This problem involves compound interest, where Mr. and Mrs. Bonifacio invest ₱15,000 for their son, with a 7% annual interest rate, compounded yearly, for 21 years. The compound interest formula is:

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

Where:

• $A$ is the amount of money in the account after interest,
• $P$ is the initial principal (₱15,000),
• $r$ is the annual interest rate (7% or 0.07),
• $n$ is the number of times the interest is compounded per year (since it's compounded annually, $n = 1$),
• $t$ is the number of years the money is invested (21 years).

Substituting the values:

$A = 15,000 \left(1 + \frac{0.07}{1}\right)^{1 \times 21}$ $A = 15,000 \left(1.07\right)^{21}$

Now, calculating the value of $(1.07)^{21}$:

$(1.07)^{21} \approx 4.016$

Now multiply this by 15,000:

$A \approx 15,000 \times 4.016 = 60,240$

So, after 21 years, Mr. and Mrs. Bonifacio will have approximately ₱60,240 in their account.

Would you like a detailed explanation of the steps or have any further questions?

Here are 5 related questions you might find interesting:

1. What happens if the interest rate changes after a few years?
2. How would the amount change if the interest is compounded monthly instead of annually?
3. What if Mr. and Mrs. Bonifacio wanted to withdraw the money after 15 years instead of 21 years?
4. How can the formula be adapted for different compounding frequencies?
5. What is the future value if they invest ₱20,000 instead of ₱15,000?

Tip: Compounding more frequently (e.g., monthly) results in more interest earned compared to annual compounding!