1. **Problem Statement:** Hakim invests money in an account with 4.1% annual interest compounded quarterly. After 8 years, the amount is 180547.26. We need to find: a) Initial investment (principal) $P$. b) Total interest earned. c) Time to reach 250000. d) Required nominal annual interest rate to reach 300000 in 5 years. 2. **Formula for compound interest:** $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ = amount after time $t$, - $P$ = principal (initial investment), - $r$ = annual nominal interest rate (decimal), - $n$ = number of compounding periods per year, - $t$ = time in years. 3. **Given:** - $A = 180547.26$ - $r = 0.041$ - $n = 4$ - $t = 8$ --- **a) Find initial investment $P$:** $$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} = \frac{180547.26}{\left(1 + \frac{0.041}{4}\right)^{4 \times 8}}$$ Calculate: $$1 + \frac{0.041}{4} = 1 + 0.01025 = 1.01025$$ $$nt = 4 \times 8 = 32$$ $$P = \frac{180547.26}{(1.01025)^{32}}$$ Calculate $(1.01025)^{32}$: $$\ln(1.01025) \approx 0.010197$$ $$32 \times 0.010197 = 0.3263$$ $$e^{0.3263} \approx 1.386$$ So, $$P = \frac{180547.26}{1.386} \approx 130260.15$$ --- **b) Total interest earned:** $$\text{Interest} = A - P = 180547.26 - 130260.15 = 50287.11$$ --- **c) Time to reach 250000:** Use formula: $$250000 = 130260.15 \times \left(1.01025\right)^{4t}$$ Divide both sides: $$\frac{250000}{130260.15} = (1.01025)^{4t}$$ Calculate ratio: $$1.918 = (1.01025)^{4t}$$ Take natural log: $$\ln(1.918) = 4t \times \ln(1.01025)$$ Calculate: $$\ln(1.918) \approx 0.650$$ $$\ln(1.01025) \approx 0.010197$$ Solve for $t$: $$t = \frac{0.650}{4 \times 0.010197} = \frac{0.650}{0.04079} \approx 15.94 \text{ years}$$ --- **d) Find nominal annual interest rate $r$ to reach 300000 in 5 years:** Given: $$A = 300000, P = 130260.15, t = 5, n = 4$$ Formula: $$300000 = 130260.15 \times \left(1 + \frac{r}{4}\right)^{4 \times 5} = 130260.15 \times \left(1 + \frac{r}{4}\right)^{20}$$ Divide: $$\frac{300000}{130260.15} = \left(1 + \frac{r}{4}\right)^{20}$$ Calculate ratio: $$2.302 = \left(1 + \frac{r}{4}\right)^{20}$$ Take natural log: $$\ln(2.302) = 20 \times \ln\left(1 + \frac{r}{4}\right)$$ Calculate: $$\ln(2.302) \approx 0.834$$ Solve for $\ln\left(1 + \frac{r}{4}\right)$: $$\ln\left(1 + \frac{r}{4}\right) = \frac{0.834}{20} = 0.0417$$ Exponentiate: $$1 + \frac{r}{4} = e^{0.0417} \approx 1.0426$$ Solve for $r$: $$\frac{r}{4} = 0.0426 \Rightarrow r = 4 \times 0.0426 = 0.1704 = 17.04\%$$ Quarterly periodic rate: $$\frac{r}{4} = 0.0426 = 4.26\%$$ --- **Final answers:** - a) Initial investment $P \approx 130260.15$ - b) Total interest earned $\approx 50287.11$ - c) Time to reach 250000 $\approx 15.94$ years - d) Required nominal annual interest rate $\approx 17.04\%$, quarterly rate $\approx 4.26\%$