Five men can check a set of exam papers in 12 days by working 6 hours daily.…

When men, days and hours together determine how much work gets done, a work-equivalence identity connects two scenarios doing possibly different amounts of the same kind of work: (M1 × D1 × H1) / W1 = (M2 × D2 × H2) / W2, where M is the number of workers, D is the number of days, H is the hours worked per day, and W is the quantity of work completed. This lets the hours needed by a different team, working a different number of days on a different amount of work, be found directly.

1. Identify the known scenario: M1 = 5 men, D1 = 12 days, H1 = 6 hours per day, checking W1 = 1 unit of papers.

2. Identify the new scenario: M2 = 4 men, D2 = 20 days, checking W2 = 2 units of papers (double the earlier amount); H2 is the unknown daily hours.

3. Apply the identity: (5 × 12 × 6) / 1 = (4 × 20 × H2) / 2.

4. Simplify the left side: 5 × 12 × 6 = 360.

5. Simplify the right side: 4 × 20 = 80, so the equation becomes 360 = (80 × H2) / 2 = 40 × H2.

6. Solve for H2: H2 = 360 / 40 = 9.

Independent check via total man-hours: the original team's total effort is 5 × 12 × 6 = 360 man-hours for 1 unit of papers, so double the papers needs 720 man-hours. The new team supplies 4 × 20 = 80 man-days, so the daily hours must be 720 / 80 = 9 — the same value.

So 4 men must work 9 hours a day to check double the papers in 20 days.