1. **Problem statement:** Hakim invests money at 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 at same rate. d) Nominal annual interest rate compounded quarterly to reach 300000 in 5 years and the quarterly rate. 2. **Formula used:** Compound interest formula: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ 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:** - $A = 180547.26$ - $r = 0.041$ - $n = 4$ (quarterly) - $t = 8$ --- ### 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 inside the parentheses: $$1 + \frac{0.041}{4} = 1 + 0.01025 = 1.01025$$ Calculate exponent: $$4 \times 8 = 32$$ Calculate denominator: $$1.01025^{32} \approx e^{32 \times \ln(1.01025)} \approx e^{32 \times 0.010197} = e^{0.3263} \approx 1.3865$$ Calculate $P$: $$P = \frac{180547.26}{1.3865} \approx 130230.15$$ --- ### b) Total interest earned: $$\text{Interest} = A - P = 180547.26 - 130230.15 = 50317.11$$ --- ### c) Time $t$ to reach $A = 250000$ with same $P$, $r$, $n$: Use formula: $$250000 = 130230.15 \times \left(1.01025\right)^{4t}$$ Divide both sides: $$\left(1.01025\right)^{4t} = \frac{250000}{130230.15} \approx 1.919$$ Take natural log: $$4t \times \ln(1.01025) = \ln(1.919)$$ Calculate logs: $$\ln(1.01025) \approx 0.010197$$ $$\ln(1.919) \approx 0.651$$ Solve for $t$: $$4t = \frac{0.651}{0.010197} = 63.87$$ $$t = \frac{63.87}{4} = 15.97 \text{ years}$$ --- ### d) Find nominal annual interest rate $r$ to reach $A=300000$ in $t=5$ years with $P=130230.15$ and $n=4$: $$300000 = 130230.15 \times \left(1 + \frac{r}{4}\right)^{4 \times 5}$$ Divide: $$\left(1 + \frac{r}{4}\right)^{20} = \frac{300000}{130230.15} \approx 2.303$$ Take natural log: $$20 \times \ln\left(1 + \frac{r}{4}\right) = \ln(2.303) \approx 0.834$$ 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 $P \approx 130230.15$ - b) Total interest earned $\approx 50317.11$ - c) Time to reach 250000 $\approx 15.97$ years - d) Nominal annual interest rate $\approx 17.04\%$, quarterly rate $\approx 4.26\%$