Question

A manufacturer has 600 L of a 12% solution of acid. How many litres of 30% acid solution must be added to it, so that acid content in the resulting mixture will be more than 15% but less than 18%?

Answer 1

Let x of 30% acid solution be added to 600 L of 12% solution of acid.
Then, totall quantity of mixture = (600 + x ) L
Total acid content in the (600 + x) L of mixture
$=600\times \frac{12}{100}+x\times \frac{30}{100}=\frac{30x+7200}{100}$
It is given that, acid content in the resulting mixture must be more than 15% and less than 18% .
$\therefore 15\%of(600+x)<\frac{30x+7200}{100}<18\%of(600+x)\Rightarrow (600+x)\times \frac{15}{100}<\frac{30x+720}{100}<(600+x)\times \frac{18}{100\left[1\right]}\Rightarrow 9000+15x<30x+7200<10800+18x\Rightarrow 9000+15x<30xj+7200\text{and}30x+7200<10800+18x\Rightarrow 15x>1800\text{and}12x<3600\Rightarrow x>120\text{and}x<300\Rightarrow 120<x<300$
Hence the 30% solution of acid must be added between 120 to 300 L.