SSC CGL 18 Aug 2021 Shift 2 Paper

Given, $\left(\mathit{A}+\mathit{B}\right)$ 's 1 day work
$=\frac{1}{20}$ . . . (i)
$\left(\mathit{B}+\mathit{C}\right)$ 's 1 day work $=\frac{1}{24}$ . . . (ii)
By Eqs. (i) and (ii), $\left(\mathit{A}-\mathit{C}\right)$ 's 1 day work
$=\frac{1}{20}-\frac{1}{24}=\frac{6-5}{120}=\frac{1}{120}$
Let $\mathit{C}$ 's 1 day work $=\frac{1}{2\mathit{x}}$
The, A's 1 day work $=\frac{1}{\mathit{x}}$
According to the question,
$\frac{1}{\mathit{x}}-\frac{1}{2\mathit{x}}=\frac{1}{120}$
$\Rightarrow \frac{2-1}{2\mathit{x}}=\frac{1}{120}$
$\Rightarrow \mathit{x}=60$
Then, A's 1 day work $=\frac{1}{60}$
and C's 1 day work $=\frac{1}{120}$
Put the value of $\mathit{A}$ in Eq. (i), we get
B's 1 day work
$=\frac{1}{20}-\frac{1}{60}=\frac{3-1}{60}$
$=\frac{2}{60}=\frac{1}{30}$
Now, $100\%$ of work completed by $\mathit{B}$
$=30$ days
$\therefore 40\%$ of work completed by B
$=\frac{40}{100}\times 30$