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Comprehensive math worksheet featuring the simple interest formula and ten practice word problems for students.

Simple interest formula and 10 word problems worksheet for math students.

Simple interest formula and 10 word problems worksheet for math students.

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Show Answer Key & Explanations Step-by-step solution for: Simple interest worksheet
Let's solve each of these simple interest word problems step by step using the formula:

$$
I = P \cdot r \cdot t
$$

Where:
- $ I $ = Interest earned or paid (in dollars)
- $ P $ = Principal (initial amount in dollars)
- $ r $ = Annual interest rate (as a decimal)
- $ t $ = Time in years

---

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?



Given:
- $ P = 5000 $
- $ r = 2.5\% = 0.025 $
- $ t = 1 $

$$
I = 5000 \cdot 0.025 \cdot 1 = 125
$$

Answer: $125 interest earned.

---

2. Jessica borrowed $15,000 for 3 years at 9% annual simple interest. How much interest will she pay? What is the total amount repaid?



Given:
- $ P = 15,000 $
- $ r = 9\% = 0.09 $
- $ t = 3 $

$$
I = 15000 \cdot 0.09 \cdot 3 = 4050
$$

Total repayment = Principal + Interest = $15,000 + $4,050 = $19,050

Answer: Interest = $4,050; Total repayment = $19,050

---

3. Nancy invested $6000 at 3% and earned $450 in interest. How long was the money invested?



Given:
- $ P = 6000 $
- $ r = 3\% = 0.03 $
- $ I = 450 $

We solve for $ t $:

$$
450 = 6000 \cdot 0.03 \cdot t \\
450 = 180t \\
t = \frac{450}{180} = 2.5
$$

Answer: 2.5 years (or 2 years and 6 months)

---

4. Mr. Johnson borrowed $8000 for 4 years and repaid $10,320. What was the interest rate?



Given:
- $ P = 8000 $
- $ t = 4 $
- Total repayment = $10,320 → so interest $ I = 10,320 - 8000 = 2320 $

Now solve for $ r $:

$$
2320 = 8000 \cdot r \cdot 4 \\
2320 = 32000r \\
r = \frac{2320}{32000} = 0.0725 = 7.25\%
$$

Answer: 7.25%

---

5. John’s parents deposited $1000 at 3.25% simple interest. How much will he have after 18 years?



Given:
- $ P = 1000 $
- $ r = 3.25\% = 0.0325 $
- $ t = 18 $

$$
I = 1000 \cdot 0.0325 \cdot 18 = 585
$$

Total amount = $1000 + $585 = $1,585

Answer: $1,585

---

6. Elaine borrowed $2,000 for 3 years at 5%. How much interest? Total repayment?



Given:
- $ P = 2000 $
- $ r = 0.05 $
- $ t = 3 $

$$
I = 2000 \cdot 0.05 \cdot 3 = 300
$$

Total repayment = $2000 + $300 = $2,300

Answer: Interest = $300; Total repayment = $2,300

---

7. TJ invested $4000 at 2% and earned $200 interest. How long was it invested?



Given:
- $ P = 4000 $
- $ r = 0.02 $
- $ I = 200 $

$$
200 = 4000 \cdot 0.02 \cdot t \\
200 = 80t \\
t = \frac{200}{80} = 2.5
$$

Answer: 2.5 years

---

8. Mr. Mogi borrowed $9000 for 10 years and repaid $20,000. What was the interest rate?



Given:
- $ P = 9000 $
- $ t = 10 $
- Total repayment = $20,000 → $ I = 20,000 - 9,000 = 11,000 $

$$
11000 = 9000 \cdot r \cdot 10 \\
11000 = 90000r \\
r = \frac{11000}{90000} = 0.1222... = 12.22\%
$$

Answer: Approximately 12.22%

---

9. Bertha deposited $1000 at 7.5% simple interest. How much will she have after 50 years?



Given:
- $ P = 1000 $
- $ r = 0.075 $
- $ t = 50 $

$$
I = 1000 \cdot 0.075 \cdot 50 = 3750
$$

Total amount = $1000 + $3750 = $4,750

Answer: $4,750

---

10. Joshua borrowed $1000 and paid back $1050 in six months. What was the annual simple interest rate?



Given:
- $ P = 1000 $
- $ I = 1050 - 1000 = 50 $
- $ t = 6 \text{ months} = 0.5 \text{ years} $

Solve for $ r $:

$$
50 = 1000 \cdot r \cdot 0.5 \\
50 = 500r \\
r = \frac{50}{500} = 0.10 = 10\%
$$

Answer: 10% annual interest rate

---

Final Answers Summary:



| Problem | Answer |
|--------|--------|
| 1 | $125 interest |
| 2 | $4,050 interest; $19,050 total |
| 3 | 2.5 years |
| 4 | 7.25% |
| 5 | $1,585 |
| 6 | $300 interest; $2,300 total |
| 7 | 2.5 years |
| 8 | ~12.22% |
| 9 | $4,750 |
| 10 | 10% annual interest |

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