dave13 wrote:
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
pushpitkc,
niks18,
generis, yes its me again
Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this
"60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following
Cost of one camera \(1.20x = 250\)
\(x\) = \(208\)
Hence Mark up = \(42\)
Now 54 were sold ---> \(Profit = 42*54 = 2,268\)
6 cameras unsold and returned at 0.50% of its cost = \(104*6 = 624\)
\(Total Profit\) = \(2268-624 = 1,644\)
Dealers Initial Cost for the 60 cameras = \(208*60 = 12,480\)
So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is \(\frac{1,644}{12,480}\) = \(0.13 %\)
Thank you
Hi
dave13Yes your method is absolutely correct
As a shortcut I would advice you to work with smart numbers. you actually don't need 250 here.
Let the CP=100, so total cost = 100*60=6000
mark-up=20, hence Profit = 20*54=1080
loss on 6 cameras = 50*6=300
Hence net profit = 1080-300=780
therefore profit % = 780/6000=0.13
The calculation becomes very simple when you chose smart numbers such as 100