Six men can complete a task in 12 days, eight women can finish the same task in 18 days, and eighteen children can complete it in 10 days. Four men, twelve women, and twenty children work together for two days. If the remaining work is to be completed by men alone in one day, how many men will be required?

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Accepted Answer

The total work is considered as one unit. The rate of work per day for one man is $$ \frac{1}{72} $$, for one woman is $$ \frac{1}{144} $$, and for one child is $$ \frac{1}{180} $$.

First, calculate the work done by four men, twelve women, and twenty children in two days:  
$$
	ext{Work done in two days} = 2 	imes \left(4 	imes \frac{1}{72} + 12 	imes \frac{1}{144} + 20 	imes \frac{1}{180}ight)
$$  
Simplify the fractions:  
$$
= 2 	imes \left(\frac{4}{72} + \frac{12}{144} + \frac{20}{180}ight) = 2 	imes \left(\frac{1}{18} + \frac{1}{12} + \frac{1}{9}ight)
$$

The least common multiple (LCM) of 18, 12, and 9 is 36. Expressing all fractions with a denominator of 36:  
$$
\frac{1}{18} = \frac{2}{36}, \quad \frac{1}{12} = \frac{3}{36}, \quad \frac{1}{9} = \frac{4}{36}
$$  
Substituting back:  
$$
= 2 	imes \left(\frac{2}{36} + \frac{3}{36} + \frac{4}{36}ight) = 2 	imes \frac{9}{36} = \frac{18}{36} = \frac{1}{2}
$$

Thus, the work completed in two days is $$ \frac{1}{2} $$, leaving $$ 1 - \frac{1}{2} = \frac{1}{2} $$ of the work to be done.

To complete the remaining $$ \frac{1}{2} $$ work in one day using men only:  
$$
	ext{Number of men required} = 	ext{Remaining work} / (	ext{Work rate of one man per day}) = \frac{\frac{1}{2}}{\frac{1}{72}} = 72 	imes \frac{1}{2} = 36
$$

Hence, **36 men are required** to complete the remaining work in one day.

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