Ans: 25

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)

Actually 150 workers work on day-1, 146 workers work on day-2, ... and work is completed in n days. Therefore,
total work = 150 + 146 + ...  (n 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)

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

Solution:

Suppose the work is completed in n days.            

Since 4 workers went away on every day except the first day.

∴ Total number of worker who worked all the n days is the sum of n terms of A.P. with first term 150 and common difference – 4.

Total number of worker who worked all the n days = n/2[2 x 150 + (n-1) x -4 ] =  n (152 – 2n)

If the workers would not have went away, then the work would have finished in (n – 8) days with 150 workers working on every day.

∴ Total number of workers who would have worked all n days = 150 (n – 8)

∴ n (152 – 2n) = 150 (n – 8)  

⇒ 152n – 2n² = 150n – 1200     

 ⇒ 2n² – 2n – 1200 = 0

⇒ n² – n – 600 = 0                 

⇒ n² – 25n + 24n – 600 = 0        

⇒ n(n – 25) + 24 (n + 25) = 0

⇒ (n – 25) (n + 24) = 0 

⇒ n – 25 = 0 or n + 24 = 0                  

⇒ n = 25 or n = – 24

⇒ n = 25 ( Number of days cannot be negative)

Thus, the work is completed in 25 days.