A diet contains 300 grams of a mixture of two foods, food X and food Y. Food X contains 10...

1. Translate the problem requirements: We have 300 grams total of two foods mixed together. Food X has \(10\%\) protein, Food Y has \(15\%\) protein. The mixture gives us exactly 38 grams of protein. We need to find how many grams of Food X are in the mix.
2. Set up the mixture relationship: Express the total weight and total protein using variables, recognizing that if we have \(\mathrm{x}\) grams of Food X, then we have \((300-\mathrm{x})\) grams of Food Y.
3. Create the protein equation: Calculate total protein by adding protein contributions from both foods, setting this equal to the target 38 grams.
4. Solve for Food X amount: Solve the linear equation to find the grams of Food X in the mixture.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have in everyday terms. Think of this like making a protein shake by mixing two different protein powders.

We have:

• A total mixture weighing 300 grams
• Food X: contains \(10\%\) protein (so out of every 100 grams of Food X, 10 grams is protein)
• Food Y: contains \(15\%\) protein (so out of every 100 grams of Food Y, 15 grams is protein)
• The final mixture gives us exactly 38 grams of protein
• We want to find: how many grams of Food X are in this 300-gram mixture

This is essentially asking: "If I mix a weaker protein food (\(10\%\)) with a stronger protein food (\(15\%\)) to get a specific amount of protein, how much of each food do I need?"

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

2. Set up the mixture relationship

Here's the key insight: if the total mixture is 300 grams, and we call the amount of Food X as '\(\mathrm{x}\) grams', then the amount of Food Y must be '\((300 - \mathrm{x})\) grams'.

Why? Because Food X + Food Y = Total mixture
So: \(\mathrm{x} + \text{Food Y} = 300\)
Therefore: \(\text{Food Y} = 300 - \mathrm{x}\)

This makes sense - if I have 300 grams total and 100 grams is Food X, then 200 grams must be Food Y.

Let's define our variable:

• Let \(\mathrm{x}\) = grams of Food X in the mixture
• Then \((300 - \mathrm{x})\) = grams of Food Y in the mixture

3. Create the protein equation

Now we need to think about the protein contribution from each food.

From Food X: If we have \(\mathrm{x}\) grams of Food X, and Food X is \(10\%\) protein, then the protein from Food X = \(10\%\) of \(\mathrm{x}\) = \(0.10\mathrm{x}\) grams

From Food Y: If we have \((300 - \mathrm{x})\) grams of Food Y, and Food Y is \(15\%\) protein, then the protein from Food Y = \(15\%\) of \((300 - \mathrm{x})\) = \(0.15(300 - \mathrm{x})\) grams

Total protein = Protein from Food X + Protein from Food Y = 38 grams

So our equation becomes:
\(0.10\mathrm{x} + 0.15(300 - \mathrm{x}) = 38\)

4. Solve for Food X amount

Let's solve this step by step:

\(0.10\mathrm{x} + 0.15(300 - \mathrm{x}) = 38\)

First, let's expand the parentheses:
\(0.10\mathrm{x} + 0.15(300) - 0.15\mathrm{x} = 38\)
\(0.10\mathrm{x} + 45 - 0.15\mathrm{x} = 38\)

Combine like terms (the \(\mathrm{x}\) terms):
\((0.10 - 0.15)\mathrm{x} + 45 = 38\)
\(-0.05\mathrm{x} + 45 = 38\)

Subtract 45 from both sides:
\(-0.05\mathrm{x} = 38 - 45\)
\(-0.05\mathrm{x} = -7\)

Divide both sides by \(-0.05\):
\(\mathrm{x} = -7 \div (-0.05)\)
\(\mathrm{x} = 140\)

Let's verify this makes sense: If we have 140 grams of Food X, then we have \(300 - 140 = 160\) grams of Food Y.
Protein from Food X: \(140 \times 0.10 = 14\) grams
Protein from Food Y: \(160 \times 0.15 = 24\) grams
Total protein: \(14 + 24 = 38\) grams ✓

Final Answer

The mixture contains 140 grams of Food X.

Looking at our answer choices, this matches choice (E) 140.

The answer is (E) 140.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting what needs to be found

Students often confuse which food they need to find. The question asks for "how many grams of food X are in the mixture" but some students might set up their equation to solve for food Y instead, especially since food Y has the higher protein percentage and might seem more important.

2. Incorrectly setting up the total constraint

Students may forget that the two foods must add up to exactly 300 grams. They might set up separate equations for each food without establishing the crucial relationship that if \(\mathrm{x}\) grams is food X, then \((300-\mathrm{x})\) grams must be food Y. This leads to having two unknowns instead of one.

3. Confusing protein percentages with actual protein amounts

Students may mix up the percentage values (\(10\%\) and \(15\%\)) with actual gram amounts in their equation setup. For example, they might write "\(\mathrm{x} + (300-\mathrm{x}) = 38\)" thinking the percentages directly give protein grams, rather than understanding they need to multiply the food amounts by their respective protein percentages.

Errors while executing the approach

1. Decimal calculation errors when expanding

When expanding \(0.15(300-\mathrm{x})\), students often make arithmetic mistakes. They might calculate \(0.15 \times 300 = 4.5\) instead of 45, or incorrectly distribute to get \(0.15(300) + 0.15\mathrm{x}\) instead of \(0.15(300) - 0.15\mathrm{x}\), missing the negative sign.

2. Sign errors when combining like terms

Students frequently make errors when combining \(0.10\mathrm{x} - 0.15\mathrm{x}\). They might get \(+0.05\mathrm{x}\) instead of \(-0.05\mathrm{x}\), or calculate the subtraction incorrectly. This type of error completely changes the final answer since they would then get \(\mathrm{x} = 60\) instead of \(\mathrm{x} = 140\).

3. Division errors with decimals

When solving \(-0.05\mathrm{x} = -7\), students may struggle with dividing by the negative decimal. Common errors include forgetting the negative signs, getting the decimal placement wrong (like getting 1400 instead of 140), or incorrectly calculating \(-7 \div (-0.05)\).

Errors while selecting the answer

1. Selecting the food Y amount instead of food X

After correctly calculating that food X = 140 grams, students might mistakenly select 160 grams (which is the amount of food Y) because they lost track of what the variable \(\mathrm{x}\) represented. Since 160 appears as choice (D), this becomes a tempting wrong answer.

2. Failing to verify the answer makes sense

Students might arrive at an answer but fail to check if it's reasonable. For instance, if they made an error and got \(\mathrm{x} = 300\) or \(\mathrm{x} = 0\), they should recognize these extreme values don't make sense for a mixture problem, but may still select such answers if they appear among the choices.