1. **State the problem:** Rebecca takes out a loan with compound interest. We know the initial amount and the values after 1 and 2 years. We 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 for compound interest:** The value of the loan after $t$ years is given by: $$ A = P(1 + r)^t $$ where: - $A$ is the amount after $t$ years, - $P$ is the initial principal, - $r$ is the annual interest rate (as a decimal), - $t$ is the time in years. 3. **Find the interest rate $r$:** From the table: - Start: $P = 2500$ - After 1 year: $A = 2630$ Using the formula for $t=1$: $$ 2630 = 2500(1 + r)^1 $$ Divide both sides by 2500: $$ \frac{2630}{2500} = \cancel{\frac{2500}{2500}}(1 + r) $$ $$ 1.052 = 1 + r $$ Subtract 1: $$ r = 1.052 - 1 = 0.052 $$ Convert to percentage: $$ r = 0.052 \times 100 = 5.2\% $$ 4. **Check with 2 years data:** Using $r=0.052$ and $t=2$: $$ A = 2500(1.052)^2 = 2500 \times 1.1067 = 2766.76 $$ This matches the table, confirming our rate. 5. **Calculate value after 10 years:** Using $r=0.052$ and $t=10$: $$ A = 2500(1.052)^{10} $$ Calculate $(1.052)^{10}$: $$ (1.052)^{10} \approx 1.677 $$ So: $$ A = 2500 \times 1.677 = 4192.5 $$ Rounded to nearest penny: $$ \boxed{4192.50} $$ **Final answers:** - a) Interest rate = 5.2% per annum - b) Value after 10 years = 4192.50 pounds