RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

(a) Let 1 man’s 1 days’ work= x & 1 boy’s 1 day’s work = y
Then, $2\mathit{x}+3\mathit{y}=\frac{1}{10}\mathit{a}\mathit{n}\mathit{d}3\mathit{x}+2\mathit{y}=\frac{1}{8}$
Solving, we get :x=$\frac{7}{200}$= and$\mathit{y}=\frac{1}{100}$
∴ (2 men + 1 boy)’s 1 day’s work
$=\left[2\times \frac{7}{200}+1\times \frac{1}{100}\right]=\frac{16}{20}=\frac{2}{25}$
So, 2 men and 1 boy together can finish the work in$12\frac{1}{2}$days.