1. **Identify the principal amount.** The principal is the initial amount borrowed or invested. Answer: 5M pesos 2. **Identify the rate of interest.** The rate is the percentage charged per time period. Answer: 0.5% per month 3. **Identify the time.** Time is the duration the money is borrowed or invested. Answer: 4 years 4. **Identify the due date.** Due date is the date when the loan must be paid back. Since loan started December 13, 2025, and duration is 4 years, due date = December 13, 2029 5. **When will be the start of the loan computation?** Loan computation starts on the date the loan is taken. Answer: December 13, 2025 6. **What will be the interest amount?** Formula for simple interest: $$I = P \times r \times t$$ Where: $P = 5,000,000$ pesos $r = 0.5\% = 0.005$ per month $t = 4$ years = $4 \times 12 = 48$ months Calculate: $$I = 5,000,000 \times 0.005 \times 48 = 1,200,000$$ pesos Answer: P 1,200,000.00 (closest option is P 1,000,000.00 but correct is 1,200,000) 7. **What will be the maturity value?** Maturity value = Principal + Interest $$M = P + I = 5,000,000 + 1,200,000 = 6,200,000$$ pesos Answer: None of the options exactly match, closest is P 5,100,000.00 but correct is 6,200,000 8. **Which will produce the lowest interest?** Calculate interest for each: $I = P \times r \times t$ - 6 months, 2%: $2500 \times 0.02 \times 6 = 300$ - 9 months, 1.5%: $2500 \times 0.015 \times 9 = 337.5$ - 3 months, 1%: $2500 \times 0.01 \times 3 = 75$ - 12 months, 0.75%: $2500 \times 0.0075 \times 12 = 225$ Lowest interest is P75 from 3 months at 1% Answer: A P2,500.00 loan payable in 3 months with 1% interest per month. 9. **Which will produce the highest interest?** From above, highest is 337.5 Answer: A P2,500.00 loan payable in 9 months with 1.5% interest per month. 10. **Conclusion based on 8 and 9:** Shortest loan term with lowest interest rate produces lowest interest. Answer: The shortest the loan term with the lowest interest rate will produce the lowest interest amount. 11. **"Interest on interest" type of interest is:** Answer: Compound Interest 12. **Method favorable to businessmen with respect to number of days:** Answer: Exact Method 13. **Find simple interest rate given P12,000 cost, P1,800 interest in 6 months:** Formula: $$I = P \times r \times t$$ Rearranged: $$r = \frac{I}{P \times t}$$ $$r = \frac{1800}{12000 \times 0.5} = 0.3 = 30\%$$ Answer: 30% 14. **Loan P50,000, interest P2,500 at 10% simple interest, find time:** $$I = P \times r \times t$$ $$t = \frac{I}{P \times r} = \frac{2500}{50000 \times 0.10} = 0.5 \text{ years} = 6 \text{ months}$$ Answer: 6 months 15. **Principal to accumulate P250,000 in 2 years at 8% simple interest:** Formula: $$A = P + I = P + P r t = P(1 + r t)$$ Rearranged: $$P = \frac{A}{1 + r t} = \frac{250000}{1 + 0.08 \times 2} = \frac{250000}{1.16} = 215517.24$$ Rounded: P215,517 Answer: P215,517.00 16. **Interest earned on P8,000 at 5% compounded semi-annually for 3 years:** Compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ Where $n=2$, $r=0.05$, $t=3$ $$A = 8000 \left(1 + \frac{0.05}{2}\right)^{2 \times 3} = 8000 (1.025)^6 = 8000 \times 1.15969 = 9277.55$$ Interest = $9277.55 - 8000 = 1277.55$ Answer: P1,277.55 17. **Future value of P9,500 at 6% compounded daily for 4 years:** $n=365$, $r=0.06$, $t=4$ $$A = 9500 \left(1 + \frac{0.06}{365}\right)^{365 \times 4} = 9500 (1.00016438)^{1460} = 9500 \times 1.26807 = 12076.63$$ Answer: P12,076.63 18. **Interest earned on P25,000 at 8% compounded monthly for 3 years:** $n=12$, $r=0.08$, $t=3$ $$A = 25000 \left(1 + \frac{0.08}{12}\right)^{12 \times 3} = 25000 (1.0066667)^{36} = 25000 \times 1.26824 = 31706$$ Interest = $31706 - 25000 = 6706$ Closest option is P31,675.00 (likely a typo, correct interest is 6706) Slug: "simple interest" Subject: "finance" Desmos: {"latex":"","features":{"intercepts":false,"extrema":false}} q_count: 18