The original price of a certain TV set is discounted by x percent, and the reduced price is then discounted...

1. Translate the problem requirements: We need to find the final price after two successive discounts - first a discount of \(\mathrm{x}\) percent, then a discount of \(\mathrm{2x}\) percent applied to the already reduced price. The key insight is that successive discounts are applied to the running total, not the original price.
2. Apply the first discount systematically: Calculate the price after the \(\mathrm{x}\) percent discount by finding what percentage remains \((100-\mathrm{x})\) percent and expressing this as a decimal multiplier.
3. Apply the second discount to the reduced price: Apply the \(\mathrm{2x}\) percent discount to the price from step 2, again using the percentage that remains \((100-\mathrm{2x})\) percent as a decimal multiplier.
4. Expand and simplify the expression: Multiply out the terms and convert all percentages to decimal form to match the answer choice format.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English. We start with a TV that costs \(\mathrm{P}\) dollars. Then two things happen in sequence:

• First, the store gives us a discount of \(\mathrm{x}\) percent off the original price
• Then, they give us another discount of \(\mathrm{2x}\) percent, but this second discount is applied to the already-reduced price, not the original price

The key insight here is understanding that successive discounts work differently than adding discounts together. When you get a second discount, it's calculated based on what you're already paying after the first discount, not on the original price.

Think of it like this: if you have a $100 item with a 10% discount, you pay $90. If there's then a 20% discount on top, you get 20% off the $90 (which is $18 off), not 20% off the original $100.

Process Skill: TRANSLATE - Converting the problem language into mathematical understanding

2. Apply the first discount systematically

Let's work through the first discount step by step. When something is discounted by \(\mathrm{x}\) percent, we pay \((100 - \mathrm{x})\) percent of the original price.

For example, if \(\mathrm{x} = 15\), then a 15% discount means we pay 85% of the original price.

In mathematical terms, if we discount \(\mathrm{P}\) by \(\mathrm{x}\) percent:

• The percentage we pay = \((100 - \mathrm{x})\%\)
• Converting to decimal form = \(\frac{(100 - \mathrm{x})}{100} = 1 - \frac{\mathrm{x}}{100} = 1 - 0.01\mathrm{x}\)
• Price after first discount = \(\mathrm{P} \times (1 - 0.01\mathrm{x})\)

3. Apply the second discount to the reduced price

Now we need to apply a \(\mathrm{2x}\) percent discount to our already-reduced price from step 2.

The price we're starting with for the second discount is: \(\mathrm{P}(1 - 0.01\mathrm{x})\)

When we discount this by \(\mathrm{2x}\) percent, we pay \((100 - \mathrm{2x})\) percent of this reduced price:

• The percentage we pay = \((100 - \mathrm{2x})\%\)
• Converting to decimal = \(\frac{(100 - \mathrm{2x})}{100} = 1 - \frac{\mathrm{2x}}{100} = 1 - 0.02\mathrm{x}\)
• Final price = [Price after first discount] × \((1 - 0.02\mathrm{x})\)
• Final price = \(\mathrm{P}(1 - 0.01\mathrm{x}) \times (1 - 0.02\mathrm{x})\)

4. Expand and simplify the expression

Now we need to multiply out the expression \(\mathrm{P}(1 - 0.01\mathrm{x})(1 - 0.02\mathrm{x})\) and simplify:

\(\mathrm{P}(1 - 0.01\mathrm{x})(1 - 0.02\mathrm{x})\)

Let's expand the terms inside the parentheses using the distributive property:

\((1 - 0.01\mathrm{x})(1 - 0.02\mathrm{x}) = 1 \times 1 + 1 \times (-0.02\mathrm{x}) + (-0.01\mathrm{x}) \times 1 + (-0.01\mathrm{x}) \times (-0.02\mathrm{x})\)

\(= 1 - 0.02\mathrm{x} - 0.01\mathrm{x} + (0.01\mathrm{x})(0.02\mathrm{x})\)

\(= 1 - 0.03\mathrm{x} + 0.0002\mathrm{x}^2\)

Therefore, our final answer is: \(\mathrm{P}(1 - 0.03\mathrm{x} + 0.0002\mathrm{x}^2)\)

Process Skill: MANIPULATE - Expanding algebraic expressions systematically

5. Final Answer

The price of the television set after the two successive discounts is \(\mathrm{P}(1 - 0.03\mathrm{x} + 0.0002\mathrm{x}^2)\)

Looking at our answer choices, this matches Answer Choice B exactly.

We can verify this makes sense: the 1 represents 100% of the original price, the \(-0.03\mathrm{x}\) term represents the combined effect of both discounts \((0.01\mathrm{x} + 0.02\mathrm{x})\), and the \(+0.0002\mathrm{x}^2\) term represents the "discount on the discount" - the small amount we save because the second discount is applied to an already-reduced price rather than the original price.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding successive discounts

Students often think that successive discounts can be simply added together. They might incorrectly assume that an \(\mathrm{x}\%\) discount followed by a \(\mathrm{2x}\%\) discount equals a total discount of \((\mathrm{x} + \mathrm{2x})\% = \mathrm{3x}\%\). This leads them to think the final price is \(\mathrm{P}(1 - 0.03\mathrm{x})\), missing the crucial interaction between the two discounts.

Faltering Point 2: Applying the second discount to the original price

Students frequently misinterpret "the reduced price is then discounted by \(\mathrm{2x}\) percent" and apply the \(\mathrm{2x}\%\) discount to the original price \(\mathrm{P}\) instead of to the already-reduced price from the first discount. This fundamental misunderstanding of the sequential nature of the problem leads to an entirely incorrect setup.

Errors while executing the approach

Faltering Point 1: Decimal conversion errors

When converting percentages to decimals, students often make mistakes such as writing \(\mathrm{x}\%\) as \(0.1\mathrm{x}\) instead of \(0.01\mathrm{x}\), or \(\mathrm{2x}\%\) as \(0.2\mathrm{x}\) instead of \(0.02\mathrm{x}\). These conversion errors propagate through the entire calculation and lead to incorrect final expressions.

Faltering Point 2: Algebraic expansion mistakes

When expanding \((1 - 0.01\mathrm{x})(1 - 0.02\mathrm{x})\), students commonly make errors in the multiplication, particularly with the quadratic term. They might incorrectly calculate \((0.01\mathrm{x})(0.02\mathrm{x})\) as \(0.02\mathrm{x}^2\) instead of \(0.0002\mathrm{x}^2\), or forget to include the positive sign for this term when combining like terms.

Faltering Point 3: Sign errors in combining terms

Students often make sign errors when combining the linear terms \(-0.01\mathrm{x}\) and \(-0.02\mathrm{x}\), sometimes getting \(-0.01\mathrm{x}\) instead of \(-0.03\mathrm{x}\), or incorrectly handling the sign of the quadratic term in the final expression.

Errors while selecting the answer

Faltering Point 1: Coefficient magnitude confusion

Students who arrive at the correct form \(\mathrm{P}(1 - 0.03\mathrm{x} + 0.0002\mathrm{x}^2)\) might select answer choice A [\(\mathrm{P}(1 - 0.03\mathrm{x} + 0.02\mathrm{x}^2)\)] because they recognize the \(-0.03\mathrm{x}\) term but fail to carefully check the coefficient of \(\mathrm{x}^2\), missing that \(0.0002 \neq 0.02\). The similar structure of the expressions can lead to hasty, incorrect selections.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for x

Let's choose \(\mathrm{x} = 10\). This gives us clean percentage calculations since 10% and 20% are easy to work with.

With \(\mathrm{x} = 10\):

• First discount = 10%
• Second discount = \(\mathrm{2x}\% = 20\%\)

Step 2: Apply the discounts to a concrete original price

Let's set the original price \(\mathrm{P} = \$100\) for easy percentage calculations.

Step 3: Calculate the price after the first discount

First discount of 10% means the customer pays 90% of the original price:

Price after first discount = \(\$100 \times 0.90 = \$90\)

Step 4: Calculate the price after the second discount

Second discount of 20% is applied to the already reduced price of $90:

Price after second discount = \(\$90 \times 0.80 = \$72\)

Step 5: Test our result against the answer choices

With \(\mathrm{P} = 100\) and \(\mathrm{x} = 10\), let's evaluate each answer choice:

Choice B: \(\mathrm{P}(1 - 0.03\mathrm{x} + 0.0002\mathrm{x}^2)\)

\(= 100(1 - 0.03(10) + 0.0002(10)^2)\)

\(= 100(1 - 0.30 + 0.0002(100))\)

\(= 100(1 - 0.30 + 0.02)\)

\(= 100(0.72) = \$72\) ✓

This matches our calculated result, confirming that Choice B is correct.

We can verify that the other choices don't yield $72 when we substitute \(\mathrm{P} = 100\) and \(\mathrm{x} = 10\), but Choice B gives us exactly the result we calculated step-by-step.