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(A) Let the efficiency of a man per hour be $E_m$ and that of a woman be $E_w$.Given $E_w = \frac{2}{3} E_m$,which implies $E_m = \frac{3}{2} E_w$.Tot

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$16$

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(A) Let the efficiency of a man per hour be $E_m$ and that of a woman be $E_w$.Given $E_w = \frac{2}{3} E_m$,which implies $E_m = \frac{3}{2} E_w$.Total work done by $10$ men and $6$ women in $18$ days:Daily work of $10$ men $= 10 	imes 9 	imes E_m = 90 E_m$.Daily work of $6$ women $= 6 	imes 7.5 	imes E_w = 45 E_w$.Total daily work $= 90 E_m + 45 E_w = 90(\frac{3}{2} E_w) + 45 E_w = 135 E_w + 45 E_w = 180 E_w$.Total work $= 180 E_w 	imes 18 = 3240 E_w$.Now,for $10$ men and $9$ women:Daily work $= 10 	imes 9 	imes E_m + 9 	imes 7.5 	imes E_w = 90 E_m + 67.5 E_w$.Substituting $E_m = \frac{3}{2} E_w$:Daily work $= 90(\frac{3}{2} E_w) + 67.5 E_w = 135 E_w + 67.5 E_w = 202.5 E_w$.Number of days required $= \frac{	ext{Total Work}}{	ext{Daily Work}} = \frac{3240 E_w}{202.5 E_w} = 16$ days.

$A$ piece of work can be completed by $10$ men and $6$ women in $18$ days. Men work $9$ hours per day while women work $7.5$ hours per day. The per hour efficiency of a woman is $\frac{2}{3}$ of a man's efficiency. In how many days will $10$ men and $9$ women complete the work?

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