150 workers were engaged to finish a pieve of work in a certain number of days.Four workers dropped the second day, four more the third day and so on. IT takes 8 more days to finish the work now. Find the number of days in which the work was completed.

Suppose 1 worker does 1 unit work in a day
Assume 150 workers can finish the work in (n-8) days, if all workers work all the days.
Then, total work =150(n−8) ⋯(1)=150(n−8) ⋯(1)
Actually 150 workers work on day-1, 146 workers work on day-2, ... and work is completed in nn days. Therefore,
total work = 150 + 146 + ...  (nn terms)

This is an arithmetic progression with a = 150, d = -4. Therefore,
total work
=n/2[2×150+(n−1)(−4)]
=n/2[300−4n+4]
=n/2[304−4n]
=n(152−2n) ⋯(2)
=n/2[2×150+(n−1)(−4)]
=n/2[300−4n+4]
=n/2[304−4n]
=n(152−2n) ⋯(2)

From(1) and (2)
150(n−8)=n(152−2n)
75(n−8)=n(76−n)
75n−600=76n−n^2
n^2−n−600=0
(n−25)(n+24)=0
n = 25
i.e., number of days in which the work was completed = 25