Dilution with added water: An initial 5 L alcohol–water solution has 40% alcohol. If 1 L of water is added, what is the percentage strength of alcohol in the new 6 L solution?

Introduction / Context:
When only water is added to an alcohol–water solution, the amount of alcohol remains constant while total volume increases. Hence the concentration (percentage strength) decreases predictably based on new total volume.

Given Data / Assumptions:

• Initial volume = 5 L.
• Alcohol fraction initially = 40% = 0.40.
• Water added = 1 L.
• Alcohol volume does not change upon adding water.

Concept / Approach:
Compute alcohol volume first: 0.40 * 5 = 2 L. New total volume = 5 + 1 = 6 L. New strength = alcohol volume / total volume * 100%.

Step-by-Step Solution:

Verification / Alternative check:
As a fraction, 2 parts alcohol out of 6 total is indeed one-third, confirming the percentage 33 1/3%.

Why Other Options Are Wrong:

• 30% and 33%: Rounding or estimation errors; exact value is 33 1/3%.
• 33 2/3%: Would require alcohol 2 of 5.94 approx; not this case.

Common Pitfalls:
Recomputing alcohol amount after adding water or using 40% of 6 L. The alcohol quantity stays 2 L; only the denominator (total volume) changes.

Final Answer: