15 men complete a work in 10 days. 15 women complete the same work in 12 days.…

Concept — rates add, days do not. If a group finishes an entire job in d days, then in one day that group finishes 1/d of the job; this fraction is its work rate. When two groups work at the same time, their one-day rates are added, and the time the combined group needs is the reciprocal of that combined rate: time = 1 ÷ (sum of the rates). Days themselves are never added or averaged.

Application. Apply this principle to the two groups described in the question.

1. The 15 men finish the whole job in 10 days, so in one day the 15 men together finish 1/10 of the job.

2. The 15 women finish the same job in 12 days, so in one day the 15 women together finish 1/12 of the job.

3. All of them work at the same time, so add the two rates: 1/10 + 1/12. Using LCM(10, 12) = 60, this becomes 6/60 + 5/60 = 11/60 of the job per day.

4. Take the reciprocal of the combined rate to get the time: time = 1 ÷ 11/60 = 60/11 days.

5. Convert to a mixed number: 60 ÷ 11 gives quotient 5 and remainder 5, so the time is 5 5/11 days.

Cross-check — the LCM (total-units) method. Let the whole job be LCM(10, 12) = 60 units. The 15 men then do 60 ÷ 10 = 6 units per day and the 15 women do 60 ÷ 12 = 5 units per day, so together they do 6 + 5 = 11 units per day, giving 60 ÷ 11 = 60/11 days by an independent route. A bound confirms it too: two groups working together must take less time than the faster group alone (under 10 days) yet more than half of that time (over 5 days), and 5 5/11 lies inside that window.

Result. Working together, the men and the women complete the job in 60/11 days, that is 5 5/11 days.