1. **Stating the problem:** We are given $x = 2 \times 10^{5}$ and $y = 3 \times 10^{-3}$, each correct to one significant figure. We need to find the greatest and least possible values of: (i) $xy$ (ii) $\frac{x}{y}$ 2. **Understanding significant figures and error bounds:** For one significant figure, the value can vary by half the unit of the last significant digit. - For $x = 2 \times 10^{5}$, the last significant digit is in the $10^{5}$ place, so the possible range is: $$1.5 \times 10^{5} \leq x \leq 2.5 \times 10^{5}$$ - For $y = 3 \times 10^{-3}$, the last significant digit is in the $10^{-3}$ place, so the possible range is: $$2.5 \times 10^{-3} \leq y \leq 3.5 \times 10^{-3}$$ 3. **(i) Finding greatest and least possible values of $xy$:** - Greatest value of $xy$ is when both $x$ and $y$ are at their greatest: $$x_{max} = 2.5 \times 10^{5}, \quad y_{max} = 3.5 \times 10^{-3}$$ $$xy_{max} = (2.5 \times 10^{5})(3.5 \times 10^{-3}) = 2.5 \times 3.5 \times 10^{5 - 3} = 8.75 \times 10^{2}$$ - Least value of $xy$ is when both $x$ and $y$ are at their least: $$x_{min} = 1.5 \times 10^{5}, \quad y_{min} = 2.5 \times 10^{-3}$$ $$xy_{min} = (1.5 \times 10^{5})(2.5 \times 10^{-3}) = 1.5 \times 2.5 \times 10^{5 - 3} = 3.75 \times 10^{2}$$ 4. **(ii) Finding greatest and least possible values of $\frac{x}{y}$:** - Greatest value of $\frac{x}{y}$ is when $x$ is greatest and $y$ is least: $$x_{max} = 2.5 \times 10^{5}, \quad y_{min} = 2.5 \times 10^{-3}$$ $$\frac{x}{y}_{max} = \frac{2.5 \times 10^{5}}{2.5 \times 10^{-3}} = \cancel{\frac{2.5}{2.5}} \times 10^{5 - (-3)} = 10^{8}$$ - Least value of $\frac{x}{y}$ is when $x$ is least and $y$ is greatest: $$x_{min} = 1.5 \times 10^{5}, \quad y_{max} = 3.5 \times 10^{-3}$$ $$\frac{x}{y}_{min} = \frac{1.5 \times 10^{5}}{3.5 \times 10^{-3}} = \frac{1.5}{3.5} \times 10^{5 - (-3)} = \frac{1.5}{3.5} \times 10^{8} \approx 0.4286 \times 10^{8} = 4.286 \times 10^{7}$$ **Final answers:** - Greatest $xy = 8.75 \times 10^{2}$ - Least $xy = 3.75 \times 10^{2}$ - Greatest $\frac{x}{y} = 10^{8}$ - Least $\frac{x}{y} \approx 4.29 \times 10^{7}$