If 3 men or 9 boys can finish a piece of work in 21 days, in how many days can 5 men and 6 boys together do the same piece of work? (Bank Recruitment, 2010)

### Concept & Work Equivalence When an "or" condition is given, it implies that the groups can complete the same work in the same amount of time, allowing us to find their relative efficiencies. $$ M_1 \times D_1 = M_2 \times D_2 $$ ### Step-by-Step Solution * **Establish Efficiency Ratio:** The problem states $3$ men or $9$ boys can do the work in $21$ days. This means the work done by $3$ men is equal to the work done by $9$ boys. $3 \text{ Men (M)} = 9 \text{ Boys (B)}$ $1 \text{ M} = 3 \text{ B}$ * **Determine Total Work:** Let's use boys as the base unit. $9$ boys can finish the work in $21$ days. Total Work = $9 \text{ B} \times 21 \text{ days} = 189 \text{ boy-days}$. * **Convert the Target Group:** The group we need to evaluate is $5$ men and $6$ boys. Convert men to boys: $5 \text{ M} = 5 \times 3 \text{ B} = 15 \text{ B}$. Total target workforce = $15 \text{ B} + 6 \text{ B} = 21 \text{ B}$. * **Calculate Time:** Time taken = Total Work / Total Workforce Time = $189 \text{ boy-days} / 21 \text{ B} = 9 \text{ days}$. * Comparing our result ($9$ days) with the given options, it does not match (a), (b), or (c). Therefore, the answer is "None of these". ### Exam Strategy & Shortcut Directly equate the efficiencies: $3M = 9B \rightarrow M = 3B$. We need the time for $5M + 6B$, which is $5(3B) + 6B = 21B$. Since $9B$ take $21$ days, the total work is $9 \times 21$. Time for $21B = (9 \times 21) / 21 = 9$ days. ### Common Pitfall A frequent mistake is interpreting "or" as "and", which would lead to adding the rates of the men and boys together initially instead of equating them. ### Final Answer Therefore, the correct answer is **None of these**.