Question

A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 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

given a manufacturer has 600 litres of 12%acid solution

therefore the acid content =$0.12\times 600=72$

let us assume another solution of x litres and 30% acid solution

therefore the acid content in this solution is $0.3\times x=0.3x$

now if we mix both total volume is $600+x$

and total acid content is $72+0.3x$

given the acid percentage is 15% to 18%

$\Rightarrow 0.15<\frac{72+0.3x}{600+x}<0.18$

$\Rightarrow 0.15(600+x)<72+0.3x\Rightarrow 90+0.15x<72+0.3x\Rightarrow 18<0.15xorx>(\frac{18}{0.15}=120)$

$\Rightarrow 72+0.3x<0.18(600+x)\Rightarrow 72+0.3x<108+0.18x\Rightarrow 0.12x<36orx<(\frac{36}{0.12}=300)$

therefore $120<x<300$