35\%$ in $2010$
Joseph projected that the revenue growth for his firm in the next $3$ years will be at least $30\%$ per year.
How much does he expect his minimum revenue to be for $2013$?


We can solve this question in two ways - by using Quantitative Reasoning and the Compound Interest Formula
Please note that for the second method - using the Compound Interest formula, we have to begin with $2011$ up until $2013$ because those are the years when he expects a steady rate of increase
So, our principal will be the amount at the end of $2010$

$ \underline{First\:\:Method - Quantitative\:\:Reasoning} \\[3ex] In \\[3ex] 2008;\:\: revenue = \$250000 \\[3ex] 2009;\:\:25\% growth = \dfrac{25}{100} * 250000 = 0.25 * 250000 = 62500 \\[5ex] New\:\:revenue = 250000 + 62500 = 312500 \\[3ex] 2010;\:\:35\% growth = \dfrac{35}{100} * 312500 = 0.35 * 312500 = 109375 \\[5ex] New\:\:revenue = 312500 + 109375 = 421875 \\[3ex] 2011;\:\:30\% growth = \dfrac{30}{100} * 421875 = 0.3 * 421875 = 126562.50 \\[5ex] New\:\:revenue = 421875 + 126562.50 = 548437.50 \\[3ex] 2012;\:\:30\% growth = \dfrac{30}{100} * 548437.50 = 0.3 * 548437.50 = 164531.25 \\[5ex] New\:\:revenue = 548437.50 + 164531.25 = 712968.75 \\[3ex] 2013;\:\:30\% growth = \dfrac{30}{100} * 712968.75 = 0.3 * 712968.75 = 213890.625 \\[5ex] New\:\:revenue = 712968.75 + 213890.625 = 926859.375 \\[3ex] $ He should expect a minimum revenue of $\$926,859.38$ in $2013$

$ \underline{Second\:\:Method - Compound\:\:Interest\:\:Formula} \\[3ex] Beginning\:\:from\:\:2011 \\[3ex] r = 30\% = \dfrac{30}{100} = 0.3 \\[5ex] P = \$421875 \\[3ex] t = 3\:years \\[3ex] Compounded\:\:per\:\:year \implies m = 1 \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = P\left(1 + \dfrac{0.3}{1}\right)^{1(3)} \\[5ex] A = 421875(1 + 0.3)^3 \\[4ex] A = 421875(1.3)^3 \\[4ex] A = 421875(2.197) \\[3ex] A = 926859.375 \\[3ex] $ He should expect a minimum revenue of $\$926,859.38$ in $2013$
(10.) JAMB If the population of a town was $240000$ in January $1998$ and it increased by $2\%$ each year, what would be the population of the town in January $2000$?

$ A.\:\: 480,000 \\[3ex] B.\:\: 249,696 \\[3ex] C.\:\: 249,600 \\[3ex] D.\:\: 244,800 \\[3ex] $

We can solve this question in two ways - by using Quantitative Reasoning and the Compound Interest Formula
Use any way that is faster for you.

$ \underline{First\:\:Method - Quantitative\:\:Reasoning} \\[3ex] From\:\:January,1998\:\:to\:\:January, 2000 \\[3ex] Population\:\:in\:\:1998 = 240000 \\[3ex] 2\%\:\:increase\:\:for\:\:1999 = \dfrac{2}{100} * 240000 = 2(2400) = 4800 \\[5ex] Population\:\:in\:\:1999 = 240000 + 4800 = 244800 \\[3ex] 2\%\:\:increase\:\:for\:\:2000 = \dfrac{2}{100} * 244800 = 2(2448) = 4896 \\[5ex] Population\:\:in\:\:2000 = 244800 + 4896 = 249696 \\[3ex] $ Ceteris paribus, the population of the town in January, $2000$ would be $249,696$ people

$ \underline{Second\:\:Method - Compound\:\:Interest\:\:Formula} \\[3ex] From\:\:January,1998\:\:to\:\:January, 2000 \implies 2\:years \\[3ex] P = 240000 \\[3ex] r = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] t = 2\:years \\[3ex] m = 1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 240000\left(1 + \dfrac{0.02}{1}\right)^{1(2)} \\[5ex] A = 240000(1 + 0.02)^2 \\[4ex] A = 240000(1.02)^2 \\[4ex] A = 240000(1.0404) \\[3ex] A = 249696 \\[3ex] $ Ceteris paribus, the population of the town in January, $2000$ would be $249,696$ people

Ask students their preferred method.
They should give reasons for their choices.
(11.) JAMB A man wishes to keep some money in a savings account at $25\%$ compound interest so that after $3$ years, he can buy a car for $₦150,000$
How much does he need to deposit now?

$ A.\:\: ₦112,000.50 \\[3ex] B.\:\: ₦96,000.00 \\[3ex] C.\:\: ₦85,714.28 \\[3ex] A.\:\: ₦76,800.00 \\[3ex] $

The question is asking us to calculate the principal.

$ r = 25\% = \dfrac{25}{100} = \dfrac{1}{4} \\[5ex] m = 1 \:\:(compounded\:\: annually) \\[3ex] t = 3 \\[3ex] A = 150000 \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] \dfrac{r}{m} = r \div m = \dfrac{1}{4} \div 1 = \dfrac{1}{4} \\[5ex] 1 + \dfrac{r}{m} = 1 + \dfrac{1}{4} = \dfrac{4}{4} + \dfrac{1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4} \\[5ex] mt = 1(3) = 3 \\[3ex] \left(1 + \dfrac{r}{m}\right)^{mt} = \left(\dfrac{5}{4}\right)^3 = \dfrac{5^3}{4^3} \\[7ex] P = \dfrac{150000}{\dfrac{5^3}{4^3}} \\[7ex] P = 150000 \div \dfrac{5^3}{4^3} \\[7ex] = 150000 * \dfrac{4^3}{5^3} \\[7ex] = \dfrac{150000 * 64}{5 * 5 * 5} \\[5ex] = \dfrac{30000 * 64}{5 * 5} \\[5ex] = \dfrac{6000 * 64}{5} \\[5ex] = 1200 * 64 \\[3ex] = 76800 \\[3ex] $ He should deposit $ ₦76,800.00$ now in order to earn $₦150,000$ in $3$ years ceteris paribus.
(12.) CSEC Faye borrowed $\$9\:600$ at $8\%$ per annum compound interest.
(i) Calculate the interest on the loan for the first year.
At the end of the first year, she repaid $\$4\:368$.
(ii) How much did she still owe at the beginning of the second year?
(iii) Calculate the interest on the remaining balance for the second year.


$ P = \$9600 \\[3ex] r = 8\% = \dfrac{8}{100} = 0.08 \\[5ex] (i) \\[3ex] t = 1\:year \\[3ex] Compounded\:\:annually \rightarrow m = 1 \\[3ex] A = ? \\[3ex] CI = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 9600\left(1 + \dfrac{0.08}{1}\right)^{1(1)} \\[5ex] = 9600(1 + 0.08)^{1} \\[4ex] = 9600(1.08)^{1} \\[4ex] = 9600(1.08) \\[3ex] A = 10368 \\[3ex] CI = A - P \\[3ex] CI = 10368 - 9600 \\[3ex] CI = \$768 \\[3ex] $ The interest on the loan for the first year is $\$768$

$ (ii) \\[3ex] \underline{End\:\:of\:\:First\:\:Year} \\[3ex] Repaid\:\: \$4368 \\[3ex] \underline{Beginning\:\:of\:\:Second\:\:Year} \\[3ex] Balance = A - 4368 \\[3ex] Balance = 10368 - 4368 \\[3ex] Balance = \$6000 \\[3ex] $ She still owes $\$6000$ at the beginning of the second year.

$ (iii) \\[3ex] \underline{Second\:\:Year} \\[3ex] Balance = P = \$6000 \\[3ex] r = 8\% = 0.08 \\[3ex] t = 1\:year \\[3ex] m = 1 \\[3ex] A = ? \\[3ex] CI = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 6000\left(1 + \dfrac{0.08}{1}\right)^{1(1)} \\[5ex] = 6000(1 + 0.08)^{1} \\[4ex] = 6000(1.08)^{1} \\[4ex] = 6000(1.08) \\[3ex] A = 6480 \\[3ex] CI = A - P \\[3ex] CI = 6480 - 6000 \\[3ex] CI = \$480 \\[3ex] $ The interest on the remaining balance on the loan for the second year is $\$480$
(13.) Mr. Williams bought a plot of land for $\$40,000$.
The value of the land appreciated by $7\%$ each year.
Calculate the value of the land after a $30-year$ period.
Compare to Question $5$.
Do you see the importance of using the Compound Interest formula?


For this question, it is much better to use the Compound Interest Formula

$ P = 40000 \\[3ex] r = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] t = 30\:years \\[3ex] m = 1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 40000\left(1 + \dfrac{0.07}{1}\right)^{1(30)} \\[5ex] A = 40000(1 + 0.07)^{30} \\[4ex] A = 40000(1.07)^{30} \\[4ex] A = 40000(7.61225504) \\[3ex] A = 304490.202 \\[3ex] A \approx \$304,490.20 \\[3ex] $ The value of the land after thirty years is $\$304,490.20$
(14.) Five and a half years ago, Peter invested $\$20,000$ which is worth $\$39,253.55$ today if the rate is compounded quarterly.
What rate of interest is used?
Compare and confirm your answer with Question $(6.)$


$ t = 5\dfrac{1}{2}\: years = 5.5\: years \\[5ex] P = \$20000 \\[3ex] A = \$39253.55 \\[3ex] Compounded\:\:Quarterly \implies m = 4 \\[3ex] r = ? \\[3ex] r = m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] \\[7ex] mt = 4(5.5) = 22 \\[3ex] \dfrac{1}{mt} = \dfrac{1}{22} = 0.0454545455 \\[5ex] \dfrac{A}{P} = \dfrac{39253.55}{20000} = 1.9626775 \\[5ex] \left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} = 1.9626775^0.0454545455 = 1.031125 \\[7ex] \left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1 = 1.031125 - 1 = 0.031125 \\[7ex] m\left[\left(\dfrac{A}{P}\right)^{\dfrac{1}{mt}} - 1\right] = 4(0.031125) = 0.1245 \\[7ex] 0.1245\:\:to\:\:\% = 0.1245(100) = 12.45\% \\[3ex] r = 12.45\% \\[3ex] $ The rate of interest is $12.45\%$
(15.) The John's family spends $\$960$ per month on food.
How much would they spend on food per month if inflation occurs at the rate of $3\%$ per year over the next $5$ years?


$ P = 960 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 5\:years \\[3ex] m = 1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 960\left(1 + \dfrac{0.03}{1}\right)^{1(5)} \\[5ex] A = 960(1 + 0.03)^{5} \\[4ex] A = 960(1.03)^{5} \\[4ex] A = 960(1.159274074) \\[3ex] A = 1112.903111 \\[3ex] A \approx \$1,112.90 $
(16.) A woman just inherited $\$250,000$.
If she invests the money at $4.5\%$ at compound interest, what can she expect to have at the end of $15$ years when she retires?


Sometimes; if the question does not state that the interest is simple interest, then assume compound interest.

If the question does not state the number of compounded periods per year for which the interest is compounded, then assume that the interest is compounded annually.

$ \underline{Compound\:\:Interest} \\[3ex] P = \$250000 \\[3ex] r = 4.5\% = \dfrac{4.5}{100} = 0.045 \\[5ex] t = 15\:years \\[3ex] Compounded\:\:annually\rightarrow m = 1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[7ex] = 250000\left(1 + \dfrac{0.045}{1}\right)^{1 * 15} \\[7ex] = 250000(1 + 0.045)^{15} \\[5ex] = 250000(1.045)^{15} \\[5ex] = 250000(1.93528244) \\[3ex] = \$483820.61 \\[3ex] $ Ceteris paribus, she would expect to have $\$483,820.61$ at the end of $15$ years when she retires.
(17.) The consumption of solar energy is expected to increase by $7\%$ per year during the next decade.
Assume Sun Systems services the City of Surprise, Arizona.
By how much will the company need to increase its capacity in order to meet demand at the end of the decade?


$ r = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] t = 10\:years \\[3ex] m = 1 \\[3ex] P = ? \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = P\left(1 + \dfrac{0.07}{1}\right)^{1(10)} \\[5ex] A = P(1 + 0.07)^{10} \\[4ex] A = P(1.07)^{10} \\[4ex] A = P(1.967151357) \\[3ex] A = 1.967151357P \\[3ex] Amount\:\:of\:\:increase = New - Initial \\[3ex] Amount\:\:of\:\:increase = A - P \\[3ex] Amount\:\:of\:\:increase = 1.967151357P - P \\[3ex] Amount\:\:of\:\:increase = 0.967151357P \\[3ex] Amount\:\:of\:\:increase\:\:as\:\:a\:\:percent \\[3ex] = 0.967151357P * 100 \\[3ex] = 96.7151357P\% \\[3ex] $ By the end of the decade, the company should increase its present capacity by about $96.72\%$
(18.) 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 = ? \\[3ex] P = 750 \: dollars \\[3ex] t = 16 \: years \\[3ex] Compounded\:\:annually\rightarrow m = 1 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[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] A \approx \$1,203.53 \\[3ex] $ Ceteris paribus, the amount in the prepaid account at the end of sixteen years will be $\$1203.53$
(19.) The inflation rates in the United States economy for $2004$ through $2007$ are $2.3\%$, $2.7\%$, $3.4\%$, and $3.2\%$ respectively.
What was the purchasing power of a dollar at the beginning of $2007$ compared to that at the beginning of $2004$?


We can solve this question in two ways.
Use any method you prefer.
First Method: Calculate the present value/principal for each year starting from $2004$

$ \underline{2004} \\[3ex] r = 2.3\% = \dfrac{2.3}{100} = 0.023 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] A = \$1 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{1}{\left(1 + \dfrac{0.023}{1}\right)^{1 * 1}} \\[7ex] P = \dfrac{1}{(1 + 0.023)^{1}} \\[5ex] P = \dfrac{1}{(1.023)^{1}} \\[5ex] P = \dfrac{1}{1.023} \\[5ex] P = 0.977517106 \\[5ex] \underline{2005} \\[3ex] r = 2.7\% = \dfrac{2.7}{100} = 0.027 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] A = \$1 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{1}{\left(1 + \dfrac{0.027}{1}\right)^{1 * 1}} \\[7ex] P = \dfrac{1}{(1 + 0.027)^{1}} \\[5ex] P = \dfrac{1}{(1.027)^{1}} \\[5ex] P = \dfrac{1}{1.027} \\[5ex] P = 0.973709834 \\[5ex] \underline{2006} \\[3ex] r = 3.4\% = \dfrac{3.4}{100} = 0.034 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] A = \$1 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{1}{\left(1 + \dfrac{0.034}{1}\right)^{1 * 1}} \\[7ex] P = \dfrac{1}{(1 + 0.034)^{1}} \\[5ex] P = \dfrac{1}{(1.034)^{1}} \\[5ex] P = \dfrac{1}{1.034} \\[5ex] P = 0.967117988 \\[5ex] \underline{2007} \\[3ex] r = 3.2\% = \dfrac{3.2}{100} = 0.032 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] A = \$1 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] P = \dfrac{1}{\left(1 + \dfrac{0.032}{1}\right)^{1 * 1}} \\[7ex] P = \dfrac{1}{(1 + 0.032)^{1}} \\[5ex] P = \dfrac{1}{(1.032)^{1}} \\[5ex] P = \dfrac{1}{1.032} \\[5ex] P = 0.968992248 \\[5ex] Purchasing\:\:power\:\:at\:\:the\:\:beginning\:\:of\:\:2007\:\:compared\:\:to\:\:the\:\:beginning\:\:of\:\:2004 \\[3ex] = 0.977517106 * 0.973709834 * 0.967117988 * 0.968992248 \\[3ex] = 0.891977061 \\[3ex] \approx \$0.89 \\[3ex] $ Second Method: Let the purchasing power of a dollar at the beginning of $2007$ compared to that at the beginning of $2004$ = $p$

$ \underline{2004} \\[3ex] r = 2.3\% = \dfrac{2.3}{100} = 0.023 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] P = \$1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 1\left(1 + \dfrac{0.023}{1}\right)^{1(1)} \\[5ex] A = 1(1 + 0.023)^{1} \\[4ex] A = 1(1.023)^{1} \\[4ex] A = 1(1.023) \\[3ex] A = 1.023 \\[5ex] \underline{2005} \\[3ex] r = 2.7\% = \dfrac{2.7}{100} = 0.027 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] P = \$1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 1\left(1 + \dfrac{0.027}{1}\right)^{1(1)} \\[5ex] A = 1(1 + 0.027)^{1} \\[4ex] A = 1(1.027)^{1} \\[4ex] A = 1(1.027) \\[3ex] A = 1.027 \\[5ex] \underline{2006} \\[3ex] r = 3.4\% = \dfrac{3.4}{100} = 0.034 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] P = \$1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 1\left(1 + \dfrac{0.034}{1}\right)^{1(1)} \\[5ex] A = 1(1 + 0.034)^{1} \\[4ex] A = 1(1.034)^{1} \\[4ex] A = 1(1.034) \\[3ex] A = 1.034 \\[5ex] \underline{2007} \\[3ex] r = 3.2\% = \dfrac{3.2}{100} = 0.032 \\[5ex] t = 1\:year \\[3ex] m = 1 \\[3ex] P = \$1 \\[3ex] A = ? \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] A = 1\left(1 + \dfrac{0.032}{1}\right)^{1(1)} \\[5ex] A = 1(1 + 0.032)^{1} \\[4ex] A = 1(1.032)^{1} \\[4ex] A = 1(1.032) \\[3ex] A = 1.032 \\[5ex] (1.023)(1.027)(1.034)(1.032) * p = 1 \\[3ex] 1.121105062 * p = 1 \\[3ex] p = \dfrac{1}{1.121105062} \\[5ex] p = 0.8919770628 \\[3ex] p \approx \$0.89 \\[3ex] $ The purchasing power of a dollar at the beginning of $2007$ compared to that at the beginning of $2004$ is approximately eighty nine cents.
(20.) If the inflation rate is $2.95\%$ compounded annually, how long will it take for prices to double?


$ r = 2.95\% = \dfrac{2.95}{100} = 0.0295 \\[5ex] Compounded\:\:annually\rightarrow m = 1 \\[3ex] P = P \\[3ex] A = 2P...double\:\:P \\[3ex] t = ? \\[3ex] t = \dfrac{\log\left(\dfrac{A}{P}\right)}{m\log\left(1 + \dfrac{r}{m}\right)} \\[7ex] t = \dfrac{\log\left(\dfrac{2P}{P}\right)}{1 * \log\left(1 + \dfrac{0.0295}{1}\right)} \\[7ex] t = \dfrac{\log 2}{1 * \log(1 + 0.0295)} \\[5ex] t = \dfrac{\log 2}{1 * \log(1.0295)} \\[5ex] t = \dfrac{\log 2}{\log(1.0295)} \\[5ex] t = \dfrac{0.3010299957}{0.01262635095} \\[5ex] t = 23.84140888 \\[3ex] t \approx 23.84\:years $




Top




(21.)

$ P = ₦400 \\[3ex] SI = ₦24 \\[3ex] t = 3\:years \\[3ex] r = ? \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{24}{400(3)} \\[5ex] r = \dfrac{24}{1200} \\[5ex] r = \dfrac{2}{100} \\[5ex] r = 2\% \\[3ex] $ The interest on $₦400$ will increase to $₦24$ in $3$ years at the rate of $2\%$
(22.) CMAT A certain amount of money invested at $10\%$ per annum compound interest for two years became $Rs\:2000$.
What is the initial investment?

$ 1.\:\: Rs.\:856 \\[3ex] 2.\:\: Rs.\:1,625 \\[3ex] 3.\:\: Rs.\:1,653 \\[3ex] 4.\:\: Rs.\:1,275 \\[3ex] $

$ r = 10\% = \dfrac{10}{100} = \dfrac{1}{10} \\[5ex] m = 1 \\[3ex] t = 2 \\[3ex] A = 2000 \\[3ex] P = \dfrac{A}{\left(1 + \dfrac{r}{m}\right)^{mt}} \\[7ex] mt = 1 * 2 = 2 \\[3ex] \dfrac{r}{m} = r \div m = \dfrac{1}{10} \div 1 = \dfrac{1}{10} \\[5ex] 1 + \dfrac{r}{m} = 1 + \dfrac{1}{10} = \dfrac{10}{10} + \dfrac{1}{10} = \dfrac{10 + 1}{10} = \dfrac{11}{10} \\[5ex] \left(1 + \dfrac{r}{m}\right)^{mt} = \left(\dfrac{11}{10}\right)^2 = \dfrac{121}{100} \\[5ex] \rightarrow P = 2000 \div \dfrac{121}{100} \\[5ex] P = 2000 * \dfrac{100}{121} \\[5ex] P \approx 1,653 $