AFCAT II 23 Aug 2025 Shift 1 Paper

Let Total Work =72 units
A's efficiency =72∕9=8 units/day
B's efficiency =72∕12=6 units/day
C's efficiency = 72∕18=4 units/day
Total efficiency for 3 days =(A+B+C)=8+6+4=18 units/day
Work done in 3 days =18×3=54 units
Remaining work =72−54=18 units
Now only B continues the work (efficiency =6 units/day)
Time to complete remaining 18 units =18∕6=3 days
Total time to complete the work =3 (initial) +3 (B alone) =6 days