The weight of an empty bucket is $25\%$ of the weight of the bucket when filled with some liquid. Some of the liquid has been removed. Then, the bucket, along with the remaining liquid, weighed three-fifths of the original weight. What percentage of the liquid has been removed?

### Concept & Strategy When a problem gives only relative percentages and fractions without absolute values, the best approach is to establish a base total (like $100$ units). Break the total down into its fixed component (bucket) and variable component (liquid) to track the changes. $$\text{Total Weight} = \text{Bucket Weight} + \text{Liquid Weight}$$ ### Step-by-step Solution **Given:** Empty bucket = $25\%$ of filled weight. New total weight = $\frac{3}{5}$ of original filled weight. **Step 1: Assign a convenient base total** Let the Original Filled Total Weight = $100$ units. **Step 2: Deduce component weights** Weight of empty bucket = $25\%$ of $100 = 25$ units. Weight of original liquid = $100 - 25 = 75$ units. **Step 3: Calculate the new total weight after removal** The new total is $\frac{3}{5}$ of the original. New Total = $\frac{3}{5} \times 100 = 60$ units. **Step 4: Find the remaining liquid** The bucket's weight ($25$) never changes. Remaining Liquid = New Total - Bucket Remaining Liquid = $60 - 25 = 35$ units. **Step 5: Calculate percentage of liquid removed** Liquid Removed = Original Liquid - Remaining Liquid Liquid Removed = $75 - 35 = 40$ units. We need the percentage of *liquid* that was removed relative to the *original liquid*: Percentage Removed $= \left(\frac{40}{75}\right) \times 100$ $= \left(\frac{8}{15}\right) \times 100 = \frac{800}{15} = \frac{160}{3} = 53 \frac{1}{3}\%$ ### Exam Strategy & Shortcut Avoid assigning variables like $x$ and $y$. Directly use $100$. Bucket is $25$, Liquid is $75$. Total drops to $60$ (since $\frac{3}{5} = 60\%$). Loss is entirely liquid: $100 \rightarrow 60$ is a loss of $40$. Percentage of liquid lost = $\frac{40}{75} = \frac{8}{15}$. Since $\frac{1}{15} \approx 6.66\%$, $8 \times 6.66 \approx 53.33\%$, which corresponds to $53 \frac{1}{3}\%$. ### Common Pitfall The most common trap is calculating the removed liquid ($40$) as a percentage of the *total* original weight ($100$) instead of the *original liquid* weight ($75$). Doing so results in $40\%$, which is prominently displayed as trap option (a). Read the final question carefully: "percentage of the **liquid** has been removed". ### Final Answer **Therefore, the correct answer is $53 \frac{1}{3}\%$.**