Question

A solution of $9\%$ acid is to be diluted by adding $3\%$ acid solution to it. The resulting mixture is to be more than $5\%$ but less than $7\%$ acid. If there is $460$ litres of the $9\%$ solution, how many litres of $3\%$ solution will have to be added?

Answer 1

Given data :

Let $x$ litres of $3\%$ solution be added
to $460$ litres of $9\%$ solution of acid.

$\Rightarrow$ Total quantity of the mixture

$=(460+x)$ litres

Total acid content in $(460+x)$ litres of mixture

$=460\times \frac{9}{100}+x+\frac{3}{100}$

It is given that the acid content in the resulting mixture must be more than $5\%$ but less than $7\%$ acid.

$\Rightarrow 5\%of(460+x)<460\times \frac{9}{100}+x+\frac{3}{100}<7\%of(460+x)$

$\Rightarrow \frac{5(460+x)}{100}<\frac{4140+3x}{100}<\frac{7(460+x}{100}$

$\Rightarrow 23000+5x<4140+3x<3220+7x$

$\Rightarrow 23000+5x<4140+3x$ and $4140+3x<3220+7x$

$\Rightarrow 2x<1840$ and $4x>920$

$\Rightarrow x<920$ and $x>230$

Drawing both on the number line and taking intersection.



$\therefore x\in (230,920)$

Hence, $3\%$ acid solution must be more than $230$ litres & less than $920$ litres.