RRB Paramedical Staff Nurse Model Paper 4

Given, tank filled by A in 2h and tank filled by B in 6h
Then, part of the tank filled in 1 h by pipe $\mathrm{A}=\frac{1}{2}$
Part of the tank filled in 1 h by pipe $\mathrm{B}=\frac{1}{6}$
According to the question,
Pipe A was opened at 10 am and till 11 am pipe $\mathit{A}$ alone worked. So, part of the tank filed between 10 am to 11 am by pipe $\mathit{A}=\frac{1}{2}$
Thus remaining work $=1-\frac{1}{2}=\frac{1}{2}$
Now, remaining work will be done by both pipes $\left(\mathit{A}+\mathit{B}\right)$
Tank filled by (A + B) working together
$=\frac{\mathit{x}\mathit{y}}{\mathit{x}+\mathit{y}}$ (where $\mathit{x}=2,\mathit{y}=6)$
$=\frac{2\times 6}{6+2}=\frac{12}{8}=\frac{3}{2}\mathit{h}$
Hence, half tank filled by $\left(\mathit{A}+\mathit{B}\right)$
$=\frac{3}{2\times 2}\mathit{h}=\frac{3}{4}\mathit{h}$
$=\frac{3}{4}\times 60=45\mathit{m}\mathit{i}\mathit{n}$
Hence. the tank will be filled at 11: 45 am