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(A) Total work = $30 	imes 38 = 1140$ man-days.Work done in $25$ days = $30 	imes 25 = 750$ man-days.Remaining work = $1140 - 750 = 390$ man-days.Af

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(A) Total work = $30 	imes 38 = 1140$ man-days.Work done in $25$ days = $30 	imes 25 = 750$ man-days.Remaining work = $1140 - 750 = 390$ man-days.After $25$ days,the number of men becomes $30 + 5 = 35$ men.These $35$ men finished the remaining work in $390 / 35$ days,which is approximately $11.14$ days. However,the problem states the work was finished one day earlier,meaning the remaining work was completed in $38 - 25 - 1 = 12$ days.If $5$ more men were not employed,the remaining $390$ man-days of work would have been done by the original $30$ men.Time taken for remaining work = $390 / 30 = 13$ days.Total time taken without extra men = $25 + 13 = 38$ days.Wait,let's re-evaluate: The original plan was $38$ days. If $30$ men work for $25$ days,$13$ days of work remain. If $5$ men are added,they finish in $12$ days. If $5$ men are $NOT$ added,the $30$ men take $13$ days. Total time = $25 + 13 = 38$ days. The delay is $38 - 38 = 0$. Re-reading the prompt: The prompt implies a delay exists. Let's assume the work was finished in $38-1=37$ days total. Remaining work $390$ man-days done by $35$ men in $390/35$ days. There is a discrepancy in the problem statement's logic. Based on standard competitive math logic for this specific problem type: $35$ men do the rest of the job in $12$ days. $30$ men would do it in $(35 	imes 12) / 30 = 14$ days. Total time = $25 + 14 = 39$ days. Delay = $39 - 38 = 1$ day.

$A$ team of $30$ men is supposed to do a work in $38$ days. After $25$ days,$5$ more men were employed and the work was finished one day earlier. How many days would it have been delayed if $5$ more men were not employed? (in days)

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