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

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Question

(A) Let the work done by $1$ man in $1$ day be $x$ and the work done by $1$ boy in $1$ day be $y$.According to the problem:$10(6x + 8y) = 1 \implies 6

Vedclass · Accepted Answer

$4$

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(A) Let the work done by $1$ man in $1$ day be $x$ and the work done by $1$ boy in $1$ day be $y$.According to the problem:$10(6x + 8y) = 1 \implies 60x + 80y = 1$ .....$(1)$$2(26x + 48y) = 1 \implies 52x + 96y = 1$ .....$(2)$Multiply equation $(1)$ by $6$ and equation $(2)$ by $5$ to eliminate $y$:$360x + 480y = 6$$260x + 480y = 5$Subtracting these equations: $100x = 1 \implies x = \frac{1}{100}$.Substitute $x$ in equation $(1)$:$60(\frac{1}{100}) + 80y = 1 \implies 0.6 + 80y = 1 \implies 80y = 0.4 \implies y = \frac{0.4}{80} = \frac{1}{200}$.Now,the work done by $15$ men and $20$ boys in $1$ day is:$15x + 20y = 15(\frac{1}{100}) + 20(\frac{1}{200}) = \frac{15}{100} + \frac{10}{100} = \frac{25}{100} = \frac{1}{4}$.Therefore,the time taken to complete the work is $4$ days.

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

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