1. The problem states that Halima owed £427.33 at the end of 2002 and £592.39 at the end of 2007. 2. We need to find how much interest the loan gathered over the next 3 years starting from 2007, so we consider the amount owed in 2007 as the principal for the next 3 years. 3. However, the problem only provides amounts for 2002 and 2007, so it is necessary to interpret carefully: The question asks for interest gathered over the next 3 years from 2007, but no amount after 2007 is given. 4. It seems the problem expects to calculate the interest accrued between 2002 and 2007 (5 years), so confirming the question: "How much interest did the loan gather over the next 3 years?" Assuming this means after 2007, but no data is given beyond 2007; hence, likely it means between 2002 and 2007 first, then to find interest for 3 years. 5. Given only amounts at 2002 and 2007, compute the annual interest rate assuming it’s compound interest. 6. The compound interest formula: $$A = P(1 + r)^t$$ where $A = 592.39$ (amount at 2007), $P = 427.33$ (amount at 2002), $t = 5$ years, $r$ is the annual interest rate. 7. Solve for $r$: $$592.39 = 427.33 (1 + r)^5$$ $$\frac{592.39}{427.33} = (1 + r)^5$$ $$1.3869 \approx (1 + r)^5$$ 8. Taking the fifth root: $$1 + r = (1.3869)^{1/5}$$ $$1 + r \approx 1.0685$$ $$r \approx 0.0685 = 6.85\%$$ 9. Now calculate how much interest accumulates over the next 3 years from 2007 using this rate. Principal at 2007: £592.39 Time: 3 years Interest rate $r = 0.0685$ Amount after 3 years: $$A = 592.39 (1 + 0.0685)^3$$ $$A = 592.39 \times (1.0685)^3$$ $$A = 592.39 \times 1.2195 = 722.00$$ (rounded to 2 decimal places) 10. Interest gathered over 3 years: $$\text{Interest} = A - P = 722.00 - 592.39 = 129.61$$ Final answer: The loan gathered £129.61 interest over the next 3 years to the nearest 1p.