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Compound Interest Practice Problems Answer Key - Solutions for calculating final amounts with different compounding frequencies.

Answer key for compound interest practice problems, showing calculations and solutions for various scenarios with interest rates, compounding periods, and time.

Answer key for compound interest practice problems, showing calculations and solutions for various scenarios with interest rates, compounding periods, and time.

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Show Answer Key & Explanations Step-by-step solution for: Compound Interest ALGEBRA Review + Practice
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:
$$
P = \frac{10000}{2.00957} \approx 4977.74
$$

#### Final Answer:
$$
\boxed{4977.74}
$$

---

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 \( (1.000219178)^{30.4167} \):
$$
(1.000219178)^{30.4167} \approx 1.00671
$$
Now multiply:
$$
A = 100 \times 1.00671 \approx 100.671
$$

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

---

Summary of Answers:


1. $ \boxed{5634.13} $
2. $ \boxed{17246.98} $
3. $ \boxed{4977.74} $
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!
Parent Tip: Review the logic above to help your child master the concept of compound interest word problems worksheet.
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