A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup...
GMAT Word Problems : (WP) Questions
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer's initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?
- Translate the problem requirements: Find the dealer's initial cost per camera, calculate total costs and revenues for sold and returned cameras, then determine profit/loss as a percentage of total initial cost
- Determine the dealer's initial cost structure: Work backwards from the selling price and markup percentage to find the cost per camera and total initial investment
- Calculate revenues from sold cameras and refunds: Find income from the 54 cameras sold at full price plus the refund received for 6 returned cameras
- Compute profit and convert to percentage: Subtract total initial cost from total revenue, then express as a percentage of the initial cost for all 60 cameras
Execution of Strategic Approach
1. Translate the problem requirements
Let's break down what we need to find step by step. The dealer is running a camera business, and we want to know whether they made money or lost money overall, expressed as a percentage.
Here's what happened:
- The dealer ordered 60 cameras
- Each camera was sold for $250 (when sold)
- This $250 represents a 20% markup over what the dealer paid
- 6 cameras were returned and the dealer got back 50% of what they originally paid
- 54 cameras were actually sold
We need to find: \(\mathrm{(Total\,money\,in - Total\,money\,out) ÷ (Total\,money\,out) × 100\%}\)
Process Skill: TRANSLATE - Converting the business scenario into clear mathematical relationships
2. Determine the dealer's initial cost structure
To find the dealer's profit or loss, we first need to know how much they spent initially.
The selling price of $250 represents the dealer's cost plus a 20% markup. In everyday terms, if the dealer paid $100 for a camera, they would sell it for $120 (that's $100 + 20% of $100).
So if \(\$250 = \mathrm{Cost} + 20\% \,\mathrm{of}\, \mathrm{Cost}\), then:
\(\$250 = \mathrm{Cost} × (1 + 0.20) = \mathrm{Cost} × 1.20\)
Therefore: \(\mathrm{Cost\,per\,camera} = \$250 ÷ 1.20 = \$208.33\) (approximately)
\(\mathrm{Total\,initial\,investment} = 60 \,\mathrm{cameras} × \$208.33 = \$12,500\)
3. Calculate revenues from sold cameras and refunds
Now let's figure out how much money the dealer received:
Revenue from sold cameras:
- 54 cameras were sold at $250 each
- \(\mathrm{Revenue\,from\,sales} = 54 × \$250 = \$13,500\)
Refund from returned cameras:
- 6 cameras were returned
- Dealer got back 50% of original cost for each
- \(\mathrm{Refund} = 6 × (\$208.33 × 0.50) = 6 × \$104.17 = \$625\)
\(\mathrm{Total\,money\,received} = \$13,500 + \$625 = \$14,125\)
4. Compute profit and convert to percentage
Now we can calculate the dealer's profit:
\(\mathrm{Profit} = \mathrm{Total\,money\,received} - \mathrm{Total\,initial\,investment}\)
\(\mathrm{Profit} = \$14,125 - \$12,500 = \$1,625\)
Profit as a percentage of initial cost:
\(\mathrm{Profit\,percentage} = (\$1,625 ÷ \$12,500) × 100\%\)
\(\mathrm{Profit\,percentage} = 0.13 × 100\% = 13\%\)
Since this is a positive number, the dealer made a 13% profit.
4. Final Answer
The dealer's approximate profit was 13% of the initial cost for the 60 cameras.
The answer is D. 13% profit
Common Faltering Points
Errors while devising the approach
1. Misunderstanding what "20% markup" means
Students often confuse markup calculations. They might think that if the selling price is $250 with a 20% markup, then the dealer's cost is $250 - 20% of $250 = $200. This is incorrect because the 20% markup is calculated on the dealer's cost, not on the selling price. The correct relationship is: \(\mathrm{Selling\,Price} = \mathrm{Cost} × (1 + \mathrm{markup\,percentage})\).
2. Confusion about what constitutes "initial cost for the 60 cameras"
The question asks for profit/loss as a percent of the dealer's initial cost for the 60 cameras. Some students might mistakenly think this refers only to the cost of the 54 cameras that were sold, forgetting that the dealer initially invested in all 60 cameras. The denominator for the percentage calculation should be the total initial investment of $12,500, not just the cost of the sold cameras.
3. Misinterpreting the refund condition
Students might misread "refund of 50 percent of the dealer's initial cost" and think the dealer gets back 50% of what they would have made in profit, or 50% of the selling price. The correct interpretation is that the dealer gets back 50% of what they originally paid for each returned camera.
Errors while executing the approach
1. Arithmetic errors in markup calculation
When calculating the dealer's cost from \(\$250 ÷ 1.20\), students often make division errors or rounding mistakes. Some might calculate \(\$250 ÷ 1.2 = \$208.33\) incorrectly, leading to wrong values for all subsequent calculations. Even small rounding errors here get magnified through the remaining calculations.
2. Calculation errors in revenue components
Students might make errors when calculating either the revenue from sold cameras \((54 × \$250)\) or the refund amount \((6 × \$208.33 × 0.5)\). Common mistakes include using the wrong number of cameras (using 60 instead of 54 for sold cameras, or using wrong numbers due to misreading) or applying the wrong percentages.
3. Errors in percentage calculation
When computing the final percentage \((\$1,625 ÷ \$12,500 × 100\%)\), students often make errors in decimal conversion. They might calculate \(1625/12500 = 0.13\) correctly but then forget to multiply by 100 to convert to a percentage, or they might make errors in the division itself.
Errors while selecting the answer
1. Confusing profit vs. loss
Even if students calculate the correct numerical value of 13%, they might select "13% loss" instead of "13% profit" if they made a sign error in their profit calculation or misunderstood whether a positive result represents a profit or loss.
2. Selecting the wrong percentage due to calculation errors
If students made errors in earlier steps (especially in determining the dealer's cost or in calculating revenues), they might arrive at values closer to 7% or 15% and select those options instead. For instance, if they incorrectly calculated the dealer's cost as $200 per camera, their final percentage would be different.