1. **Problem statement:** Two capitals of 300000F and 700000F are invested for the same duration at an interest rate of 7% per annum. The total interest earned from both investments is 105000F. We need to find the common duration of the investment in months. 2. **Formula used:** The simple interest formula is $$I = P \times r \times t$$ where $I$ is the interest, $P$ is the principal (capital), $r$ is the annual interest rate (in decimal), and $t$ is the time in years. 3. **Important notes:** - Since the interest rate is annual, the time $t$ must be expressed in years. - The total interest is the sum of interests from both capitals. 4. **Set up the equation:** Let $t$ be the duration in years. Interest from first capital: $$I_1 = 300000 \times 0.07 \times t$$ Interest from second capital: $$I_2 = 700000 \times 0.07 \times t$$ Total interest: $$I_1 + I_2 = 105000$$ Substitute: $$300000 \times 0.07 \times t + 700000 \times 0.07 \times t = 105000$$ 5. **Simplify the equation:** $$0.07 t (300000 + 700000) = 105000$$ $$0.07 t \times 1000000 = 105000$$ 6. **Solve for $t$:** $$0.07 \times 1000000 \times t = 105000$$ $$70000 t = 105000$$ $$t = \frac{105000}{70000}$$ $$t = 1.5$$ 7. **Convert $t$ to months:** Since $t$ is in years, $$t = 1.5 \text{ years} = 1.5 \times 12 = 18 \text{ months}$$ **Final answer:** The common duration of the investment is **18 months**.