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Let's go through the questions one by one, calculating the simple interest where needed.

### Formula for Simple Interest:
\[
\text{Interest (I)} = \text{Principal (P)} \times \text{Rate (R)} \times \text{Time (T)}
\]

Where:
- \( I \) is the interest.
- \( P \) is the principal (initial amount).
- \( R \) is the interest rate (in decimal form).
- \( T \) is the time period (in years).

Now, let’s apply the formula to each problem.

---

### 1. How much interest is earned on ₱6,200 at 4% for one year?
- **Given**: \( P = ₱6,200 \), \( R = 4\% = 0.04 \), \( T = 1 \text{ year} \)
\[
I = 6200 \times 0.04 \times 1 = ₱248
\]
**Answer**: ₱248 is earned in interest.

---

### 2. What was the interest rate if your balance on an investment of ₱27,750 at the end of six years is ₱41,070?
- **Given**: \( P = ₱27,750 \), \( A = ₱41,070 \), \( T = 6 \text{ years} \), find \( R \).

First, find the interest \( I \):
\[
I = A - P = 41,070 - 27,750 = ₱13,320
\]
Now, solve for \( R \):
\[
I = P \times R \times T \implies 13,320 = 27,750 \times R \times 6
\]
\[
R = \frac{13,320}{27,750 \times 6} \approx 0.08
\]
**Answer**: The interest rate is 8%.

---

### 3. At what rate was an investment made that obtains ₱2,520 on ₱10,500 over four years?
- **Given**: \( P = ₱10,500 \), \( I = ₱2,520 \), \( T = 4 \text{ years} \), find \( R \).
\[
I = P \times R \times T \implies 2,520 = 10,500 \times R \times 4
\]
\[
R = \frac{2,520}{10,500 \times 4} \approx 0.06
\]
**Answer**: The interest rate is 6%.

---

### 4. If a loan is taken out for ₱15,900 at 9% and costs ₱8,586, how long was the loan for?
- **Given**: \( P = ₱15,900 \), \( I = ₱8,586 \), \( R = 9\% = 0.09 \), find \( T \).
\[
I = P \times R \times T \implies 8,586 = 15,900 \times 0.09 \times T
\]
\[
T = \frac{8,586}{15,900 \times 0.09} \approx 6
\]
**Answer**: The loan was for 6 years.

---

### 5. A bank is offering 3.5% simple interest on a savings account. If you deposit ₱10,555, how much interest will you earn in two years?
- **Given**: \( P = ₱10,555 \), \( R = 3.5\% = 0.035 \), \( T = 2 \text{ years} \)
\[
I = 10,555 \times 0.035 \times 2 = ₱739
\]
**Answer**: ₱739 interest will be earned.

---

### 6. If you borrow ₱75,000 for two and half years at an interest rate of 9.5%, what is the maturity value of the loan?
- **Given**: \( P = ₱75,000 \), \( R = 9.5\% = 0.095 \), \( T = 2.5 \text{ years} \)
\[
I = 75,000 \times 0.095 \times 2.5 = ₱17,812.50
\]
The maturity value \( A \) is:
\[
A = P + I = 75,000 + 17,812.50 = ₱92,812.50
\]
**Answer**: The maturity value of the loan is ₱92,812.50.

---

Would you like more details on any of the steps? Here are five related questions:

1. How do you calculate the maturity amount for an investment after interest?
2. How would the interest formula change if the rate were compounded instead of simple?
3. What factors influence the time it takes for a loan to double?
4. How do changes in the interest rate affect the total interest earned?
5. What are the key differences between simple and compound interest?

**Tip**: Always check the interest type (simple or compound) when analyzing investments or loans, as the formulas differ significantly.

How much interest is earned on ₱6,200 at 4% for one year?

Calculate the simple interest earned on ₱6,200 at an interest rate of 4% over one year. This problem explores simple interest concepts, interest rate, and time, suitable for Grades 7-9 students.

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