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A dealer originally bought 100 identical batteries at a total cost of \(\mathrm{q}\) dollars. If each battery was sold at \(50\) percent above the original cost per battery, then, in terms of \(\mathrm{q}\), for how many dollars was each battery sold?
Let's start by understanding what we know and what we need to find in plain English.
What we know:
What we need to find:
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
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!
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 = \(\frac{3}{2} \times \frac{\mathrm{q}}{100} = \frac{3\mathrm{q}}{200}\)
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:
Looking at our answer choices, \(\frac{3\mathrm{q}}{200}\) matches choice (A) exactly.
The selling price per battery is \(\frac{3\mathrm{q}}{200}\) dollars, which corresponds to answer choice (A).
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."
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 \(\frac{1.5\mathrm{q}}{100}\).
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.
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.