You need a 55% alcohol solution. On hand, you have a 50 mL of a 25% alcohol mixture. You also have 60% alcohol mixture. How much of the 60% mixture will you need to add to obtain the desired solution?

There is a simple diagrammatic way to solve mixture problems, illustrated in the attachment. In the left column are the available solution percentages. In the center is the percentage of the desired mix. In the column to the right of that are the diagonal differences between the desired value and the constituent values. Those differences represent the proportions of the constituents on the same horizontal line.

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Here, that means 30 parts of 60% solution are needed for 5 parts of 25% solution. Since we have 50 mL of 25% solution, we can multiply these ratio values by 10 mL to find the quantities needed. That has been done here in the right-most column.

300 mL of 60% solution are needed to obtain the desired mixture.

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Additional comment

If you want to write an equation, you can let x represent the quantity of high-concentration solution needed. Then the amount of alcohol in the mix is ...

0.60x +0.25(50) = 0.55(x +50)

0.05x = 0.30(50) . . . . . . . subtract 0.55x+0.25(50)

x = 300 . . . . . . . . . . . . mL of 60% solution needed