- 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.
- 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.
- 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.
- 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.
