It takes 10 men and 20 women to complete a project in 48 days. After the first day, a woman was replaced by a man and it took them 45 more days to complete the work. If 10 men and 20 women had worked on the same project, and after 2 days, if 2 of the women had been replaced by 2 men, how many more days would it have taken to finish the project? Round up to the nearest integer.
I have a doubt in my approach for this question , please let me know what is the flaw in it ;
Let's take the work done by 1 man in a day to m (i.e. the efficiency is m units/day) ; similarly for 1 woman I took it as w units/day
Now it is given that after the first day one woman was replaced by 1 man and due to this they finished 45 days more than the scheduled 48 days
Writing this in the form of equation :-
The work that had to be done by 1 woman in 47 days is now being done by 11 men and 19 women in those extra 45 days over the scheduled time
OR,
47*W=(11M+19W)*45
Now while finding the ratios of M and W , I am getting negative values , what is wrong in the equation ?
You have misinterpreted the question. The 45 additional days to complete the job are being counted after the first day, when the switch occurred, not after the originally scheduled 48 days. So what the problem is saying is that after the work done during Day 1, one woman is switched out for a man, and then with this new team composition, the work is finished on day 46. This is the only reasonable interpretation of the question because, if by switching out one person for another, it nearly doubles the working time, then the logical conclusion would be that the group that the new worker belongs to would have to have negative productivity.
So, with this corrected interpretation in mind, we use the same notation as you adopted. If a man completes $m$ units of work (expressed as some proportion of the total job) in one day and a woman completes $w$ units per day, then the total number of units completed in the first day is $10m + 20w$ units, and over the originally projected completion time of $48$ days, we have $$48(10m + 20w) = 1. \tag{1}$$ Beginning on the second day, we have $11$ men and $19$ women, and it takes $45$ days plus the initial first day to complete the work; i.e., $$(10m + 20w) + 45(11m + 19w) = 1. \tag{2}$$ Equivalently, we may write this as $$(10m + 20w) + 45(10m + 20w + m - w) = 46(10m + 20w) + 45(m-w) = 1. \tag{3}$$ Then substitution of $(1)$ into $(3)$ yields $$45(m-w) = \frac{1}{24}. \tag{4}$$ Now we are asked to find the quantity $d$ satisfying $$2(10m + 20w) + d(12m + 18w) = 1, \tag{5}$$ where $d$ represents the number of additional days after the second working day, that a team of $2$ more men and $2$ fewer women, would need to complete the job. But we can express $(5)$ as $$2(10m + 20w) + d(10m + 20w + 2(m-w)) = (2+d)(10m + 20w) + 2d(m-w) = 1, \tag{6}$$ and since we can determine from $(1)$ and $(4)$ the quantities $10m + 20w$ and $m-w$, it is straightforward to substitute and solve for $d$. Then, to check that your answer is correct, we should intuitively note that $d$ must be less than $45$, since if one replacement after one day shortens the project completion timeline by $3$ days, two replacements after two days should further decrease the timeline below $45$.
As a further exercise, consider the following modifications to the problem: