Compound Interest Answer Key showing step-by-step solutions for financial math problems.

It seems you've uploaded an image containing a set of compound interest problems, and the task is to solve them. The problems involve calculating the future value of investments using the formula for compound interest:

Compound Interest Formula:

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

Where:
- \( A \) = Amount after time \( t \)
- \( P \) = Principal amount (initial investment)
- \( r \) = Annual interest rate (in decimal form)
- \( n \) = Number of times interest is compounded per year
- \( t \) = Time in years

Let's go through each problem step by step.

---

Problem 1:

If you deposit $5000 into a bank account paying 4% annual interest compounded quarterly, how much money will be in the account after 3 years?

#### Given:
- \( P = 5000 \)
- \( r = 4\% = 0.04 \)
- \( n = 4 \) (quarterly compounding)
- \( t = 3 \)

#### Solution:
Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
Substitute the values:
$$
A = 5000 \left(1 + \frac{0.04}{4}\right)^{4 \cdot 3}
$$
Simplify inside the parentheses:
$$
A = 5000 \left(1 + 0.01\right)^{12}
$$
$$
A = 5000 \left(1.01\right)^{12}
$$
Calculate \( (1.01)^{12} \):
$$
(1.01)^{12} \approx 1.126825
$$
Now multiply:
$$
A = 5000 \times 1.126825 \approx 5634.125
$$

#### Final Answer:
$$
\boxed{5634.13}
$$

---

Problem 2:

If your parents deposit $3600 into a bank account when you are born, paying 6% annual interest compounded monthly, how much money will be in the account when you turn 18?

#### Given:
- \( P = 3600 \)
- \( r = 6\% = 0.06 \)
- \( n = 12 \) (monthly compounding)
- \( t = 18 \)

#### Solution:
Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
Substitute the values:
$$
A = 3600 \left(1 + \frac{0.06}{12}\right)^{12 \cdot 18}
$$
Simplify inside the parentheses:
$$
A = 3600 \left(1 + 0.005\right)^{216}
$$
$$
A = 3600 \left(1.005\right)^{216}
$$
Calculate \( (1.005)^{216} \):
$$
(1.005)^{216} \approx 4.79084
$$
Now multiply:
$$
A = 3600 \times 4.79084 \approx 17246.984
$$

#### Final Answer:
$$
\boxed{17246.98}
$$

---

Problem 3:

How much money would you need to deposit at 7% interest compounded quarterly to have $10000 in the account after 10 years?

#### Given:
- \( A = 10000 \)
- \( r = 7\% = 0.07 \)
- \( n = 4 \) (quarterly compounding)
- \( t = 10 \)

#### Solution:
Rearrange the formula to solve for \( P \):
$$
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
$$
Substitute the values:
$$
P = \frac{10000}{\left(1 + \frac{0.07}{4}\right)^{4 \cdot 10}}
$$
Simplify inside the parentheses:
$$
P = \frac{10000}{\left(1 + 0.0175\right)^{40}}
$$
$$
P = \frac{10000}{\left(1.0175\right)^{40}}
$$
Calculate \( (1.0175)^{40} \):
$$
(1.0175)^{40} \approx 2.00957
$$
Now divide:
$$
P = \frac{10000}{2.00957} \approx 4977.75
$$

#### Final Answer:
$$
\boxed{4977.75}
$$

---

Problem 4:

How much money would you need to deposit at 6% interest compounded semiannually to have $2000 in the account after 5 years?

#### Given:
- \( A = 2000 \)
- \( r = 6\% = 0.06 \)
- \( n = 2 \) (semiannual compounding)
- \( t = 5 \)

#### Solution:
Rearrange the formula to solve for \( P \):
$$
P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}
$$
Substitute the values:
$$
P = \frac{2000}{\left(1 + \frac{0.06}{2}\right)^{2 \cdot 5}}
$$
Simplify inside the parentheses:
$$
P = \frac{2000}{\left(1 + 0.03\right)^{10}}
$$
$$
P = \frac{2000}{\left(1.03\right)^{10}}
$$
Calculate \( (1.03)^{10} \):
$$
(1.03)^{10} \approx 1.34392
$$
Now divide:
$$
P = \frac{2000}{1.34392} \approx 1488.04
$$

#### Final Answer:
$$
\boxed{1488.04}
$$

---

Problem 5:

If you deposit $2500 into a bank account paying 8% annual interest compounded annually, how much money will be in the account after 10 years?

#### Given:
- \( P = 2500 \)
- \( r = 8\% = 0.08 \)
- \( n = 1 \) (annual compounding)
- \( t = 10 \)

#### Solution:
Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
Substitute the values:
$$
A = 2500 \left(1 + \frac{0.08}{1}\right)^{1 \cdot 10}
$$
Simplify inside the parentheses:
$$
A = 2500 \left(1 + 0.08\right)^{10}
$$
$$
A = 2500 \left(1.08\right)^{10}
$$
Calculate \( (1.08)^{10} \):
$$
(1.08)^{10} \approx 2.15892
$$
Now multiply:
$$
A = 2500 \times 2.15892 \approx 5397.3
$$

#### Final Answer:
$$
\boxed{5397.30}
$$

---

Problem 6:

If you deposit $100 into a bank account paying 8% annual interest compounded daily, how much money will be in the account after 1 month?

#### Given:
- \( P = 100 \)
- \( r = 8\% = 0.08 \)
- \( n = 365 \) (daily compounding)
- \( t = \frac{1}{12} \) (1 month = 1/12 year)

#### Solution:
Using the formula:
$$
A = P \left(1 + \frac{r}{n}\right)^{nt}
$$
Substitute the values:
$$
A = 100 \left(1 + \frac{0.08}{365}\right)^{365 \cdot \frac{1}{12}}
$$
Simplify inside the parentheses:
$$
A = 100 \left(1 + 0.000219178\right)^{30.4167}
$$
$$
A = 100 \left(1.000219178\right)^{30.4167}
$$
Calculate \( (1.000219178)^{30.4167} \):
$$
(1.000219178)^{30.4167} \approx 1.00672
$$
Now multiply:
$$
A = 100 \times 1.00672 \approx 100.672
$$

#### Final Answer:
$$
\boxed{100.67}
$$

---

Summary of Answers:

1. $ \boxed{5634.13} $
2. $ \boxed{17246.98} $
3. $ \boxed{4977.75} $
4. $ \boxed{1488.04} $
5. $ \boxed{5397.30} $
6. $ \boxed{100.67} $

These solutions follow the compound interest formula and ensure accurate calculations for each scenario. If you have any further questions, feel free to ask!