AMC8 2025

Let the total amount of cheese be $1$. We will track the amount of cheese Sarika eats throughout the process. First Round: Sarika eats half of the total cheese, so she eats: \[\frac{1}{2}.\] Second Round: Dev eats half of what remains after Sarika's turn, which is: \[\frac{1}{4}.\] Third Round: Rajiv eats half of the remaining cheese after Dev’s turn, which is: \[\frac{1}{8}.\] At the end of the first round, the total cheese eaten is: \[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}.\] We observe that Sarika’s consumption follows a geometric sequence. In the first round, she eats $\frac{1}{2}$, and in subsequent rounds, she eats half of what remains from the previous round. This gives the following series for Sarika’s total consumption: \[\frac{1}{2} + \frac{1}{16} + \frac{1}{128} + \cdots\] This is a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{8}$. The sum $S$ of this infinite geometric series is given by the formula: \[S = \frac{a}{1 - r},\] where $a$ is the first term and $r$ is the common ratio. Substituting $a = \frac{1}{2}$ and $r = \frac{1}{8}$: \[S = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{1}{2} \times \frac{8}{7} = \frac{4}{7}.\] Thus, Sarika eats $\frac{4}{7}$ of the original block of cheese. The correct answer is: \[\boxed{\textbf{(A) } \frac{4}{7}}.\]