Question

$8$ men and $12$ boys can finish apiece 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.

Answer 1

Given,

$8$ men and $12$ days can finish a place of work in $10$ days

$6$ men and $8$ days can finish it in $14$ days

also given,

to find the time taken by $14$ man alone and by $1$ boy alone to finish the work

Let us now assume,

wotk done by $1$ man in $1$ day $=x$

work done by $1$ boy in $1$ day $=y$

then, $8$ men for $1$ day will be $8x$ and $12$ boys for $1$ day will be $12y$

$\iff 8x+12y=\frac{1}{10}\to \left(1\right)$ equation

Similarly for $6$ men for $1$ day $=6x$ and

$8$ boys for $1$ day $=8y$

$\iff 6x+8y=\frac{1}{14}\to \left(2\right)$ equation

Now, let us Solve equation $\left(1\right)$ & equation $\left(2\right)$ to find the value of ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$

$3X\left(1\right)=24x+3by=3/10\to$ equation $\left(3\right)$

$4X\left(2\right)=24x+32y=4/14\to$ equation $\left(4\right)$

we are multiplying equation $\left(1\right)$ with $3$ and equation $\left(2\right)$ with $4$ just to from equation to solve easily.

equation $\left(3\right)$-equation $\left(4\right)\Rightarrow \frac{24x+36y=3/1024x+32y=4/14---}{4y=\frac{3}{10}-\frac{4}{14}}$

$\Rightarrow 4y=\frac{3}{10}-\frac{2}{7}$

$\Rightarrow 4y=\frac{3\times 7-2\times 10}{70}\iff 4y=\frac{21-20}{70}=\frac{1}{70}$

$\therefore y=\frac{1}{70\times 4}=\frac{1}{280}$

Now. let us apply ${}^{\prime }{y}^{\prime }$ value in equation $\left(1\right)$ we get,

$8x+12\times \frac{1}{80}=\frac{1}{10}\iff 560x+3=\frac{70}{10}$

$\Rightarrow 560x=7-3\Rightarrow 4$

$\iff x=\frac{41}{560}\iff x=1/140$

Naw, to find the time taken by $1$ man alone and $1$ boy alone we get

$\iff 1$ man $=x=\frac{1}{140}$, so, here ${}^{\prime }{1}^{\prime }$ man taken $140$ days to complete the work alone,

and $1$ boy $=y=\frac{1}{280}$ so, here ${}^{\prime }{1}^{\prime }$ boy taken $280$ days to complete the work alone.