Bob expected to spend a total of $9.00 to buy a given amount of pasta salad at a fixed price...

1. Translate the problem requirements: Bob planned to spend $9 for pasta salad at an expected price per pound. The actual price was $0.20 more per pound than expected. With $9, he bought 0.5 pounds less than planned. Find the actual price per pound.
2. Set up variables for expected vs actual scenarios: Define the expected price per pound and use it to express both the expected quantity and actual quantity Bob could buy with $9.
3. Create equation from the quantity difference: Since Bob bought 0.5 pounds less than expected, set up an equation comparing the expected quantity minus actual quantity equals 0.5.
4. Solve for the expected price and calculate actual price: Solve the equation to find Bob's expected price per pound, then add $0.20 to get the actual price he paid.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what happened to Bob in everyday language:

• Bob had a plan: spend exactly $9.00 on pasta salad at what he thought was the regular price
• Reality check: the pasta salad cost $0.20 more per pound than he expected
• The result: he still spent his $9.00, but got 0.5 pounds less pasta salad than he planned

We need to find the actual price Bob paid per pound (which was $0.20 more than expected).

Process Skill: TRANSLATE - Converting the story into clear mathematical relationships

2. Set up variables for expected vs actual scenarios

Let's use simple language to set up our unknowns:

• Let the expected price per pound = \(\mathrm{x}\) dollars
• Then the actual price per pound = \(\mathrm{x} + 0.20\) dollars

Now, since Bob had $9.00 to spend:

• Expected quantity he could buy = $9.00 ÷ \(\mathrm{x}\) = \(\frac{9}{\mathrm{x}}\) pounds
• Actual quantity he bought = $9.00 ÷ \(\(\mathrm{x} + 0.20\)\) = \(\frac{9}{\mathrm{x} + 0.20}\) pounds

Both scenarios involve the same $9.00, but different prices per pound lead to different quantities.

3. Create equation from the quantity difference

The key insight is that Bob got 0.5 pounds less than he planned:

In plain English: Expected quantity - Actual quantity = 0.5 pounds

Translating this: \(\frac{9}{\mathrm{x}} - \frac{9}{\mathrm{x} + 0.20} = 0.5\)

To solve this, let's find a common denominator for the fractions:

• \(\frac{9}{\mathrm{x}} - \frac{9}{\mathrm{x} + 0.20} = 0.5\)
• \(\frac{9(\mathrm{x} + 0.20) - 9\mathrm{x}}{\mathrm{x}(\mathrm{x} + 0.20)} = 0.5\)
• \(\frac{9\mathrm{x} + 1.8 - 9\mathrm{x}}{\mathrm{x}(\mathrm{x} + 0.20)} = 0.5\)
• \(\frac{1.8}{\mathrm{x}(\mathrm{x} + 0.20)} = 0.5\)

4. Solve for the expected price and calculate actual price

Continuing from our equation: \(\frac{1.8}{\mathrm{x}(\mathrm{x} + 0.20)} = 0.5\)

Cross multiply: \(1.8 = 0.5 \times \mathrm{x}(\mathrm{x} + 0.20)\)
\(1.8 = 0.5\mathrm{x}(\mathrm{x} + 0.20)\)
\(3.6 = \mathrm{x}(\mathrm{x} + 0.20)\)
\(3.6 = \mathrm{x}^2 + 0.20\mathrm{x}\)
\(\mathrm{x}^2 + 0.20\mathrm{x} - 3.6 = 0\)

Using the quadratic formula or factoring:
\(\mathrm{x}^2 + 0.20\mathrm{x} - 3.6 = 0\)

Multiplying by 100 to avoid decimals: \(100\mathrm{x}^2 + 20\mathrm{x} - 360 = 0\)
Dividing by 20: \(5\mathrm{x}^2 + \mathrm{x} - 18 = 0\)

Factoring: \((5\mathrm{x} - 9)(\mathrm{x} + 2) = 0\)
So \(\mathrm{x} = \frac{9}{5} = 1.8\) or \(\mathrm{x} = -2\)

Since price must be positive, the expected price was \(\mathrm{x} = \$1.80\) per pound.

Therefore, the actual price Bob paid = $1.80 + $0.20 = $2.00 per pound

Final Answer

Bob spent $2.00 per pound for the salad.

Verification:

• Expected: $9.00 ÷ $1.80 = 5 pounds
• Actual: $9.00 ÷ $2.00 = 4.5 pounds
• Difference: 5 - 4.5 = 0.5 pounds ✓

The answer is E: $2.00

Common Faltering Points

Errors while devising the approach

• Misinterpreting "12 pound less": Students might read this as "1/2 pound less" instead of "0.5 pounds less," leading to an incorrect constraint equation. The fraction 1/2 in the problem statement refers to half a pound, which equals 0.5 pounds.
• Setting up the wrong relationship: Students may confuse which price to solve for, setting up their variable as the actual price instead of the expected price, making it harder to track the $0.20 difference correctly.
• Incorrect quantity relationship: Students might set up the equation as actual quantity - expected quantity = 0.5 instead of expected quantity - actual quantity = 0.5, since Bob got less than he expected.

Errors while executing the approach

• Algebraic manipulation errors with fractions: When combining the fractions \(\frac{9}{\mathrm{x}} - \frac{9}{\mathrm{x} + 0.20}\), students often make errors in finding the common denominator or simplifying the numerator, particularly missing that 9x terms cancel out.
• Quadratic equation solving mistakes: Students may make arithmetic errors when cross-multiplying \((1.8 = 0.5\mathrm{x}(\mathrm{x} + 0.20))\) or when factoring the resulting quadratic equation \(5\mathrm{x}^2 + \mathrm{x} - 18 = 0\).
• Sign errors in the constraint equation: Students might incorrectly set up \(3.6 = \mathrm{x}^2 - 0.20\mathrm{x}\) instead of \(3.6 = \mathrm{x}^2 + 0.20\mathrm{x}\), leading to a completely different quadratic equation.

Errors while selecting the answer

• Selecting the expected price instead of actual price: After correctly solving for \(\mathrm{x} = \$1.80\) (the expected price), students might choose answer A ($1.80) instead of remembering that the question asks for the actual price Bob paid, which is $1.80 + $0.20 = $2.00.