Question: 31-35. Stephen paid a total of ₱538,000 on the money he borrowed 2.5 years ago * at 10% simple annual interest. How much did he borrow originally?
1. **State the problem:** Stephen paid a total of 538000 after 2.5 years with 10% simple annual interest. We need to find the original amount borrowed (principal $P$).
2. **Formula for simple interest:**
$$ A = P + I = P + P \times r \times t = P(1 + rt) $$
where $A$ is the total amount paid, $P$ is the principal, $r$ is the annual interest rate (in decimal), and $t$ is the time in years.
3. **Given values:**
- $A = 538000$
- $r = 10\% = 0.10$
- $t = 2.5$
4. **Substitute into formula:**
$$ 538000 = P(1 + 0.10 \times 2.5) $$
5. **Calculate inside parentheses:**
$$ 1 + 0.10 \times 2.5 = 1 + 0.25 = 1.25 $$
6. **Rewrite equation:**
$$ 538000 = 1.25P $$
7. **Solve for $P$ by dividing both sides by 1.25:**
$$ P = \frac{538000}{1.25} $$
8. **Show cancellation:**
$$ P = \frac{\cancel{538000}}{\cancel{1.25}} \text{ (division)} $$
9. **Calculate:**
$$ P = 430400 $$
**Final answer:** Stephen originally borrowed $\boxed{430400}$.