First, calculate how much water is in the tank after the technician empties 25% of it.

1. Initial water amount: 20 gallons
2. Amount emptied: $25\% \times 20 = \frac{25}{100} \times 20 = 5$ gallons
3. Water left after removing 25%: $20 - 5 = 15$ gallons

Next, determine how much water corresponds to $\frac{2}{5}$ of the tank's original capacity.

4. Water needed for $\frac{2}{5}$: $\frac{2}{5} \times 20 = 8$ gallons

Now calculate how much more water must be removed to reach 8 gallons:

5. Additional water to remove: $15 - 8 = 7$ gallons

Final Answer:

The technician must remove 7 more gallons of water.

Would you like further clarification?

1. How is the 25% converted to gallons?
2. Why is $\frac{2}{5}$ of the tank’s capacity equal to 8 gallons?
3. Could the remaining water be calculated differently?
4. How would this change if the tank held more than 20 gallons?
5. How do percentages and fractions interplay in this problem?

Tip: Always verify percentages by converting them to fractions or decimals for accuracy.