AB work 3 days; C is faster than (A+B) together by a factor of time A can do a job in 12 days, B in 8 days. C can do the same job in four-fifths of the time required by (A and B together). A and B work together for 3 days, then C alone completes the remainder. For how many days did C work?

Introduction / Context:
The original stem was broken. By the Recovery-First Policy, we minimally repair it to reflect a common variant: C’s solo time is four-fifths of the joint time of A and B. We then compute how long C needs to finish after A and B have worked some initial days.

Given Data / Assumptions:

• A = 12 days ⇒ rate 1/12 per day.
• B = 8 days ⇒ rate 1/8 per day.
• (A+B) rate = 1/12 + 1/8 = 5/24 per day ⇒ time = 24/5 days.
• C’s time = (4/5)*(24/5) = 96/25 days ⇒ C’s rate = 25/96 per day.
• A and B work jointly for 3 days first.

Concept / Approach:
Compute initial work by A and B, find the leftover fraction, then divide by C’s rate.

Step-by-Step Solution:

Verification / Alternative check:
Multiply C’s rate by 36/25: (25/96)*(36/25) = 36/96 = 3/8, which matches the remainder.

Why Other Options Are Wrong:
8, 6, 3 are far from 1.44 days; the only valid choice is “None of these.”

Common Pitfalls:
Interpreting “four-fifths” against A or B instead of the joint A+B time; or adding times rather than rates.

Final Answer:
None of these