A sum of money at a simple rate of interest $i_1$, doubles in 5 years. At another simple rate of interest $i_2$, it becomes three times in 12 years. Then, the two rates of interest $i_1$ and $i_2$ respectively are

### Concept & Simple Interest Let principal be $P$. Case 1: Amount $= 2P \Rightarrow S.I. = 2P - P = P$ Case 2: Amount $= 3P \Rightarrow S.I. = 3P - P = 2P$ $$R = \frac{100 \times S.I.}{P \times T}$$ ### Step-by-Step Solution * **Case 1:** $S.I. = P$, $T = 5$ years. $i_1 = \left(\frac{100 \times P}{P \times 5}\right)$% $= 20$% * **Case 2:** $S.I. = 2P$, $T = 12$ years. $i_2 = \left(\frac{100 \times 2P}{P \times 12}\right)$% $= \frac{200}{12}$% $= \frac{50}{3}$% $= 16\frac{2}{3}$% ### Exam Strategy & Shortcut Use formula $R = \frac{100(n - 1)}{T}$. $i_1 = \frac{100(2 - 1)}{5} = 20$%. $i_2 = \frac{100(3 - 1)}{12} = \frac{200}{12} = 16\frac{2}{3}$%. ### Common Pitfall Mixing up the formulas for simple and compound interest. The problem specifies simple interest. ### Final Answer Therefore, the correct answer is **20%, $16\\frac{2}{3}$%**.