Question

Eight men and twelve boys can finish a piece of work in 10 days while six men and eight boys can finish the same work in 14 days. Find the number of days taken by one man alone to complete the work and also one boy alone to complete the work.

Answer 1

Let x denote the number of days needed for one man to finish the work and y denote the number of days needed for one boy to finish the work. Clearly, $x\ne 0$ and $y\ne 0$.
So, one man can complete $\frac{1}{x}$ part of the work in one day and one boy can complete $\frac{1}{y}$ part of the work in one day.
The amount of work done by 8 men and 12 boys in one day is $\frac{1}{10}$.
Thus, we have $\frac{8}{x}+\frac{12}{y}=\frac{1}{10}$.
The amount of work done by 6 men and 8 boys in one day is $\frac{1}{14}$.
Thus, we have $\frac{6}{x}+\frac{8}{y}=\frac{1}{14}$
Let $a=$\frac{1}{x}$and$b = \frac{1}{y}$. Then (1) and (2) give, respectively,
$8a+12b=\frac{1}{10}\to 4a+6b-\frac{1}{20}=0$ (3)
$6a+8b=\frac{1}{14}\to 3a+4b-\frac{1}{28}=0$ (4)
Writing the coefficients of (3) and (4) for the cross multiplication, we have
Thus, we have $\frac{a}{-\frac{3}{14}+\frac{1}{5}}=\frac{b}{-\frac{3}{20}+\frac{1}{7}}=\frac{1}{16-18}$. i.e., $\frac{a}{-\frac{1}{70}}=\frac{b}{-\frac{1}{140}}=\frac{1}{-2}$.
That is, $a=\frac{1}{140},b=\frac{1}{280}$
Thus, we have $x=\frac{1}{a}=140,y=1b=280$.
Hence, one man can finish the work individually in 140 days and one boy can finish the work individually in 280 days.