1. **State the problem:** We have a principal amount $P = 18300$ that grows to an amount $A = 89265$ at a compound interest rate of 5% per annum. We need to find the time $t$ in years. 2. **Recall the compound interest formula:** $$A = P \left(1 + \frac{r}{100}\right)^t$$ where $r$ is the annual interest rate and $t$ is the time in years. 3. **Substitute the known values:** $$89265 = 18300 \left(1 + \frac{5}{100}\right)^t = 18300 \times (1.05)^t$$ 4. **Divide both sides by 18300:** $$\frac{89265}{18300} = (1.05)^t$$ $$4.875 = (1.05)^t$$ 5. **Take the natural logarithm on both sides:** $$\ln(4.875) = \ln\left((1.05)^t\right) = t \ln(1.05)$$ 6. **Solve for $t$:** $$t = \frac{\ln(4.875)}{\ln(1.05)}$$ Calculate the values: $$\ln(4.875) \approx 1.5847$$ $$\ln(1.05) \approx 0.04879$$ $$t \approx \frac{1.5847}{0.04879} \approx 32.47 \text{ years}$$ 7. **Convert years to months:** $$32.47 \times 12 \approx 389.64 \text{ months}$$ Since none of the options match 389.64 months, let's check if the interest is compounded more frequently or if the time is in months directly. **Alternative approach assuming time $t$ is in years but interest compounded monthly:** If interest is compounded monthly, the formula is: $$A = P \left(1 + \frac{r}{12 \times 100}\right)^{12t}$$ Substitute values: $$89265 = 18300 \left(1 + \frac{5}{1200}\right)^{12t} = 18300 \times (1.0041667)^{12t}$$ Divide: $$4.875 = (1.0041667)^{12t}$$ Take natural log: $$\ln(4.875) = 12t \ln(1.0041667)$$ Solve for $t$: $$t = \frac{\ln(4.875)}{12 \times \ln(1.0041667)}$$ Calculate: $$\ln(1.0041667) \approx 0.004158$$ $$t = \frac{1.5847}{12 \times 0.004158} = \frac{1.5847}{0.0499} \approx 31.75 \text{ years}$$ Convert to months: $$31.75 \times 12 = 381 \text{ months}$$ Still no match. **Check if time is in months directly with annual rate:** If $t$ is in months, then convert rate to monthly rate: $$r_{monthly} = \frac{5}{12} \approx 0.4167\%$$ Formula: $$A = P \left(1 + \frac{r_{monthly}}{100}\right)^t = 18300 \times (1.004167)^t$$ Solve: $$4.875 = (1.004167)^t$$ Take ln: $$\ln(4.875) = t \ln(1.004167)$$ $$t = \frac{1.5847}{0.004158} \approx 381 \text{ months}$$ No match with options. **Try to check options by plugging back:** - For 24 months (2 years): $$A = 18300 \times (1.05)^2 = 18300 \times 1.1025 = 20117.25$$ - For 36 months (3 years): $$A = 18300 \times (1.05)^3 = 18300 \times 1.157625 = 21172.24$$ - For 48 months (4 years): $$A = 18300 \times (1.05)^4 = 18300 \times 1.215506 = 22224.9$$ - For 72 months (6 years): $$A = 18300 \times (1.05)^6 = 18300 \times 1.3401 = 24522.83$$ None of these match 89265. **Conclusion:** The problem likely assumes the time is in months and the rate is monthly compounded at 5% per annum, but the amount is much larger, so the time must be longer. **Final step:** Since the amount is approximately 4.875 times the principal, and the annual compound rate is 5%, the time in years is approximately 32.47 years, which is about 389.64 months. Since none of the options match, the closest is 36 months (3 years), but it is far off. **Answer:** None of the given options is correct based on the compound interest formula and given data. If forced to choose, the problem might expect 72 months (6 years) as the answer, but mathematically it does not fit. **Summary:** The time $t$ is approximately $32.47$ years or $389.64$ months. **Final answer:** $$\boxed{t \approx 32.47 \text{ years}}$$