If 6 men and 8 boys can do a piece of work in | vedclass

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(B) Let the efficiency of one man be $M$ and one boy be $B$.According to the problem:$(6M + 8B) 	imes 10 = (26M + 48B) 	imes 2$Dividing both sides b

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$4$

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(B) Let the efficiency of one man be $M$ and one boy be $B$.According to the problem:$(6M + 8B) 	imes 10 = (26M + 48B) 	imes 2$Dividing both sides by $2$:$(6M + 8B) 	imes 5 = 26M + 48B$$30M + 40B = 26M + 48B$$4M = 8B \implies M = 2B$.Now,express the total work in terms of boys:Total work $= (6M + 8B) 	imes 10 = (6(2B) + 8B) 	imes 10 = (12B + 8B) 	imes 10 = 20B 	imes 10 = 200B$ units.We need to find the time taken by $15$ men and $20$ boys:$15M + 20B = 15(2B) + 20B = 30B + 20B = 50B$.Time taken $= \frac{	ext{Total Work}}{	ext{Efficiency}} = \frac{200B}{50B} = 4$ days.

If $6$ men and $8$ boys can do a piece of work in $10$ days and $26$ men and $48$ boys can do the same work in $2$ days,the time taken by $15$ men and $20$ boys to do the same type of work will be (in days)

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