A certain number of men can finish a job in 100 days. If there were 10 men fewer, the job would take 10 days longer (110 days). How many men were there originally?

Introduction:
This classic manpower-time problem uses the fact that total work (in man-days) remains constant. When the number of men changes, the time adjusts inversely. By writing two expressions for the same total work, we can solve for the unknown original headcount directly.

Given Data / Assumptions:

• Let original men = m; original time = 100 days.
• With (m − 10) men, time = 110 days.
• Total work W is constant and measured in man-days.

Concept / Approach:
Write W in two ways: W = m * 100 and W = (m − 10) * 110. Equate them, then solve the simple linear equation in m. This is a direct application of the unitary method with work conservation.

Step-by-Step Solution:

Verification / Alternative check:
With 110 men: time = W / 110 = (110 * 100)/110 = 100 days. With 100 men: time = (110 * 100)/100 = 110 days. Both match the conditions precisely.

Why Other Options Are Wrong:
75, 92, 105, 120 do not satisfy the equality 100m = 110(m − 10) and produce inconsistent times for the modified headcount case.

Common Pitfalls:
Averaging the times or men, or assuming proportional difference in days equals difference in men without using the exact equation. Always equate man-days for accuracy.

Final Answer:
110