Question

If $8$ men and $12$ boys can finish a piece of work in $10$ days while $6$ men and $8$ boys can finish it in $14$ days. Find the time taken by one man alone and that by one boy alone to finish the work.

Answer 1

Suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
One man's one day's work $=\frac{1}{x}$

One boy's one day's work $=\frac{1}{y}$

$\therefore$ Eight men's one day's work $=\frac{8}{x}$

$12$ boy's one day's work $=\frac{12}{y}$

Since $8$ men and $12$ boys can finish the work in $10$ days

$10(\frac{8}{x}+\frac{12}{y})=1\Rightarrow \frac{80}{x}+\frac{120}{y}=1$ ..(i)

Again, $6$ men and $8$ boys can finish the work in $14$ days.
$\therefore 14(\frac{6}{x}+\frac{8}{y})=1\Rightarrow \frac{84}{x}+\frac{112}{y}=1$ (ii)

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in equations (i) and (ii), we get

$80u+120v-1=0$
$84u+112v-1=0$

By using cross-multiplication method, we have
$\frac{u}{-120+112}=\frac{-v}{-80+84}=\frac{1}{80\times 112-120\times 84}$

$\Rightarrow \frac{u}{-8}=\frac{v}{-4}=\frac{1}{-1120}$

$\Rightarrow u=\frac{-8}{-1120}=\frac{1}{140}$ and $v=\frac{-4}{-1120}=\frac{1}{280}$

Now, $u=\frac{1}{140}\Rightarrow \frac{1}{x}=\frac{1}{140}\Rightarrow x=140$

and, $v=\frac{1}{280}\Rightarrow \frac{1}{y}=\frac{1}{280}\Rightarrow y=280$.

Thus, one man alone can finish the work in $140$ days and one boy alone can finish the work in $280$ days.