A dealer originally bought 100 identical batteries at a total cost of q dollars. If each battery was sold at...

1. Translate the problem requirements: We need to find the selling price per battery when 100 batteries cost q dollars total, and each is sold at 50% above its original cost
2. Calculate the original cost per battery: Determine what each battery cost the dealer initially
3. Apply the markup to find selling price: Add 50% to the original cost per battery to get the selling price
4. Express the final answer in terms of q: Simplify the expression to match one of the given answer choices

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we know and what we need to find in plain English.

What we know:

• The dealer bought 100 batteries
• The total cost was q dollars
• Each battery is sold at \(50\%\) above its original cost

What we need to find:

• The selling price of each individual battery (in terms of q)

Think of it this way: if you bought a bulk pack of 100 batteries for some total amount, then decided to sell each one with a 50% markup, how much would you charge for each battery?

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Calculate the original cost per battery

Since the dealer paid q dollars for 100 batteries, let's figure out what each battery cost originally.

In everyday terms: if 100 items cost q dollars total, then each item costs \(\mathrm{q} \div 100\) dollars.

So the original cost per battery = \(\frac{\mathrm{q}}{100}\) dollars

Let's check this makes sense: if we had 100 batteries costing $50 total (q = 50), then each battery would cost \(\$50 \div 100 = \$0.50\). That's reasonable!

3. Apply the markup to find selling price

Now we need to add a 50% markup to the original cost per battery.

When we say "50% above the original cost," this means:
Selling price = Original cost + (50% of original cost)
Selling price = Original cost + \(0.5 \times \text{Original cost}\)
Selling price = \(\text{Original cost} \times (1 + 0.5)\)
Selling price = \(\text{Original cost} \times 1.5\)

Since the original cost per battery is \(\frac{\mathrm{q}}{100}\), the selling price per battery is:
Selling price = \(\left(\frac{\mathrm{q}}{100}\right) \times 1.5\)
Selling price = \(\frac{1.5\mathrm{q}}{100}\)

We can simplify this fraction by writing 1.5 as \(\frac{3}{2}\):
Selling price = \(\frac{3}{2} \times \frac{\mathrm{q}}{100} = \frac{3\mathrm{q}}{200}\)

4. Express the final answer in terms of q

Our final expression is \(\frac{3\mathrm{q}}{200}\).

Let's verify this makes sense with our earlier example: if q = 50 (total cost $50 for 100 batteries), then:

• Original cost per battery = \(\frac{\$50}{100} = \$0.50\)
• Selling price per battery = \(\frac{3(50)}{200} = \frac{150}{200} = \$0.75\)
• Markup check: \(\$0.75 = \$0.50 \times 1.5\) ✓

Looking at our answer choices, \(\frac{3\mathrm{q}}{200}\) matches choice (A) exactly.

Final Answer

The selling price per battery is \(\frac{3\mathrm{q}}{200}\) dollars, which corresponds to answer choice (A).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "50 percent above the original cost"
Students often confuse "50% above" with "50% of" the original cost. They might think the selling price is just 50% of the original cost \((0.5 \times \text{original cost})\) instead of 150% of the original cost \((1.5 \times \text{original cost})\). The phrase "above" means in addition to the original amount.

2. Confusion about what needs to be calculated
Students may try to find the total revenue from selling all 100 batteries instead of focusing on the selling price per individual battery. The question specifically asks "for how many dollars was each battery sold" - the key word being "each."

Errors while executing the approach

1. Arithmetic errors with fractions
When calculating \(\left(\frac{\mathrm{q}}{100}\right) \times 1.5\), students often make mistakes converting 1.5 to \(\frac{3}{2}\) or multiplying fractions. They might write \(\frac{1.5\mathrm{q}}{100}\) as \(\frac{15\mathrm{q}}{100}\) instead of \(\frac{3\mathrm{q}}{200}\), forgetting that \(1.5 = \frac{3}{2}\), not \(\frac{15}{10}\).

2. Incorrect order of operations
Students may apply the 50% markup incorrectly by calculating 50% of q first, then dividing by 100, instead of first finding the cost per battery \(\left(\frac{\mathrm{q}}{100}\right)\) and then applying the markup. This leads to expressions like \(\frac{0.5\mathrm{q}}{100}\) instead of \(\frac{1.5\mathrm{q}}{100}\).

Errors while selecting the answer

1. Not simplifying the final expression properly
Students may arrive at \(\frac{1.5\mathrm{q}}{100}\) but fail to convert it to the form \(\frac{3\mathrm{q}}{200}\) that appears in the answer choices. They might select answer choice (D) \(\frac{\mathrm{q}}{100}\) thinking it's close enough, or get confused because their expression doesn't exactly match any option at first glance.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for q

Let's set q = 100 dollars (chosen to make the arithmetic clean since we have 100 batteries)

Step 2: Calculate original cost per battery

With q = 100, the total cost is $100 for 100 batteries

Original cost per battery = \(\frac{\$100}{100} = \$1\)

Step 3: Apply the 50% markup

Selling price per battery = Original cost + 50% of original cost

Selling price per battery = \(\$1 + 0.50 \times \$1 = \$1 + \$0.50 = \$1.50\)

Step 4: Express in terms of q

We found that each battery sells for $1.50 when q = 100

So selling price = \(\$1.50 = \frac{3}{2}\) dollars

Since q = 100, we can write: \(\frac{3}{2} = \frac{3 \times 100}{200} = \frac{3\mathrm{q}}{200}\)

Step 5: Verify with answer choices

Our result \(\frac{3\mathrm{q}}{200}\) matches answer choice (A)

This smart numbers approach confirms that the selling price per battery is \(\frac{3\mathrm{q}}{200}\) dollars.