SSC CHSL 6-Dec-2015 Shift 1 Paper

Let 1 worker does 1 unit work in a day.
Let 150 workers can finish the work in (n - 8 ) days, if all workers work all the days.
Then, total work = 150(n - 8) -------(i)
Also, 150 workers work on day 1, 146 workers work on day 2, ... and so on. Work is completed in n days.
Thus, total work = 150 + 146 +....(n terms)
This is an arithmetic progression with first term,a = 150, d = - 4.
Thus, total work =

n

2

[2a+(n−1)d]
=

n

2

[2(150)+(n−1)(−4)]
=

n

2

[300−4n+4]
=

n

2

[304−4n]=n(152−2n) ----(ii)
Comparing equations (i) and (ii),
⇒ 150(n−8)=n(152−2n)
⇒ 75(n−8)=n(76−n)
⇒ 75n−600=76n−n2
⇒ n2−n−600=0
⇒ (n−25)(n+24)=0
⇒ n=25,−24
∵ n cannot be negative , ⇒ n = 25
⇒ Number of days in which the work was completed = 25