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.