$10$ mens can do a work in $15$ days and $15$ women can do a work in $12$ day, if both $10$ men and $15$ women, so the work how many day the to complete the work.

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Solution

Work done by $10$ men in $1$ day $= \frac{W}{15}$
Work done by $15$ women in $41$ day $= \frac{W}{10}$
If all these together do the work.
$(\frac{W}{15} + \frac{W}{12}) \times d = W \frac{3}{20} \times d = 1 d = \frac{20}{3} = 6.67 \approx 7 d a y s$

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10 mens can do a work in 15 days and 15 women can do a work in 12 day, if both 10 men and 15 women, so the work how many day the to complete the work.

Work done by 10 men in 1 day=W15Work done by 15 women in 41 day=W10If all these together do the work.(W15+W12)×d=W320×d=1d=203=6.67≈7days

$10$ mens can do a work in $15$ days and $15$ women can do a work in $12$ day, if both $10$ men and $15$ women, so the work how many day the to complete the work.

Work done by $10$ men in $1$ day $= \frac{W}{15}$
Work done by $15$ women in $41$ day $= \frac{W}{10}$
If all these together do the work.
$(\frac{W}{15} + \frac{W}{12}) \times d = W \frac{3}{20} \times d = 1 d = \frac{20}{3} = 6.67 \approx 7 d a y s$