pgmat wrote:
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?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Approach 1:
54 cameras yield a 20% profit, while 6 cameras suffer a 50% loss.
Average for all 60 cameras \(= \frac{(54*20 - (6*50)}{60} = \frac{780}{60} = 13\)
Approach 2:
Since the question stem asks for a PERCENTAGE, ignore the $250 selling price.
Plug in values that satisfy the given conditions and make the math easy.
Fraction of cameras not sold \(= \frac{6}{60} = \frac{1}{10}\)
Let the number of cameras ordered \(= 10\)
Let the cost per camera \(= 10\)
Total cost \(= 10*10 = 100\)
Number not sold \(= \frac{1}{10}*10 = 1\)
For this one camera, the refund received \(= 5*10 =5\)
Number of cameras sold \(= 9\)
With a markup of 20%, the selling price per camera \(= 12\)
Total revenue \(= 9*12 = 108\)
Refund + revenue = 5+108 = 113, a profit of 13%.
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