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?

Solution:

A man can alone finish the work in $x$ days and one boy alone can finish it in $y$days then

One mans one days work =\frac{1}{x}

One boys one days work= \frac{1}{y}

2men one day work=\frac{2}{x}

7boys one day work=$\frac{7}{y}$

Since 2 men and 7 boys can finish the work in 4 days

$4\left(\frac{2}{x}+\frac{7}{y}\right)=1$

$\frac{8}{x}+\frac{28}{y}=1 \cdots(i)$

Again 4 men and 4 boys can finish the work in 3 days

$3\left(\frac{4}{x}+\frac{4}{y}\right)=1$

 

$\frac{12}{x}+\frac{12}{y}=1 \cdots(i i)$

$3\left(\frac{4}{x}+\frac{4}{y}\right)=1$

 

$\frac{12}{x}+\frac{12}{y}=1 \cdots(i i)$

Putting $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in equation $(i)$ and $(i i)$ we get

$8 u+28 v=1$

$12 u+12 v=1$

$8 u+28 v-1 \cdots(i i i)$

 

$12 u+12 v-1 \cdots(i v)$

By using cross multiplication we have

Now,

$u=\frac{1}{15}$

$\frac{1}{x}=\frac{1}{15}$

$x=15$

and

$\begin{aligned} v &=\frac{1}{60} \\ \frac{1}{y} &=\frac{1}{60} \\ y &=60 \end{aligned}$

Hence, one man alone can finish the work in 15 days and one boy alone can finish the work in 60 days