Discussion

Explanation:

Let efficiency of each worker by 1 unit/day.

Case 1: When a worked leaves every day.

Since n workers start on day 1 and on last day only 1 worker will remain, hence, the work will last n days.

Day 1 2 3 ... n
Number of workers n n - 1 n - 2 ... n - 3
Workdone/day × 1 (n - 1) × 1 (n - 2) × 1 ... (n - 3) × 1

⇒ Total work done = n × 1 + (n - 1) × 1 + (n - 2) × 1 + … + n × 1 = n(n+1)/2   …(1)

Case 2: When no workers leave

Time taken is 2/3rd of n days.

⇒ Total work done = n × 2/3 n = 2/3 × n2   …(2)

Since same work is done in both the cases, we can equate (1) and (2).

n(n + 1)/2 = 2/3 × n2

⇒ n = 3.

Hence, option (b).

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