Problem 23 - Algebra and Trigonometry with Analytic Geometry

Question

Mixing acid solutions Three solutions contain a certain acid. The first contains \(10 \%\) acid, the second \(30 \%\), and the third \(50 \%\). A chemist wishes to use all three solutions to obtain a 50 -liter mixture containing \(32 \%\) acid. If the chemist wants to use twice as much of the \(50 \%\) solution as of the \(30 \%\) solution, how many liters of each solution should be used?

Step-by-Step Solution

Answer

Use 17 liters of 10% solution, 11 liters of 30% solution, and 22 liters of 50% solution.

Step 1: Define the Variables

Let \(x\) be the amount of the 10% solution in liters, \(y\) be the amount of the 30% solution, and \(z\) be the amount of the 50% solution to be used in the mixture.

Step 2: Set up the Volume Equation

Because the total mixture is to be 50 liters, we have the equation: \[ x + y + z = 50 \]

Step 3: Set up the Acid Percentage Equation

The target acid concentration of the mixture is 32%, so the amount of acid in the mix is \(0.32 \times 50 = 16\) liters. This gives us the equation: \[ 0.10x + 0.30y + 0.50z = 16 \]

Step 4: Relate 30% and 50% Solution Quantities

The chemist wants to use twice as much of the 50% solution as the 30% solution, giving the equation: \[ z = 2y \]

Step 5: Solve the Equations

Substitute \(z = 2y\) into the equations from Step 2 and Step 3. Solve the system: 1. \( x + y + 2y = 50 \) simplifies to \( x + 3y = 50 \)2. \( 0.10x + 0.30y + 0.50(2y) = 16 \) simplifies to \( 0.10x + 1.3y = 16 \) Solve for \(x\) and \(y\):- From \(x + 3y = 50\), \( x = 50 - 3y \)- Substitute \(x = 50 - 3y\) into the second equation: \( 0.10(50 - 3y) + 1.3y = 16 \) \( 5 - 0.3y + 1.3y = 16 \) \( 1.0y = 11 \) \( y = 11 \)Find \(x\):\(x = 50 - 3\times11\)\(x = 17\)Find \(z\):\(z = 2 \times 11 = 22\)

Step 6: Verify the Solution

Check the solution by verifying each condition. The volumes sum to 50: \(17 + 11 + 22 = 50\). The acid amount is correct: \(0.10(17) + 0.30(11) + 0.50(22) = 1.7 + 3.3 + 11 = 16\). The ratio \(z = 2y\) holds: \(22 = 2\times11\). Thus, the solution is verified.

Key Concepts

Acid Solution

In chemistry, an acid solution is a liquid mixture containing a certain percentage of acid mixed with other components, usually a solvent like water. In this problem, we deal with three such solutions:

• 10% acid solution
• 30% acid solution
• 50% acid solution

This info is essential in many real-life applications. Scientists often tweak these mixtures to get just the right concentration for a desired purpose. Here, the chemist wants to create a 50-liter mixture with a specific concentration of 32% acid. This illustrates the juggling act of using solutions with different acid strengths to achieve an exact mixture concentration.

System of Equations

Solving mixture problems involves setting up a system of equations, a fundamental algebraic strategy. It allows us to deal with multiple equations at once and find unknown quantities.

• Equation 1: Represents the total volume of the solution as 50 liters: \( x + y + z = 50 \)
• Equation 2: Accounts for the acid concentration requirement: \( 0.10x + 0.30y + 0.50z = 16 \)
• Equation 3: Reflects the relationship between the volumes of the 30% and 50% solutions: \( z = 2y \)

These three equations form a system that can be solved simultaneously to find the amount of each solution needed. Understanding how to formulate and work with such systems is key to tackling more complicated word problems in algebra.

Percentage Concentration

Percentage concentration tells us how much of a substance is present in a solution compared to the entire solution. In mixtures, it's a way to measure the strength or potency of the solution, for example, the amount of acid present in an acid solution.
In this problem, the chemist aims for a 32% concentration of acid in a 50-liter mixture. That translates to having exactly 16 liters of pure acid in the mixture (because 32% of 50 liters is 16 liters).
Each solution contributes a different percentage of acid, which is why we see different coefficient values in the equations. The understanding of these percentages and how they add up helps ensure the final mixture meets the desired specifications. This fundamental idea applies not only in chemistry but also in cooking, pharmacology, and other practical fields.

Word Problems in Algebra

Word problems often intimidate students, but they're just translations of real-life situations into mathematical language. They require critical thinking and logical reasoning to break down what the problem is asking and how to solve it.
In this exercise, the situation is about mixing acid solutions, which is a common type of word problem in algebra that models real-world scenarios. First, start by identifying all the essential pieces of information given. Then, establish relationships among the quantities involved, often expressed through equations.

• Determining the right amounts of each solution involves interpreting the word problem correctly to set up accurate equations.
• Realize that the problem constraints, like total volume or concentration, form medium for creating these equations.

By practicing more word problems, students enhance their ability to translate spoken or written scenarios into mathematical equations, an invaluable skill in both education and everyday life.