Aimathic – Simple interest – Free Math Worksheets

Simple interest

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Difficulty: 1 – 5

5127367

Two banks advertise one-year savings accounts. Bank A offers an annual interest rate of $2\%$ on a deposit of $\$2500$. Bank B pays exactly $\$60$ in interest after one year on a deposit of $\$2000$. a) Find the annual interest earned at Bank A. b) Find Bank B’s annual interest rate. c) Which bank offers the higher interest rate?

Hints

- What formula relates principal, interest rate, and interest? - How do you write a decimal as a percent? - In each part, decide whether you are finding a dollar amount or a percent.

Solution

1. For Bank A, the interest is $\$2500 \cdot 0.02 = \$50$. 2. For Bank B, the interest rate is $\frac{60}{2000} = 0.03 = 3\%$. 3. Since $3\% > 2\%$, Bank B offers the higher interest rate.

Answer

a) Bank A pays $\$50$ in annual interest. b) Bank B’s annual interest rate is $3\%$. c) Bank B offers the higher interest rate.

5127577

Ms. Miller deposits $\$4500$ in a savings account that pays $1.8\%$ simple interest per year. How much interest will she earn after one full year? How much interest would she earn if she kept the money in the account for only $4$ months at the same rate?

Hints

- The stated rate is an annual rate. - What fraction of a year is four months? - Scale the annual interest by that fraction.

Solution

1. The interest for one year is $I = Prt = \$4500 \cdot 0.018 \cdot 1 = \$81$. 2. Four months is $\frac{4}{12} = \frac{1}{3}$ of a year. 3. The interest for four months is $\$81 \cdot \frac{1}{3} = \$27$.

Answer

She will earn $\$81$ after one year and $\$27$ after four months.

5127727

An investment of $\$1200$ earns $1.5\%$ simple interest per year for four years. The interest is paid out at the end of each year and is not added to the principal. Find the total amount consisting of the original principal plus all interest received after four years.

Hints

- Find the interest earned in one year. - How many times is that amount paid during four years? - Add the total interest to the original principal.

Solution

1. The interest earned each year is $\$1200 \cdot 0.015 = \$18$. 2. The total interest over four years is $\$18 \cdot 4 = \$72$. 3. The original principal plus all interest is $\$1200 + \$72 = \$1272$.

Answer

The total amount is $\$1272$.

5127757

Ms. Meyer uses a $\$1200$ line of credit for $45$ days to buy an electric bike. The bank charges $9.5\%$ annual simple interest. How much interest must she pay? Use a $360$-day year.

Hints

- Use the simple interest formula for a fraction of a year. - How many days are in the bank year stated in the problem? - Simplify the fraction of the year before multiplying.

Solution

1. Use $I = Prt$, with $P = 1200$, $r = 0.095$, and $t = \frac{45}{360}$. 2. The interest is $\$1200 \cdot 0.095 \cdot \frac{45}{360} = \$114 \cdot \frac{1}{8} = \$14.25$.

Answer

Ms. Meyer must pay $\$14.25$ in interest.

5127817

Ms. Miller deposits $\$2500$ in a savings account that pays $1.2\%$ simple interest for one year. Find the interest earned and the total balance at the end of the year.

Hints

- Identify the principal and the interest rate. - Write the percent as a decimal before multiplying. - The ending balance includes both the principal and the interest.

Solution

1. The interest is $\$2500 \cdot 0.012 = \$30$. 2. The ending balance is $\$2500 + \$30 = \$2530$.

Answer

The account earns $\$30$ in interest, and the ending balance is $\$2530$.

5127827

A sports club borrows $\$12{,}000$ to buy training equipment. After exactly one year, the club has paid $\$540$ in simple interest. Find the annual interest rate charged by the bank.

Hints

- Which formula relates principal, interest, rate, and time? - Rearrange the formula to isolate the rate. - Convert the resulting decimal to a percent.

Solution

1. Use $I = Prt$, with $I = 540$, $P = 12{,}000$, and $t = 1$. 2. Solve for the rate: $r = \frac{I}{Pt} = \frac{540}{12000} = 0.045$. 3. Write the decimal as a percent: $0.045 = 4.5\%$.

Answer

The annual interest rate is $4.5\%$.

5127507

A family compares two one-year loans for a kitchen renovation. - Offer A: Borrow $\$5000$ and repay a total of $\$5325$ after one year. - Offer B: Borrow $\$4500$ at an annual simple interest rate of $6\%$. Find the interest charged in dollars for each offer, and determine which offer has the lower interest rate.

Hints

- For Offer A, subtract the amount borrowed from the total repayment. - Use the interest and principal to find Offer A’s rate. - Use the given rate to find Offer B’s interest. - Compare the rates, not just the dollar amounts.

Solution

1. For Offer A, the interest is $\$5325 - \$5000 = \$325$. 2. Offer A’s interest rate is $\frac{325}{5000} \cdot 100\% = 6.5\%$. 3. For Offer B, the interest is $\$4500 \cdot 0.06 = \$270$. 4. Since $6\% < 6.5\%$, Offer B has the lower interest rate.

Answer

Offer A charges $\$325$ in interest at a rate of $6.5\%$. Offer B charges $\$270$ in interest at a rate of $6\%$. Offer B has the lower interest rate.

5127527

A youth group wants its savings to earn exactly $\$180$ in simple interest over one year. a) The bank currently offers an annual interest rate of $1.5\%$. How much principal must the group deposit to earn exactly $\$180$? b) Suppose the group has only $\$10{,}000$ to deposit. What annual interest rate would be needed to earn $\$180$ in one year?

Hints

- Identify whether each part asks for the principal or the interest rate. - Start with the simple interest formula $I = Prt$. Here, $t = 1$ year. - For part b, compare the interest with the principal.

Solution

1. For part a, use $I = Pr$. Solving for the principal gives $P = \frac{I}{r}$. 2. Substitute the values: $P = \frac{180}{0.015} = 12{,}000$. The required principal is $\$12{,}000$. 3. For part b, solve $I = Pr$ for the rate: $r = \frac{I}{P}$. 4. Substitute the values: $r = \frac{180}{10000} = 0.018 = 1.8\%$.

Answer

a) The group must deposit $\$12{,}000$. b) The required annual interest rate is $1.8\%$.

5127537

Mr. Miller deposits $\$5000$ in a savings account. During the first year, the account earns exactly $\$100$ in simple interest. a) Find the annual interest rate for the first year. b) In the second year, the bank raises the interest rate by $0.5$ percentage point. How much interest will $\$5000$ earn during the second year? c) At the new interest rate, how much total principal would be needed to earn exactly $\$200$ in one year?

Hints

- A percentage-point increase is added directly to the original percent. - Find the original rate before finding the new rate. - For part c, decide whether the required principal should be greater or less than $\$5000$.

Solution

1. For part a, the first-year rate is $\frac{100}{5000} = 0.02 = 2\%$. 2. For part b, the new rate is $2\% + 0.5$ percentage point $= 2.5\%$. The second-year interest is $\$5000 \cdot 0.025 = \$125$. 3. For part c, let $P$ be the required principal. From $200 = 0.025P$, $P = \frac{200}{0.025} = 8000$.

Answer

a) The first-year interest rate is $2\%$. b) The second-year interest is $\$125$. c) The required principal is $\$8000$.

5127547

Lucas plans to buy a new bike in three years. He wants the simple interest from his savings account to total exactly $\$108$ over those three years. The account pays a fixed annual interest rate of $1.2\%$, and the interest is paid out each year instead of being added to the principal. How much money must Lucas deposit at the beginning?

Hints

- How much interest must be earned in one year? - Which quantity is unknown in the simple interest formula? - Rearrange the formula so the principal is isolated.

Solution

1. Because the total interest over three years is $\$108$, the interest earned each year must be $\$108 \div 3 = \$36$. 2. Using $I = Prt$ for one year, $36 = P(0.012)(1)$. 3. Solving for $P$ gives $P = \frac{36}{0.012} = 3000$.

Answer

Lucas must deposit $\$3000$.

5127557

Ms. Weber wants to invest $\$2500$ for five years. Under both options, the interest is paid out at the end of each year and is not added to the principal. Option A: A fixed annual interest rate of $2.1\%$ for all five years. Option B: An annual rate of $1\%$ in the first year that increases by $0.5$ percentage point each year. Find the difference in total interest earned after five years. Which option is better?

Hints

- Find the total interest for each option separately. - List all five annual rates for Option B. - Compare the two total interest amounts.

Solution

1. For Option A, the total interest is $\$2500 \cdot 0.021 \cdot 5 = \$262.50$. 2. The five rates for Option B are $1\%$, $1.5\%$, $2\%$, $2.5\%$, and $3\%$. 3. The total interest for Option B is $\$2500(0.01 + 0.015 + 0.02 + 0.025 + 0.03) = \$2500(0.10) = \$250$. 4. The difference is $\$262.50 - \$250 = \$12.50$, so Option A earns more.

Answer

Option A is better because it earns $\$12.50$ more in total interest.

5127587

A sports club needs a $\$15{,}000$ loan to buy new equipment. Two banks offer one-year loans: Bank A charges $6.4\%$ simple interest for the year. Bank B charges a fixed $\$75$ in interest each month. Which offer has the lower total interest cost for one year? Justify your answer with calculations.

Hints

- Convert Bank B’s monthly charge to a yearly total. - Use the principal and rate to find Bank A’s interest. - Compare the two annual dollar amounts.

Solution

1. Bank A’s annual interest is $\$15{,}000 \cdot 0.064 = \$960$. 2. Bank B’s annual interest is $\$75 \cdot 12 = \$900$. 3. Since $\$900 < \$960$, Bank B has the lower interest cost.

Answer

Bank B is less expensive. Bank A charges $\$960$ in interest for the year, while Bank B charges $\$900$.

5127597

A family is planning a renovation and can spend at most $\$40$ per month on interest. A bank offers a loan with an annual simple interest rate of $7.5\%$. What is the greatest loan amount the family can borrow without exceeding its monthly interest limit?

Hints

- Convert the monthly interest limit to an annual limit. - Which quantity is unknown in the simple interest formula? - Rearrange the formula to solve for the principal.

Solution

1. The maximum interest for one year is $\$40 \cdot 12 = \$480$. 2. Using $I = Prt$ with $t = 1$, solve for the principal: $P = \frac{I}{r} = \frac{480}{0.075} = 6400$.

Answer

The greatest loan amount is $\$6400$.

5127637

A gaming laptop costs $\$1200$. Lucas pays $\$400$ now and delays paying the remaining $\$800$ for three months. The store charges an additional $\$24$ for this delay. Using simple interest and proportional annualization, what annual interest rate does the $\$24$ charge represent?

Hints

- Which part of the purchase price is paid later? - What percent of the financed amount is the three-month charge? - How many three-month periods are in one year? - Use proportional annualization, as stated in the problem.

Solution

1. The amount financed for three months is $\$1200 - \$400 = \$800$. 2. The three-month rate is $\frac{24}{800} = 0.03 = 3\%$. 3. There are four three-month periods in one year. With simple interest, the annual rate is $3\% \cdot 4 = 12\%$.

Answer

The charge represents an annual simple interest rate of $12\%$.

5127647

Ms. Meyer wants to invest $\$2500$ for exactly nine months. A local bank offers an annual simple interest rate of $1.5\%$. An online bank instead offers a fixed $\$25$ in interest for the entire nine-month term. Which offer gives the greater return? Justify your decision with a calculation.

Hints

- What information is given for the local bank’s offer? - Write nine months as a fraction of a year. - Compare the calculated interest with the online bank’s fixed payment.

Solution

1. Nine months is $\frac{9}{12} = 0.75$ year. 2. The local bank pays $I = Prt = \$2500 \cdot 0.015 \cdot 0.75 = \$28.125 \approx \$28.13$. 3. Since $\$28.13 > \$25$, the local bank gives the greater return.

Answer

The local bank is the better offer because it pays approximately $\$28.13$, compared with $\$25$ from the online bank.

5127667

Ms. Miller invests $\$4500$ for three months at an annual simple interest rate of $1.2\%$. Mr. Smith invests the same amount for $100$ days at an annual simple interest rate of $1.1\%$. Who earns more interest? Use ordinary simple interest, with $30$ days per month and $360$ days per year.

Hints

- Convert three months to interest days using the stated convention. - Use the simple interest formula for each investment. - Compare the two dollar amounts.

Solution

1. Three months is $3 \cdot 30 = 90$ days. Ms. Miller earns $\$4500 \cdot 0.012 \cdot \frac{90}{360} = \$13.50$. 2. Mr. Smith earns $\$4500 \cdot 0.011 \cdot \frac{100}{360} = \$13.75$. 3. Since $\$13.75 > \$13.50$, Mr. Smith earns more interest.

Answer

Mr. Smith earns more interest: $\$13.75$, compared with Ms. Miller’s $\$13.50$.

5127677

An investment of $\$12{,}000$ earns $\$112.50$ in simple interest over $75$ days. What annual interest rate does the investment pay? Use a $360$-day year.

Hints

- Identify the known quantities in the simple interest formula. - Write $75$ days as a fraction of a $360$-day year. - Rearrange the formula to isolate the interest rate.

Solution

1. Use $I = Prt$, where $P = 12{,}000$, $I = 112.50$, and $t = \frac{75}{360}$. 2. Solve for the rate: $r = \frac{I}{Pt} = \frac{112.50}{12000\left(\frac{75}{360}\right)} = 0.045$. 3. Write the decimal as a percent: $0.045 = 4.5\%$.

Answer

The annual interest rate is $4.5\%$.

5127737

Ms. Miller wants to invest $\$2500$ for exactly nine months. She compares two simple-interest offers: Offer A: An annual interest rate of $2.4\%$. Offer B: A monthly interest rate of $0.25\%$ applied to the original principal. Which offer pays more interest after nine months? Justify your answer with calculations.

Hints

- One offer gives an annual rate, while the other gives a monthly rate. - Write nine months as a fraction of a year for Offer A. - For Offer B, determine how many monthly interest periods occur.

Solution

1. For Offer A, nine months is $\frac{9}{12}$ year, so the interest is $\$2500 \cdot 0.024 \cdot \frac{9}{12} = \$45$. 2. For Offer B, the interest is $\$2500 \cdot 0.0025 \cdot 9 = \$56.25$. 3. Since $\$56.25 > \$45$, Offer B pays more.

Answer

Offer B pays more interest: $\$56.25$, compared with $\$45$ from Offer A.

5127767

A short-term savings account pays $1.8\%$ annual simple interest. An investor deposits $\$5000$ and earns exactly $\$15$ in interest. For how many days was the money invested? Use a $360$-day year.

Hints

- Which quantity is unknown: principal, rate, or time? - How much interest would the account earn in a full $360$-day year? - Compare the actual interest with the full-year interest.

Solution

1. The interest for a full year would be $\$5000 \cdot 0.018 = \$90$. 2. The actual interest is $\frac{15}{90} = \frac{1}{6}$ of the full-year interest, so the investment lasted $\frac{1}{6}$ of a year. 3. The number of days is $\frac{1}{6} \cdot 360 = 60$.

Answer

The money was invested for $60$ days.

5127777

Lucas needs to borrow $\$1500$ for exactly three months, or $90$ days. He compares two options: Offer A: A line of credit charging $11\%$ annual simple interest. Offer B: A small loan charging $6\%$ annual simple interest plus a one-time $\$15$ processing fee. Use a $360$-day year. Which option costs less, and by how much?

Hints

- Find the simple-interest charge for each offer. - Add the one-time fee to Offer B’s interest. - Compare the two total costs.

Solution

1. Offer A’s interest is $\$1500 \cdot 0.11 \cdot \frac{90}{360} = \$41.25$. 2. Offer B’s interest is $\$1500 \cdot 0.06 \cdot \frac{90}{360} = \$22.50$. 3. Including the fee, Offer B costs $\$22.50 + \$15 = \$37.50$. 4. Offer B is cheaper by $\$41.25 - \$37.50 = \$3.75$.

Answer

Offer B costs less. It saves Lucas $\$3.75$ compared with Offer A.

5127837

A family compares two one-year renovation loans. Bank A charges $6.5\%$ simple interest. At that rate, the family would pay $\$455$ in interest for the year. Bank B charges $7.2\%$ simple interest. First find the amount of the loan. Then determine how much more interest the family would pay at Bank B for the same loan amount.

Hints

- Use Bank A’s rate and interest to find the loan principal. - Once you know the principal, find Bank B’s interest. - Subtract the two interest amounts.

Solution

1. For Bank A, $455 = 0.065P$, so the loan amount is $P = \frac{455}{0.065} = 7000$. 2. At Bank B, the annual interest would be $\$7000 \cdot 0.072 = \$504$. 3. The additional interest is $\$504 - \$455 = \$49$.

Answer

The loan amount is $\$7000$. The family would pay $\$49$ more in interest at Bank B.

5127387

An investor has $\$12{,}000$ in an account earning $1.2\%$ simple interest per year. The annual interest rate drops to $0.8\%$ for the entire balance. How much additional money must the investor deposit immediately so that the account earns the same amount of interest for the year as it would have earned at the original rate?

Hints

- First find the interest the original balance would earn at the original rate. - Use that interest amount as the target under the new rate. - The question asks for the additional deposit, not the new total balance.

Solution

1. At the original rate, the annual interest would be $\$12{,}000 \cdot 0.012 = \$144$. 2. Let $P$ be the total principal needed at the lower rate. Then $0.008P = 144$, so $P = \frac{144}{0.008} = 18{,}000$. 3. The additional deposit is $\$18{,}000 - \$12{,}000 = \$6000$.

Answer

The investor must deposit an additional $\$6000$.

5127687

An account contains $\$8000$ and pays $2\%$ annual simple interest. Use a $360$-day year. a) How many days will it take the account to earn exactly $\$40$ in interest? b) At the same principal and with the same interest goal, how would the time change if the annual rate were $4\%$? Explain without completely repeating the calculation from part a).

Hints

- For part a, isolate the time in the simple interest equation. - For part b, examine how the rate and time must change when the principal and interest stay fixed. - Think about inverse proportionality.

Solution

1. For part a, use $I = Prt$, with time measured in years: $40 = 8000(0.02)\frac{d}{360}$. 2. Solving gives $d = \frac{40 \cdot 360}{8000 \cdot 0.02} = 90$ days. 3. For part b, when the principal and target interest stay fixed, the interest rate and time are inversely proportional. 4. Doubling the rate from $2\%$ to $4\%$ halves the required time from $90$ days to $45$ days.

Answer

a) It will take $90$ days. b) The time will be cut in half, to $45$ days, because the rate doubles while the principal and target interest remain fixed.

5127927

An investment earns exactly $\$36$ in simple interest over $240$ days at an annual rate of $1.5\%$. Use a $360$-day year. a) How much money was originally invested? b) How much interest would the same principal earn at the same rate over a full $360$-day year?

Hints

- Identify the unknown quantity in the simple interest formula. - Compare $240$ days with a full $360$-day year. - Part b can be solved either from the principal or by scaling the interest. - Check that the full-year interest is greater than the $240$-day interest.

Solution

1. For part a, use $I = Prt$, where $I = 36$, $r = 0.015$, and $t = \frac{240}{360}$. 2. Solve for the principal: $P = \frac{36}{0.015\left(\frac{240}{360}\right)} = 3600$. 3. For part b, the full-year interest is $\$3600 \cdot 0.015 = \$54$. 4. Equivalently, $360$ days is $1.5$ times $240$ days, so the interest is $\$36 \cdot 1.5 = \$54$.

Answer

a) The original principal was $\$3600$. b) The full-year interest would be $\$54$.

5142627

Two friends compare short-term loans. Lucas borrows $\$1200$ at an annual simple interest rate of $10\%$ and repays the loan after $90$ days. Leah borrows $\$1500$ at an annual simple interest rate of $6\%$. After how many days must Leah repay her loan to pay exactly the same amount of interest as Lucas? Use a $360$-day year.

Hints

- First find the amount of interest Lucas pays. - Use that interest amount as the target for Leah’s loan. - Identify the unknown quantity in Leah’s simple interest equation. - Rearrange the equation to solve for the number of days.

Solution

1. Lucas pays $\$1200 \cdot 0.10 \cdot \frac{90}{360} = \$30$ in interest. 2. Let $d$ be Leah’s loan length in days. Her interest is $\$1500 \cdot 0.06 \cdot \frac{d}{360}$. 3. Set the interest amounts equal: $1500(0.06)\frac{d}{360} = 30$. 4. Solving gives $d = \frac{30 \cdot 360}{1500 \cdot 0.06} = 120$.

Answer

Leah must repay her loan after $120$ days.

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