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 Dominic Chukwuemeka

Solved Examples: Simple Interest

Pre-requisites: Measurements and Units
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.
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Interpret your solutions.

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.) Dorcas deposited $3700 in an account with an annual simple interest of 3.4% for 15 years.
How much will she have at the end of 15 years?

$ P = \$3700 \\[3ex] r = 3.4\% = \dfrac{3.4}{100} = 0.034 \\[5ex] t = 15\:years \\[3ex] SI = P * r * t \\[3ex] A = P(1 + rt) \\[3ex] A = 3700(1 + 0.034 * 15) \\[3ex] A = 3700(1 + 0.51) \\[3ex] A = 3700(1.51) \\[3ex] A = 5587.00 \\[3ex] $ Ceteris paribus, she will have $5587.00 at the end of 15 years.

(2.) How much should Jonah invest at 3.7% simple interest in order to earn 85 interest in 12 months?

$ P = ? \\[3ex] r = 3.7\% = \dfrac{3.7}{100} = 0.037 \\[5ex] SI = \$85 \\[3ex] t = 12\:months = 1\:year \\[3ex] P = \dfrac{SI}{rt} \\[5ex] P = \dfrac{85}{0.037 * 1} \\[5ex] P = \dfrac{85}{0.037} \\[5ex] P = 2297.297297 \\[3ex] P \approx \$2,297.30 $

(3.) CSEC Mr. Adams invested $5000 at the credit union and received $5810 inclusive of simple interest, after 3 years.
Determine
(i) the simple interest earned

(ii) the annual interest rate paid by the credit union

(iii) the length of time it will take for Mr. Adams' investment to be doubled, at the same rate of interest.

$ P = \$5000 \\[3ex] A = \$5810 \\[3ex] t = 3\:years \\[3ex] (i) \\[3ex] A = P + SI \\[3ex] SI = A - P \\[3ex] SI = 5810 - 5000 \\[3ex] SI = \$810 \\[3ex] $ The simple interest earned from an investment of $\$5000$ at the credit union will be $\$810$ after $3$ years.

$ (ii) \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{810}{5000(3)} \\[5ex] r = \dfrac{810}{15000} \\[5ex] r = 0.054 \\[3ex] r = 5.4\% \\[3ex] $ The annual or yearly interest rate paid by the credit union is $5.4\%$

$ (iii) \\[3ex] P = \$5000 \\[3ex] A = 2(5000) = \$10000 \\[3ex] r = 0.054 \\[3ex] t = ? \\[3ex] t = \dfrac{A - P}{Pr} \\[5ex] t = \dfrac{10000 - 5000}{5000(0.054)} \\[5ex] t = \dfrac{5000}{270} \\[5ex] t = 18.5185185 \\[3ex] t \approx 18.5\:years \\[3ex] $ It will take approximately $18.5\:years$ to double the investment at $5.4\%$ interest rate.

(4.) JAMB A sum of money was invested at $8\%$ per annum simple interest.
If after $4$ years, the money amounts to $₦330.00$, find the amount originally invested.

$ F.\:\: ₦180.00 \\[3ex] G.\:\: ₦165.00 \\[3ex] H.\:\: ₦150.00 \\[3ex] J.\:\: ₦200.00 \\[3ex] K.\:\: ₦250.00 \\[3ex] $

$ r = 18\% = \dfrac{8}{100} = 0.08 \\[5ex] t = 4\:years \\[3ex] A = ₦330 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{1 + rt} \\[5ex] P = \dfrac{330}{1 + (0.08)(4)} \\[5ex] P = \dfrac{330}{1 + 0.32} \\[5ex] P = \dfrac{330}{1.32} \\[5ex] P = ₦250.00 \\[3ex] $ $₦250.00$ was initially invested.

(5.) CSEC Mr Mitchell deposited $\$40,000$ in a bank and earned simple interest at $7\%$ per annum for two years.
Calculate the amount he will receive at the end of the two-year period.

$ P = \$40000 \\[3ex] r = 7\% = \dfrac{7}{100} = 0.07 \\[5ex] t = 2\:years \\[3ex] A = ? \\[3ex] A = P(1 + rt) \\[3ex] A = 40000[1 + (0.07)(2)] \\[3ex] A = 40000(1 + 0.14) \\[3ex] A = 40000(1.14) \\[3ex] A = \$45,600 \\[3ex] $ Mr Mitchell will receive $\$45,600.00$ at the end of the two-year period.

(6.) Paul purchased $\$16,000$ worth of $52-week$ treasury bills (T-bills) for $\$15,750$.
What is the rate of return on his investment?

$ t = 52\:weeks = 1\:year \\[3ex] A = \$16000 \\[3ex] P = \$15750 \\[3ex] r = ? \\[3ex] r = \dfrac{A - P}{Pt} \\[5ex] r = \dfrac{16000 - 15750}{15750(1)} \\[5ex] r = \dfrac{250}{15750} \\[5ex] r = 0.0158730159 \\[3ex] r \approx 1.59\% \\[3ex] $ The rate of return on Paul's investment is $1.59\%$

(7.) SouthEast Bank pays simple interest at the rate of $6\%$ per year.
Deborah deposited a sum of money which grew to $\$1300$ in $8$ months.
How much did she deposit?

$ r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] P = ? \\[3ex] A = \$1300 \\[3ex] t = 8\:months = \dfrac{8}{12} \\[5ex] P = \dfrac{A}{1 + rt} \\[5ex] rt = 0.06 * \dfrac{8}{12} = 0.04 \\[5ex] 1 + rt = 1 + 0.04 = 1.04 \\[3ex] P = \dfrac{1300}{1.04} \\[5ex] P = 1250 \\[3ex] $ Deborah deposited a sum of $\$1250.00$

(8.) How many days would it take for a sum of $\$1300$ to earn $\$26$ interest if it is invested at an interest rate of $4\%$ per annum?
Use a $360-day$ year.

Per annum means Per year

$ P = \$1300 \\[3ex] SI = \$26 \\[3ex] r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = ? \\[3ex] t = \dfrac{SI}{Pr} \\[5ex] t = \dfrac{26}{1300(0.04)} \\[5ex] t = \dfrac{26}{52} \\[5ex] t = 0.5\:year \\[3ex] 0.5\:year\:\:to\:\:days = 0.5(360) = 180\:\:days \\[3ex] t = 180\:\:days \\[3ex] $ It would take $180$ days for a sum of $\$1300$ to earn $\$26$ interest if it is invested at an interest rate of $4\%$ per year

(9.) CSEC A loan of $\$12,000$ was borrowed from a bank at $14\%$ per annum.

Calculate
(i) the interest on the loan at the end of the first year

(ii) the total amount owing at the end of the first year.

A repayment of $\$7,800$ was made at the start of the second year.

Calculate
(iii) the amount still outstanding at the start of the second year.

(iv) the interest on the outstanding amount at the end of the second year.

$ P = \$12000 \\[3ex] r = 14\% = \dfrac{14}{100} = 0.14 \\[5ex] (i) \\[3ex] End\:\:of\:\:first\:\:year \implies t = 1\:year \\[3ex] SI = Prt \\[3ex] SI = 12000 * 0.14 * 1 \\[3ex] SI = \$1680 \\[3ex] $ The interest on the loan at the end of the first year is $\$1680.00$

$ (ii) \\[3ex] A = P + SI \\[3ex] A = 12000 + 1680 \\[3ex] A = \$13680 \\[3ex] $ The total amount owing at the end of the first year is $\$13,680.00$

$ (iii) \\[3ex] \underline{Beginning\:\:of\:\:second\:\:year} \\[3ex] Amount\:\:owed = 13680 \\[3ex] Repayment = 7800 \\[3ex] Balance = 13680 - 7800 \\[3ex] Balance = \$5880 \\[3ex] $ The amount still outstanding at the start of the second year is $\$5,880.00$
This balance is now the new principal.
Interest is still being charged on this balance.

$ (iv) \\[3ex] \underline{End\:\:second\:\:year} \\[3ex] P = \$5880 \\[3ex] r = 14\% = \dfrac{14}{100} = 0.14 \\[5ex] End\:\:of\:\:first\:\:year \implies t = 1\:year \\[3ex] SI = Prt \\[3ex] SI = 5880 * 0.14 * 1 \\[3ex] SI = \$823.20 \\[3ex] $ The interest on the outstanding amount at the end of second year is $\$823.20$

(10.) ACT Jamal invested $\$1,000$ on January $1$.
At the end of $9$ months, during which time Jamal made no withdrawals and no other deposits, the investment has earned $\$75$ in interest.
Jamal's $\$1,000$ investment returned an annual percentage yield closest to which of the following percents?
(Note: Interest can be estimated using $I = Prt$, where $I$ is the amount of interest earned; $P$ is the amount of money initially invested; $r$ is the annual percentage yield that the money returned; and $t$ is the time, in years, the money is invested.)

$ F.\:\: 12\% \\[3ex] G.\:\: 11\% \\[3ex] H.\:\: 10\% \\[3ex] J.\:\: 8\% \\[3ex] K.\:\: 7\% \\[3ex] $

Recall what we discussed about Word Problems, and how you should not skip them because they are lengthy, but rather read to understand, paraphrase, and shorten them.

Ask your students to paraphrase and shorten that question.
Some of the responses may be:

(1.) Determine the rate if the principal is $\$1000$, time is $9$ months, and simple interest is $\$75$

(2.) Determine $r$ if $P = 1000, t = 9\:months, SI = 75$

(3.) If an investment of a thousand dollars yields seventy five dollars as interest over a period of nine months, what interest rate was used?...among others

$ P = \$1000 \\[3ex] t = 9\:months = \dfrac{9}{12}\:years = 0.75\:year \\[5ex] SI = \$75 \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{75}{1000(0.75)} \\[5ex] r = \dfrac{75}{750} \\[5ex] r = 0.1 \\[3ex] Convert\:\:to\:\:percent \\[3ex] 0.1 * 100 = 10\% \\[5ex] r = 10\% \\[3ex] $ The annual percentage yield is $10\%$

(11.) WASCCE The interest to be paid by a borrower was reduced from $2\dfrac{1}{2}\%$ to $2\%$.

If he paid $₦500.00$ less, calculate the amount borrowed.

The time was not given.
But, we do know that the time is the same in both cases (what he should have paid and what he actually paid)
Assume the time to be 1 year

$ \underline{What\:\:he\:\:should\:\:have\:\:paid} \\[3ex] r = 2\dfrac{1}{2}\% = 2.5\% = \dfrac{2.5}{100} = 0.025 \\[5ex] t = 1\:year \\[3ex] P = ? \\[3ex] SI = P * r * 1 \\[3ex] SI = P * 0.025 * 1 \\[3ex] SI = 0.025P \\[3ex] \underline{What\:\:he\:\:paid} \\[3ex] r = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] t = 1\:year \\[3ex] P = ? \\[3ex] SI = P * r * 1 \\[3ex] SI = P * 0.02 * 1 \\[3ex] SI = 0.02P \\[3ex] $ But, the interest he paid is $₦500.00$ less than what he should have paid

$ \therefore 0.025P - 500 = 0.02P \\[3ex] 0.025P - 0.02P = 500 \\[3ex] 0.005P = 500 \\[3ex] P = \dfrac{500}{0.005} \\[5ex] P = ₦100,000 \\[3ex] $ He borrowed ₦100,000.00

$ \underline{Check} \\[3ex] Supposed\:\:to\:\:pay \implies 2.5\% \:\:of\:\: 100000 = 0.025 * 100000 = ₦2500 \\[3ex] Actually\:\:paid \implies 2\% \:\:of\:\: 100000 = 0.02 * 100000 = ₦2000 \\[3ex] ₦2500- ₦2000 = ₦500 $

(12.) Luke owns $\$20,000$ worth of $10-year$ bonds of the City of Truth or Consequences, New Mexico
The bonds pay interest every $6$ months at the simple interest rate of $2\%$ per year.
(a.) How much interest will Luke receive every $6$ months?
(b.) How much will he receive over the life of the bonds?

$ P = \$20000 \\[3ex] r = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] (a.) \\[3ex] t = 6\:months = \dfrac{6}{12} = 0.5\: year \\[5ex] I = ? \\[3ex] I = P * r * t \\[3ex] I = 20000 * 0.02 * 0.5 \\[3ex] I = \$200 \\[3ex] $ Luke will receive $\$200.00$ every $6$ months

$ (b.) \\[3ex] t = 10\:years \\[3ex] I = P * r * t \\[3ex] I = 20000 * 0.02 * 10 \\[3ex] I = \$4000 \\[3ex] $ Luke will receive $\$4,000.00$ over the life of the bonds.

(13.) JAMB A man invested a total of $₦50,000$ in two companies.
If these companies pay dividend of $6\%$ and $8\%$ respectively, how much did he invest at $8\%$ if the total yield is $₦3,700$?

$ A.\:\: ₦15,000 \\[3ex] B.\:\: ₦29,600 \\[3ex] C.\:\: ₦21,400 \\[3ex] D.\:\: ₦27,800 \\[3ex] E.\:\: ₦35,000 \\[3ex] $

$(1.)$ Let the investment (Principal) at the $6\%$ rate be $x$
$(2.)$ Let the investment (Principal) at the $8\%$ rate be $y$

$ x + y = 50000 \\[3ex] \rightarrow y = 50000 - x \\[3ex] $ $(2.)$ So, the investment on the $8\%$ rate is $50000 - x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:6\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:8\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] 8\% = \dfrac{8}{100} = 0.08 \\[5ex] Let\:\:t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$x$	$0.06$	$1$	$0.06x$
$B$	$50000 - x$	$0.08$	$1$	$0.08(50000 - x)$
$Total$				$3700$

$ \rightarrow 0.06x + 0.08(50000 - x) = 3700 \\[3ex] 0.06x + 4000 - 0.08x = 3700 \\[3ex] -0.02x = 3700 - 4000 \\[3ex] -0.02x = -300 \\[3ex] x = \dfrac{-300}{-0.02} \\[5ex] x = ₦15000.00 \\[3ex] y = 50000 - x \\[3ex] y = 50000 - 15000 \\[3ex] y = ₦35000.00 \\[3ex] $ He invested $₦15,000.00$ at $6\%$ interest rate and $₦35,000.00$ at $8\%$ interest rate in order to earn $₦3,700.00$ interest.

(14.) JAMB A man invests a sum of money at $4\%$ per annum simple interest.
After $3$ years, the principal amounts to $₦7,000.00$.
Find the sum invested.

$ A.\:\: ₦7,840.00 \\[3ex] B.\:\: ₦6,250.00 \\[3ex] C.\:\: ₦6,160.00 \\[3ex] D.\:\: ₦5,833.33 \\[3ex] $

$ r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = 3\: years \\[3ex] A = ₦7000 \\[3ex] P = ? \\[3ex] P = \dfrac{A}{1 + rt} \\[5ex] P = \dfrac{7000}{1 + (0.04)(3)} \\[5ex] P = \dfrac{7000}{1 + 0.12} \\[5ex] P = \dfrac{7000}{1.12} \\[5ex] P = ₦6250 \\[3ex] $ The sum invested was $₦6,250.00$

(15.) Sarah has $\$2250$ in her savings account at the end of a certain period of time.
She invested $\$1525$ at a $3.75\%$ simple annual interest rate.
How long, in years, was her money invested?

$ A = \$2250 \\[3ex] P = \$1525 \\[3ex] r = 3.75\% = \dfrac{3.75}{100} = 0.0375 \\[5ex] t = ? \\[3ex] t = \dfrac{A - P}{Pr} \\[5ex] t = \dfrac{2250 - 1525}{1525(0.0375)} \\[5ex] t = \dfrac{725}{57.1875} \\[5ex] t = 12.67759563 \\[3ex] t \approx 12.68\:years $

(16.) A retail store in the town of Neversink, New York charges a $23\%$ annual rate for overdue accounts.
How much interest is owed on a $\$912$ account that is $2$ months overdue?

$ r = 23\% = \dfrac{23}{100} = 0.23 \\[5ex] P = \$912 \\[3ex] t = 2\:months = \dfrac{2}{12} = \dfrac{1}{6} \\[5ex] I = P * r * t \\[3ex] I = 912 * 0.23 * \dfrac{1}{6} \\[5ex] I = \dfrac{209.76 * 1}{6} \\[5ex] I = \$34.96 \\[3ex] $ The interest owed on a $\$912$ account that is $2$ months overdue is $\$34.96$

(17.) Hosea borrowed $\$2000$ and has to pay $289.05 for 7 months to repay the loan.
(a.) How much simple interest is paid on the loan?
(b.) What APR, rounded to the nearest hundredth of a percent, would give this interest with a principal of $2000 and a term of 7 months?

$ P = \$2000 \\[3ex] A = 289.05\:\:per\:\:month\:\:for\:\:7\:\:months = 289.05(7) = \$2023.35 \\[3ex] t = 7\:months = \dfrac{7}{12}\:years \\[5ex] (a.) \\[3ex] SI = ? \\[3ex] SI = A - P \\[3ex] SI = 2023.35 - 2000 \\[3ex] SI = \$23.35 \\[3ex] (b.) \\[3ex] r = ? \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = SI \div Pt \\[3ex] r = 23.35 \div \left(2000 * \dfrac{7}{12}\right) \\[5ex] r = 23.35 \div \dfrac{14000}{12} \\[5ex] r = 23.35 * \dfrac{12}{14000} \\[5ex] r = \dfrac{280.2}{14000} \\[5ex] r = 0.02001428571 \\[3ex] to\:\:percent = 0.02001428571(100) \\[3ex] r = 2.001428571\% \\[3ex] r \approx 2.00\% \\[3ex] r \approx 2\% $

(18.) A check for $\$4270$ was used to pay up a $10-month$ $\$4000$ loan.
What annual rate of interest was charged?

$ A = \$4270 \\[3ex] t = 10\:months = \dfrac{10}{12} = \dfrac{5}{6}\:year \\[5ex] P = \$4000 \\[3ex] r = \dfrac{A - P}{Pt} \\[5ex] Pt = 4000 * \dfrac{5}{6} = 2000 * \dfrac{5}{3} = \dfrac{10000}{3} \\[5ex] A - P = 4270 - 4000 = 270 \\[3ex] r = 270 \div \dfrac{10000}{3} \\[5ex] r = 270 * \dfrac{3}{10000} \\[5ex] r = \dfrac{27 * 3}{1000} \\[5ex] r = \dfrac{81}{1000} \\[5ex] Convert\:\:to\:\:percent \\[3ex] \dfrac{81}{1000} * 100 = \dfrac{81}{10} = 8.1\% \\[5ex] $ The interest rate that was charged is $8.1\%$

(19.) WASSCE If the simple interest on a certain amount of money saved in a bank for $5$ years at $2\dfrac{1}{2}\%$ per annum is $₦500.00$, calculate the total amount due after $6$ years at the same rate.

$ A.\:\: ₦2,500.00 \\[3ex] B.\:\: ₦2,600.00 \\[3ex] C.\:\: ₦4,500.00 \\[3ex] D.\:\: ₦4,600.00 \\[3ex] $

This question has two cases - two time periods.
First: We need to find the principal - the sum of money that was initially deposited.
We shall do so based on the $5$ years that was given to us.
Second: We will calculate the amount due after $6$ years

$ \underline{First\:\:time\:\:period} \\[3ex] t = 5\:years \\[3ex] r = 2\dfrac{1}{2}\% = 2.5\% = \dfrac{2.5}{100} = 0.025 \\[5ex] SI = ₦500 \\[3ex] P = \dfrac{SI}{rt} \\[5ex] P = \dfrac{500}{0.025(5)} \\[5ex] P = \dfrac{500}{0.125} \\[5ex] P = ₦4000 \\[3ex] \underline{Second\:\:time\:\:period} \\[3ex] t = 6\:years \\[3ex] Same\:\:rate \\[3ex] A = P(1 + rt) \\[3ex] A = 4000[1 + (0.025)(6)] \\[3ex] A = 4000(1 + 0.15) \\[3ex] A = 4000(1.15) \\[3ex] A = ₦4600 \\[3ex] $ The total amount due after $6$ years at the same rate is $₦4,600.00$

(20.) JAMB If the interest on $₦150.00$ for $2\dfrac{1}{2}$ years is $₦4.50$, find the interest on $₦250.00$ for $6$ months at the same rate.

$ A.\:\: ₦1.50 \\[3ex] B.\:\: ₦7.50 \\[3ex] C.\:\: ₦15.00 \\[3ex] D.\:\: ₦18.00 \\[3ex] $

$ \underline{First\:\:case} \\[3ex] P = ₦150 \\[3ex] t = 2\dfrac{1}{2}\:years = 2.5\: years \\[5ex] SI = ₦4.50 \\[3ex] r = ? \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{4.50}{150(2.5)} \\[5ex] r = \dfrac{4.5}{375} \\[5ex] r = 0.012 \\[3ex] \underline{Second\:\:case} \\[3ex] P = ₦250 \\[3ex] t = 6\:months = \dfrac{6}{12}\:years = 0.5\:years \\[5ex] Same\:\:rate \implies r = 0.012 \\[3ex] SI = ? \\[3ex] SI = P * r * t \\[3ex] SI = 250(0.012)(0.5) \\[3ex] SI = 1.5 \\[3ex] $ The simple interest on $₦250.00$ for $6$ months at the same rate is $₦1.50$

Top

(21.) JAMB At what rate will the interest on $₦400$ increase to $₦24$ in $3$ years reckoning in simple interest?

$ A.\:\: 3\% \\[3ex] B.\:\: 2\% \\[3ex] C.\:\: 5\% \\[3ex] D.\:\: 4\% \\[3ex] $

$ 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.) JAMB A man invested $₦5000$ for $9$ months at $4\%$.
What is the simple interest?

$ A.\:\: ₦220 \\[3ex] B.\:\: ₦150 \\[3ex] C.\:\: ₦130 \\[3ex] D.\:\: ₦250 \\[3ex] $

$ P = ₦5000 \\[3ex] t = 9\:months = \dfrac{9}{12}\:year = 0.75\:year \\[5ex] r = 4\% = \dfrac{4}{100} = 0.04 \\[5ex] SI = P * r * t \\[3ex] SI = 5000 * 0.75 * 0.04 \\[3ex] SI = ₦150 \\[3ex] $ The simple interest on an investment of $₦5000$ for $9$ months at $4\%$ is $₦150$

(23.) JAMB Calculate the time taken for $₦3,000$ to earn $₦600$ if invested at $8\%$ simple interest.

$ A.\:\: 3\:\:years \\[3ex] B.\:\: 3\dfrac{1}{2}\:\:years \\[5ex] C.\:\: 1\dfrac{1}{2}\:\:years \\[5ex] D.\:\: 2\dfrac{1}{2}\:\:years \\[5ex] $

$ P = ₦3000 \\[3ex] SI = ₦600 \\[3ex] r = 8\% = \dfrac{8}{100} = \dfrac{2}{25} \\[5ex] t = ? \\[3ex] t = \dfrac{SI}{Pr} \\[5ex] Pr = 3000 * \dfrac{2}{25} = 600 * \dfrac{2}{5} = 120 * 2 = 240 \\[5ex] t = \dfrac{600}{240} = \dfrac{60}{24} = \dfrac{10}{4} = \dfrac{5}{2} \\[5ex] t = 2\dfrac{1}{2} \\[3ex] $ The time taken for $₦3,000$ to earn $₦600$ if invested at $8\%$ simple interest. is $2\dfrac{1}{2}\:\:years$

(24.) JAMB Musa borrows $₦10.00$ at $2\%$ per month interest and repays $₦8.00$ after $4$ months.
How much does he still owe?

$ A.\:\: ₦10.80 \\[3ex] B.\:\: ₦10.67 \\[3ex] C.\:\: ₦2.80 \\[3ex] D.\:\: ₦2.67 \\[3ex] $

The interest is $2\%$ per month (rather than $2\%$ per year)
The time is $4$ months (rather than $4$ years)
It is still in the same units (in months) so we are good
We do not need to convert to years

$ P = ₦10 \\[3ex] r = 2\% = \dfrac{2}{100} = 0.02 \\[5ex] t = 4\:months \\[3ex] SI = ? \\[3ex] SI = P * r * t \\[3ex] SI = (10)(0.02)(4) \\[3ex] SI = 0.8 \\[3ex] He\:\:repaid\:\:₦8 \\[3ex] Remaining:\:\: 10 - 8 = 2 \\[3ex] Total\:\:Balance\:\:Amount = Remaining + Simple\:\:Interest \\[3ex] Total\:\:Balance\:\:Amount = 2 + 0.8 \\[3ex] Total\:\:Balance\:\:Amount = 2.8 \\[3ex] $ Musa still owes ₦2.80

(25.) JAMB Oke deposited $₦800.00$ in the bank at the rate of $12\dfrac{1}{2}\%$ simple interest.
After some time, the total amount was one and half times the principal.
For how many years was the money left in the bank?

$ A.\:\: 2 \\[3ex] B.\:\: 4 \\[3ex] C.\:\: 5\dfrac{1}{2} \\[5ex] D.\:\: 8 \\[3ex] $

$ P = ₦800 \\[3ex] r = 12\dfrac{1}{2}\% = 12.5\% = \dfrac{12.5}{100} = 0.125 \\[5ex] A = 1\dfrac{1}{2} * 800 = \dfrac{3}{2} * 800 = 3(400) \\[5ex] A = ₦1200 \\[3ex] t = \dfrac{A - P}{Pr} \\[5ex] t = \dfrac{1200 - 800}{800(0.125)} \\[5ex] t = \dfrac{400}{100} \\[5ex] t = 4 \\[3ex] $ The money was left in the bank for $4$ years

(26.) JAMB At what rate would a sum of $₦100.00$ deposited for $5$ years raise an interest of $₦7.50$?

$ A.\:\: 1\dfrac{1}{2}\% \\[5ex] B.\:\: 2\dfrac{1}{2}\% \\[5ex] C.\:\: 15\% \\[3ex] D.\:\: 25\% \\[3ex] $

$ P = ₦100 \\[3ex] t = 5\:years \\[3ex] I = ₦7.50 \\[3ex] r = ? \\[3ex] r = \dfrac{I}{Pt} \\[5ex] r = \dfrac{7.50}{100(5)} \\[5ex] r = \dfrac{1.5}{100} \\[5ex] r = 1.5\% \\[3ex] Convert\:\:the\:\:decimal\:\:to\:\:fraction \\[3ex] 1.5 = \dfrac{1.5}{10} = \dfrac{3}{2} = 1\dfrac{1}{2} \\[5ex] \therefore r = 1\dfrac{1}{2}\% \\[5ex] $ A rate of $1\dfrac{1}{2}\%$ is needed for a sum of $₦100.00$ deposited for $5$ years to raise an interest of $₦7.50$

(27.) JAMB A man invested a sum of $₦280.00$ partly at $5\%$ and partly at $4\%$.
if the total interest is $₦12.80$ per annum, find the amount invested at $5\%$

$ A.\:\: ₦14.00 \\[3ex] B.\:\: ₦120.00 \\[3ex] C.\:\: ₦140.00 \\[3ex] D.\:\: ₦160.00 \\[3ex] $

$(1.)$ Let the investment (Principal) at the $5\%$ rate be $x$
$(2.)$ Let the investment (Principal) at the $4\%$ rate be $y$

$ x + y = 280 \\[3ex] \rightarrow y = 280 - x \\[3ex] $ $(2.)$ So, the investment on the $8\%$ rate is $280 - x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:5\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 5\% = \dfrac{5}{100} = 0.05 \\[5ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] Per\:\:annum\:\:means\:\:t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$x$	$0.05$	$1$	$0.05x$
$B$	$280 - x$	$0.04$	$1$	$0.04(280 - x)$
$Total$				$12.80$

$ \rightarrow 0.05x + 0.04(280 - x) = 12.8 \\[3ex] 0.05x + 11.2 - 0.04x = 12.8 \\[3ex] 0.01x = 12.8 - 11.2 \\[3ex] 0.01x = 1.6 \\[3ex] x = \dfrac{1.6}{0.01} \\[5ex] x = 160 \\[3ex] y = 280 - x \\[3ex] y = 280 - 160 \\[3ex] y = 120 \\[3ex] $ He invested $₦160.00$ at $5\%$ interest rate and $₦120.00$ at $4\%$ interest rate in order to earn $₦12.80$ interest.

(28.) JAMB If $₦225.00$ yields $₦27.00$ in $x$ years simple interest at the rate of $4\%$ per annum, find $x$

$ A.\:\: 3 \\[3ex] B.\:\: 4 \\[3ex] C.\:\: 12 \\[3ex] D.\:\: 27 \\[3ex] $

$ P = ₦225 \\[3ex] SI = ₦27 \\[3ex] t = x\:years \\[3ex] r = 4\% = \dfrac{4}{100} = \dfrac{1}{25} \\[5ex] t = \dfrac{SI}{Pr} \\[5ex] Pr = 225 * \dfrac{1}{25} = \dfrac{45}{5} = 9 \\[5ex] t = \dfrac{27}{9} \\[5ex] t = 3 \\[3ex] $ $x$ is $3$ years

(29.) How many days would it take for a sum of $\$1600$ to earn $\$7.00$ interest if it is invested at an interest rate of $3\%$ per year?
Use a $365-day$ year.

$ P = \$1600 \\[3ex] SI = \$7.00 \\[3ex] r = 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = ? \\[3ex] t = \dfrac{SI}{Pr} \\[5ex] t = \dfrac{7}{1600(0.03)} \\[5ex] t = \dfrac{7}{48} \\[5ex] t = 0.145833333\:year \\[3ex] 0.145833333\:year\:\:to\:\:days = 0.145833333(365) = 53.2291665\:\:days \\[3ex] t \approx 53.23\:\:days \\[3ex] $ It would take approximately $53.23$ days for a sum of $\$1600$ to earn $\$7$ interest if it is invested at an interest rate of $3\%$ per year

(30.) JAMB Find the simple interest rate percent per annum at which $₦1000$ accumulates to $₦1240$ in $3$ years.

$ A.\:\: 6\% \\[3ex] B.\:\: 8\% \\[3ex] C.\:\: 10\% \\[3ex] D.\:\: 12\% \\[3ex] $

$ P = ₦1000 \\[3ex] A = ₦1240 \\[3ex] t = 3\:years \\[3ex] r = \dfrac{A - P}{Pt} \\[5ex] r = \dfrac{1240 - 1000}{1000(3)} \\[5ex] r = \dfrac{240}{1000(3)} \\[5ex] r = \dfrac{8}{100} \\[5ex] r = 8\% \\[3ex] $ The simple interest rate percent per annum at which $₦1000$ accumulates to $₦1240$ in $3$ years is $8\%$

(31.) CSEC Thomas invested $\$1498$ at $6\%$ simple interest per annum.
Calculate:
(i) The interest, in dollars, earned after six months.
(ii) The TOTAL amount of money in his account after $3$ years.
(iii) How long it will be before his investment earns $\$449.40$

$ P = \$1498 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] (i) \\[3ex] t = 6\:months = \dfrac{6}{12} = 0.5\:year \\[5ex] SI = ? \\[3ex] SI = P * r * t \\[3ex] SI = 1498 * 0.06 * 0.5 \\[3ex] SI = \$44.94 \\[3ex] (ii) \\[3ex] t = 3\:years \\[3ex] A = ? \\[3ex] A = P(1 + rt) \\[3ex] A = 1498(1 + 0.06 * 3) \\[3ex] A = 1498(1 + 0.18) \\[3ex] A = 1498(1.18) \\[3ex] A = \$1767.64 \\[3ex] (iii) \\[3ex] SI = \$449.40 \\[3ex] t = ? \\[3ex] t = \dfrac{SI}{Pr} \\[5ex] t = \dfrac{449.40}{1498 * 0.06} \\[5ex] t = \dfrac{449.4}{89.88} \\[5ex] t = 5\:years \\[3ex] $ (i) The interest earned after six months is $\$44.94$
(ii) The total amount of money in his account after $3$ years is $\$1,767.64$
(iii) Ceteris paribus, it will take $5$ years before his investment earns $\$449.40$

(32.) ACT The simple interest for 1 year for an investment was $372
If the interest rate had been 1% higher for this investment, the simple interest for 1 year would have been $434
How much money was invested?

$ A.\;\; \$ 62 \\[3ex] B.\;\; \$ 520 \\[3ex] C.\;\; \$ 620 \\[3ex] D.\;\; \$5,200 \\[3ex] E.\;\; \$6,200 \\[3ex] $

$ \underline{First\;\;Case} \\[3ex] t = 1\;year \\[3ex] SI = \$372 \\[3ex] r = ? \\[3ex] P = ? \\[3ex] SI = P * r * t \\[3ex] 372 = P * r * 1 \\[3ex] 372 = Pr...eqn.(1) \\[3ex] \underline{Second\;\;Case} \\[3ex] rate\;\;is\;\;1\%\;\;higher \\[3ex] t = 1\; year \\[3ex] SI = \$434 \\[3ex] r = r + 1\% = r + \dfrac{1}{100} = r + 0.01 \\[5ex] P = P = ? \\[3ex] SI = P * r * t \\[3ex] 434 = P * (r + 0.01) * 1 \\[3ex] 434 = Pr + 0.01P \\[3ex] Substitute\;\;372\;\;for\;\;Pr...eqn.(1) \\[3ex] 434 = 372 + 0.01P \\[3ex] 434 - 372 = 0.01P \\[3ex] 62 = 0.01P \\[3ex] 0.01P = 62 \\[3ex] P = \dfrac{62}{0.01} \\[5ex] P = \$6200.00 \\[3ex] \underline{Check}...if\;\;you\;\;have\;\;time \\[3ex] P = \$6200 \\[3ex] t = 1\; year \\[3ex] SI = \$372 \\[3ex] r = ? \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{372}{6200(1)} \\[5ex] r = 0.06 = 6\% \\[3ex] Increase\;\;rate\;\;by\;\;1\% \\[3ex] r = 6\% + 1\% = 7\% = 0.07 \\[3ex] P = \$6200 \\[3ex] t = 1\; year \\[3ex] SI = ? \\[3ex] SI = P * r * t \\[3ex] SI = 6200 * 0.07 * 1 \\[3ex] SI = \$434 $

(33.) Philemon invested a total of $\$5000.00$, part at $4\%$ simple interest and part at $5\%$ simple interest.
At the end of $1$ year, the investments had earned $\$226.00$ interest.
How much was invested at each rate?

What are we looking for?
Two things
$(1.)$ Let the investment (Principal) at the $4\%$ rate be $x$
$(2.)$ Let the investment (Principal) at the $5\%$ rate be $y$

$ x + y = 5000 \\[3ex] \rightarrow y = 5000 - x \\[3ex] $ $(2.)$ So, the investment on the $5\%$ rate is $5000 - x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:5\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] 5\% = \dfrac{5}{100} = 0.05 \\[5ex] t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$x$	$0.04$	$1$	$0.04x$
$B$	$5000 - x$	$0.05$	$1$	$0.05(5000 - x)$
$Total$				$226$

$ \rightarrow 0.04x + 0.05(5000 - x) = 226 \\[3ex] 0.04x + 250 - 0.05x = 226 \\[3ex] -0.01x = 226 - 250 \\[3ex] -0.01x = -24 \\[3ex] x = \dfrac{-24}{-0.01} \\[5ex] x = \$2400.00 \\[3ex] y = 5000 - x \\[3ex] y = 5000 - 2400 \\[3ex] y = \$2600.00 \\[3ex] $ Philemon invested $\$2,400.00$ at $4\%$ interest rate and $\$2,600.00$ at $5\%$ interest rate in order to earn $\$226.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2400 * 0.04 * 1 = \$96 ...Bank\:\:A \\[3ex] SI = 2600 * 0.05 * 1 = \$130 ...Bank\:\:B \\[3ex] \$96 + \$130 = \$226 $

(34.) Phoebe invests her savings in two accounts, one paying $6\%$ and the other paying $10\%$ simple interest per year.
She puts twice as much in the lower-yielding account because it is less risky.
Her annual interest is $\$4950$.
How much did she invest at each rate?

Let the investment (Principal) at the $10\%$ rate (the higher-yielding account) be $x$
This means that the investment at the $6\%$ rate (the lower-yielding account) = $2x$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:6\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:10\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 6\% = \dfrac{6}{100} = 0.06 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$2x$	$0.06$	$1$	$0.12x$
$B$	$x$	$0.1$	$1$	$0.1x$
$Total$				$4950$

$ \rightarrow 0.12x + 0.1x = 4950 \\[3ex] 0.22x = 4950 \\[3ex] x = \dfrac{4950}{0.22} \\[5ex] x = \$22500.00 \\[3ex] 2x = 2(22500) = \$45000 \\[3ex] $ Phoebe invested $\$45,000.00$ at $6\%$ interest rate and $\$22,500.00$ at $10\%$ interest rate in order to earn $\$4,950.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 45000 * 0.06 * 1 = \$2700 ...Bank\:\:A \\[3ex] SI = 22500 * 0.1 * 1 = \$2250 ...Bank\:\:B \\[3ex] \$2700 + \$2250 = \$4950 $

(35.) Ruth invested some money at $9\%$, and $\$1600$ less than that amount at $4\%$.
The investments produced a total of $\$274$ interest in $1$ year.
How much did she invest at each rate?

What are we looking for?
Two things
$(1.)$ Let the investment (Principal) at the $9\%$ rate be $x$
$(2.)$ The investment (Principal) at the $4\%$ rate be $x - 1600$

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:9\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:4\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 9\% = \dfrac{9}{100} = 0.09 \\[5ex] 4\% = \dfrac{4}{100} = 0.04 \\[5ex] t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$x$	$0.09$	$1$	$0.09x$
$B$	$x - 1600$	$0.04$	$1$	$0.04(x - 1600)$
$Total$				$274$

$ \rightarrow 0.09x + 0.04(x - 1600) = 274 \\[3ex] 0.09x + 0.04x - 64 = 274 \\[3ex] 0.13x = 274 + 64 \\[3ex] 0.13x = 338 \\[3ex] x = \dfrac{338}{0.13} \\[5ex] x = \$2600.00 \\[3ex] x - 1600 = 2600 - 1600 = \$1000.00 \\[3ex] $ Ruth invested $\$2,600.00$ at $9\%$ interest rate and $\$1,000.00$ at $4\%$ interest rate in order to earn $\$274.00$ interest in one year

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 2600 * 0.09 * 1 = \$234 ...Bank\:\:A \\[3ex] SI = 1000 * 0.04 * 1 = \$40 ...Bank\:\:B \\[3ex] \$234 + \$40 = \$274 $

(36.) Samson invested a total of $\$12,900$ in two accounts.
The first account earned an annual rate of return of $12\%$.
However, the second account suffered a $3\%$ loss in the same period.
At the end of one year, the total amount of money gained was $\$243$
How much was invested into each account?

What are we looking for?
Two things
$(1.)$ Let the investment (Principal) at the $12\%$ rate be $x$
$(2.)$ Let the investment (Principal) at the $3\%$ rate be $y$

$ x + y = 12900 \\[3ex] \rightarrow y = 12900 - x \\[3ex] $ $(2.)$ So, the investment on the $3\%$ rate is $12900 - x$

What other thing should we note?
The interest rate in the account that earned a gain (the 12\% rate account) is positive.
It is positive because of the gain.

The interest rate in the account that suffered a loss (the 3\% account) is negative.
It is negative because of the loss.

$ P-r-t-I \\[3ex] I = P * r * t \\[3ex] Let\:\:the\:\:12\%\:\:investment\:\:rate = Bank\:\:A \\[3ex] Let\:\:the\:\:3\%\:\:investment\:\:rate = Bank\:\:B \\[3ex] 12\% = \dfrac{12}{100} = 0.12 \\[5ex] 3\% = \dfrac{3}{100} = 0.03 \\[5ex] t = 1\:\:year \\[3ex] $

Bank	$P$	$r$	$t$	$I = P * r * t$
$A$	$x$	$0.12$	$1$	$0.12x$
$B$	$12900 - x$	$-0.03$	$1$	$-0.03(12900 - x)$
$Total$				$243$

$ \rightarrow 0.12x + -0.03(12900 - x) = 243 \\[3ex] 0.12x - 0.03(12900 - x) = 243 \\[3ex] 0.12x - 387 + 0.03x = 243 \\[3ex] 0.12x + 0.03x = 243 + 387 \\[3ex] 0.15x = 630 \\[3ex] x = \dfrac{630}{0.15} \\[5ex] x = \$4200.00 \\[3ex] y = 12900 - x \\[3ex] y = 12900 - 4200 \\[3ex] y = \$8700.00 \\[3ex] $ Samson invested $\$4,200.00$ at $12\%$ interest rate and $\$8,700.00$ at negative $3\%$ interest rate in order to earn $\$243.00$ interest in one year.

$ \underline{Check} \\[3ex] SI = Prt \\[3ex] SI = 4200 * 0.12 * 1 = \$504 ...Bank\:\:A \\[3ex] SI = 8700 * -0.03 * 1 = \$-261 ...Bank\:\:B \\[3ex] \$504 + -\$261 = \$504 - \$261 = \$243 $

(37.) Often, when you take out a loan, some additional fees are included in the principal of the loan that are not part of the loan proceeds (the amount you actually borrow).
You need a little extra money to pay your tuition and your school has agreed to loan you the money.
You need to borrow $\$400$
Your school charges an interest rate of $9\%$ over $4$ months (you have to pay the loan before the end of the semester.)
There is a also a $\$12$ charge for the loan, which is included in the principal.
(a.) What is the principal used to calculate the interest for this loan?
(b.) What is the total interest you pay on the loan?
(c.) What is the total cost of credit?
(d.) What is the total repayment for the loan?
(e.) What will your monthly payments be?
(f.) What is the actual APR for the loan?

$ (a.) \\[3ex] Principal = Amount\:\:borrowed + fee\:\:charged\:\:for\:\:borrowing \\[3ex] P = 400 + 12 \\[3ex] P = \$412.00 \\[3ex] (b.) \\[3ex] Total\:\:interest\:\;on\:\:loan = SI \\[3ex] P = \$412 \\[3ex] r = 9\% = \dfrac{9}{100} = 0.09 \\[5ex] t = 4\:months = \dfrac{4}{12} = \dfrac{1}{3} \\[5ex] SI = ? \\[3ex] SI = P * r * t \\[3ex] SI = 412 * 0.09 * \dfrac{1}{3} \\[5ex] SI = \dfrac{37.08}{3} \\[5ex] SI = \$12.36 \\[3ex] (c.) \\[3ex] Total\:\:cost\:\:of\:\:credit = fee\:\:charged\:\:for\:\:borrowing + total\:\:interest\:\;on\:\:loan \\[3ex] Total\:\:cost\:\:of\:\:credit = 12 + 12.36 \\[3ex] Total\:\:cost\:\:of\:\:credit = \$24.36 \\[3ex] (d.) \\[3ex] Total\:\:repayment\:\:for\:\:the\:\:loan = A \\[3ex] A = P + SI \\[3ex] A = 412 + 12.36 \\[3ex] A = \$424.36 \\[3ex] (e.) \\[3ex] Monthly\:\:payments = \dfrac{Total\:\:repayment\:\:for\:\:the\:\:loan}{4} \\[3ex] Monthly\:\:payments = \dfrac{424.36}{4} \\[5ex] Monthly\:\:payments = \$106.09 \\[3ex] (f.) \\[3ex] Actual\:\:APR\:\:for\:\:the\:\:loan = r \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = SI \div Pt \\[3ex] r = 12.36 \div \left(412 * \dfrac{4}{12}\right) \\[5ex] r = 12.36 \div \dfrac{1648}{12} \\[5ex] r = 12.36 * \dfrac{12}{1648} \\[5ex] r = \dfrac{148.32}{1648} \\[5ex] r = 0.09 \\[3ex] to\:\:percent = 0.09(100) \\[3ex] r = 9\% $

(38.) JAMB Udoh deposited $₦150.00$ in the bank.
At the end of $5$ years, the simple interest on the principal was $₦55.00$
At what rate per annum was the interest paid?

$ A.\:\: 11\% \\[3ex] B.\:\: 7\dfrac{1}{3}\% \\[5ex] C.\:\: 5\% \\[3ex] D.\:\: 3\dfrac{1}{2}\% \\[5ex] $

$ P = ₦150 \\[3ex] t = 5\:years \\[3ex] SI = ₦55 \\[3ex] r = \dfrac{SI}{Pt} \\[5ex] r = \dfrac{55}{150(5)} \\[5ex] r = \dfrac{11}{150} \\[5ex] Convert\:\:to\:\:percent \\[3ex] \dfrac{11}{150} * 100 = \dfrac{11}{3} * 2 = \dfrac{22}{3}\% \\[5ex] \dfrac{22}{3} = 7\dfrac{1}{3} \\[5ex] \therefore r = 7\dfrac{1}{3}\% \\[5ex] $ The interest rate per annum is $7\dfrac{1}{3}\%$

(41.) JAMB Find the principal which amounts to $₦5,500$ at simple interest in $5$ years at $2\%$ per annum.

$ A.\:\: ₦4,900 \\[3ex] B.\:\: ₦5,000 \\[3ex] C.\:\: ₦4,700 \\[3ex] D.\:\: ₦4,800 \\[3ex] $

$ A = 5500 \\[3ex] t = 5 \\[3ex] r = 2\% = \dfrac{2}{100} \\[5ex] P = \dfrac{A}{1 + rt} \\[5ex] rt = \dfrac{2}{100} * 5 = \dfrac{10}{100} = \dfrac{1}{10} \\[5ex] 1 + rt = 1 + \dfrac{1}{10} = \dfrac{10}{10} + \dfrac{1}{10} = \dfrac{10 + 1}{10} = \dfrac{11}{10} \\[5ex] \rightarrow P = 5500 \div \dfrac{11}{10} \\[5ex] P = 5500 * \dfrac{10}{11} \\[5ex] P = 500 * 10 \\[3ex] P = ₦5000 \\[3ex] $ A deposit of a principal sum of ₦5000 at $2\%$ per annum interest rate in $5$ years will amount to $₦5,500$ at simple interest.

(42.) CMAT In how many years will $Rs.\:2\:lakh$ double itself at 11.5% per annum simple interest?

$ 1.\:\: Less\:\:than\:\:8 \\[3ex] 2.\:\: Between\:\:8\:\:and\:\:9 \\[3ex] 3.\:\: 9.3 \\[3ex] 4.\:\: 10.5 \\[3ex] $

$ P = 2 \\[3ex] A = 4 \:(double\:2) \\[3ex] r = 11.5\% = \dfrac{11.5}{100} \\[3ex] t = \dfrac{A - P}{Pr} \\[5ex] Pr = 2 * \dfrac{11.5}{100} = \dfrac{23}{100} \\[5ex] A - P = 4 - 2 = 2 \\[3ex] \rightarrow t = 2 \div \dfrac{23}{100} \\[5ex] t = 2 * \dfrac{100}{23} = \dfrac{200}{23} \\[5ex] t \:\:is\:\: Between\:\:8\:\:and\:\:9 \\[3ex] $ $Rs.\:2\:lakh$ will double itself at $11.5\%$ per annum simple interest in about $8.7$ years

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