40 men can complete a work in 40 days. They s | vedclass

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(A) Total work = $40 	imes 40 = 1600$ man-days.Work done in first $10$ days by $40$ men = $40 	imes 10 = 400$ units.Work done in next $10$ days by $

Vedclass · Accepted Answer

$56 \frac{2}{3}$

Vedclass · Answer

(A) Total work = $40 	imes 40 = 1600$ man-days.Work done in first $10$ days by $40$ men = $40 	imes 10 = 400$ units.Work done in next $10$ days by $35$ men = $35 	imes 10 = 350$ units.Work done in next $10$ days by $30$ men = $30 	imes 10 = 300$ units.Work done in next $10$ days by $25$ men = $25 	imes 10 = 250$ units.Work done in next $10$ days by $20$ men = $20 	imes 10 = 200$ units.Total work done in $50$ days = $400 + 350 + 300 + 250 + 200 = 1500$ units.Remaining work = $1600 - 1500 = 100$ units.Now,$15$ men are left. Time taken to complete the remaining work = $\frac{100}{15} = \frac{20}{3} = 6 \frac{2}{3}$ days.Total time = $50 + 6 \frac{2}{3} = 56 \frac{2}{3}$ days.

$40$ men can complete a work in $40$ days. They started the work together. But at the end of each $10^{th}$ day,$5$ men left the job. The work would have been completed in (in days):

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