A group of workers was engaged to complete a job. From the second day onwards, one worker was withdrawn each day. The job was finished on the day when the last remaining worker was withdrawn. If no worker had been withdrawn at any stage, the group would have finished the job in only 55% of the actual time taken. How many workers were originally in the group?

Introduction / Context: This is a challenging time and work problem involving a varying number of workers. The workforce decreases by one worker each day until the job ends. You are also told how long the work would take if the original full group remained constant. From this information you must find the initial number of workers. Given Data / Assumptions:

• On the first day, there are N workers (unknown).
• From the second day onwards, one worker is withdrawn each day.
• The job finishes exactly on the day when the last worker is withdrawn, so the number of days worked = N.
• If no worker had been withdrawn, the group would finish the job in only 55% of the actual time taken.
• All workers are identical with constant efficiency.

Concept / Approach: Compute total work in terms of worker-days. With decreasing workers, total work is the sum of workers each day: N + (N - 1) + ... + 1. If nobody were withdrawn, daily workers would always be N, and the work would be completed in T constant days. The problem states T is 55% of the actual time, which is N days. Use these relationships to form an equation and solve for N. Step-by-Step Solution: Let each worker do 1 unit of work per day. With decreasing workers, the number of workers each day is: N, N - 1, N - 2, ..., 2, 1. Number of days until the last worker is withdrawn = N days. Total work W = N + (N - 1) + ... + 1 = N(N + 1) / 2 worker-days. Now, suppose no worker is withdrawn. Then N workers work every day, and let the work finish in T days. Total work W in this scenario = N * T worker-days. So N * T = N(N + 1) / 2, which implies T = (N + 1) / 2. Given T is 55% of the actual time. Actual time = N days, so T = 55% of N = 0.55N = 55N / 100 = 11N / 20. Set the two expressions for T equal: (N + 1) / 2 = 11N / 20. Multiply both sides by 20: 10(N + 1) = 11N. 10N + 10 = 11N, so N = 10. Verification / Alternative check: With N = 10, actual time with decreasing workers is 10 days. Constant workers time T = (N + 1) / 2 = 11 / 2 = 5.5 days. Percentage: T / actual time = 5.5 / 10 = 0.55 = 55%, which matches the condition. Why Other Options Are Wrong: N = 40, 45 or 50 would not satisfy the equation (N + 1) / 2 = 11N / 20. Each would give a different fraction for T / actual time, not equal to 55%. Common Pitfalls: Learners may misinterpret "finished when the last worker was withdrawn" and assign wrong number of days. Another error is to forget that the sequence of workers per day is an arithmetic series and not a fixed number. Always express total work in both scenarios and equate them to solve such problems. Final Answer: The original group had 10 workers.