150 workers were engaged to finish a piece of | vedclass

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(C) Let the original number of days be $n$. The total work is $150n$ man-days.The number of workers on each day forms an Arithmetic Progression $(AP)$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Let the original number of days be $n$. The total work is $150n$ man-days.The number of workers on each day forms an Arithmetic Progression $(AP)$ with first term $a = 150$ and common difference $d = -4$.The work is completed in $n + 8$ days.The sum of workers over $n + 8$ days is given by the formula $S_k = \frac{k}{2} [2a + (k - 1)d]$,where $k = n + 8$.$S_{n+8} = \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$$-n^2 - 15n + 544 = 0$$n^2 + 15n - 544 = 0$$(n + 32)(n - 17) = 0$Since $n > 0$,we have $n = 17$.The total number of days taken is $n + 8 = 17 + 8 = 25$.

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

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