1. Problem: Find the exact interest on PHP 65,250.00 at 9.5% from Mar. 12, 2020, to Sept. 15, 2020, using Approximate Time. - Approximate Time counts each month as 30 days. - Time period: Mar 12 to Sept 15 = 6 months and 3 days approx. - Approximate time in years: $$\frac{6 \times 30 + 3}{360} = \frac{183}{360} = 0.5083$$ - Interest formula: $$I = P \times r \times t$$ - Calculate: $$I = 65250 \times 0.095 \times 0.5083 = 3151.03$$ 2. Problem: Find the ordinary interest on PHP 65,250.00 at 9.5% from Mar. 12, 2020, to Sept. 15, 2020, using Actual Time. - Ordinary interest uses 360 days per year. - Actual days from Mar 12 to Sept 15: 187 days. - Time in years: $$\frac{187}{360} = 0.5194$$ - Calculate: $$I = 65250 \times 0.095 \times 0.5194 = 3219.91$$ 3. Problem: Calculate compound amount for Php 8000 at 8% interest, compounded quarterly, for 5 years. - Compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ - Here, $$P=8000$$, $$r=0.08$$, $$n=4$$, $$t=5$$ - Calculate: $$A = 8000 \left(1 + \frac{0.08}{4}\right)^{4 \times 5} = 8000 \times (1.02)^{20} = 11887.58$$ 4. Problem: Find exact interest on PHP 65,250.00 at 9.5% from Mar. 12, 2020, to Sept. 15, 2020, using Actual Time. - Actual days: 187 days. - Time in years: $$\frac{187}{365} = 0.5123$$ - Calculate: $$I = 65250 \times 0.095 \times 0.5123 = 3167.12$$ 5. Problem: Calculate simple interest on a 120-day loan of 7000 at 5.25% annual rate. - Time in years: $$\frac{120}{365} = 0.3288$$ - Calculate: $$I = 7000 \times 0.0525 \times 0.3288 = 120.50$$ (closest option is 122.50) 6. Problem: ₱10,000 deposited for 2 years at 8% interest compounded semiannually. - Compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ - $$P=10000$$, $$r=0.08$$, $$n=2$$, $$t=2$$ - Calculate: $$A = 10000 \times (1 + 0.04)^4 = 10000 \times 1.16985856 = 11698.59$$ - Interest earned: $$11698.59 - 10000 = 1698.59$$ 7. Problem: Find ordinary interest on PHP 65,250.00 at 9.5% from Mar. 12, 2020, to Sept. 15, 2020, using Approximate Time. - Same as problem 1: $$3151.03$$ 8. Problem: Calculate simple interest on a 3-month loan of 2000 at 6.5%. - Time in years: $$\frac{3}{12} = 0.25$$ - Calculate: $$I = 2000 \times 0.065 \times 0.25 = 32.50$$ 9. Problem: Future value of Php 4000 at 6% interest compounded monthly for 6 years. - $$P=4000$$, $$r=0.06$$, $$n=12$$, $$t=6$$ - Calculate: $$A = 4000 \times (1 + \frac{0.06}{12})^{12 \times 6} = 4000 \times (1.005)^{72} = 5674.08$$ 10. Problem: Simple interest on a 4-month loan of 1500 at 5.25%. - Time in years: $$\frac{4}{12} = 0.3333$$ - Calculate: $$I = 1500 \times 0.0525 \times 0.3333 = 26.25$$ 11. Problem: ₱10,000 deposited for 2 years at 8% interest. - Interest compounded quarterly vs semiannually. - Quarterly: $$A_q = 10000 \times (1 + 0.02)^8 = 11716.59$$ - Semiannually: $$A_s = 10000 \times (1 + 0.04)^4 = 11698.59$$ - Difference in interest: $$11716.59 - 11698.59 = 18$$ 12. Problem: ₱10,000 deposited for 2 years at 8% interest compounded quarterly. - Same as above quarterly amount: Interest = $$11716.59 - 10000 = 1716.59$$