A2. Julia had eaten a box of candies in 4 days. On the first day she ate $\frac{1}{5}$ of the total number of candies. On the second day she ate half of what was left after the first day. On the third day she ate half of what was left after the second day. What portion of the candies initially contained in the box did she eat on the fourth day? Express your answer as an irreducible fraction.
Solution 1: We organize the information in the form of a table.
\begin{tabular}{|c|c|c|c|c|}
\hline day \# & 1 & 2 & 3 & 4 \\
\hline ate of the total & $\frac{1}{5}$ & $\frac{1}{2} \times \frac{4}{5}=\frac{2}{5}$ & $\frac{1}{2} \times \frac{2}{5}=\frac{1}{5}$ & $\frac{1}{5}$ \\
\hline left of the total & $1-\frac{1}{5}=\frac{4}{5}$ & $\frac{4}{5}-\frac{2}{5}=\frac{2}{5}$ & $\frac{2}{5}-\frac{1}{5}=\frac{1}{5}$ & 0 \\
\hline
\end{tabular}

Solution 2: Assume that box contains $5 n$ (for some $n \geq 1$ ) candies and count the numbers of candies (instead of fractions of the box).
\begin{tabular}{|c|c|c|c|c|}
\hline day \# & 1 & 2 & 3 & 4 \\
\hline ate & $n$ & $\frac{1}{2} \times 4 n=2 n$ & $\frac{1}{2} \times 2 n=n$ & $n$ \\
\hline left & $5 n-n=4 n$ & $4 n-2 n=2 n$ & $2 n-n=n$ & 0 \\
\hline
\end{tabular}

The portion she ate on the 4 th day is $\frac{n}{5 n}=\frac{1}{5}$.
Answer: $1 / 5$.