Man-hours balancing: 10 men working 6 h/day complete a job in 18 days. How many hours/day must 15 men work to finish the job in 12 days?

Introduction / Context:
For constant efficiency, work done equals total man-hours. We use the first scenario to get the total man-hours required, and then solve for the daily hours in the second scenario so that the man-hours match the same total work.

Given Data / Assumptions:

• Scenario A: 10 men, 6 h/day, 18 days
• Scenario B: 15 men, H h/day, 12 days
• Same job; efficiencies unchanged

Concept / Approach:
Equate man-hours required: 10*6*18 = 15*H*12. Solve for H (hours/day) algebraically. This is the standard unitary-method application in work-rate problems.

Step-by-Step Solution:

Verification / Alternative check:
Back-check: 15 * 6 * 12 = 1080, same as Scenario A, so the job can be finished in the new schedule with 6 hours/day.

Why Other Options Are Wrong:

• 4, 3, 5, 8: These produce total man-hours different from 1080, so the job would be under- or over-scheduled.

Common Pitfalls:
Forgetting to multiply by days, or mis-canceling factors when solving the proportion. Keep all factors visible until the final step.

Final Answer:
6