A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind the schedule the work should have been finished? (G.B.O., 2007)

### Concept & Formula This problem is solved using the fundamental equation of work: $$ \text{Total Work} = \sum (\text{Number of Men} \times \text{Number of Days}) $$ We calculate the actual total work done in terms of "man-days" and then find out how long the original workforce would have taken to complete the same amount of work. ### Step-by-Step Solution * **Actual Work Done:** $100$ men worked for the first $35$ days. Then, $100$ more men joined ($200$ total men), and they worked for the remaining $5$ days (since the work was completed in the stipulated $40$ days). * Total work = $(100 \text{ men} \times 35 \text{ days}) + (200 \text{ men} \times 5 \text{ days})$ * Total work = $3500 + 1000 = 4500$ man-days. * **Hypothetical Scenario:** If no additional men were engaged, only $100$ men would have worked the entire time. * Time taken by $100$ men = Total Work / $100$ = $4500 / 100 = 45$ days. * **Delay Calculation:** The stipulated time was $40$ days. The hypothetical time is $45$ days. * Days behind schedule = $45 - 40 = 5$ days. ### Exam Strategy & Shortcut Focus only on the extra work done. The $100$ extra men worked for $5$ days, doing $500$ man-days of work. If those $100$ extra men hadn't been there, the original $100$ men would have had to do that $500$ man-days of work. It would take them $500 / 100 = 5$ extra days. ### Common Pitfall A frequent error is calculating the work for the second part as $100$ men for $5$ days instead of $200$ men (the original $100$ plus the $100$ new ones). ### Final Answer Therefore, the correct answer is **5**.