To calculate the balance after 10 years in a savings account with compounded interest, we use the compound interest formula:

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

Where:

• $A$ is the final amount (balance after interest),
• $P$ is the principal amount (initial deposit),
• $r$ is the annual interest rate (in decimal form),
• $n$ is the number of times interest is compounded per year,
• $t$ is the time in years.

For this problem:

• $P = 100$,
• $r = 10\% = 0.10$,
• $n = 4$ (since interest is compounded quarterly),
• $t = 10$ years.

Now, plug these values into the formula:

$A = 100 \left(1 + \frac{0.10}{4}\right)^{4 \times 10}$ $A = 100 \left(1 + 0.025\right)^{40}$ $A = 100 \left(1.025\right)^{40}$

Now, calculate:

$A = 100 \times 2.685$ $A \approx 268.50$

So, the balance after 10 years is approximately $268.50.

Would you like a more detailed breakdown, or do you have any other questions?

Here are 5 related questions:

1. How would the final amount change if the interest was compounded monthly instead of quarterly?
2. What would happen to the balance if the interest rate were 5% instead of 10%?
3. How does the frequency of compounding affect the overall growth of an investment?
4. Can you calculate the balance after 5 years with the same interest rate and compounding period?
5. How does the compound interest formula differ from simple interest?

Tip: The more frequently interest is compounded, the greater the final balance will be, even if the interest rate stays the same!