Question

$8$ men and $10$ boys do a work in the same days as $4$ men and $16$ boys do. $1$ man and $3$ boys start a work to finish it in $40$ days. After eighteen days, $\frac{9}{20}$ th part of work had been finished. How many boys should be increased to make the work complete in exact time ?

Answer 1

According to given condition,

$8M+10B=4M+16B$

$\therefore 4M=6B$

$\therefore 2M=3B$

$M=3$ units/day

$B=2$ units/day

In second condition, it is given that,

$1$ man and $3$ boys complete the same work in $40$ days

After eighteen days,

$1$ man and $3$boys complete$\frac{9}{20}$ th part of work

i.e. $(1\times 3+3\times 2)\times 18=162$ units of work in $18$ days

$162=\frac{9}{20}\times$total work

$\therefore$ total work = $360$ units

So. remaining work = $360-162=198$units

let $x$ no. of boys are added to complete the work in 22 remaining days

$\therefore 198$ units of work in $22$ days $=22\times (1\times 3+[3+x]\times 2)$

$\therefore 198=22\times (3+6+2x)$

$\therefore 9=3+6+2x$

$\therefore x=0$

Hence no additional boys are required.