Mr. X can do a job in 10 days, Mr. Y in 15 days and Mr. Z in 20 days. They work together for 2 days. Then Mr. X leaves the job. Mr. Y leaves the job a day earlier to the completion of the work. The job was completed in how many days ?
My attempt:-
Let total work be 60 units
Efficiency of X = 6 units / day
Efficiency of Y = 4 units / day
Efficiency of Z = 3 units / day
Work completed in first 2 days = (6+4+3) * 2 = 26 units
Work Remaining = 34 units
Y and Z can complete this remaining work at a combined efficiency of 7 units/ day in 34/7 days or on the partial 5th day of them working together, to be precise 4+(6/7) days , since Y leaves a day prior, therefore Y and Z together worked for 4 days and completed 4*7=28 units
Finally, the remaining work of 34-28 =6 units would be completed by Z alone in 2 days Therefore total days required to complete the job = 2+4+2=8 days
However my answer is not matching with that of author. Official answer given is 52/7 days
Your use of efficiency for an assumed units of work (usually the L C M) is a very good strategy to avoid reciprocals.
The other simplifying technique you can use is to compute from both ends, viz
In first two days, $13*2 = 26$ units of work is completed, $34$ units remain.
Looking from the end, on the last day, $Z$ completes $3$ units of work, $31$ units of work left for middle period
Time needed for the balance work by Y and Z can be now easily determined as $\dfrac{31}7 +3$ and the three to give the book answer of $\boxed{\dfrac{52}7}$
Note for future reference that whenever in such problems, it is said that someone(s) left $x$ days before completion of work, it always means exactly $x$ days before completion of work.