Question

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

Answer 1

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man's one day's work = $\frac{1}{x}$
And, one boy's one day's work = $\frac{1}{y}$
2 men and 5 boys can finish the work in 4 days.
∴ (2 men's one day's work) + (5 boys' one day's work) = $\frac{1}{4}$
⇒ $\frac{2}{x}+\frac{5}{y}=\frac{1}{4}$
⇒ $2u+5v=\frac{1}{4}$ ...(i) Here, $\frac{1}{x}=u\mathrm{and}\frac{1}{\mathrm{y}}=\mathrm{v}$
Again, 3 men and 6 boys can finish the work in 3 days.
∴ (3 men's one day's work) + (6 boys' one day's work) = $\frac{1}{3}$
⇒ $\frac{3}{x}+\frac{6}{y}=\frac{1}{3}$
⇒ $3u+6v=\frac{1}{3}$ ....(ii) Here, $\frac{1}{x}=u\mathrm{and}\frac{1}{\mathrm{y}}=\mathrm{v}$
On multiplying (i) by 6 and (ii) by 5, we get:
$12u+30v=\frac{6}{4}$ ....(iii)
$15u+30v=\frac{5}{3}$ ....(iv)
On subtracting (iii) from (iv), we get:
$3u=\left(\frac{5}{3}-\frac{6}{4}\right)=\frac{2}{12}=\frac{1}{6}$
⇒ $u=\frac{1}{6\times 3}=\frac{1}{18}\Rightarrow \frac{1}{x}=\frac{1}{18}\Rightarrow x=18$
On substituting $u=\frac{1}{18}$ in (i), we get:
$2\times \frac{1}{18}+5v=\frac{1}{4}\Rightarrow 5v=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
⇒ $v=\left(\frac{5}{36}\times \frac{1}{5}\right)=\frac{1}{36}\Rightarrow \frac{1}{y}=\frac{1}{36}\Rightarrow y=36$
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.