Work-rate with unequal efficiencies: 10 men and 8 women together finish a job in 5 days. Given that one woman's daily work equals half of one man's daily work, in how many days will 4 men and 6 women complete the same job?

Introduction / Context:
This unitary-method problem tests conversion of mixed-efficiency teams to a common unit (e.g., “man-equivalents”) and then using rate * time = work. By expressing every worker's contribution in comparable terms, we can scale the time for any new team composition.

Given Data / Assumptions:

• 10 men + 8 women finish the work in 5 days.
• 1 woman does half the daily work of 1 man.
• Find days needed by 4 men + 6 women to finish the same job.

Concept / Approach:
Let a man's daily rate be m units/day. Then a woman's daily rate is m/2. Convert each team to man-equivalents, compute the total work once, and reuse it to find the time for the second team.

Step-by-Step Solution:

Verification / Alternative check:
Because 4 men + 6 women is exactly half of the original team in man-equivalents (7m vs 14m), their time must double from 5 to 10 days. This proportionality check confirms the result.

Why Other Options Are Wrong:
9, 8 2/3, 11, 12 days do not preserve the rate-time inverse relationship between a 14m team and a 7m team for the same job size.

Common Pitfalls:
Forgetting that a woman works at m/2, or adding women as if they worked at rate m like men. Mixing up proportional reasoning is another frequent error.

Final Answer:
10 days