A 10, text{hectare} field is reaped by 2, tex | vedclass

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(D) Let the efficiency of a man, a woman, and a child be $5k, 4k,$ and $2k$ respectively.Total work done by $(2\, 	ext{men} + 3\, 	ext{women} + 4\,

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$8$

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(D) Let the efficiency of a man, a woman, and a child be $5k, 4k,$ and $2k$ respectively.Total work done by $(2\, 	ext{men} + 3\, 	ext{women} + 4\, 	ext{children})$ in $10\, 	ext{days}$ is:$	ext{Work} = (2 	imes 5k + 3 	imes 4k + 4 	imes 2k) 	imes 10 = (10k + 12k + 8k) 	imes 10 = 30k 	imes 10 = 300k$.Given that this work corresponds to $10\, 	ext{hectares},$ so $10\, 	ext{hectares} = 300k,$ which means $1\, 	ext{hectare} = 30k$.Now, we need to reap $16\, 	ext{hectares},$ which is equivalent to $16 	imes 30k = 480k$ units of work.The group of $(6\, 	ext{men} + 4\, 	ext{women} + 7\, 	ext{children})$ has a daily efficiency of:$	ext{Efficiency} = (6 	imes 5k + 4 	imes 4k + 7 	imes 2k) = (30k + 16k + 14k) = 60k$.Let the number of days required be $x$.$	ext{Total Work} = 	ext{Efficiency} 	imes 	ext{Days} ightarrow 480k = 60k 	imes x$.$x = \frac{480k}{60k} = 8\, 	ext{days}$.

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