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

**
Pre-requisites: Measurements and Units
Formulas Used: Mathematics of Finance
Verify Answers with Calculator: Financial Mathematics 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 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.) **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.

$ 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.

(2.) **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

$ 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

(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.

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

$ 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.

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.

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 $1.59\%$

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 $\$1250.00$

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

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 $\$823.20$

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 $10\%$

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.

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 $

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.

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] $

$ \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.

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 $₦6,250.00$

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 $

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 $\$34.96$

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 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\% $

(a.) How much 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 $8.1\%$

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 $₦4,600.00$

$ 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 $₦1.50$

$ 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 $₦1.50$

(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 $2\%$

$ 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 $₦150$

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 $2\dfrac{1}{2}\:\:years$

$ 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

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

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 $₦7.50$

$ 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] $

$ \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.

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

$ 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

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 $8\%$

$ 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 $\$449.40$

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.) 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 $

$ 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 $

(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] $

$ \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 $

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] $

$ \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 $

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] $

$ \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 $

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] $

$ \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 $

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\% $

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 $7\dfrac{1}{3}\%$

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 $7\dfrac{1}{3}\%$