Six men can finish a work in 12 days; eight women can finish it in 18 days; and eighteen children can finish it in 10 days. First, 4 men, 12 women, and 20 children work together for 2 days. If only men must finish the remaining work in 1 additional day, how many men are required?

Introduction / Context:Convert each group's capability into an equivalent daily rate per individual, then add contributions over the first phase. The remainder is completed by men in a final 1-day push.

Given Data / Assumptions:

• 6 men → 12 days → 72 man-days → 1 man = 1/72 per day.
• 8 women → 18 days → 144 woman-days → 1 woman = 1/144 per day.
• 18 children → 10 days → 180 child-days → 1 child = 1/180 per day.

Concept / Approach:Compute the fraction of work done in first 2 days, then set n*(1/72) = remaining work (to be done in 1 day) to find n.

Step-by-Step Solution:

Verification / Alternative check:Cross-check equivalences: woman = 1/144 = half of man; child ≈ 0.4 of man; totals align to 1/4 per day.

Why Other Options Are Wrong:Undershoot or overshoot the exact half-work remaining within a single day.

Common Pitfalls:Arithmetic slips when summing heterogeneous rates; forgetting that the final phase is only 1 day.

Final Answer:36