Compound Interest Cbfe1A

1. **State the problem:** Calculate the interest on a loan of $345,000$ that matures after $3$ years and $9$ months with an interest rate of $2.5\%$ compounded quarterly. 2. **Formula used:** The compound interest formula is: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after interest - $P$ is the principal amount ($345,000$) - $r$ is the annual interest rate (as a decimal, so $0.025$) - $n$ is the number of compounding periods per year ($4$ for quarterly) - $t$ is the time in years 3. **Convert time to years:** $3$ years and $9$ months = $3 + \frac{9}{12} = 3.75$ years 4. **Calculate the amount $A$:** $$A = 345000 \left(1 + \frac{0.025}{4}\right)^{4 \times 3.75}$$ 5. **Simplify inside the parentheses:** $$1 + \frac{0.025}{4} = 1 + 0.00625 = 1.00625$$ 6. **Calculate the exponent:** $$4 \times 3.75 = 15$$ 7. **Calculate $A$:** $$A = 345000 \times (1.00625)^{15}$$ 8. **Calculate $(1.00625)^{15}$:** $$ (1.00625)^{15} \approx 1.09744 $$ 9. **Calculate $A$:** $$ A \approx 345000 \times 1.09744 = 378,016.80 $$ 10. **Calculate the interest $I$:** $$ I = A - P = 378,016.80 - 345,000 = 33,016.80 $$ **Final answer:** The interest on the loan is approximately $33,016.80$ rounded to two decimal places.