If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be

### Concept & Proportional Workforce Comparison Before diving into complex algebra, check if the target workforce is a direct multiple of one of the given workforces. If team B is exactly $k$ times the size of team A, they will take $1/k$ the amount of time. $$M_1 \times D_1 = M_2 \times D_2$$ ### Step-by-Step Solution * **Establish the equivalence:** Equate the total work from both given groups. * $10 \times (6m + 8b) = 2 \times (26m + 48b)$. * $60m + 80b = 52m + 96b$. * $8m = 16b$. * $1m = 2b$. * **Determine total work:** Substitute the relation ($1m = 2b$) into the first group to find total work in terms of boys. * First group = $6m + 8b = 6(2b) + 8b = 12b + 8b = 20$ boys. * Total work = $20 \text{ boys} \times 10 \text{ days} = 200 \text{ boy-days}$. * **Calculate target workforce capacity:** The target is 15 men and 20 boys. * Convert to boys: $15(2b) + 20b = 30b + 20b = 50$ boys. * **Find required time:** Time = Total Work / Target Workforce. * Time = $200 \text{ boy-days} / 50 \text{ boys} = 4 \text{ days}$. ### Exam Strategy & Shortcut Observe the target group carefully: $15m + 20b$. Notice that this is exactly $2.5$ times the first group: $2.5 \times (6m + 8b) = 15m + 20b$. Since the workforce is $2.5$ times larger, the time required will be divided by $2.5$. Original time = $10$ days. New time = $10 / 2.5 = 4$ days. You don't even need the second scenario! ### Common Pitfall A common mistake is solving the entire system the long way when a direct scaling factor exists, burning valuable exam time. Always inspect the target workforce for proportionality to the given conditions first. ### Final Answer Therefore, the correct answer is **4 days**.