We have a mixture of 300 ml with 15% alcohol content, this means that in this mixture we have an alcohol total of:
![\text{alcohol}=300\cdot(15)/(100)=45\text{ ml}](https://img.qammunity.org/2023/formulas/mathematics/college/hmbm259av6mtbzkfyk3naurb7djsi9kgu4.png)
And we want to obtain a 60% alcohol solution by mixing a 90% mixture with the 15% one we have.
![\begin{gathered} \text{volume 90 \%}=v \\ \text{ alcohool content}=0.9\cdot v \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/td8o6k340loxd9krr0ermxbbchb9otxao0.png)
When we add this volume to what we already have, we will obtain:
![\begin{gathered} \text{ final mixture}=300+v \\ \text{alcohool content}=45+0.9\cdot v \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ardsxb6g3jvltw9hljzwv7yab57fzv5a7b.png)
And what we want is that the alcohol content must be equal to 60% of the final mixture, so we have:
![\begin{gathered} 60\text{ \%}\cdot\text{ final mixture}=\text{ alcohol content} \\ (60)/(100)\cdot(300+v)=45+0.9\cdot v \\ 0.6\cdot(300+v)=45+0.9\cdot v \\ 180+0.6\cdot v=45+0.9\cdot v \\ \text{0}.9\cdot v-0.6\cdot v=180-45 \\ 0.3\cdot v=135 \\ v=(135)/(0.3)=450\text{ ml} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fbu99gtexclxc0rbfeqfq2ignq4o91f6fo.png)
This means that we need to add 450 ml of the 90% solution to obtain the desired final mixture.