1. **State the problem:** We are given the initial loan amount and its value after 1 and 2 years. We need to find: a) The annual rate of interest 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 loan grows according to compound interest: $$ A = P(1 + r)^t $$ 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. **Find the rate of interest $r$:** From the data: - Start: $P = 7500$ - After 1 year: $A = 7920$ Using the formula for $t=1$: $$ 7920 = 7500(1 + r)^1 $$ Divide both sides by 7500: $$ 1 + r = \frac{7920}{7500} = 1.056 $$ So, $$ r = 1.056 - 1 = 0.056 $$ Convert to percentage: $$ r = 0.056 \times 100 = 5.6\% $$ 4. **Check with year 2 data:** $$ A = 7500(1.056)^2 = 7500 \times 1.115136 = 8363.52 $$ Matches given data, so $r=5.6\%$ is correct. 5. **Calculate loan value after 10 years:** Using $r=0.056$ and $t=10$: $$ A = 7500(1.056)^{10} $$ Calculate: $$ (1.056)^{10} \approx 1.74494 $$ So, $$ A = 7500 \times 1.74494 = 13087.05 $$ Rounded to nearest penny: $$ 13087.05 $$ **Final answers:** - a) Rate of interest per annum = **5.6%** - b) Value of loan after 10 years = **13087.05** pounds