Question

$4$ men and $6$ women can finish work in $8$ days while $3$ men and $7$ women can do that same work in $10$ days. In how many days will $10$ women do the same work.

Answer 1

Step 1: Form a pair of linear equations in two variables

$4$ men and $6$ women can finish work in $8$ days

Let work done by one man in a day $=\mathrm{x}$

and work done by one woman in a day $=y$

$\Rightarrow$Work done by $4$ men in a day $=4\mathrm{x}$

$\Rightarrow$Work done by $6$ women in a day $=6y$

According to first condition of the question, $4$ men and $6$ women can finish work in $8$ days

$$\begin{gathered} \Rightarrow 4\mathrm{x}+6\mathrm{y}=\frac{1}{8} \\ \Rightarrow 32x+48y=1................\left(\mathrm{i}\right) \end{gathered}$$

$3$ men and $7$ women can do that same work in $10$ days

$\Rightarrow$Work done by $3$ men in a day $=3\mathrm{x}$

$\Rightarrow$Work done by $7$ women in a day $=7y$

According to the second condition of the question, $3$ men and $7$ women can do that same work in $10$ days

$$\begin{gathered} \Rightarrow 3\mathrm{x}+7\mathrm{y}=\frac{1}{10} \\ \Rightarrow 30\mathrm{x}+70\mathrm{y}=1................\left(\mathrm{ii}\right) \end{gathered}$$

Step 2: Find the value of $y$

Write $x$ in terms of $y$

From $\left(\mathrm{i}\right)$, we get

$x=\frac{1-48y}{32}$

Substitute the value of $x$ in $\left(\mathrm{ii}\right)$, we get

$$\begin{gathered} 30\left(\frac{1-48y}{32}\right)+70y=1 \\ \Rightarrow 30-1440y+2240y=32 \\ \Rightarrow 800y=2 \\ \Rightarrow y=\frac{2}{800} \\ \Rightarrow y=\frac{1}{400} \end{gathered}$$

Step 3: Finding the work done by ten women

Work done by ten women in a day $=10y=10\times \frac{1}{400}$

$=\frac{1}{40}$

Hence, ten women will complete the work in $40\mathrm{days}.$