One worker leaves midway; other completes: A can do a job in 60 days and B in 75 days. They begin together, but after some days A leaves and B finishes the remaining work in 30 days. After how many days from the start did A leave?

Introduction / Context:
When one worker departs partway through, split the timeline into the joint phase and the solo phase. Convert solo phase duration into completed fraction of work, then deduce how much was done during the joint phase. From that, compute the length of the joint phase by dividing its fraction by the combined rate.

Given Data / Assumptions:

• A alone time = 60 days ⇒ rate(A) = 1/60.
• B alone time = 75 days ⇒ rate(B) = 1/75.
• B finishes the remainder in 30 days after A leaves.

Concept / Approach:
Work done by B alone in 30 days = 30 * 1/75 = 2/5. Hence, the fraction completed during the joint phase is 1 − 2/5 = 3/5. Since the combined rate is 1/60 + 1/75 = 9/300 = 3/100, the joint-phase duration t satisfies (3/100)*t = 3/5. Solve for t to get when A left.

Step-by-Step Solution:

Verification / Alternative check:
In 20 days together they do (3/100)*20 = 3/5 of the work; remaining 2/5 is done by B in 30 days, consistent.

Why Other Options Are Wrong:
25, 21, 24, or 18 days would not split the work into 3/5 and 2/5 consistent with B’s 30-day solo effort.

Common Pitfalls:
Forgetting to convert the solo phase into a fraction of work or misadding the combined rate of A and B.

Final Answer:
20 days