Given:
Efficiency of Rohit = 2.5 times efficiency of Pankaj
Rohit takes 6 days less than Pankaj to complete a work.
Rohit and Pankaj work on alternate days, Rohit works on first day.
Concept used:
If A completes a work in x days, then work done by A in one day = 1 / x.
Efficiency is the work done by a person in one day.
Solution:
Let the number of days taken by Rohit and Pankaj working alone to complete the job be x and (x + 6).
Part of work done by Rohit in 1 day = 1 / x.
Part of work done by Pankaj in 1 day = 1 / (x + 6).
As efficiency of Rohit = 2.5 times efficiency of Pankaj
(1 / x) = 2.5 × (1 / (x + 6))
x + 6 = 2.5x
1.5x = 6
x = 4.
Thus, number of days taken by Rohit to complete the work = 4.
And number of days taken by Pankaj to complete the work = 10.
Let total work be 1 unit.
Work done by Rohit on the first day = 1 / 4.
Work done by Pankaj on second day = 1 / 10.
Sum of work done by Rohit and Pankaj in first two days
= (1 / 4) + (1 / 10)
= (5 + 2) / 20
= 7 / 20.
Sum of work done by Rohit and Pankaj in first four days
= (7 / 20) × 2
= 14 / 20
= 7 / 10.
Remaining work = 1 - (7 / 10) = 3 / 10.
Work done by Rohit on 5th day = 1 / 4.
Remaining work
= 3 / 10 - 1 / 4
= (6 - 5) / 20
= 1 / 20.
Time taken by Pankaj to complete 1 / 20 of work on the sixth day:
= (1 / 20) / (1 / 10)
= 10 / 20
= 0.5 days.
Thus, total number of days taken by Rohit and Pankaj to complete the work working alternatively is 5.5 days.
∴ The work will be completed in 5.5 days.

Alternate Method
Given:
Efficiency of Rohit = 2.5 times efficiency of Pankaj
Rohit takes 6 days less than Pankaj to complete a work.
Rohit and Pankaj work on alternate days, Rohit works on first day.
Concept used:
Efficiency is inversely proportional to number of days taken by a person to complete a work.
In LCM method we find LCM of quantities and use their efficiencies to solve the problem.
Solution:
As efficiencies are inversely proportional to number of days and efficiency of Rohit = 2.5 times efficiency of Pankaj, let us assume Rohit takes x days and Pankaj takes (x + 6) days to complete the work.
1 / x = 2.5 × 1 / (x + 6)
x + 6 = 2.5x
1.5x = 6
x = 4.
Number of days taken by Rohit to complete the work alone = 4 days.
Number of days taken by Pankaj to complete the work alone = 10 days.
Thus, we assume total work = 20 (LCM of 4 and 10).

The numbers above the names indicate the number of days taken by Rohit and Pankaj to complete the work alone while the numbers on the line indicate their efficiencies and total work is represented by the number on the right.
Efficiency of Rohit = 20 / 4 = 5.
Efficiency of Pankaj = 20 / 10 = 2.
Work done by Rohit on 1st day = 5.
Work done by Pankaj on 2nd day = 2
Total work done by Rohit and Pankaj in first 2 days = 5 + 2 = 7.
Total work done by Rohit and Pankaj in first 4 days = 7 × 2 = 14.
Remaining work = 20 - 14 = 6.
Work done by Rohit on 5th day = 5.
Remaining work = 6 - 5 = 1.
Time taken by Pankaj to complete remaining work on 6th day
= 1 / 2
= 0.5 days.
Thus, total number of days taken by Rohit and Pankaj to complete the work working alternatively is 5.5 days.
∴ The work will be completed in 5.5 days.