1. The problem asks to find the annual interest rate and the value of the loan after 10 years based on the given amounts at different times. 2. Let the rate of interest per annum be $r$ (as a decimal). The initial loan is £7500, after 1 year it becomes £7905, and after 2 years it is £8331.87. 3. Using the compound interest formula: $$A = P(1 + r)^t$$ where $A$ is the amount after time $t$, $P$ is the principal, and $r$ the rate. 4. From start to 1 year: $$7905 = 7500(1 + r)^1 \implies 1 + r = \frac{7905}{7500} = 1.054$$ 5. Therefore, the interest rate is $$r = 1.054 - 1 = 0.054 = 5.4\%$$ (to 1 decimal place). 6. Checking for 2 years: $$8331.87 \approx 7500(1.054)^2 = 7500 \times 1.110916 = 8331.87$$ confirming $r=0.054$ is correct. 7. To find the value after 10 years: $$A = 7500(1.054)^{10}$$ 8. Calculating: $$ (1.054)^{10} \approx 1.70158 $$ 9. So, $$A \approx 7500 \times 1.70158 = 12761.87$$ 10. Rounded to the nearest penny, the loan value after 10 years is £12761.87. **Final answers:** a) Rate of interest per annum = 5.4%. b) Value of loan after 10 years = £12761.87.