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Ryan, Engineer
Category: Pre-Calculus
Satisfied Customers: 9064
Experience:  B.S. in Civil Engineering
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You need 375 mL of a 60% alcohol solution. On hand, you have

Customer Question

You need 375 mL of a 60% alcohol solution. On hand, you have a 75% alcohol mixture. How much of the 75% alcohol mixture and pure water will you need to obtain the desired solution?
Submitted: 7 months ago.
Category: Pre-Calculus
Expert:  Ryan replied 7 months ago.


Thank you for using the site. I'll be happy to help you with this problem.

To solve this, you need to account for both the amount of alcohol in each component of the mixture, and the total volume.

Let x represent the amount of 75% alcohol, and let y represent the amount of pure water used.

Since the total volume of the mixture is 375 ml, you would write:

x + y = 375

The amount of alcohol in x ml of the 75% mixture is 0.75x, and the amount of alcohol in the pure water will be 0y = 0. These must add up to the amount of alcohol that is in the final 60% solution, which would be 0.6(375). This gives you another equation:

0.75x + 0y = 0.6(375)

Solving the second equation for x gives:

x = 0.6(375)/0.75 = 300 ml

Using the first equation to solve for y, after substituting the value 300 for x gives:

300 + y = 375

y = 375 - 300 = 75

Solution: The mixture requires 300 ml of the 75% alcohol solution, and 75 ml of pure water.

Please feel free to ask if you have any questions about this solution.