Question

A group of workers starts a job. From Day 2 onward, 1 worker is removed each day. The job ends when the last worker is withdrawn. If all had worked full time, the work would’ve been done in 2/3 time. How many workers were initially in the group?

• A. 2
• B. 3
• C. 5
• D. 11

Hint:

Use arithmetic series to model cumulative work done over time with changing workforce.

Answer 1

The Correct Option is B

Let \( n \) be the number of workers. Each day from Day 2, one worker is withdrawn, so total working days = \( n \).
Work done under actual condition: Day 1: \( n \) workers
Day 2: \( n-1 \) workers
...
Day \( n \): 1 worker
So, total work = \( n + (n-1) + ... + 1 = \frac{n(n+1)}{2} \) Work if all worked all days: Each of \( n \) workers works for \( n \) days = \( n^2 \) Now, it's given: \[ \frac{n(n+1)}{2} = \frac{2}{3} n^2 3n(n+1) = 4n^2 3n^2 + 3n = 4n^2 n^2 - 3n = 0 n(n - 3) = 0 n = 0 \text{ or } 3 \] Ignore \( n = 0 \). So, \[ \boxed{3 \text{ workers}} \]