A machine will be replaced in 10 years from now at a cost of Php750,000. How much should the owner save at the beginning of each quarter in order to replace the old machine if his special savings earns interest at 8.75% compounded quarterly?

To solve this problem, we will use the **future value of an ordinary annuity formula**, which accounts for regular savings (deposits) earning compound interest. 

The formula for the future value of an ordinary annuity is:

\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]

Where:
- \(FV\) is the future value (the amount needed to replace the machine),
- \(P\) is the quarterly payment (the amount the owner needs to save each quarter),
- \(r\) is the interest rate per period (per quarter),
- \(n\) is the total number of periods (total number of quarters).

### Given dataTo solve this problem, we will use the **future value of an ordinary annuity formula**, which accounts for regular savings (deposits) earning compound interest. 

The formula for the future value of an ordinary annuity is:

\[
FV = P \times \frac{(1 + r)^n - 1}{r}
\]

Where:
- \(FV\) is the future value (the amount needed to replace the machine),
- \(P\) is the quarterly payment (the amount the owner needs to save each quarter),
- \(r\) is the interest rate per period (per quarter),
- \(n\) is the total number of periods (total number of quarters).

### Given 
- \(FV = \text{Php } 750,000\) (amount needed after 10 years),
- Annual interest rate = 8.75%,
- Compounding is quarterly, so there are 4 periods per year,
- Time = 10 years.

### Step-by-step solution:

1. **Convert the annual interest rate to a quarterly interest rate:**
   \[
   r = \frac{8.75\%}{4} = \frac{0.0875}{4} = 0.021875
   \]

2. **Calculate the total number of quarters over 10 years:**
   \[
   n = 10 \times 4 = 40 \text{ quarters}
   \]

3. **Substitute these values into the future value formula and solve for \(P\):**

\[
750,000 = P \times \frac{(1 + 0.021875)^{40} - 1}{0.021875}
\]

Now let's solve for \(P\).The owner should save approximately **Php 11,920.09** at the beginning of each quarter to replace the machine in 10 years, assuming the savings earn interest at 8.75% compounded quarterly.

Would you like more details or have any other questions?

Here are five related questions for further exploration:
1. What happens if the interest rate changes during the 10-year period?
2. How much total interest will be earned over the 10 years?
3. What would the required quarterly saving be if the machine cost increases?
4. How does the formula change if the savings were compounded monthly?
5. What is the difference between an ordinary annuity and an annuity due?

**Tip:** Regularly reviewing the interest rate on your savings ensures that you are maximizing the benefits of compounding interest.

Learn how to calculate the quarterly savings required to replace a machine costing Php 750,000 in 10 years. This finance problem uses the future value of an ordinary annuity formula with an interest rate of 8.75% compounded quarterly. Suitable for Grades 10-12 students.

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