Printable worksheet for practicing simple interest calculations, featuring 12 problems with given values and spaces to show work.

Let's solve each problem step by step using the simple interest formula:

$$
I = P \times R \times T
$$

Where:
- $ I $ = Interest
- $ P $ = Principal (initial amount)
- $ R $ = Annual interest rate (as a decimal)
- $ T $ = Time in years

We'll use four decimal places for fractions of time and round answers to the nearest cent or tenth of a percent, as instructed.

---

Part 1: Calculate Interest (I)

#### 1a.
Given:
- $ P = 500 $
- $ R = 8\% = 0.08 $
- $ T = 3 \text{ months} = \frac{3}{12} = 0.25 \text{ years} $

$$
I = 500 \times 0.08 \times 0.25 = 500 \times 0.02 = \boxed{\$10.00}
$$

#### 1b.
- $ P = 50 $
- $ R = 12\% = 0.12 $
- $ T = 1 \text{ month} = \frac{1}{12} \approx 0.0833 \text{ years} $

$$
I = 50 \times 0.12 \times \frac{1}{12} = 50 \times 0.01 = \boxed{\$0.50}
$$

#### 1c.
- $ P = 1,000 $
- $ R = 18\% = 0.18 $
- $ T = 24 \text{ months} = \frac{24}{12} = 2 \text{ years} $

$$
I = 1000 \times 0.18 \times 2 = 1000 \times 0.36 = \boxed{\$360.00}
$$

#### 1d.
- $ P = 600 $
- $ R = 15\% = 0.15 $
- $ T = 60 \text{ days} = \frac{60}{360} = \frac{1}{6} \approx 0.1667 \text{ years} $

$$
I = 600 \times 0.15 \times \frac{60}{360} = 600 \times 0.15 \times 0.1667 = 600 \times 0.025 = \boxed{\$15.00}
$$

---

Part 2: Find Principal (P)

Use:
$$
P = \frac{I}{R \times T}
$$

#### 2a.
- $ I = 6 $
- $ R = 12\% = 0.12 $
- $ T = 3 \text{ months} = \frac{3}{12} = 0.25 $

$$
P = \frac{6}{0.12 \times 0.25} = \frac{6}{0.03} = \boxed{\$200.00}
$$

#### 2b.
- $ I = 15 $
- $ R = 15\% = 0.15 $
- $ T = 90 \text{ days} = \frac{90}{360} = 0.25 $

$$
P = \frac{15}{0.15 \times 0.25} = \frac{15}{0.0375} = \boxed{\$400.00}
$$

#### 2c.
- $ I = 300 $
- $ R = 12\% = 0.12 $
- $ T = 6 \text{ months} = \frac{6}{12} = 0.5 $

$$
P = \frac{300}{0.12 \times 0.5} = \frac{300}{0.06} = \boxed{\$5,000.00}
$$

#### 2d.
- $ I = 90 $
- $ R = 6\% = 0.06 $
- $ T = 60 \text{ days} = \frac{60}{360} = \frac{1}{6} \approx 0.1667 $

$$
P = \frac{90}{0.06 \times \frac{60}{360}} = \frac{90}{0.06 \times 0.1667} = \frac{90}{0.01} = \boxed{\$9,000.00}
$$

---

Part 3: Find Rate (R)

Use:
$$
R = \frac{I}{P \times T}
$$

#### 3a.
- $ I = 12 $
- $ P = 200 $
- $ T = 6 \text{ months} = \frac{6}{12} = 0.5 $

$$
R = \frac{12}{200 \times 0.5} = \frac{12}{100} = 0.12 = \boxed{12\%}
$$

#### 3b.
- $ I = 390 $
- $ P = 2,000 $
- $ T = 18 \text{ months} = \frac{18}{12} = 1.5 $

$$
R = \frac{390}{2000 \times 1.5} = \frac{390}{3000} = 0.13 = \boxed{13\%}
$$

#### 3c.
- $ I = 50 $
- $ P = 25 $
- $ T = 6 \text{ weeks} = \frac{6}{52} \approx 0.1154 \text{ years} $

But note: The worksheet says T = 42/360, so we use that.

$$
T = \frac{42}{360} \approx 0.1167 \text{ years}
$$

$$
R = \frac{50}{25 \times \frac{42}{360}} = \frac{50}{25 \times 0.1167} = \frac{50}{2.9167} \approx 17.1429
$$

Convert to percentage:
$$
R \approx 17.1429 \Rightarrow \boxed{1714.29\%}
$$

Wait — that seems extremely high. Let’s double-check.

Wait! $ I = \$50 $, $ P = \$25 $ → So you're earning \$50 on \$25 over 6 weeks?

That’s double the principal in just 6 weeks!

Let’s compute:

$$
R = \frac{50}{25 \times \frac{42}{360}} = \frac{50}{25 \times 0.116666...} = \frac{50}{2.916666...} \approx 17.142857
$$

So $ R = 17.142857 $ → That means 1714.29% annual rate.

Yes, that’s correct mathematically, though unrealistic. But based on data:

$$
\boxed{1714.3\%} \quad \text{(rounded to nearest tenth of a percent)}
$$

#### 3d.
- $ I = 3 $
- $ P = 50 $
- $ T = 180 \text{ days} = \frac{180}{360} = 0.5 $

$$
R = \frac{3}{50 \times 0.5} = \frac{3}{25} = 0.12 = \boxed{12\%}
$$

---

Part 4: Word Problems

#### 4a.
You borrow $50 and pay back $58 after 6 months.

- Amount paid back = $58
- Principal = $50
- Interest = $58 - $50 = $8

Time = 6 months = $ \frac{6}{12} = 0.5 $ years

Now find interest rate:

$$
I = P \times R \times T \Rightarrow 8 = 50 \times R \times 0.5
$$

$$
8 = 25R \Rightarrow R = \frac{8}{25} = 0.32 = \boxed{32\%}
$$

Answer:
- Interest paid: $8
- Interest rate: 32%

#### 4b.
Borrow $800 at 18% for 11 months.

- $ P = 800 $
- $ R = 18\% = 0.18 $
- $ T = \frac{11}{12} \approx 0.9167 $ years

$$
I = 800 \times 0.18 \times \frac{11}{12} = 800 \times 0.18 \times 0.9167
$$

First: $ 800 \times 0.18 = 144 $

Then: $ 144 \times \frac{11}{12} = 144 \times 0.9167 = 131.999 \approx \boxed{\$132.00} $

---

✔ Final Answers:

---

#### 1.
- 1a: $ I = \$10.00 $
- 1b: $ I = \$0.50 $
- 1c: $ I = \$360.00 $
- 1d: $ I = \$15.00 $

#### 2.
- 2a: $ P = \$200.00 $
- 2b: $ P = \$400.00 $
- 2c: $ P = \$5,000.00 $
- 2d: $ P = \$9,000.00 $

#### 3.
- 3a: $ R = 12\% $
- 3b: $ R = 13\% $
- 3c: $ R = 1714.3\% $
- 3d: $ R = 12\% $

#### 4a.
- Interest paid: \$8
- Interest rate: 32%

#### 4b.
- Total interest: \$132.00

---

Let me know if you'd like this formatted as a printable answer sheet!