Question

How many litres of water will have to be added to $1125$ litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?

Answer 1

Let $x$ litre of water is required to be added
Then, total mixture $=(x+1125)$ litres
It is evident that the amount of acid contained in the resulting mixture is $45\%$ of $1125$ litres.
This resulting mixture will contains more than $25\%$ but less than $30\%$ acid content.
$\therefore 30\%$ of $(1125+x)>45\%$ of $1125$
And, $25\%$ of $(1125+x)<45\%$ of $1125$
$\Rightarrow \frac{30}{100}(1125+x)>\frac{45}{100}\times 1125$
$\Rightarrow 30(1125+x)>45\times 1125$
$\Rightarrow 30\times 1125+30x>45\times 1125$
$\Rightarrow 30>45\times 1125-30\times 1125$
$\Rightarrow 30x>(45-30)\times 1125$
$\Rightarrow x>\frac{15\times 1125}{30}=562.5$
$25\%$ of $(1125\times x)<45\%$ of $1125$
$\Rightarrow \frac{25}{100}(1125+x)<\frac{45}{100}\times 1125$
$\Rightarrow 25(1125+x)>45\times 1125$
$\Rightarrow 25\times 1125+25x>45\times 1125$
$\Rightarrow 25x>45\times 1125-25\times 1125$
$\Rightarrow 25x>(45-25)\times 1125$
$\Rightarrow x>\frac{20\times 1125}{25}=900$
$\therefore 562.5<x<900$
Thus, the required number of litres of water that is to be added will have to be more than $562.5$ but less than $900$.