2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?
Given:
2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys.
To do:
We have to find the time taken by one man and one boy to do the work.
Solution:
Let the number of days taken by one man to finish the work alone be $x$.
This implies,
The amount of work done by one man in a day $=\frac{1}{x}$.
Let the number of days taken by one boy to finish the work alone be $y$.
This implies,
The amount of work done by one boy in a day $=\frac{1}{y}$.
In the first case, 2 men and 7 boys finish the work in 4 days.
The amount of work done by two men in 1 day $=\frac{2}{x}$.
The amount of work done by 7 boys in 1 day $=\frac{7}{y}$.
According to the question,
$4(\frac{2}{x}+\frac{7}{y})=1$
$\frac{8}{x}+\frac{28}{y}=1$....(i)
In the second case, 4 men and 4 boys finish the work in 3 days.
The amount of work done by 4 men in 1 day $=\frac{4}{x}$.
The amount of work done by 4 boys in 1 day $=\frac{4}{y}$.
According to the question,
$3(\frac{4}{x}+\frac{4}{y})=1$
$\frac{12}{x}+\frac{12}{y}=1$....(ii)
Multiplying equation (i) by 3 and equation (ii) by 2, we get,
$3(\frac{8}{x}+\frac{28}{y})=3(1)$
$\frac{24}{x}+\frac{84}{y}=3$.....(iii)
$2(\frac{12}{x}+\frac{12}{y})=2(1)$
$\frac{24}{x}+\frac{24}{y}=2$.....(iv)
Subtracting (iv) from (iii), we get,
$\frac{24}{x}+\frac{84}{y}-(\frac{24}{x}+\frac{24}{y})=3-2$
$\frac{84-24}{y}=1$
$y=1(60)$
$y=60$
Substituting $y=60$ in (i), we get,
$\frac{8}{x}+\frac{28}{60}=1$
$\frac{8}{x}=1-\frac{7}{15}$
$\frac{8}{x}=\frac{15-7}{15}$
$\frac{8}{x}=\frac{8}{15}$
$x=\frac{8\times15}{8}$
$x=15$
Therefore, one man takes 15 days to finish the work alone and one boy takes 60 days to finish the work alone.
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