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(C) Let the original number of days be $n$. The total work is $150n$ man-days.According to the problem,the workers follow an arithmetic progression: $

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Let the original number of days be $n$. The total work is $150n$ man-days.According to the problem,the workers follow an arithmetic progression: $150, 146, 142, \dots$ for $(n+8)$ days.The sum of this arithmetic progression is given by $S = \frac{N}{2} [2a + (N-1)d]$,where $N = n+8$,$a = 150$,and $d = -4$.$S = \frac{n+8}{2} [2(150) + (n+8-1)(-4)] = 150n$$\frac{n+8}{2} [300 - 4n - 28] = 150n$$(n+8)(272 - 4n) = 300n$$(n+8)(68 - n) = 75n$$68n - n^2 + 544 - 8n = 75n$$-n^2 + 60n + 544 = 75n$$n^2 + 15n - 544 = 0$Solving the quadratic equation: $(n + 32)(n - 17) = 0$.Since $n$ must be positive,$n = 17$.The total number of days taken to complete the work is $n + 8 = 17 + 8 = 25$ days.

$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped out on the second day,$4$ more workers dropped out on the third day,and so on. It takes $8$ more days to finish the work now. The number of days in which the work was completed is:

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