Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixture Y is 25% ryegrass and 75% fescue....

1. Translate the problem requirements: We have two seed mixtures with known ryegrass percentages (X has \(40\%\) ryegrass, Y has \(25\%\) ryegrass), and we need to find what percentage of a combined mixture must be X to achieve \(30\%\) ryegrass overall.
2. Set up the weighted average relationship: The final ryegrass percentage (\(30\%\)) must be a weighted average of the ryegrass percentages in X (\(40\%\)) and Y (\(25\%\)), where the weights are the proportions of X and Y in the mixture.
3. Create the mixture equation: Express that the total ryegrass comes from ryegrass contributed by mixture X plus ryegrass contributed by mixture Y, equaling \(30\%\) of the total mixture.
4. Solve for the proportion of X: Use algebra to find what fraction of the total mixture must be X to achieve the target \(30\%\) ryegrass content.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have in plain English:

• Mixture X contains \(40\%\) ryegrass (and \(60\%\) bluegrass)
• Mixture Y contains \(25\%\) ryegrass (and \(75\%\) fescue)
• We want to combine X and Y to create a new mixture that has exactly \(30\%\) ryegrass
• We need to find what percentage of this combined mixture should be X

Think of it like mixing two different coffee blends with different caffeine strengths to get a specific target strength. We're doing the same thing but with ryegrass percentages.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Set up the weighted average relationship

Here's the key insight in everyday language: when we mix two things together, the final percentage of any component will be somewhere between the percentages in the original mixtures.

Since X has \(40\%\) ryegrass and Y has \(25\%\) ryegrass, our final mixture with \(30\%\) ryegrass makes sense because \(30\%\) is between \(25\%\) and \(40\%\).

The closer our final percentage is to one of the original mixtures, the more of that mixture we need. Since \(30\%\) is closer to \(25\%\) (Y's percentage) than to \(40\%\) (X's percentage), we should expect to need more of mixture Y than mixture X.

This weighted average relationship means:
Final percentage = (Fraction of X) × (X's percentage) + (Fraction of Y) × (Y's percentage)

3. Create the mixture equation

Let's say X makes up some fraction of our final mixture. If we call this fraction 'p', then Y makes up the remaining fraction, which is (1-p).

In plain English: if X is \(30\%\) of our mixture, then Y must be the remaining \(70\%\) of our mixture.

Now we can write our equation using the weighted average principle:
\(30\% = \mathrm{p} \times 40\% + (1-\mathrm{p}) \times 25\%\)

This equation captures the logic that the ryegrass in our final mixture comes from two sources:
• The ryegrass contributed by mixture X: \(\mathrm{p} \times 40\%\)
• The ryegrass contributed by mixture Y: \((1-\mathrm{p}) \times 25\%\)

And these two contributions must add up to exactly \(30\%\) ryegrass in the final mixture.

4. Solve for the proportion of X

Now let's solve our equation step by step:

\(30 = \mathrm{p} \times 40 + (1-\mathrm{p}) \times 25\)

First, let's distribute the 25:
\(30 = 40\mathrm{p} + 25 - 25\mathrm{p}\)

Combine like terms:
\(30 = 15\mathrm{p} + 25\)

Subtract 25 from both sides:
\(5 = 15\mathrm{p}\)

Divide both sides by 15:
\(\mathrm{p} = \frac{5}{15} = \frac{1}{3}\)

Converting to a percentage: \(\frac{1}{3} = 33\frac{1}{3}\%\)

Let's verify this makes sense: if X is \(\frac{1}{3}\) of the mixture and Y is \(\frac{2}{3}\), then:
Ryegrass percentage = \(\frac{1}{3} \times 40\% + \frac{2}{3} \times 25\% = \frac{40}{3} + \frac{50}{3} = \frac{90}{3} = 30\%\) ✓

5. Final Answer

The mixture must contain \(33\frac{1}{3}\%\) of mixture X by weight.

This corresponds to answer choice B. \(33\frac{1}{3}\%\).

Our answer makes intuitive sense: since we wanted \(30\%\) ryegrass (closer to Y's \(25\%\) than to X's \(40\%\)), we needed more of mixture Y (\(66\frac{2}{3}\%\)) than mixture X (\(33\frac{1}{3}\%\)).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what the question is asking for

Students often confuse whether they need to find the percentage of mixture X in the final mixture OR the percentage of ryegrass that comes from mixture X. The question asks "what percent of the weight of the mixture is X?" but students might mistakenly try to find what percentage of the total ryegrass comes from X.

2. Setting up the wrong baseline for percentages

Students may get confused about whether the ryegrass percentages (\(40\%\), \(25\%\), \(30\%\)) refer to percentages within each individual mixture or percentages of the total final mixture weight. This confusion can lead to setting up incorrect equations where they mix absolute weights with relative percentages.

3. Incorrectly assuming equal amounts of X and Y

Some students jump to the conclusion that since we're mixing two substances, we must use equal amounts (\(50\%\) each). They fail to recognize that this is a weighted average problem where the proportions need to be calculated, not assumed.

Errors while executing the approach

1. Algebraic manipulation errors when expanding (1-p)

When expanding the term \((1-\mathrm{p}) \times 25\), students commonly make sign errors or forget to distribute properly. For example, writing \((1-\mathrm{p}) \times 25 = 25 - \mathrm{p}\) instead of \(25 - 25\mathrm{p}\), which leads to completely wrong equations.

2. Fraction and percentage conversion mistakes

Students often struggle with converting between fractions and percentages. When they get \(\mathrm{p} = \frac{1}{3}\), they might incorrectly convert this to \(30\%\) instead of \(33\frac{1}{3}\%\), or they might write \(33\%\) instead of \(33\frac{1}{3}\%\).

3. Verification calculation errors

Even when students get the right answer (\(\frac{1}{3}\)), they may make arithmetic errors during verification. For instance, calculating \(\frac{1}{3} \times 40\% + \frac{2}{3} \times 25\%\) incorrectly, which could make them doubt their correct answer and change it.

Errors while selecting the answer

1. Selecting the wrong percentage format

Students who correctly calculate \(\mathrm{p} = \frac{1}{3} \approx 0.333...\) might select answer choice A (\(10\%\)) or D (\(50\%\)) because they misread the decimal or made a last-minute conversion error between the fraction and percentage forms.

2. Choosing the complement instead of the answer

Since the problem involves two mixtures, students might correctly calculate that X makes up \(33\frac{1}{3}\%\) but then mistakenly select E (\(66\frac{2}{3}\%\)) thinking the question asked for the percentage of mixture Y instead of mixture X.

Alternate Solutions

Smart Numbers Approach

This mixture problem is well-suited for smart numbers because we can assign a convenient total weight to our final mixture and work with concrete amounts rather than percentages.

Step 1: Choose a smart number for total mixture weight
Let's say our final mixture weighs 100 units total. This makes percentage calculations straightforward since each unit represents \(1\%\).

Step 2: Set up what we know about the final mixture
Since the final mixture is \(30\%\) ryegrass and weighs 100 units:
Total ryegrass in final mixture = 30 units

Step 3: Define variables using our smart number
Let \(\mathrm{x}\) = weight of mixture X in our final mixture
Then \((100 - \mathrm{x})\) = weight of mixture Y in our final mixture

Step 4: Calculate ryegrass contribution from each mixture
Ryegrass from mixture X = \(40\%\) of \(\mathrm{x} = 0.4\mathrm{x}\) units
Ryegrass from mixture Y = \(25\%\) of \((100 - \mathrm{x}) = 0.25(100 - \mathrm{x})\) units

Step 5: Set up the equation
Total ryegrass = Ryegrass from X + Ryegrass from Y
\(30 = 0.4\mathrm{x} + 0.25(100 - \mathrm{x})\)

Step 6: Solve for x
\(30 = 0.4\mathrm{x} + 25 - 0.25\mathrm{x}\)
\(30 = 0.15\mathrm{x} + 25\)
\(5 = 0.15\mathrm{x}\)
\(\mathrm{x} = 5 \div 0.15 = 5 \div \frac{3}{20} = 5 \times \frac{20}{3} = \frac{100}{3} = 33\frac{1}{3}\)

Step 7: Interpret the result
Since \(\mathrm{x} = 33\frac{1}{3}\) units and our total mixture is 100 units, mixture X represents \(33\frac{1}{3}\%\) of the total weight.

Verification:
Mixture X: \(33\frac{1}{3}\) units with \(40\%\) ryegrass = \(33\frac{1}{3} \times 0.4 = 13\frac{1}{3}\) units of ryegrass
Mixture Y: \(66\frac{2}{3}\) units with \(25\%\) ryegrass = \(66\frac{2}{3} \times 0.25 = 16\frac{2}{3}\) units of ryegrass
Total ryegrass: \(13\frac{1}{3} + 16\frac{2}{3} = 30\) units ✓
Percentage: \(\frac{30}{100} = 30\%\) ✓

The answer is B. \(33\frac{1}{3}\%\)