Let the total work done be 48 units, 48 being a common multiple of 8, 12 and 6 which are the respective times taken by the different groups of people.
Let rate of A = p units per day, rate of B = q units per day and rate of C = r units per day.
Rate of working of A and B = \(\frac{48 }{ 8}\) = 6 units per day.
Therefore, p + q = 6 ----- (1)
Rate of working of B and C = \(\frac{48 }{ 12}\) = 4 units per day.
Therefore, q + r = 4 ---- (2)
Rate of working of A, B and C = \(\frac{48 }{ 6}\) = 8 units per day.
Therefore, p + q + r = 8 ---- (3)
Adding equations (1) and (2) , we have p + 2q + r = 10 ---- (4)
Subtracting equation (3) from (4), we have, q = 2. Substituting the value of q in equation (1) and (2) we have p = 4 and r = 2.
Therefore, rate of working of A and C = p + r = 6 units per day.
Time taken by A and C to do the work = \(\frac{48 }{ 6 }\)= 8 days
The correct answer option is C.