Simple interest worksheet - Free Printable

Problem: Solving Simple Interest Word Problems

The formula for simple interest is given as:
\[
I = P \cdot r \cdot t
\]
where:
- \( I \) = Interest Paid or Earned in dollars
- \( P \) = Principal (initial amount of money) in dollars
- \( r \) = Interest Rate (as a decimal)
- \( t \) = Time in years

We will solve each problem step by step using this formula.

---

Problem 1:

A bank is offering 2.5% simple interest on a savings account. If you deposit $5000, how much interest will you earn in one year?

#### Solution:
- Principal (\( P \)) = $5000
- Interest Rate (\( r \)) = 2.5% = 0.025 (as a decimal)
- Time (\( t \)) = 1 year

Using the formula:
\[
I = P \cdot r \cdot t = 5000 \cdot 0.025 \cdot 1 = 125
\]

Answer: You will earn $125 in interest in one year.

---

Problem 2:

To buy a car, Jessica borrowed $15,000 for 3 years at an annual simple interest rate of 9%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay?

#### Solution:
- Principal (\( P \)) = $15,000
- Interest Rate (\( r \)) = 9% = 0.09 (as a decimal)
- Time (\( t \)) = 3 years

First, calculate the interest (\( I \)):
\[
I = P \cdot r \cdot t = 15000 \cdot 0.09 \cdot 3 = 4050
\]

Next, calculate the total amount to be repaid:
\[
\text{Total Amount} = P + I = 15000 + 4050 = 19050
\]

Answer: Jessica will pay $4050 in interest and a total of $19,050.

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Problem 3:

Nancy invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested?

#### Solution:
- Principal (\( P \)) = $6000
- Interest Rate (\( r \)) = 3% = 0.03 (as a decimal)
- Interest (\( I \)) = $450

We need to find the time (\( t \)). Using the formula:
\[
I = P \cdot r \cdot t
\]
Rearrange to solve for \( t \):
\[
t = \frac{I}{P \cdot r} = \frac{450}{6000 \cdot 0.03} = \frac{450}{180} = 2.5
\]

Answer: The money was invested for 2.5 years.

---

Problem 4:

Mr. Johnson borrowed $8000 for 4 years to make home improvements. If he repaid a total of $10,320, at what interest rate did he borrow the money?

#### Solution:
- Principal (\( P \)) = $8000
- Time (\( t \)) = 4 years
- Total Repayment = $10,320

First, calculate the interest (\( I \)):
\[
I = \text{Total Repayment} - P = 10320 - 8000 = 2320
\]

Now, use the formula \( I = P \cdot r \cdot t \) to solve for \( r \):
\[
r = \frac{I}{P \cdot t} = \frac{2320}{8000 \cdot 4} = \frac{2320}{32000} = 0.0725
\]

Convert \( r \) to a percentage:
\[
r = 0.0725 \times 100 = 7.25\%
\]

Answer: The interest rate was 7.25%.

---

Problem 5:

John’s parents deposited $1000 into a savings account as a college fund when he was born. How much will John have in this account after 18 years at a yearly simple interest rate of 3.25%?

#### Solution:
- Principal (\( P \)) = $1000
- Interest Rate (\( r \)) = 3.25% = 0.0325 (as a decimal)
- Time (\( t \)) = 18 years

First, calculate the interest (\( I \)):
\[
I = P \cdot r \cdot t = 1000 \cdot 0.0325 \cdot 18 = 585
\]

Next, calculate the total amount in the account:
\[
\text{Total Amount} = P + I = 1000 + 585 = 1585
\]

Answer: John will have $1585 in the account after 18 years.

---

Problem 6:

To buy a laptop computer, Elaine borrowed $2,000 for 3 years at an annual simple interest rate of 5%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay?

#### Solution:
- Principal (\( P \)) = $2000
- Interest Rate (\( r \)) = 5% = 0.05 (as a decimal)
- Time (\( t \)) = 3 years

First, calculate the interest (\( I \)):
\[
I = P \cdot r \cdot t = 2000 \cdot 0.05 \cdot 3 = 300
\]

Next, calculate the total amount to be repaid:
\[
\text{Total Amount} = P + I = 2000 + 300 = 2300
\]

Answer: Elaine will pay $300 in interest and a total of $2300.

---

Problem 7:

TJ invested $4000 in a bond at a yearly rate of 2%. He earned $200 in interest. How long was the money invested?

#### Solution:
- Principal (\( P \)) = $4000
- Interest Rate (\( r \)) = 2% = 0.02 (as a decimal)
- Interest (\( I \)) = $200

We need to find the time (\( t \)). Using the formula:
\[
I = P \cdot r \cdot t
\]
Rearrange to solve for \( t \):
\[
t = \frac{I}{P \cdot r} = \frac{200}{4000 \cdot 0.02} = \frac{200}{80} = 2.5
\]

Answer: The money was invested for 2.5 years.

---

Problem 8:

Mr. Mogi borrowed $9000 for 10 years to make home improvements. If he repaid a total of $20,000, at what interest rate did he borrow the money?

#### Solution:
- Principal (\( P \)) = $9000
- Time (\( t \)) = 10 years
- Total Repayment = $20,000

First, calculate the interest (\( I \)):
\[
I = \text{Total Repayment} - P = 20000 - 9000 = 11000
\]

Now, use the formula \( I = P \cdot r \cdot t \) to solve for \( r \):
\[
r = \frac{I}{P \cdot t} = \frac{11000}{9000 \cdot 10} = \frac{11000}{90000} = 0.1222
\]

Convert \( r \) to a percentage:
\[
r = 0.1222 \times 100 \approx 12.22\%
\]

Answer: The interest rate was approximately 12.22%.

---

Problem 9:

Bertha deposited $1000 into a retirement account when she was 18. How much will Bertha have in this account after 50 years at a yearly simple interest rate of 7.5%?

#### Solution:
- Principal (\( P \)) = $1000
- Interest Rate (\( r \)) = 7.5% = 0.075 (as a decimal)
- Time (\( t \)) = 50 years

First, calculate the interest (\( I \)):
\[
I = P \cdot r \cdot t = 1000 \cdot 0.075 \cdot 50 = 3750
\]

Next, calculate the total amount in the account:
\[
\text{Total Amount} = P + I = 1000 + 3750 = 4750
\]

Answer: Bertha will have $4750 in the account after 50 years.

---

Problem 10:

Joshua borrowed $1000 from his friend and paid him back $1050 in six months. What simple annual interest did Joshua pay his friend?

#### Solution:
- Principal (\( P \)) = $1000
- Total Repayment = $1050
- Time (\( t \)) = 6 months = 0.5 years

First, calculate the interest (\( I \)):
\[
I = \text{Total Repayment} - P = 1050 - 1000 = 50
\]

Now, use the formula \( I = P \cdot r \cdot t \) to solve for \( r \):
\[
r = \frac{I}{P \cdot t} = \frac{50}{1000 \cdot 0.5} = \frac{50}{500} = 0.1
\]

Convert \( r \) to a percentage:
\[
r = 0.1 \times 100 = 10\%
\]

Answer: Joshua paid an annual interest rate of 10%.

---

Final Answers:

1. $125
2. $4050, $19,050
3. 2.5 years
4. 7.25%
5. $1585
6. $300, $2300
7. 2.5 years
8. 12.22%
9. $4750
10. 10%

\boxed{125, 4050, 19050, 2.5, 7.25\%, 1585, 300, 2300, 2.5, 12.22\%, 4750, 10\%}