Compound Interest worksheet with problems to find interest and amount, rounded to the nearest cent.

Problem: Compound Interest Calculations

We are tasked with solving compound interest problems using the formula for compound interest:

$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$

Where:
- \( A \) is the amount of money accumulated after \( n \) years, including interest.
- \( P \) is the principal amount (the initial amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times that interest is compounded per year.
- \( t \) is the time the money is invested or borrowed for, in years.

The compound interest (\( I \)) can be calculated as:
$$
I = A - P
$$

Let's solve each part step by step.

---

Part A: Find the compound interest (\( I \)) and amount (\( A \))

#### 1) Principal = \$9,400, Time = 4 years, Rate = 10% compounded annually
- \( P = 9400 \)
- \( r = 10\% = 0.10 \)
- \( n = 1 \) (compounded annually)
- \( t = 4 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 9400 \left(1 + \frac{0.10}{1}\right)^{1 \cdot 4}
$$
$$
A = 9400 \left(1 + 0.10\right)^4
$$
$$
A = 9400 \left(1.10\right)^4
$$

Calculate \( (1.10)^4 \):
$$
(1.10)^4 = 1.4641
$$

Now calculate \( A \):
$$
A = 9400 \times 1.4641 = 13780.54
$$

The amount (\( A \)) is \$13,780.54.

To find the compound interest (\( I \)):
$$
I = A - P
$$
$$
I = 13780.54 - 9400 = 4380.54
$$

So, the answers are:
$$
A = 13780.54, \quad I = 4380.54
$$

#### 2) Principal = \$12,900, Time = 2 years, Rate = 13% compounded annually
- \( P = 12900 \)
- \( r = 13\% = 0.13 \)
- \( n = 1 \) (compounded annually)
- \( t = 2 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 12900 \left(1 + \frac{0.13}{1}\right)^{1 \cdot 2}
$$
$$
A = 12900 \left(1 + 0.13\right)^2
$$
$$
A = 12900 \left(1.13\right)^2
$$

Calculate \( (1.13)^2 \):
$$
(1.13)^2 = 1.2769
$$

Now calculate \( A \):
$$
A = 12900 \times 1.2769 = 16525.01
$$

The amount (\( A \)) is \$16,525.01.

To find the compound interest (\( I \)):
$$
I = A - P
$$
$$
I = 16525.01 - 12900 = 3625.01
$$

So, the answers are:
$$
A = 16525.01, \quad I = 3625.01
$$

#### 3) Principal = \$800, Time = 9 years, Rate = 25% compounded annually
- \( P = 800 \)
- \( r = 25\% = 0.25 \)
- \( n = 1 \) (compounded annually)
- \( t = 9 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 800 \left(1 + \frac{0.25}{1}\right)^{1 \cdot 9}
$$
$$
A = 800 \left(1 + 0.25\right)^9
$$
$$
A = 800 \left(1.25\right)^9
$$

Calculate \( (1.25)^9 \):
$$
(1.25)^9 \approx 7.4505817
$$

Now calculate \( A \):
$$
A = 800 \times 7.4505817 = 5960.47
$$

The amount (\( A \)) is \$5,960.47.

To find the compound interest (\( I \)):
$$
I = A - P
$$
$$
I = 5960.47 - 800 = 5160.47
$$

So, the answers are:
$$
A = 5960.47, \quad I = 5160.47
$$

#### 4) Principal = \$4,125, Time = 5 years, Rate = 7% compounded annually
- \( P = 4125 \)
- \( r = 7\% = 0.07 \)
- \( n = 1 \) (compounded annually)
- \( t = 5 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 4125 \left(1 + \frac{0.07}{1}\right)^{1 \cdot 5}
$$
$$
A = 4125 \left(1 + 0.07\right)^5
$$
$$
A = 4125 \left(1.07\right)^5
$$

Calculate \( (1.07)^5 \):
$$
(1.07)^5 \approx 1.4025517
$$

Now calculate \( A \):
$$
A = 4125 \times 1.4025517 = 5785.24
$$

The amount (\( A \)) is \$5,785.24.

To find the compound interest (\( I \)):
$$
I = A - P
$$
$$
I = 5785.24 - 4125 = 1660.24
$$

So, the answers are:
$$
A = 5785.24, \quad I = 1660.24
$$

---

Part B: Find the amount payable on \$300 after 8 years

#### a) If it is compounded annually at 9%
- \( P = 300 \)
- \( r = 9\% = 0.09 \)
- \( n = 1 \) (compounded annually)
- \( t = 8 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 300 \left(1 + \frac{0.09}{1}\right)^{1 \cdot 8}
$$
$$
A = 300 \left(1 + 0.09\right)^8
$$
$$
A = 300 \left(1.09\right)^8
$$

Calculate \( (1.09)^8 \):
$$
(1.09)^8 \approx 1.9925626
$$

Now calculate \( A \):
$$
A = 300 \times 1.9925626 = 597.77
$$

The amount payable is \$597.77.

#### b) If it is compounded annually at 35%
- \( P = 300 \)
- \( r = 35\% = 0.35 \)
- \( n = 1 \) (compounded annually)
- \( t = 8 \)

Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
$$
A = 300 \left(1 + \frac{0.35}{1}\right)^{1 \cdot 8}
$$
$$
A = 300 \left(1 + 0.35\right)^8
$$
$$
A = 300 \left(1.35\right)^8
$$

Calculate \( (1.35)^8 \):
$$
(1.35)^8 \approx 10.418354
$$

Now calculate \( A \):
$$
A = 300 \times 10.418354 = 3125.51
$$

The amount payable is \$3,125.51.

---

Final Answers:

#### Part A:
1. \( A = 13780.54, \quad I = 4380.54 \)
2. \( A = 16525.01, \quad I = 3625.01 \)
3. \( A = 5960.47, \quad I = 5160.47 \)
4. \( A = 5785.24, \quad I = 1660.24 \)

#### Part B:
a. \( 597.77 \)
b. \( 3125.51 \)

Boxed Final Answer:

$$
\boxed{
\begin{array}{ll}
\text{Part A:} & \\
1. & A = 13780.54, \quad I = 4380.54 \\
2. & A = 16525.01, \quad I = 3625.01 \\
3. & A = 5960.47, \quad I = 5160.47 \\
4. & A = 5785.24, \quad I = 1660.24 \\
\text{Part B:} & \\
a. & 597.77 \\
b. & 3125.51 \\
\end{array}
}
$$