Question

$2$ women and $5$ men can together finish an embroidery work in $4$ days, while $3$ women and $6$ men can finish it in $3$ days. Find the time taken by $1$ women alone to finish the work, and also that taken by $1$ men alone.

Answer 1

Hint: If a person finishes a work in $n$ days, then the work done by him in one day is $\dfrac{1}{n}$. This $\dfrac{1}{n}$ is the rate of work done by the person. Using this logic, we will assume the rate of doing work by one woman and one man with the given information. Then, we use that rate of both of them according to the cases given in the problem. Multiply the rate with the number of people and equate it to the total time taken. You will get two equations solving them will give the required answer.

Complete step-by-step answer:
In the above problem, it is given that $2$ women and $5$ men finish work in $4$ days and $3$ women and $6$ men can finish it in $3$ days. And with that information, we need to find the time taken by a woman alone and time taken by a man alone to finish the same work.
Let’s go simple and step by step with the problem.
Assuming that a woman alone can finish her work in $'w'$ days
$ \Rightarrow $ Fraction or part of work done by her in one day$ = \dfrac{1}{w}$
Similarly, assuming that a man alone can finish the work in $'m'$ days
$ \Rightarrow $ Fraction or part of work done by him in one day $ = \dfrac{1}{m}$
Now, in our first case, it is given that $2$ women and $5$ men finish work in $4$days.
$ \Rightarrow $ Fraction or part of work done by $2$ women and $5$ men together in one day $ = \dfrac{1}{4}$
If we combine all these three relations, then we can substitute the rate of work
$ \Rightarrow 2 \times $ Part of work done by one woman in one day $ + 5 \times $Part of work done by one man in one day $ = \dfrac{1}{4}$
$ \Rightarrow \left( {2 \times \dfrac{1}{w}} \right) + \left( {5 \times \dfrac{1}{m}} \right) = \dfrac{1}{4}$
$ \Rightarrow \dfrac{2}{w} + \dfrac{5}{m} = \dfrac{1}{4}...........(1)$
In the second case, it is given that $3$ women and $6$ men can finish the work in $3$ days
Similarly, we can write in form of an equation as:
$ \Rightarrow 3 \times $ Part of work done by one woman in one day $ + 6 \times $Part of work done by one man in one day $ = \dfrac{1}{3}$
$ \Rightarrow \left( {3 \times \dfrac{1}{w}} \right) + \left( {6 \times \dfrac{1}{m}} \right) = \dfrac{1}{3}$
$ \Rightarrow \dfrac{3}{w} + \dfrac{6}{m} = \dfrac{1}{3}.............(2)$
Now, we got two equations (1) and (2) with two variables, which can be solved to get the two unknowns.
Let’s transform (2) in a simpler form
$ \Rightarrow \dfrac{3}{w} + \dfrac{6}{m} = \dfrac{1}{3} \Rightarrow \dfrac{1}{w} + \dfrac{2}{m} = \dfrac{1}{9}$
$ \Rightarrow \dfrac{1}{w} = \dfrac{1}{9} - \dfrac{2}{m}$
Let’s now substitute this value of $\dfrac{1}{w}$ in equation (1), we will see that the equation will change into a single variable equation
$ \Rightarrow \dfrac{2}{w} + \dfrac{5}{m} = \dfrac{1}{4} \Rightarrow 2\left( {\dfrac{1}{9} - \dfrac{2}{m}} \right) + \dfrac{5}{m} = \dfrac{1}{4}$
Now, we can solve this further to get the value of $'m'$
$ \Rightarrow \dfrac{2}{9} - \dfrac{4}{m} + \dfrac{5}{m} = \dfrac{1}{4}$
$ \Rightarrow \dfrac{1}{m} = \dfrac{1}{4} - \dfrac{2}{9}$
$ \Rightarrow \dfrac{1}{m} = \dfrac{{9 - 8}}{{4 \times 9}} = \dfrac{1}{{36}}$
Therefore, we get $m = 36$
And, using (2), $\dfrac{1}{w} = \dfrac{1}{9} - \dfrac{2}{m} = \dfrac{1}{9} - \dfrac{2}{{36}} = \dfrac{{4 - 2}}{{36}} = \dfrac{2}{{36}} = \dfrac{1}{{18}}$
Thus, we found the value $w = 18$
Hence, the time taken by a woman alone to complete a work is $w = 18$ days and the time taken by a man alone to complete the work is $m = 36$ days

Note: Try to assume a rate of doing work of a woman and a man. Always go step by step in these work-time questions to avoid complications. An alternative approach can be to find the combined rate and then multiply it to the total number of days and equate it to$1$. This is a way in which you will be equating the total work done with the different rate with the completely finished part of work, i.e. $1$.