Given:
Time taken by A to complete the work = 10 days
Time taken by B to complete the work = 12 days
Time taken by C to complete the work = 15 days
Concept used:
Total work = LCM
Efficiency = (Total work)/(Total time)
Calculations:
Total work = LCM(10, 12, 15) = 60
| Person | Time | Total work | Efficiency |
|---|
| A | 10 | 60 units | 60/10 = 6 |
| B | 12 | 60 units | 60/12 = 5 |
| C | 15 | 60 units | 60/15 = 4 |
Total efficiency of A, B and C together = 6 + 5 + 4 = 15
Work done by A, B and C in 2 days = 15 × 2 = 30 units
Remaining work = 60 – 30 = 30 units
According to the question, B left the job 3 days before the work was completed.
So, in the last 3 days, only C was working
So, work done by C in 3 days = 4 × 3 = 12 units
Remaining work = 30 – 12 = 18 units
These 18 unit work is completed by B and C together
Total efficiency of B and C = 5 + 4 = 9
So, time taken by B and C to complete 18 units of work = 18/9 = 2 days
Total time to complete the work = 2 + 2 + 3 = 7 days
Total work = LCM(10, 12, 15) = 60
| Person | Time | Total work | Efficiency |
|---|
| A | 10 | 60 units | 60/10 = 6 |
| B | 12 | 60 units | 60/12 = 5 |
| C | 15 | 60 units | 60/15 = 4 |
Let the total number of days taken to finish the work be x.
C worked for all X days.
B worked for (X - 3) days
A worked for 2 days.
Total work by all together = 60 untis
4 × X + 5 × (X - 3) + 6 × 2 = 60
Or, 9X = 63
Or, X = 7
∴ Total time to complete the work is 7 days