Solution:
Given data:
Time taken by A to complete the work = 42 days
Time taken by B to complete the work = 56 days
Time taken by C to complete the work = 63 days
A left the work 10 days before completion
B left the work 12 days before completion
The concept:
LCM (Least Common Multiple) and work done concept
Find LCM of 42, 56, and 63 to find the total units of work.
Use work = efficiency × time to find the work done by each individual.
Calculation:
LCM of 42, 56, 63 = 504 units
⇒ Daily work done by A = 504 / 42 = 12 units/day
⇒ Daily work done by B = 504 / 56 = 9 units/day
⇒ Daily work done by C = 504 / 63 = 8 units/day
Let's assume the work is completed in D days.
⇒ Total work done by A = 12 × (D - 10) (as A left 10 days before completion)
⇒ Total work done by B = 9 × (D - 12) (as B left 12 days before completion)
⇒ Total work done by C = 8 × D
Summing up work done by A, B, and C should be equal to total units of work (504).
⇒ 12(D - 10) + 9(D - 12) + 8D = 504
⇒ 12D - 120 + 9D - 108 + 8D = 504
⇒ (12 + 9 + 8)D = 504 + 120 + 108
⇒ 29D = 732
⇒ D = 732 / 29
⇒ D ≈ 25 days
Therefore, the work will be completed in approximately 25 days.
