RRB Group D Model Paper 3 with solutions

(2) Let (A + C) complete the work in ‘a’ days. So B will complete the work in 3a days.
$\frac{1}{\mathit{a}}+\frac{1}{3\mathit{a}}=\frac{1}{40}$ → 3a = 40 ⇒ a = $\frac{40}{3}$
Let (A + B) complete the work in n days. So C will complete the work in 2n days.
$\frac{1}{\mathit{n}}+\frac{1}{2\mathit{n}}=\frac{1}{10}$ ⇒ n = $\frac{30}{2}$ = 15
(A + B + C)'s1 day's work = $\frac{1}{10}$
$\frac{1}{\mathit{A}}+\frac{1}{40}+\frac{1}{30}=\frac{1}{10}$
⇒ $\frac{1}{\mathit{A}}=\frac{12-3-4}{120}$ = $\frac{5}{120}=\frac{1}{24}$
Therefore, A will complete the work in 24 days.