SSC CGL 20 Aug 2021 Shift 2 Paper

$\ln 1\mathit{h}$, part filled by pipe $\mathit{A}=\frac{1}{10}$
In $1\mathit{h}$, part filled by pipe $\mathit{B}=\frac{1}{16}$
Now, In 2 h, part filled $=\frac{1}{10}+\frac{1}{16}=\frac{13}{80}$
In $12\mathit{h}$, part filled $=\frac{13\times 6}{80}=\frac{78}{80}$
Remaining part $=1-\frac{78}{80}=\frac{2}{80}=\frac{1}{40}$
$\therefore \frac{1}{40}$ part filled by pipe $\mathit{B}=\frac{\frac{1}{\frac{40}{1}}}{\frac{16}{5}}=\frac{2}{5}\mathit{h}$
Hence, total time take by pipe $\mathit{A}$ and $\mathit{B}$
$=\left(12+\frac{2}{5}\right)\mathit{h}=12\frac{2}{5}\mathit{h}$