1. **Problem Statement:** We are given the ratio of the difference between Compound Interest (C.I) and Simple Interest (S.I) for 3 years and 2 years as 10:3. We need to find the rate of interest per annum. 2. **Formulas and Important Rules:** - Simple Interest (S.I) for $n$ years: $$SI = \frac{P \times R \times n}{100}$$ - Compound Interest (C.I) for $n$ years: $$CI = P \left(1 + \frac{R}{100}\right)^n - P$$ - The difference between C.I and S.I for $n$ years is: $$D_n = CI - SI = P \left(1 + \frac{R}{100}\right)^n - P - \frac{P \times R \times n}{100}$$ 3. **Given:** $$\frac{D_3}{D_2} = \frac{10}{3}$$ 4. **Calculate $D_3$ and $D_2$:** $$D_3 = P \left(1 + \frac{R}{100}\right)^3 - P - \frac{3 P R}{100}$$ $$D_2 = P \left(1 + \frac{R}{100}\right)^2 - P - \frac{2 P R}{100}$$ 5. **Divide $D_3$ by $D_2$ and simplify:** $$\frac{D_3}{D_2} = \frac{\left(1 + \frac{R}{100}\right)^3 - 1 - \frac{3 R}{100}}{\left(1 + \frac{R}{100}\right)^2 - 1 - \frac{2 R}{100}} = \frac{10}{3}$$ 6. Let $x = \frac{R}{100}$. Then: $$\frac{(1+x)^3 - 1 - 3x}{(1+x)^2 - 1 - 2x} = \frac{10}{3}$$ 7. Expand and simplify numerator and denominator: - Numerator: $$(1+x)^3 - 1 - 3x = (1 + 3x + 3x^2 + x^3) - 1 - 3x = 3x^2 + x^3$$ - Denominator: $$(1+x)^2 - 1 - 2x = (1 + 2x + x^2) - 1 - 2x = x^2$$ 8. Substitute back: $$\frac{3x^2 + x^3}{x^2} = \frac{10}{3}$$ 9. Simplify the fraction: $$3 + x = \frac{10}{3}$$ 10. Solve for $x$: $$x = \frac{10}{3} - 3 = \frac{10 - 9}{3} = \frac{1}{3}$$ 11. Recall $x = \frac{R}{100}$, so: $$\frac{R}{100} = \frac{1}{3} \implies R = \frac{100}{3} = 33.33\%$$ **Final Answer:** The rate of interest per annum is **33.33%**.