NOTE :-
Two operations are performed in sequence means than after adding m%, from that total amount n% is subtracted.
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Let the Amount in the tank at any point be 100 liters.
(100 + m% of 100) - { n% of (100 + m% of 100) }
$\Rightarrow$$(100 + \frac{m}{100} * 100$ ) - { $\frac{n}{100} * (100 + \frac{m}{100} * 100$ ) }
$\Rightarrow$$(100 + m$ ) - { $n + \frac{nm}{100}$}
$\Rightarrow$ $100 + m - n - \frac{nm}{100}$
$\therefore$ $m - n - \frac{nm}{100}$ is the net increase or decrease in the original value.
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If the net increase is 0 after each hour then the value will remain 100 liters after each iteration.
$\Rightarrow$ m - n - $\frac{nm}{100}$ = 0
$\Rightarrow$ m - n = $\frac{nm}{100}$
$\Rightarrow$ 100m - 100n = nm
$\Rightarrow$ 100m - 100n = positive number ($\because$ n and m are positive numbers so n *m will also be positive)
$\Rightarrow$ m>n.
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$\therefore$ Option $B.$ is the correct answer.