Question

$6$ men and $10$ women can finish making pots in $8$ days, while the $4$ men and $6$ women can finish it in $12$ days. Find the time taken by the one man alone from that of one woman alone to finish the work.

Answer 1

Hint: We need to express the work done by one man and the work done by one woman algebraically. Then we will find the equations relating the work done by men and women and the total work. After that, we will solve the equation to find the time taken by the one man alone and that of the one woman alone.

Complete step by step solution:
Let us denote the work done by one man in one day and the work done by one woman in one day algebraically.
Let $\dfrac{1}{x}$ be the work done by one man in one day and $\dfrac{1}{y}$ the work done by one woman in one day.
In the question, it is given that $6$ men and $10$ women can finish making pots in $8$ days.
Since the work done by one man in one day is $\dfrac{1}{x}$ and the work done by one women in one day is $\dfrac{1}{y},$ we can say that the following equation can correspond the work done by $6$ men and $10$ women in $8$ days, $\dfrac{6}{x}+\dfrac{10}{y}=\dfrac{1}{8}.......\left( 1 \right)$
In that case, we can represent the work done by $4$ men and $6$ women in $12$ days as the following equation $\dfrac{4}{x}+\dfrac{6}{y}=\dfrac{1}{12}.......\left( 2 \right)$
Let us put $a=\dfrac{1}{x}$ and $b=\dfrac{1}{y}.$
Now the equations $\left( 1 \right)$ and $\left( 2 \right)$ will, respectively, become $6a+10b=\dfrac{1}{8}.......\left( 3 \right)$ and $4a+6b=\dfrac{1}{12}.......\left( 4 \right)$
We will solve the equations to find the values of the unknowns $a$ and $b.$
Let us multiply the equation $\left( 3 \right)$ with $4$ and the equation $\left( 4 \right)$ with $6$ and subtract the equation $\left( 4 \right)$ from the equation $\left( 3 \right).$
Now, we will get $4\times \left( 3 \right)\Rightarrow 24a+40b=\dfrac{4}{8}=\dfrac{1}{2}.$
We will get $6\times \left( 4 \right)\Rightarrow 24a+36b=\dfrac{6}{12}=\dfrac{1}{2}.$
When we find the difference of these equations, we will obtain $4b=0\Rightarrow b=0.$
Now, we will substitute this value in the equation $\left( 3 \right)$ to get $6a+10\times 0=\dfrac{1}{8}\Rightarrow 6a=\dfrac{1}{8}\Rightarrow a=\dfrac{1}{48}.$
Hence the time taken by the one man alone from that of the one woman alone to finish the work is $48$ days.

Note: Instead of multiplying the equation $\left( 1 \right)$ with $4$ and the equation $\left( 2 \right)$ with $6,$ we can transpose the values on the RHS of both the equations to the LHS and then we can find the difference. Equation $\left( 1 \right)$ will become $48a+80b=1$ and equation $\left( 2 \right)$ will become $48a+72b=1.$ Now we will get $b=0.$