Let the principal be \(p\) rupees and the annual rate of interest be \(r\%\). So, the simple interest for 2 years = \(\cfrac{p \times 2 \times r}{100}\) rupees According to the question: \(\cfrac{p \times 2 \times r}{100} = 800\) ⇒ \(pr = 40000\) — (i) Now, the compound amount after 2 years = \(p\left(1 + \cfrac{r}{100}\right)^2\) rupees So, compound interest = \(p\left(1 + \cfrac{r}{100}\right)^2 - p = 820\) ⇒ \(p\left\{\left(1 + \cfrac{r}{100}\right)^2 - 1\right\} = 820\) ⇒ \(p\left\{\cancel{1} + \cfrac{2r}{100} + \cfrac{r^2}{10000} - \cancel{1}\right\} = 820\) ⇒ \(\cfrac{pr}{100} \left(2 + \cfrac{r}{100}\right) = 820\) Substituting \(pr = 40000\) from equation (i): \(\cfrac{40000}{100} \left(2 + \cfrac{r}{100}\right) = 820\) ⇒ \(2 + \cfrac{r}{100} = \cfrac{820}{400}\) ⇒ \(\cfrac{r}{100} = \cfrac{41}{20} - 2 = \cfrac{1}{20}\) ⇒ \(r = \cfrac{1}{20} \times 100 = 5\) Now, substituting \(r = 5\) in equation (i): \(5p = 40000\) ⇒ \(p = 8000\) Therefore, the required principal is ₹8000 and the rate of interest is 5%.