RRB NTPC 29 April 2016 Paper

(2) A and B do $\frac{1}{12}$part of work in 1day

B and C do $\frac{1}{15}$part of work in 1day

C and A do $\frac{1}{20}$part of work in 1day

Adding all, we have
$2\mathrm{l}\left[\frac{1}{\mathit{A}}+\mathrm{}\frac{1}{\mathit{B}}+\frac{1}{\mathit{c}}\right]=\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$$=\frac{12}{60}=\frac{1}{5}$
$\mathrm{}\frac{1}{\mathit{A}}+\mathrm{}\frac{1}{\mathit{B}}+\mathrm{}\frac{1}{\mathit{C}}=\mathrm{}110$
A does in 1 day
$=\frac{1}{\mathit{A}}+\frac{1}{\mathit{B}}+\frac{1}{\mathit{C}}-\frac{1}{\mathit{B}}-\frac{1}{\mathit{C}}$$=110-115=3-230$
= $\frac{1}{30}$ part of wok
So, A will compete the work in 30days.