1. **Problem statement:** Jack opens an account with an initial deposit of 500000 Frw and deposits the same amount at the beginning of every year. The bank pays compound interest at 13.5% per annum. We need to find: - The amount he will receive at the end of 9 years. - The time it takes for the money to accumulate to 3320000 Frw. 2. **Formula used:** This is a problem of compound interest with regular annual deposits (an annuity). The future value $A$ after $n$ years with annual deposit $P$, interest rate $r$, is given by: $$A = P \times \frac{(1+r)^n - 1}{r} \times (1+r)$$ This formula accounts for deposits made at the beginning of each year. 3. **Calculate amount after 9 years:** Given: $P = 500000$ $r = 0.135$ $n = 9$ Calculate: $$A = 500000 \times \frac{(1+0.135)^9 - 1}{0.135} \times (1+0.135)$$ First, calculate $(1+0.135)^9$: $$1.135^9 \approx 3.1821$$ Then: $$\frac{3.1821 - 1}{0.135} = \frac{2.1821}{0.135} \approx 16.1637$$ Multiply by $(1+0.135) = 1.135$: $$16.1637 \times 1.135 \approx 18.345$$ Finally multiply by $P$: $$500000 \times 18.345 = 9,172,500$$ So, at the end of 9 years, Jack will have approximately 9,172,500 Frw. 4. **Calculate time to accumulate 3,320,000 Frw:** We want to find $n$ such that: $$3320000 = 500000 \times \frac{(1+0.135)^n - 1}{0.135} \times (1+0.135)$$ Divide both sides by $500000 \times 1.135$: $$\frac{3320000}{500000 \times 1.135} = \frac{(1.135)^n - 1}{0.135}$$ Calculate left side: $$\frac{3320000}{567500} \approx 5.85$$ Multiply both sides by 0.135: $$5.85 \times 0.135 = (1.135)^n - 1$$ $$0.78975 = (1.135)^n - 1$$ Add 1: $$(1.135)^n = 1.78975$$ Take natural logarithm: $$n \ln(1.135) = \ln(1.78975)$$ Calculate logarithms: $$\ln(1.135) \approx 0.1267$$ $$\ln(1.78975) \approx 0.582$$ Solve for $n$: $$n = \frac{0.582}{0.1267} \approx 4.59$$ So, it will take approximately 4.59 years for the money to accumulate to 3,320,000 Frw. **Final answers:** - Amount after 9 years: approximately 9,172,500 Frw. - Time to reach 3,320,000 Frw: approximately 4.59 years.