1. **Stating the problem:** An investor receives 31000 in interest per annum from a sum of money invested partly at 10% and partly at 7% simple interest. If the amounts invested at these rates are interchanged, the interest increases by 1000. We need to find the total amount invested. 2. **Define variables:** Let the amount invested at 10% be $x$ and at 7% be $y$. 3. **Write equations from the problem:** The total interest from the original investment is: $$0.10x + 0.07y = 31000$$ If the amounts are interchanged, the interest becomes: $$0.10y + 0.07x = 31000 + 1000 = 32000$$ 4. **Set up the system of equations:** $$\begin{cases} 0.10x + 0.07y = 31000 \\ 0.07x + 0.10y = 32000 \end{cases}$$ 5. **Solve the system:** Multiply the first equation by 10 and the second by 7 to clear decimals: $$\begin{cases} x + 0.7y = 310000 \\ 0.49x + 0.7y = 224000 \end{cases}$$ 6. **Subtract second from first:** $$x + 0.7y - (0.49x + 0.7y) = 310000 - 224000$$ $$x - 0.49x + 0.7y - 0.7y = 86000$$ $$0.51x = 86000$$ 7. **Solve for $x$:** $$x = \frac{86000}{0.51}$$ $$x = 168627.45$$ 8. **Substitute $x$ back into first equation:** $$168627.45 + 0.7y = 310000$$ $$0.7y = 310000 - 168627.45 = 141372.55$$ 9. **Solve for $y$:** $$y = \frac{141372.55}{0.7} = 201960.79$$ 10. **Calculate total amount invested:** $$x + y = 168627.45 + 201960.79 = 370588.24$$ **Final answer:** The total amount invested is approximately $370588.24$.