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# $150$ workers were engaged to finish a job in certain no. of days. $4$ workers dropped out on second day. $4$ more workers dropped out on third day and so on. It took $8$ more days to finish the work. Find the no. of days in which the work was completed.

$\begin{array}{1 1}25\;days \\ 24\;days \\ 27\;days \\ 28\;days \end{array}$

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• Sum of $n$ terms of an A.P. $=\large\frac{n}{2}$$\big[2a+(n-1)d\big] Let the no. of days the work was to be completed be n Given: No. of workers on 1^{st} day =150 Each day 4 workers dropped on the subsequent days. Also given that it took 8 more days to finish the work. \therefore Total number of days the work went on =n+8 Step 1 If 150 men work for n days the work will be finished by n days. \therefore The work done by 1 man on one day =\large\frac{1}{n\times 150} On first day no. of workers = 150 \therefore The work done on first day =\large\frac{1}{n\times 150}$$\times 150$
Step 2
the no. of workers on second day $= 146$
$\therefore$ The work done on second day $=\large\frac{1}{n\times 150}$$\times 146 Similarly the no. of workers on third day 142 \therefore The work done on third day =\large\frac{1}{n\times 150}$$\times 142$
and so on
the work goes on for $n+8$ days
Step 3
Thus the if we assume the total work to be completed as $1$, then
$1=\large\frac{1}{n\times 150}$$\times 150$$+\large\frac{1}{n\times 150}$$\times 146$$+\large\frac{1}{n\times 150}$$\times 142$$+........$(upto $n+8$ terms.)
Taking $\large\frac{1}{150n}$ common we get
$1=\large\frac{1}{150 n}$$\big(150+146+142+....... upto (n+8) terms\big) \big(150+146+142+....... upto (n+8) terms\big) is an A.P. with first term a=150 and common difference d=-4 We know that sum of n terms of an A.P. =\large\frac{n}{2}$$\big[2a+(n-1)d\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[2\times 150+(n+8-1)(-4)\big] \Rightarrow\:150n=\large\frac{n+8}{2}$$\big[300-(n+7)(4)\big]$
$\Rightarrow\:150n=\large\frac{n+8}{2}$$\big[300-4n-28\big] \Rightarrow\:150n=\large\frac{n+8}{2}$$\big[272-4n\big]$
$\Rightarrow\:150n=(n+8)(-2n+136)$
$\Rightarrow\:2n^2+30n-1088=0$
$\Rightarrow\:n^2+15n-544=0$
$\Rightarrow\:n^2+32n-17n-544=0$
$\Rightarrow\:(n-17)(n(+32)=0$
$\Rightarrow\:n=17\:\:or\:\:-32$
But since $n$ cannot be $-ve$, $n=17$
$\therefore$ The no. of days the work was completed $=17+8=25\:days.$
answered Apr 12, 2014
edited Apr 12, 2014