Question

Two men and $7$ children complete a certain piece of work in $4$ days while $4$ men and $4$ children complete the same work in only $3$ days. The number of days required by $1$ man to complete the work is

• A. $60$ days
• B. $15$ days
• C. $6$ days
• D. $51$ days

Answer 1

Answer: (B)

$15$ days

Let work done by $1$ man in $1$ day $=$ $x$
Let work done by $1$ child in $1$ day $=$ $y$
Given,
$2x+7y=\frac{1}{4}$ ....(i) and
$4x+4y=\frac{1}{3}$ ....(ii)
Multiplying equation (i) by $2$, we get
$2x+7y=\frac{1}{4}$
$\Rightarrow 4x+14y=\frac{1}{2}$ ....(iii)
Subtracting equation (ii) by (iii), we get
$4x+14y=\frac{1}{2}$
$4x+4y=\frac{1}{3}$

$10y=\frac{1}{2}-\frac{1}{3}$
$\Rightarrow 10y=\frac{3-2}{6}$
$\Rightarrow 10y=\frac{1}{6}$
$\Rightarrow y=\frac{1}{60}$
Putting the value of $y$ in equation (i), we get
$2x+7y=\frac{1}{4}$
$\Rightarrow 2x+7\left(\frac{1}{60}\right)=\frac{1}{4}$
$\Rightarrow 2x=\frac{1}{4}-\frac{7}{60}$
$\Rightarrow 2x=\frac{15-7}{60}$
$\Rightarrow 2x=\frac{8}{60}$
$\Rightarrow 2x=\frac{2}{15}$
$\Rightarrow x=\frac{1}{15}$
$\therefore$ Work done by $1$ man in $1$ day $\frac{1}{15}$.
$\therefore$ days taken to complete the job is $15$.