1. **State the problem:** Andrew and Aisha each invest 7500. Andrew's account pays 7 3/4 % interest compounded daily, and Aisha's pays 7 1/4 % compounded continuously. We want to find how much longer it takes for Aisha's money to triple compared to Andrew's, rounded to the nearest hundredth of a year. 2. **Convert interest rates to decimals:** Andrew's rate: 7 3/4 % = 7.75% = 0.0775 Aisha's rate: 7 1/4 % = 7.25% = 0.0725 3. **Formulas:** - For compound interest compounded daily: $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where $n=365$ (days per year), $r$ is annual rate, $t$ is time in years. - For continuous compounding: $$A = Pe^{rt}$$ 4. **Set final amount to triple the principal:** $$A = 3P$$ 5. **Solve for Andrew's time $t_A$:** $$3P = P\left(1 + \frac{0.0775}{365}\right)^{365t_A}$$ Divide both sides by $P$: $$3 = \left(1 + \frac{0.0775}{365}\right)^{365t_A}$$ Take natural log: $$\ln(3) = 365t_A \ln\left(1 + \frac{0.0775}{365}\right)$$ Solve for $t_A$: $$t_A = \frac{\ln(3)}{365 \ln\left(1 + \frac{0.0775}{365}\right)}$$ 6. **Solve for Aisha's time $t_B$:** $$3P = Pe^{0.0725 t_B}$$ Divide both sides by $P$: $$3 = e^{0.0725 t_B}$$ Take natural log: $$\ln(3) = 0.0725 t_B$$ Solve for $t_B$: $$t_B = \frac{\ln(3)}{0.0725}$$ 7. **Calculate values:** Calculate $t_A$: $$t_A = \frac{\ln(3)}{365 \ln\left(1 + \frac{0.0775}{365}\right)} \approx \frac{1.0986}{365 \times 0.000211} = \frac{1.0986}{0.077015} \approx 14.26 \text{ years}$$ Calculate $t_B$: $$t_B = \frac{1.0986}{0.0725} \approx 15.15 \text{ years}$$ 8. **Find difference:** $$t_B - t_A = 15.15 - 14.26 = 0.89 \text{ years}$$ **Final answer:** It takes Aisha approximately **0.89 years longer** than Andrew for her money to triple.