Mixture problems and their solutions are presented along with their solutions. Percentages are also used to solve these types of problems.
How many liters of \( 20\% \) alcohol solution should be added to 40 liters of a \( 50\% \) alcohol solution to make a \( 30\% \) solution?
Let \( x \) be the number of liters of the 20% alcohol solution to be added. Let \( y \) be the total quantity of the resulting 30% solution. Since we're adding \( x \) liters to 40 liters, we have: \[ x + 40 = y \] Express the total amount of alcohol before and after mixing. The alcohol in the 20% solution is \( 0.20x \), and in the 50% solution, it's \( 0.50 \times 40 \). The total alcohol in the final mixture is \( 0.30y \). So: \[ 0.20x + 0.50 \times 40 = 0.30y \] Substitute \( y = x + 40 \) into the equation: \[ 0.20x + 0.50 \times 40 = 0.30(x + 40) \] Simplify: \[ 0.20x + 20 = 0.30x + 12 \] Bring all terms to one side: \[ 0.20x - 0.30x + 20 - 12 = 0 \] \[ -0.10x + 8 = 0 \] \[ x = 80 \] 80 liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% alcohol solution.
John wants to make 100 ml of a 5% alcohol solution by mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?
Let \( x \) and \( y \) be the quantities (in ml) of the 2% and 7% alcohol solutions, respectively. Then: \[ x + y = 100 \tag{1} \] The amount of alcohol in \( x \) ml of 2% solution is \( 0.02x \), and the amount in \( y \) ml of 7% solution is \( 0.07y \). The total amount of alcohol in the final mixture is: \[ 0.02x + 0.07y = 0.05 \times 100 = 5 \tag{2} \] From equation (1), solve for \( y \): \[ y = 100 - x \] Substitute into equation (2): \[ 0.02x + 0.07(100 - x) = 5 \] \[ 0.02x + 7 - 0.07x = 5 \] \[ -0.05x + 7 = 5 \] \[ -0.05x = -2 \] \[ x = 40 \] Now substitute \( x = 40 \) into equation (1) to find \( y \): \[ y = 100 - 40 = 60 \] John should mix 40 ml of the 2% solution with 60 ml of the 7% solution to obtain 100 ml of a 5% alcohol solution.
Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed with a 90% Silver alloy to obtain 500 g of a 91% Silver alloy?
Let \( x \) be the weight (in grams) of Sterling Silver (92.5%), and \( y \) be the weight (in grams) of the 90% silver alloy. We want to mix these to get 500 grams of a 91% silver alloy. So, the total mass equation is: \[ x + y = 500 \] The pure silver content equation is: \[ 0.925x + 0.90y = 0.91 \times 500 \] Substitute \( y = 500 - x \) into the equation: \[ 0.925x + 0.90(500 - x) = 455 \] Now simplify: \[ 0.925x + 450 - 0.90x = 455 \] \[ (0.925 - 0.90)x + 450 = 455 \] \[ 0.025x = 5 \] \[ x = 200 \] \( 200 \) grams of Sterling Silver is needed to make the 91% silver alloy.
How many kilograms of pure water must be added to 100 kilograms of a 30% saline solution to obtain a 10% saline solution?
Let \( x \) be the amount of pure water (in kilograms) to be added. Let \( y \) be the final weight of the 10% saline solution. So, \[ x + 100 = y \] The salt content in the pure water is 0. The salt content in the original 30% solution is: \[ 0.30 \times 100 = 30 \text{ kg of salt} \] The final solution is a 10% saline solution, so the total amount of salt is also: \[ 0.10 \times y \] Equating the amount of salt: \[ 30 = 0.10 \times y \] Substitute \( y = x + 100 \) into the equation: \[ 30 = 0.10(x + 100) \] Multiply both sides by 10 to eliminate the decimal: \[ 300 = x + 100 \] Solve for \( x \): \[ x = 200 \] \( 200 \text{ kilograms} \) of pure water must be added.
A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?
The total volume of the final mixture is: \[ 50\ \text{ml} + 30\ \text{ml} = 80\ \text{ml} \] The amount of alcohol in pure water is 0 ml. The amount of alcohol in the after-shave lotion is: \[ 30\% \text{ of } 50\ \text{ml} = 0.30 \times 50 = 15\ \text{ml} \] Let \( x \) be the percentage of alcohol in the final solution. Then: \[ x \times 80 = 15 \] Solving for \( x \): \[ x = \frac{15}{80} = 0.1875 = 18.75\% \] The percentage of alcohol in the new solution is 18.75%.
You add \( x \) ml of a 25% alcohol solution to 200 ml of a 10% alcohol solution to obtain another solution.
a) Find the amount of alcohol in the final solution in terms of \( x \).
b) Find the ratio, in terms of \( x \), of the alcohol in the final solution to the total amount of the solution.
c) What do you think will happen if \( x \) is very large?
d) Find \( x \) so that the final solution has a percentage of 15%.
a) The amount of alcohol in the 200 ml of 10% solution is: \[ 200 \times 0.10 = 20 \text{ ml} \] The amount of alcohol in \( x \) ml of 25% solution is: \[ 0.25x \text{ ml} \] The total amount of alcohol in the final solution is: \[ 20 + 0.25x \] b) The total volume of the solution is: \[ x + 200 \] - The ratio of alcohol in the final solution to the total amount of the solution is: \[ r = \frac{20 + 0.25x}{x + 200} \] c) - As \( x \) becomes very large, the expression \[ r = \frac{20 + 0.25x}{x + 200} \] \( r \) approaches 0.25 or 25%. This is because the 25% solution dominates, and the solution behaves like a 25% alcohol solution. (The horizontal asymptote of the rational function is 0.25.)
d) To find \( x \) such that the final solution has 15% alcohol, set: \[ \frac{20 + 0.25x}{x + 200} = 0.15 \] Solve the equation: \[ 20 + 0.25x = 0.15(x + 200) \] \[ 20 + 0.25x = 0.15x + 30 \] \[ 0.25x - 0.15x = 30 - 20 \] \[ 0.10x = 10 \] \[ x = 100 \] \( x = 100 \) ml so that the final solution has a percentage of 15%.