14. What is the principal amount invested on June 16, 2003 to accumulate to P4,500 at 10% interest on: a) September 14, 2003 b) October 14, 2003 c) November 2, 2005

To determine the principal amount (\( P \)) that would accumulate to ₱4,500 at a 10% interest rate over different time periods, we use the simple interest formula. 

### Simple Interest Formula:
\[
A = P \left(1 + \frac{r \times t}{n}\right)
\]
Where:
- \( A \) is the accumulated amount (₱4,500)
- \( P \) is the principal amount (unknown)
- \( r \) is the annual interest rate (10% or 0.10)
- \( t \) is the time period in days
- \( n \) is the number of days in a year (365 for exact interest)

We rearrange this formula to solve for \( P \):
\[
P = \frac{A}{1 + \frac{r \times t}{n}}
\]

### Step 1: Calculate the Number of Days for Each Scenario

#### a) June 16, 2003 to September 14, 2003

- **From June 16, 2003, to September 14, 2003:**
  - June 16 to June 30: 14 days
  - July: 31 days
  - August: 31 days
  - September 1 to September 14: 14 days

Total days \( t_a \):
\[
t_a = 14 + 31 + 31 + 14 = 90 \text{ days}
\]

#### b) June 16, 2003 to October 14, 2003

- **From June 16, 2003, to October 14, 2003:**
  - June 16 to June 30: 14 days
  - July: 31 days
  - August: 31 days
  - September: 30 days
  - October 1 to October 14: 14 days

Total days \( t_b \):
\[
t_b = 14 + 31 + 31 + 30 + 14 = 120 \text{ days}
\]

#### c) June 16, 2003 to November 2, 2005

- **From June 16, 2003, to June 16, 2004:** 1 year = 365 days
- **From June 16, 2004, to June 16, 2005:** 1 year = 365 days
- **From June 16, 2005, to November 2, 2005:**
  - June 16 to June 30: 14 days
  - July: 31 days
  - August: 31 days
  - September: 30 days
  - October: 31 days
  - November 1 to November 2: 2 days

Total days \( t_c \):
\[
t_c = 365 + 365 + 14 + 31 + 31 + 30 + 31 + 2 = 869 \text{ days}
\]

### Step 2: Calculate the Principal Amount \( P \)

#### a) Principal for September 14, 2003

Using \( t_a = 90 \) days and \( n = 365 \):

\[
P_a = \frac{4,500}{1 + \frac{0.10 \times 90}{365}}
\]

\[
P_a = \frac{4,500}{1 + 0.02466} = \frac{4,500}{1.02466} = 4,391.88 \text{ (rounded to the nearest centavo)}
\]

#### b) Principal for October 14, 2003

Using \( t_b = 120 \) days and \( n = 365 \):

\[
P_b = \frac{4,500}{1 + \frac{0.10 \times 120}{365}}
\]

\[
P_b = \frac{4,500}{1 + 0.03288} = \frac{4,500}{1.03288} = 4,355.57 \text{ (rounded to the nearest centavo)}
\]

#### c) Principal for November 2, 2005

Using \( t_c = 869 \) days and \( n = 365 \):

\[
P_c = \frac{4,500}{1 + \frac{0.10 \times 869}{365}}
\]

\[
P_c = \frac{4,500}{1 + 0.23808} = \frac{4,500}{1.23808} = 3,635.04 \text{ (rounded to the nearest centavo)}
\]

### Summary of Results:

- **a)** Principal on June 16, 2003 to accumulate to ₱4,500 by September 14, 2003: **₱4,391.88**
- **b)** Principal on June 16, 2003 to accumulate to ₱4,500 by October 14, 2003: **₱4,355.57**
- **c)** Principal on June 16, 2003 to accumulate to ₱4,500 by November 2, 2005: **₱3,635.04**

Let me know if you have further questions or need additional clarifications!

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**Related Questions:**
1. How does the length of time affect the principal required to reach a specific future value?
2. What would be the principal if the interest rate was different?
3. How does using different interest calculation methods affect the principal?
4. How does compound interest differ from simple interest in calculating future values?
5. What are the practical applications of these types of calculations?

**Tip:** The longer the investment period, the lower the initial principal needed to reach a specific future amount due to the accrual of interest over time.

Learn how to calculate the principal amount needed to accumulate ₱4,500 at 10% interest for different time periods. Includes step-by-step solutions for dates ranging from June 16, 2003, to November 2, 2005. Suitable for advanced high school students studying simple interest problems.

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