3 votes

A tank has $100$ liters of water$.$ At the end of every hour, the following two operations are performed in sequence$:$ $i)$ water equal to $m\%$ of the current contents of the tank is added to the tank $, ii)$ water equal to $n\%$ of the current contents of the tank is removed from the tank$.$ At the end of $5$ hours, the tank contains exactly $100$ liters of water $.$ The relation between $m$ and $n$ is $:$

- $m=n$
- $m>n$
- $m<n$
- None of the previous

3 votes

Best answer

Initially the tank is having $100$ liters of water.

Every hour, the capacity changes as $C' = C \left(1+\frac{m}{100}\right)\left(1 - \frac{n}{100}\right)$

So, after $5$ hours we get $C ' = 100 = 100 \left(1+\frac{m}{100}\right)^5\left(1 - \frac{n}{100}\right)^5$

$\implies \left(1+\frac{m}{100}\right)^5\left(1 - \frac{n}{100}\right)^5 = 1$

$\implies \left(1+\frac{m}{100}\right)\left(1 - \frac{n}{100}\right) = 1$

$\implies 1+\frac{m}{100}-\frac{n}{100} -\frac{mn}{10000}= 1$

$\implies100(m - n) = mn$

Since, $m,n > 0, m - n > 0 \implies m > n.$

Every hour, the capacity changes as $C' = C \left(1+\frac{m}{100}\right)\left(1 - \frac{n}{100}\right)$

So, after $5$ hours we get $C ' = 100 = 100 \left(1+\frac{m}{100}\right)^5\left(1 - \frac{n}{100}\right)^5$

$\implies \left(1+\frac{m}{100}\right)^5\left(1 - \frac{n}{100}\right)^5 = 1$

$\implies \left(1+\frac{m}{100}\right)\left(1 - \frac{n}{100}\right) = 1$

$\implies 1+\frac{m}{100}-\frac{n}{100} -\frac{mn}{10000}= 1$

$\implies100(m - n) = mn$

Since, $m,n > 0, m - n > 0 \implies m > n.$

2 votes

NOTE :-

Two operations are performed in sequence means than after adding m%, from that total amount n% is subtracted.

--------------------------------------------------------------------------------------------------------------------------------------------

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.

------------------------------------------------------------------------------------------------------------------------------------------------

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.

------------------------------------------------------------------------------------------------------------------------------------------------

$\therefore$ Option $B.$ is the correct answer.