10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all the 10 men and 15 women work together, in how many days will the work get completed ?

### Concept & Group Efficiency When given the time taken by specific groups to complete a job, treat each entire group as a single entity to find their combined daily work rate, bypassing the need to calculate individual worker efficiencies. ### Step-by-Step Solution * **Treat Groups as Entities:** * Let Group 1 = 10 men. * Let Group 2 = 15 women. * **Determine Daily Work Rates for the Groups:** * Group 1 (10 men) completes the work in 15 days. * Group 1's 1-day work = $\frac{1}{15}$. * Group 2 (15 women) completes the work in 12 days. * Group 2's 1-day work = $\frac{1}{12}$. * **Calculate Combined Daily Rate:** * When all 10 men and 15 women work together, their combined 1-day work is: $$ \frac{1}{15} + \frac{1}{12} $$ * Find the lowest common multiple of 15 and 12, which is 60. $$ \frac{4}{60} + \frac{5}{60} = \frac{9}{60} $$ * Simplify the fraction: $\frac{9}{60} = \frac{3}{20}$. * **Determine Total Time:** * Total time = Inverse of combined daily work = $\frac{20}{3}$ days. * Convert to a mixed fraction: $\frac{20}{3} = 6\frac{2}{3}$ days. ### Exam Strategy & Shortcut Use the standard "A and B working together" formula: $\frac{xy}{x+y}$. Treat the "10 men" as worker A (15 days) and the "15 women" as worker B (12 days). Time = $\frac{15 \times 12}{15 + 12} = \frac{180}{27} = \frac{20}{3} = 6\frac{2}{3}$ days. ### Common Pitfall A major time-wasting pitfall is trying to find the efficiency of 1 man and 1 woman, setting up ratios, and converting everything, which adds completely unnecessary steps since the final group size exactly matches the given group sizes. ### Final Answer Therefore, the correct answer is **$6\frac{2}{3}$**.