A small disc of radius $r$ is cut out from a disc of radius $R$. The weight of the disc which now has a hole in it, is reduced to $\frac{24}{25}$ of the original weight. If $R = xr$, what is the value of $x$?

### Concept & Proportionality of Weight and Area Assuming uniform density and thickness, the weight of a disc is directly proportional to its surface area. $$ \text{Area} = \pi r^2 $$ ### Step-by-Step Solution * **Define initial state:** Original area of the large disc = $\pi R^2$. Original weight is proportional to this area. * **Define final state:** Area of the cut-out hole = $\pi r^2$. The remaining area of the disc with the hole = $\pi R^2 - \pi r^2$. * **Set up the weight proportion:** The remaining weight is $\frac{24}{25}$ of the original weight. $\pi R^2 - \pi r^2 = \frac{24}{25} \pi R^2$ * **Simplify the equation:** Divide the entire equation by $\pi$. $R^2 - r^2 = \frac{24}{25} R^2$ $R^2 - \frac{24}{25} R^2 = r^2$ $\frac{1}{25} R^2 = r^2$ * **Solve for R:** Take the square root of both sides. $\frac{1}{5} R = r \Rightarrow R = 5r$. * **Find x:** The problem states $R = xr$. Substituting $R = 5r$, we get $5r = xr$, so $x = 5$. * **Evaluate options:** The calculated value $x = 5$ is not among the options (a, b, c, d). ### Exam Strategy & Shortcut You can interpret the problem as: The removed piece accounts for $1 - \frac{24}{25} = \frac{1}{25}$ of the total weight/area. So, $\frac{\text{Area of small disc}}{\text{Area of large disc}} = \frac{1}{25}$. This means $\left(\frac{r}{R}\right)^2 = \frac{1}{25} \Rightarrow \frac{r}{R} = \frac{1}{5} \Rightarrow R = 5r$. ### Common Pitfall A common error is equating the remaining weight fraction directly to the ratio of radii, rather than the ratio of the squares of the radii (since weight is proportional to area, not linear length). ### Final Answer Therefore, the correct answer is **None of these**.