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Arithmetic Percents

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Arithmetic Percents

by datonman » Fri Nov 27, 2015 11:44 am

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A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20% 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% 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
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by Brent@GMATPrepNow » Fri Nov 27, 2015 11:49 am
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
It's important to recognize that we really don't need to use the information about the cameras selling for $250 each. The question boils down to . . .

54 cameras were sold at a 20% markup, and 6 cameras were (essentially) sold at a 50% markdown. What was the approximate profit or loss as a percent of the dealer's initial cost for all 60 cameras?

So, we can assign A NICE value of $100 to the initial cost per camera.
This means the 60 cameras cost $6000 to buy.

54 cameras were sold at a 20% markup and 6 cameras were sold at a 50% markdown.
So, 54 cameras were sold for $120, and 6 cameras were sold for $50.
(54)($120) + (6)($50) = $6780
So, the cameras were sold for $6780

This represents a profit of $780 (eliminate A and B)

If the initial cost was $6000, we must determine the percentage equivalent to $780/$6000

$780/$6000 = 78/600 = 13/100 = 13%

Answer = D

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by GMATGuruNY » Fri Nov 27, 2015 1:57 pm
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
54 cameras yield a 20% profit, while 6 cameras suffer a 50% loss.
Average for all 60 cameras = (54*20 - 6*50)/60 = 780/60 = 13.

The correct answer is D.
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by jain2016 » Sat Jan 02, 2016 4:07 am
Hi Experts ,

Can you please check and advise what will be the next.

S.P.( Selling Price)= $250

Profit %= 20%

Profit in $ = 20/100*250 = $50

so that means C.P.( Cost rice) = $200

Total S.P. of 54 camera = 54*250 = 13500

Now I don't understand this line 50 percent of the dealer's initial cost

Please advise .

Many thanks in advance.

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by eitijan » Mon Apr 25, 2016 5:07 am
Hi Mitch

Your method seems to be very quick.
But I am not able to understand the logic on this. Could you please elaborate.
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by GMATGuruNY » Mon Apr 25, 2016 5:31 am
eitijan wrote:Hi Mitch

Your method seems to be very quick.
But I am not able to understand the logic on this. Could you please elaborate.
This is a MIXTURE problem.

The 54 sold cameras each yield a profit of 20%.
Total profit for the 54 sold cameras = (54)(20).

Since the dealer receives only a 50% refund for the 6 unsold cameras, the 6 unsold cameras each suffer a loss of -50%.
Total loss for the 6 unsold cameras = (6)(-50).

The total profit for the 54 sold cameras is combined with the total loss for the 6 unsold cameras to form a MIXTURE of 60 cameras.
Total profit or loss for the mixture of 60 cameras = 54*20 - 6*50.

To determine the dealer's profit or loss, we calculate the AVERAGE profit or loss for the MIXTURE of 60 cameras.
Average profit or loss for the mixture of 60 cameras = (total profit or loss for the mixture)/(number of cameras) = (54*20 - 6*50)/60 = 13.
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by Matt@VeritasPrep » Wed Apr 27, 2016 1:52 pm
jain2016 wrote:
Now I don't understand this line 50 percent of the dealer's initial cost
Suppose the dealer paid $250 for each of these six cameras. Unable to sell them, the dealer asks the supplier for a refund. The supplier agrees to refund 50% of the cost, or $125 each, for the six cameras.

So the dealer LOSES $125 for each camera (... which is presumably better than just leaving the cameras in storage somewhere and losing the full $250 on each of them.)
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