One liquid contains 25% benzene and another liquid contains 30% benzene. A can is filled by mixing 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the resulting mixture (assume parts are by volume and additive).

Introduction / Context:
This is a weighted-average concentration problem. When you mix two liquids with different benzene percentages, the final percentage depends on how much of each liquid is used. Because 6 parts of the first and 4 parts of the second are mixed, the mixture is not an equal average; it is a weighted mean based on parts.

Given Data / Assumptions:

• Liquid 1 benzene concentration = 25%
• Liquid 2 benzene concentration = 30%
• Parts mixed: 6 parts of liquid 1, 4 parts of liquid 2
• Total parts = 10

Concept / Approach:
Final % = (sum of benzene amounts) / (total mixture) * 100. Since parts are proportional volumes, use parts directly in the weighted calculation.

Step-by-Step Solution:

Verification / Alternative check:
The result must lie between 25% and 30%. Since more quantity is from 25% liquid (6 parts vs 4 parts), the final should be closer to 25 than to 30. 27% fits that expectation.

Why Other Options Are Wrong:

Common Pitfalls:
Many learners take a simple average (25 + 30)/2 = 27.5% and ignore the 6:4 weighting. Another mistake is to treat 6 and 4 as percentages rather than parts and miscompute the denominator.

Final Answer:
27%