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If \(\mathrm{p}\) pounds of beef cost \(\mathrm{d}\) dollars, how many cents would \(\mathrm{k}\) pounds of beef cost after the price had been reduced by \(\mathrm{c}\) cents per pound?
Let's start by understanding what we're given and what we need to find in everyday terms.
We know that p pounds of beef cost d dollars. This tells us the relationship between weight and cost for beef at the original price.
We want to find how much k pounds of beef would cost after the price has been reduced by c cents per pound. Notice two important things:
This means we'll need to be careful about converting between dollars and cents throughout our solution.
Process Skill: TRANSLATE
To find the cost of any amount of beef, we first need to figure out the cost per pound at the original price.
If p pounds cost d dollars, then 1 pound costs \(\frac{\mathrm{d}}{\mathrm{p}}\) dollars.
For example, if 5 pounds cost $20, then 1 pound costs \(\$20/5 = \$4\).
So our original price per pound is \(\frac{\mathrm{d}}{\mathrm{p}}\) dollars per pound.
Since we'll eventually need our answer in cents, let's convert this to cents per pound:
Original price per pound = \(\frac{\mathrm{d}}{\mathrm{p}}\) dollars per pound = \(\frac{\mathrm{d}}{\mathrm{p}} \times 100\) cents per pound = \(\frac{100\mathrm{d}}{\mathrm{p}}\) cents per pound
Now we need to subtract the price reduction from our original price per pound.
The reduction is c cents per pound, and our original price is \(\frac{100\mathrm{d}}{\mathrm{p}}\) cents per pound.
New price per pound = Original price per pound - Reduction
New price per pound = \(\frac{100\mathrm{d}}{\mathrm{p}} - \mathrm{c}\) cents per pound
To combine these terms with a common denominator:
New price per pound = \(\frac{100\mathrm{d} - \mathrm{cp}}{\mathrm{p}}\) cents per pound
Now that we know the new price per pound in cents, we can find the total cost for k pounds.
Total cost = (Price per pound) × (Number of pounds)
Total cost = \(\frac{100\mathrm{d} - \mathrm{cp}}{\mathrm{p}} \times \mathrm{k}\) cents
Total cost = \(\frac{(100\mathrm{d} - \mathrm{cp})\mathrm{k}}{\mathrm{p}}\) cents
This can be written as: \(\frac{100\mathrm{d} - \mathrm{cp}}{\mathrm{p}} \times \mathrm{k}\) cents
The cost of k pounds of beef after the price reduction is \(\frac{(100\mathrm{d} - \mathrm{cp})\mathrm{k}}{\mathrm{p}}\) cents.
Looking at our answer choices, this matches choice D: \(\frac{100\mathrm{d} - \mathrm{cp}}{\mathrm{p}} \times \mathrm{k}\).
Let's verify with a quick example: If 2 pounds cost $10, and the price is reduced by 50 cents per pound, what would 3 pounds cost?
Using our formula: \(\frac{(100 \times 10 - 50 \times 2)}{2} \times 3 = \frac{(1000 - 100)}{2} \times 3 = \frac{900}{2} \times 3 = 450 \times 3 = 1350\) cents ✓
The answer is D.
Students often miss that the price reduction is given in cents per pound while the original price is in dollars per pound. They may treat both as the same unit, leading to incorrect setup. For example, they might directly subtract c from \(\frac{\mathrm{d}}{\mathrm{p}}\) without converting to common units.
2. Misunderstanding what the final answer should representStudents may not notice that the question asks for the answer in cents, not dollars. This leads them to work entirely in dollars and select an answer choice that gives a dollar amount rather than converting their final result to cents.
3. Incorrect interpretation of 'reduced by c cents per pound'Some students may interpret the price reduction as a total reduction of c cents for all k pounds, rather than c cents per pound. This would lead them to subtract c once instead of multiplying c by the weight.
When converting the original price from dollars per pound to cents per pound, students may forget to multiply by 100 or make calculation mistakes. For example, writing the original price as \(\frac{\mathrm{d}}{\mathrm{p}}\) cents per pound instead of \(\frac{100\mathrm{d}}{\mathrm{p}}\) cents per pound.
2. Incorrect algebraic manipulation when finding common denominatorsWhen combining \(\frac{100\mathrm{d}}{\mathrm{p}} - \mathrm{c}\) to get a single fraction, students often make errors. They might write it as \(\frac{100\mathrm{d} - \mathrm{c}}{\mathrm{p}}\) instead of the correct \(\frac{100\mathrm{d} - \mathrm{cp}}{\mathrm{p}}\), forgetting that c needs to be multiplied by p to have the same denominator.
3. Order of operations mistakesStudents may incorrectly place parentheses or multiply k at the wrong step, leading to expressions like \(\frac{(100\mathrm{d} - \mathrm{c})\mathrm{k}}{\mathrm{p}}\) or \(100\mathrm{d} - \frac{\mathrm{cp}}{\mathrm{p}} \times \mathrm{k}\) instead of the correct \(\frac{(100\mathrm{d} - \mathrm{cp})\mathrm{k}}{\mathrm{p}}\).
Several answer choices look very similar, differing only in the placement of terms or factors of 100. Students who have the right approach but made minor calculation errors might select choice A (missing the 100 factor) or choice E (having an extra 100 factor with cp).
2. Not verifying units in the final answerStudents may select an answer that gives a result in dollars rather than cents, not double-checking that their selected choice will yield the correct units as requested in the question.
Step 1: Choose convenient smart numbers
Let's select values that make calculations clean:
Step 2: Calculate original price per pound
Original price per pound = \(\frac{\mathrm{d}}{\mathrm{p}} = \frac{\$20}{10}\) pounds = $2 per pound
Converting to cents: $2 = 200 cents per pound
Step 3: Apply the price reduction
Reduction = 50 cents per pound
New price per pound = 200 - 50 = 150 cents per pound
Step 4: Calculate cost for k pounds
Cost for 5 pounds = 150 cents/pound × 5 pounds = 750 cents
Step 5: Verify with answer choices
Let's check which formula gives us 750:
Choice D: \(\frac{(100\mathrm{d} - \mathrm{cp})\mathrm{k}}{\mathrm{p}} = \frac{(100 \times 20 - 50 \times 10) \times 5}{10} = \frac{(2000 - 500) \times 5}{10} = \frac{1500 \times 5}{10} = \frac{7500}{10} = 750\) ✓
The smart numbers confirm that choice D is correct.