Headcount from two schedules: A group can finish a job in 100 days. With 10 fewer people, the job would take 110 days. How many people were in the original group?

Introduction / Context:
This is a classic two-schedule unitary-method problem. Express total work in person-days using one schedule, then equate it to the other schedule with a different headcount to solve for the original number of people directly and exactly.

Given Data / Assumptions:

• Original team finishes in 100 days.
• With 10 fewer people, it takes 110 days.
• Per-person productivity is constant.

Concept / Approach:
Let the original headcount be n and total work be W. Then W = 100n (person-days). With 10 fewer, W = 110(n − 10). Equate and solve for n algebraically.

Step-by-Step Solution:

Verification / Alternative check:
Total work W = 110 * 100 = 11000 person-days. With 100 people (110 − 10), W / 100 = 110 days, as stated.

Why Other Options Are Wrong:
180, 190, 196, 120 do not satisfy 100n = 110(n − 10).

Common Pitfalls:
Setting up 100(n − 10) = 110n (incorrect inversion) or cancelling 100s prematurely and losing track of the equation's balance.

Final Answer:
110