If p pounds of beef cost d dollars, how many cents would k pounds of beef cost after the price...

1. Translate the problem requirements: Convert the given information into unit rates and clarify that we need the final answer in cents, not dollars
2. Establish the original unit price: Determine the cost per pound in dollars from the given information
3. Apply the price reduction: Calculate the new price per pound after the reduction in cents
4. Calculate the total cost for k pounds: Multiply the reduced price per pound by k pounds to get the final cost in cents

Execution of Strategic Approach

1. Translate the problem requirements

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:

• The price reduction is given in cents per pound
• Our final answer needs to be in cents, not dollars

This means we'll need to be careful about converting between dollars and cents throughout our solution.

Process Skill: TRANSLATE

2. Establish the original unit price

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

3. Apply the price reduction

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

4. Calculate the total cost for k pounds

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

Final Answer

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?

• Original price: \(\$10/2 = \$5\) per pound = 500 cents per pound
• New price: \(500 - 50 = 450\) cents per pound
• Cost of 3 pounds: \(450 \times 3 = 1350\) cents

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.

Common Faltering Points

Errors while devising the approach

1. Unit confusion between dollars and cents

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 represent

Students 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.

Errors while executing the approach

1. Arithmetic errors in unit conversion

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 denominators

When 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 mistakes

Students 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 \(\frac{(100\mathrm{d} - \mathrm{cp})\mathrm{k}}{\mathrm{p}}\).

Errors while selecting the answer

1. Misreading answer choice formats

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 answer

Students 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.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient smart numbers

Let's select values that make calculations clean:

• p = 10 pounds (convenient base)
• d = 20 dollars (so price per pound is $2, easy to work with)
• c = 50 cents reduction per pound
• k = 5 pounds (what we want to buy)

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.