If 5 men and 3 boys can reap 23 text{ hectare | vedclass

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(A) Let $M$ be the work done by $1$ man per day and $B$ be the work done by $1$ boy per day.From the given data,the work done is proportional to the n

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$2$

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(A) Let $M$ be the work done by $1$ man per day and $B$ be the work done by $1$ boy per day.From the given data,the work done is proportional to the number of workers and time.$(5M + 3B) 	imes 4 / 23 = (3M + 2B) 	imes 2 / 7$$28(5M + 3B) = 46(3M + 2B)$$140M + 84B = 138M + 92B$$2M = 8B \Rightarrow M = 4B$.Now,substitute $M = 4B$ into the first condition:$5(4B) + 3B = 23B$ can reap $23 	ext{ hectares}$ in $4 	ext{ days}$.This means $23B$ can reap $23 	ext{ hectares}$ in $4 	ext{ days}$,so $1B$ can reap $1 	ext{ hectare}$ in $4 	ext{ days}$.Therefore,$1$ boy can reap $1/4 	ext{ hectare}$ in $1 	ext{ day}$.We need to find the number of boys $x$ such that $(7M + xB)$ can reap $45 	ext{ hectares}$ in $6 	ext{ days}$.$(7(4B) + xB) 	imes 6 / 45 = 1 	ext{ (work unit)}$$(28B + xB) 	imes 6 = 45 	imes 4B$$(28 + x) 	imes 6 = 180$$28 + x = 30$$x = 2$.Thus,$2$ boys must assist $7$ men.

If $5$ men and $3$ boys can reap $23 \text{ hectares}$ in $4 \text{ days}$ and if $3$ men and $2$ boys can reap $7 \text{ hectares}$ in $2 \text{ days}$,how many boys must assist $7$ men in order that they may reap $45 \text{ hectares}$ in $6 \text{ days}$?

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