How much pure alcohol must be added to 400 mL of a 15% solution to make its strength 32%.

$\begin{array}{l}\text{Let the volume of the pure alcohol be}x\mathrm{ml}.\\ \text{Initial concentration=}15\%\\ \text{So, initial amount of alcohol in the solution will be}=\frac{15}{100}\times \text{400= 60 ml}\\ \text{To make the strength of the solution 32\%, we will keep the amount of water constant a}\mathrm{nd\; add}x\text{volume of pure alcohol}\text{.}\\ \mathrm{On\; adding\; pure\; alcohol},the\; volume\; of\text{the solution increases to 400 +}x.\\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question},\mathrm{we}\mathrm{have}:\\ \frac{x+60}{400+x}=\frac{32}{100}\\ \text{⇒100}x+6000=12800+32x\\ \Rightarrow \text{100}x-32x=1\text{2800}-6000\\ \Rightarrow \text{68}x=6800\\ \Rightarrow x=100\\ \text{So, amount of pure alcohol to be added=100 ml}\end{array}$