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Question:

A work is done by 30 workers, not all of whom are equally efficient. Every day exactly 2 workers come for work, with no pair of workers working together twice. Even after all possible pairs have worked once, the work isn’t complete. Thus, all the 30 workers work together for three more days to finish the work. Find the number of days in which the whole work would finish, if all the workers worked together.

Explanation:

Here let us focus on work done by each of these 30 workers.

Let the efficiency of first worked by w₁.
He would work in pairs with each of the other 29 workers, hence he would work for 29 days. Additionally he would also work for another 3 days when everyone works.

∴ Worker 1 works for a total of 29 + 3 = 32 days
Hence, work done by worker 1 = 32 × w₁.

Similarly work done by nth worker would be 32 × w_n.

⇒ Total work done by them = 32w₁ + 32w₂ + 32w₃ + … + 32w₃₀ = 32 × (sum of their efficiencies) …(1)

Now when all of them work together let’s say they take d days to complete the work.
∴ Work done by all of them together in d days = d × (sum of their efficiencies) …(2)

Now (1) = (2)
⇒ d × (sum of their efficiencies) = 32 × (sum of their efficiencies)
⇒ d = 32 days.

Hence, option (d).

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