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 = $\frac{\mathit{n}}{2}\left[2\mathit{a}+\left(\mathit{n}-1\right)\mathit{d}\right]$
= $\frac{\mathit{n}}{2}\left[2\left(150\right)+\left(\mathit{n}-1\right)\left(-4\right)\right]$
= $\frac{\mathit{n}}{2}\left[300-4\mathit{n}+4\right]$
= $\frac{\mathit{n}}{2}\left[304-4\mathit{n}\right]=\mathit{n}\left(152-2\mathit{n}\right)$ ----(ii)
Comparing equations (i) and (ii),
⇒ $150\left(\mathit{n}-8\right)=\mathit{n}\left(152-2\mathit{n}\right)$
⇒ $75\left(\mathit{n}-8\right)=\mathit{n}\left(76-\mathit{n}\right)$
⇒ $75\mathit{n}-600=76\mathit{n}-{\mathit{n}}^2$
⇒ ${\mathit{n}}^2-\mathit{n}-600=0$
⇒ $\left(\mathit{n}-25\right)\left(\mathit{n}+24\right)=0$
⇒ $\mathit{n}=25,-24$
∵ n cannot be negative , ⇒ n = 25
⇒ Number of days in which the work was completed = 25