Question

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:

Answer 1

Explanation : Let $1$ man's $1$ day's work = $x$ and $1$ boy's $1$ day's work =$y$

Then $6x+8y$ =$\frac{1}{10}$ and $26x+48y$ = $\frac{1}{2}$

$6x+8y$ =$\frac{1}{10}$ ........(1) $\times 6$

$26x+48y$ = $\frac{1}{2}$ .........(2)

Solving these two equations we get $:x=$ $\frac{1}{100}$ and $y=$ $\frac{1}{200}$

$(15men+20boy{)}^{\prime }s$ $1$ day's work =$\frac{15}{100}$+$\frac{20}{200}$ = $\frac{1}{4}$

$15$ men and $20$ boys can do the work in $4$$6x+8y$ =$\frac{1}{10}$ days.