Bruce wants to make 50 ml of an alcohol solution with a $12 \%$ concentration. He has a $10 \%$ alcohol solution and $15 \%$ alcohol solution. The equation $0.10 x+0.15(50-x)=0.12(50)$ can be used to find the amount of $10 \%$ alco solution Bruce should use.
How much of the $10 \%$ alcohol solution should Bruce use?
$\square$ mL
How much of the $15 \%$ alcohol solution should Bruce use?
$\square$ mL
Answer (1)
Define x as the amount of the 10% solution, so the amount of the 15% solution is 50 − x .
Set up the equation: 0.10 x + 0.15 ( 50 − x ) = 0.12 ( 50 ) .
Solve for x : x = 30 .
Calculate the amount of the 15% solution: 50 − 30 = 20 . The amount of the 10% alcohol solution is 30 mL and the amount of the 15% alcohol solution is 20 mL. 30 mL and 20 mL.
Explanation
Problem Analysis Let's analyze the problem. We are given that Bruce wants to create 50 ml of a 12% alcohol solution using a 10% alcohol solution and a 15% alcohol solution. We are also given the equation 0.10 x + 0.15 ( 50 − x ) = 0.12 ( 50 ) , which represents the total amount of alcohol in the final solution. Our goal is to find the amount of each solution Bruce needs to use.
Define Variables Let x be the amount (in ml) of the 10% alcohol solution Bruce should use. Then, the amount of the 15% alcohol solution Bruce should use is 50 − x ml, since the total volume of the mixture is 50 ml.
Solve for x Now, let's solve the given equation for x :
0.10 x + 0.15 ( 50 − x ) = 0.12 ( 50 ) 0.10 x + 7.5 − 0.15 x = 6 − 0.05 x = 6 − 7.5 − 0.05 x = − 1.5 x = − 0.05 − 1.5 x = 30 So, Bruce should use 30 ml of the 10% alcohol solution.
Calculate the amount of 15% solution Now we can find the amount of the 15% alcohol solution: 50 − x = 50 − 30 = 20 Therefore, Bruce should use 20 ml of the 15% alcohol solution.
Verification To verify our solution, we can plug the values back into the original equation: 0.10 ( 30 ) + 0.15 ( 20 ) = 3 + 3 = 6 0.12 ( 50 ) = 6 Since both sides of the equation are equal, our solution is correct.
Final Answer In conclusion, Bruce should use 30 ml of the 10% alcohol solution and 20 ml of the 15% alcohol solution.
Examples
Mixture problems like this are common in chemistry, pharmacy, and even cooking. For example, a pharmacist might need to mix different concentrations of a drug to achieve a specific dosage. Similarly, a chef might mix different types of vinegar to get a desired acidity level in a salad dressing. Understanding how to solve these problems allows for precise control over the final product.
