When Manisha and Seema work together, they can finish a project in 60 days. In…

Concept — In work-and-time problems the reciprocal of the time taken is the fraction of work done per day, and the rates of people working together simply add. A useful consequence: if the three possible pair rates are added, every individual appears in exactly two of those pairs, so their total equals twice the combined rate of all three people working together.

Application to this problem:

1. Take the whole project as 1 unit of work, and let M, S and R be the one-day work rates of Manisha, Seema and Reena.

2. Manisha and Seema finish in 60 days, so M + S = 1/60.

3. Seema and Reena finish in 80 days, so S + R = 1/80.

4. Reena and Manisha finish in 48 days, so R + M = 1/48.

5. Add the three equations. Each of M, S and R appears exactly twice on the left, giving 2(M + S + R) = 1/60 + 1/80 + 1/48.

6. Use the LCM 240 on the right: 1/60 = 4/240, 1/80 = 3/240 and 1/48 = 5/240, so the sum is (4 + 3 + 5)/240 = 12/240 = 1/20.

7. Hence 2(M + S + R) = 1/20, so M + S + R = 1/40; the three together do 1/40 of the project in one day.

8. The time needed is the reciprocal of that combined rate: 1 divided by 1/40 = 40 days.

Cross-check — the same result in LCM (unit-work) form. Take the whole project as 240 units:

The three pair rates add to 12 units per day, and that total counts every person twice, so the three of them working together do 12 / 2 = 6 units per day. The project of 240 units therefore takes 240 / 6 = 40 days. As a sanity check, this is faster than the quickest pair on its own (48 days), which is exactly what must happen once a third worker joins.