Question

A chemist has two different alcohol solutions available. one solution contains 5% alcohol and the other 12% how much of each should be mixed to obtain 1250 gal of a solution containing 10% alcohol?

Answer 1

To obtain 1250 gallons of a 10% alcohol solution, one would need to mix approximately 1071.43 gallons of the 5% alcohol solution with approximately 178.57 gallons of the 12% alcohol solution.

Step-by-step explanation:

To determine how much of each alcohol solution should be mixed to obtain 1250 gallons of a 10% alcohol solution, we can use a system of equations. Let x be the amount of 5% alcohol solution, and y be the amount of 12% alcohol solution required.

Our two equations based on the percentage of alcohol and the total volume will be:

• 0.05x + 0.12y = 0.10(1250)
• x + y = 1250

We can multiply the first equation by 100 to make calculations easier:

• 5x + 12y = 10×1250
• x + y = 1250

To solve this system, we can multiply the second equation by 5:

• 5x + 5y = 5×1250

And subtract this from the first equation:

• (5x + 12y) - (5x + 5y) = 10×1250 - 5×1250
• 7y = 1250
• y = 1250 / 7
• y = 178.57 (approximately)

Now we find x by substituting y back into the second equation:

• x + 178.57 = 1250
• x = 1250 - 178.57
• x = 1071.43 (approximately)

To mix 1250 gallons of a 10% alcohol solution, you would need approximately 1071.43 gallons of 5% alcohol and 178.57 gallons of 12% alcohol.