Correct option is C
Solution:
Let total work = 30 units
(LCM of 10, 15, and 6)
Step 1: Find efficiencies
1.
A + B finish work in 10 days → Efficiency of A + B = 30 / 10 =
3 units/day … (i.)
2.
B + C finish work in 15 days → Efficiency of B + C = 30 / 15 =
2 units/day … (ii.)
3.
A + B + C finish work in 6 days → Efficiency of A + B + C = 30 / 6 =
5 units/day … (iii.)
Step 2: Find individual efficiencies
From (iii.) − (ii.): A = 5 − 2 =
3 units/day
From (i.): B = (A + B) − A = 3 − 3 =
0 units/day
From (ii.): C = (B + C) − B = 2 − 0 =
2 units/day
Step 3: Efficiency of A + C
A + C = 3 + 2 =
5 units/day
Step 4: Time taken by A and C together
Time = Total Work ÷ Efficiency = 30 ÷ 5 =
6 days
Final Answer: 6 days
