1. **Stating the problem:** George takes out a loan of £7500 that gathers compound interest. We are given the loan values after 1 and 2 years and need to find: a) The annual interest rate as a percentage to 1 decimal place. b) The value of the loan after 10 years, rounded to the nearest penny. 2. **Formula used:** The compound interest formula is: $$ A = P(1 + r)^t $$ where: - $A$ is the amount after $t$ years, - $P$ is the principal (initial amount), - $r$ is the annual interest rate (as a decimal), - $t$ is the time in years. 3. **Finding the interest rate $r$:** From the table: - $P = 7500$ - After 1 year, $A = 7905$ Using the formula for $t=1$: $$ 7905 = 7500(1 + r)^1 $$ Divide both sides by 7500: $$ 1 + r = \frac{7905}{7500} = 1.054 $$ So, $$ r = 1.054 - 1 = 0.054 $$ Convert to percentage: $$ r = 0.054 \times 100 = 5.4\% $$ 4. **Check with year 2 value:** Using $r=0.054$ and $t=2$: $$ A = 7500(1.054)^2 = 7500 \times 1.110916 = 8331.87 $$ This matches the given value, confirming our rate. 5. **Calculate the loan value after 10 years:** Using $r=0.054$ and $t=10$: $$ A = 7500(1.054)^{10} $$ Calculate $(1.054)^{10}$: $$ (1.054)^{10} \approx 1.708 \quad \text{(using a calculator)} $$ So, $$ A = 7500 \times 1.708 = 12810.00 $$ Rounded to the nearest penny, the loan value after 10 years is £12810.00. **Final answers:** - a) Interest rate = 5.4% per annum - b) Loan value after 10 years = 12810.00 pounds