6 men or 8 women can do a work in 25 days. The number of days required by 8…

When two groups can each finish the very same job in the very same number of days, their total daily outputs must be equal — this equality is what fixes the ratio between one individual's daily work in one group and one individual's daily work in the other group. That per-person efficiency ratio, combined with the job's total size, is what then decides how long any differently composed team needs for the same job.

1. Let one man's one-day work be m and one woman's one-day work be w.

2. Since 6 men finish the job in 25 days and 8 women finish the same job in 25 days, both groups produce the same total work: 6m × 25 = 8w × 25, so 6m = 8w, i.e. m : w = 4 : 3.

3. Write the job as 24 units (the LCM of 6 and 8). Then 6 men completing 24 units in 25 days gives 1 man's daily work = 24 ÷ (6 × 25) = 4/25 unit; 8 women completing 24 units in 25 days gives 1 woman's daily work = 24 ÷ (8 × 25) = 3/25 unit — consistent with the 4 : 3 ratio.

4. The new team of 8 men and 6 women therefore does 8 × 4/25 + 6 × 3/25 = 32/25 + 18/25 = 50/25 = 2 units per day.

5. Days required = total work ÷ daily rate = 24 ÷ 2 = 12 days.

Cross-checking with the man-equivalent route: since 1 woman = 3/4 man (from the 4 : 3 ratio), 8 men and 6 women equal 8 + 6 × 3/4 = 12.5 man-equivalents. The whole job equals 6 men's 25-day output, i.e. 6 × 25 = 150 man-days. So the new team needs 150 ÷ 12.5 = 12 days — the same result.

So the work is completed in 12 days.