Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the...

1. Translate the problem requirements: We need to find x (gallons of 1% milk) in terms of y and z when mixing three milk grades to create a 1.5% fat mixture. The key insight is that fat content is preserved - total fat going in equals total fat in the final mixture.
2. Set up the fat conservation equation: Calculate the total fat content from each milk grade and set it equal to the fat content in the final mixture.
3. Simplify the equation systematically: Remove common factors and rearrange terms to isolate x on one side of the equation.
4. Express x in terms of y and z: Solve for x to get the final answer that matches one of the given choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're dealing with in everyday terms. Think of this like mixing different types of cream for coffee - some have more fat content than others, and when you mix them, you get something in between.

We have three types of milk:

• Type 1: 1% fat (x gallons)
• Type 2: 2% fat (y gallons)
• Type 3: 3% fat (z gallons)

When we mix all these together, we get \((\mathrm{x} + \mathrm{y} + \mathrm{z})\) gallons of milk that is 1.5% fat.

The key insight here is simple: the total amount of fat going into the mixture must equal the total amount of fat in the final mixture. Nothing disappears or gets created - fat is conserved.

Process Skill: TRANSLATE - Converting the mixing scenario into a mathematical relationship based on conservation of fat content

2. Set up the fat conservation equation

Now let's calculate the actual fat content from each component:

Fat from 1% milk: If we have x gallons of 1% milk, the actual fat content is \(\mathrm{x} \times 0.01 = 0.01\mathrm{x}\) gallons of pure fat

Fat from 2% milk: If we have y gallons of 2% milk, the actual fat content is \(\mathrm{y} \times 0.02 = 0.02\mathrm{y}\) gallons of pure fat

Fat from 3% milk: If we have z gallons of 3% milk, the actual fat content is \(\mathrm{z} \times 0.03 = 0.03\mathrm{z}\) gallons of pure fat

Fat in final mixture: We have \((\mathrm{x} + \mathrm{y} + \mathrm{z})\) gallons of 1.5% milk, so the fat content is \((\mathrm{x} + \mathrm{y} + \mathrm{z}) \times 0.015 = 0.015(\mathrm{x} + \mathrm{y} + \mathrm{z})\) gallons of pure fat

Since fat is conserved, we can write:

\(0.01\mathrm{x} + 0.02\mathrm{y} + 0.03\mathrm{z} = 0.015(\mathrm{x} + \mathrm{y} + \mathrm{z})\)

3. Simplify the equation systematically

Let's clean up this equation step by step. First, let's multiply everything by 1000 to get rid of the decimals (this makes our arithmetic much cleaner):

\(10\mathrm{x} + 20\mathrm{y} + 30\mathrm{z} = 15(\mathrm{x} + \mathrm{y} + \mathrm{z})\)

Now let's expand the right side:

\(10\mathrm{x} + 20\mathrm{y} + 30\mathrm{z} = 15\mathrm{x} + 15\mathrm{y} + 15\mathrm{z}\)

Let's move all terms with x to one side by subtracting \(10\mathrm{x}\) from both sides:

\(20\mathrm{y} + 30\mathrm{z} = 15\mathrm{x} - 10\mathrm{x} + 15\mathrm{y} + 15\mathrm{z}\)

\(20\mathrm{y} + 30\mathrm{z} = 5\mathrm{x} + 15\mathrm{y} + 15\mathrm{z}\)

Now subtract \(15\mathrm{y}\) and \(15\mathrm{z}\) from both sides:

\(20\mathrm{y} - 15\mathrm{y} + 30\mathrm{z} - 15\mathrm{z} = 5\mathrm{x}\)

\(5\mathrm{y} + 15\mathrm{z} = 5\mathrm{x}\)

Process Skill: SIMPLIFY - Systematically eliminating decimal complications and rearranging terms to isolate the target variable

4. Express x in terms of y and z

We have: \(5\mathrm{y} + 15\mathrm{z} = 5\mathrm{x}\)

Divide everything by 5:

\(\mathrm{y} + 3\mathrm{z} = \mathrm{x}\)

Therefore: \(\mathrm{x} = \mathrm{y} + 3\mathrm{z}\)

Looking at our answer choices, this matches option A: \(\mathrm{y} + 3\mathrm{z}\)

Let's verify this makes intuitive sense: Since we want a 1.5% mixture, and our 1% milk is below this target, we need the higher-fat milks (2% and 3%) to pull the average up. The 3% milk has a bigger impact (coefficient of 3) compared to the 2% milk (coefficient of 1), which makes sense since 3% is further from our 1% baseline.

Final Answer

\(\mathrm{x} = \mathrm{y} + 3\mathrm{z}\)

This corresponds to Answer Choice A.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what needs to be conserved
Students often focus on conserving volume (which is obvious since we're just adding gallons) rather than recognizing that the key insight is conserving the actual fat content. They might set up equations like \(\mathrm{x} + \mathrm{y} + \mathrm{z} = \mathrm{x} + \mathrm{y} + \mathrm{z}\), which doesn't help solve the problem. The critical realization is that the total amount of pure fat going into the mixture must equal the total amount of pure fat in the final mixture.

Faltering Point 2: Setting up percentage equations incorrectly
Students frequently mix up the percentage calculation, writing something like \(0.01 + 0.02 + 0.03 = 0.015\), treating percentages as if they can be directly added without considering the volumes. They fail to multiply each percentage by its corresponding volume to get the actual fat content in gallons.

Faltering Point 3: Misinterpreting the target
Some students might think they need to find specific values for x, y, and z rather than understanding that they need to express x in terms of y and z. This leads them to look for numerical solutions instead of setting up the algebraic relationship requested in the question.

Errors while executing the approach

Faltering Point 1: Decimal arithmetic errors
When working with 0.01, 0.02, 0.03, and 0.015, students often make calculation mistakes. They might incorrectly expand \(0.015(\mathrm{x} + \mathrm{y} + \mathrm{z})\) or make errors when subtracting decimal terms. The solution wisely multiplies by 1000 to avoid decimals, but students working directly with decimals are prone to computational errors.

Faltering Point 2: Algebraic manipulation mistakes
When rearranging the equation \(10\mathrm{x} + 20\mathrm{y} + 30\mathrm{z} = 15\mathrm{x} + 15\mathrm{y} + 15\mathrm{z}\), students commonly make sign errors or forget to apply operations to all terms. For example, they might correctly subtract \(10\mathrm{x}\) from the left side but forget to subtract it from the right side, or make errors when collecting like terms.

Faltering Point 3: Incorrect factoring or simplification
When reaching \(5\mathrm{y} + 15\mathrm{z} = 5\mathrm{x}\), students might fail to factor out the 5, or they might factor incorrectly. Some might write the final answer as \(5\mathrm{y} + 15\mathrm{z}\) instead of properly isolating x, or they might make errors in the division step.

Errors while selecting the answer

Faltering Point 1: Selecting expressions that look similar but are algebraically different
After correctly finding \(\mathrm{x} = \mathrm{y} + 3\mathrm{z}\), students might second-guess themselves and select answer choice E (\(3\mathrm{y} + 4.5\mathrm{z}\)) because it contains similar terms, or choice C (\(2\mathrm{y} + 3\mathrm{z}\)) because it has the "\(3\mathrm{z}\)" term they expect to see. They don't verify their answer by substitution or logical checking.

Faltering Point 2: Forgetting to solve for the requested variable
Some students might arrive at a correct equation like \(5\mathrm{y} + 15\mathrm{z} = 5\mathrm{x}\) but then select an answer choice that represents this intermediate form rather than the final simplified form \(\mathrm{x} = \mathrm{y} + 3\mathrm{z}\). They don't complete the final step of expressing x explicitly.

Alternate Solutions

Smart Numbers Approach

This mixture problem can be effectively solved using smart numbers by choosing convenient values for y and z, then calculating the required value of x.

Step 1: Choose smart values for y and z
Let's choose values that will make our calculations clean:
• \(\mathrm{y} = 2\) gallons (2% milk)
• \(\mathrm{z} = 2\) gallons (3% milk)

Step 2: Set up the fat conservation equation
Total fat from individual components = Fat in final mixture
• Fat from x gallons of 1% milk: \(0.01\mathrm{x}\)
• Fat from 2 gallons of 2% milk: \(0.02(2) = 0.04\)
• Fat from 2 gallons of 3% milk: \(0.03(2) = 0.06\)
• Fat in final mixture: \(0.015(\mathrm{x} + 2 + 2) = 0.015(\mathrm{x} + 4)\)

Step 3: Solve for x
\(0.01\mathrm{x} + 0.04 + 0.06 = 0.015(\mathrm{x} + 4)\)
\(0.01\mathrm{x} + 0.10 = 0.015\mathrm{x} + 0.06\)
\(0.10 - 0.06 = 0.015\mathrm{x} - 0.01\mathrm{x}\)
\(0.04 = 0.005\mathrm{x}\)
\(\mathrm{x} = 8\)

Step 4: Verify using answer choices
With \(\mathrm{y} = 2\) and \(\mathrm{z} = 2\), let's check each answer choice:

• Choice A: \(\mathrm{y} + 3\mathrm{z} = 2 + 3(2) = 2 + 6 = 8\) ✓
• Choice B: \((\mathrm{y} + \mathrm{z})/4 = (2 + 2)/4 = 1\) ✗
• Choice C: \(2\mathrm{y} + 3\mathrm{z} = 2(2) + 3(2) = 10\) ✗
• Choice D: \(3\mathrm{y} + \mathrm{z} = 3(2) + 2 = 8\) ✓
• Choice E: \(3\mathrm{y} + 4.5\mathrm{z} = 3(2) + 4.5(2) = 15\) ✗

Step 5: Test with different smart numbers
To distinguish between A and D, let's try \(\mathrm{y} = 1, \mathrm{z} = 2\):
\(0.01\mathrm{x} + 0.02(1) + 0.03(2) = 0.015(\mathrm{x} + 3)\)
\(0.01\mathrm{x} + 0.08 = 0.015\mathrm{x} + 0.045\)
\(0.035 = 0.005\mathrm{x}\)
\(\mathrm{x} = 7\)

Checking answer choices with \(\mathrm{y} = 1, \mathrm{z} = 2\):

• Choice A: \(\mathrm{y} + 3\mathrm{z} = 1 + 3(2) = 7\) ✓
• Choice D: \(3\mathrm{y} + \mathrm{z} = 3(1) + 2 = 5\) ✗

Conclusion: The answer is A: \(\mathrm{x} = \mathrm{y} + 3\mathrm{z}\)