A piece of work was planned to finish in 40 days. The initial team managed to complete only half the work in 24 days. Then, 16 more men were added to the team, and the whole work was still finished exactly in the scheduled 40 days. How many men were employed at first?

Introduction / Context:
When team size changes mid-project, convert to rate equations relative to the total work W. Here half the work is done by the initial team in 24 days; the rest is done by a larger team in the final 16 days. We solve for the original team size.

Given Data / Assumptions:

• Total planned duration = 40 days.
• Half of W completed in first 24 days by x men.
• 16 more men added after day 24; remaining half finished in 16 days.
• Each man works at the same constant rate r (work units per day).

Concept / Approach:
Let W be total work. Then x*r*24 = W/2 and (x + 16)*r*16 = W/2. Eliminate W and r to solve for x, the initial number of men.

Step-by-Step Solution:

Verification / Alternative check:
First half: 32*r*24 = W/2. Second half: 48*r*16 = W/2. Both equal W/2 ⇒ consistent. Total time = 24 + 16 = 40 as required.

Why Other Options Are Wrong:
16/24/48 men yield equations that do not balance to halves over 24 and 16 days. Only 32 satisfies the equal half-work condition under a constant per-man rate.

Common Pitfalls:
Assuming linearity without setting equations; mixing total duration with per-phase work; or forgetting both phases are exactly half of W.

Final Answer:
32 men