Simple Interest Word Problems 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

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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?

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

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

✔ Answer: You will earn $125 in interest.

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Problem 2

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

- $ P = 15000 $
- $ r = 9\% = 0.09 $
- $ t = 3 $

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

Total repayment = Principal + Interest
$$
= 15000 + 4050 = 19050
$$

✔ Answer: She will pay $4,050 in interest and repay a total of $19,050.

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Problem 3

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

- $ P = 6000 $
- $ r = 3\% = 0.03 $
- $ I = 450 $
- Find $ 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.

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Problem 4

Mr. Johnson borrowed $8000 for 4 years and repaid $10,320. What interest rate did he borrow at?

- $ P = 8000 $
- $ t = 4 $
- Total repayment = $10,320 → So interest $ I = 10320 - 8000 = 2320 $

Now find $ r $:

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

✔ Answer: The interest rate was 7.25%.

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Problem 5

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

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

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

Total amount = $1000 + 585 = 1585$

✔ Answer: John will have $1,585.

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Problem 6

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

- $ P = 2000 $
- $ r = 5\% = 0.05 $
- $ t = 3 $

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

Total repayment = $2000 + 300 = 2300$

✔ Answer: Interest = $300, Total repayment = $2,300.

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Problem 7

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

- $ P = 4000 $
- $ r = 2\% = 0.02 $
- $ I = 200 $

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

✔ Answer: The money was invested for 2.5 years.

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Problem 8

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

- $ P = 9000 $
- $ t = 10 $
- Total repayment = $20,000 → Interest $ I = 20000 - 9000 = 11000 $

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

✔ Answer: The interest rate was approximately 12.22%.

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Problem 9

Bertha deposited $1000 at age 18 at 7.5% simple interest. How much will she have in 50 years?

- $ P = 1000 $
- $ r = 7.5\% = 0.075 $
- $ t = 50 $

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

Total = $1000 + 3750 = 4750$

✔ Answer: Bertha will have $4,750.

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Problem 10

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

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

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

✔ Answer: The annual interest rate was 10%.

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $125 |
| 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% |

Let me know if you'd like these formatted into a printable answer sheet!