2 men and 7 boys can do a piece of work in 14 days; 3 men and 8 boys can do the same in 11 days. Then, 8 men and 6 boys can do three times the amount of this work in :

### Concept & Simultaneous Equations for Efficiency When given two distinct scenarios of mixed groups completing the same work, equate the total "man-days" to find the relative efficiency between the worker types. $$ (M_1 \times D_1) = (M_2 \times D_2) $$ ### Step-by-Step Solution * **Equate Total Work:** * $(2M + 7B) \times 14 = (3M + 8B) \times 11$ * **Solve for Efficiency Ratio:** * $28M + 98B = 33M + 88B$ * $98B - 88B = 33M - 28M$ * $10B = 5M \implies 1M = 2B$ (One man is twice as efficient as one boy). * **Calculate Baseline Work:** Substitute $1M = 2B$ into the first group to find total work in terms of boys. * Group 1 = $2(2B) + 7B = 4B + 7B = 11B$. * Total Work = $11B \times 14 \text{ days} = 154$ boy-days. * **Target Work:** We need to do "three times the amount of this work". * Target Work = $154 \times 3 = 462$ boy-days. * **Evaluate New Group:** 8 men and 6 boys. * New Group = $8(2B) + 6B = 16B + 6B = 22B$. * **Calculate Time:** Time = Target Work / New Group Size. * Days = $462 / 22 = 21$ days. ### Exam Strategy & Shortcut Once you find $1M = 2B$, evaluate the group strengths mentally: Group 1 acts like 11 boys taking 14 days ($11 \times 14$). The target group (8 men + 6 boys) acts like 22 boys. Since the target group has exactly double the manpower ($22$ vs $11$), they would take half the time ($7$ days) for normal work. For $3\times$ the work, it takes $7 \times 3 = 21$ days. ### Common Pitfall Failing to multiply the final required work by 3, missing the "three times the amount" clause in the question, resulting in an answer of 7 days (which isn't an option, but would cause panic). ### Final Answer Therefore, the correct answer is **21 days**.