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Problem solving Q182 OG 13

by verma.kumarrishikesh » Tue Sep 02, 2014 11:18 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
(0 7% profit
(D) 13% profit
(E) 15% profit
The Solution in OG is lengthy
Possible Alternatives
Answer
Target Sales =250x60= 15000
Dealers Initial Cost= 15000X100/120=12500
Actual Sales =54x250= 13500
Actual Profit/LOSS =13500-12500+Refund =13500-12500+ (6x12500/60) =1000+1250=2250
Profit % on Initial cost =2250/12500= 13%
Alternatively
10% of Cameras were returned (6 put of 60) hence
90% sold at 20% profit (Planned markup)
10% sold at 50% loss (Recovered Half of invested amount)
What is profit or loss%?
Now assume Initial Cost is 10€. And he bought 10 Camera
Total Investment = 100 €
Total Sales = 9 (90%) at 12 €. (120% of 10r€) + 1(10%) at 5€ (loss of 50% on 10€) =108+5=113
Profit =113-100=13 on Rs.100=13%
Ans is D
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by GMATGuruNY » Tue Sep 02, 2014 11:25 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
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 Brent@GMATPrepNow » Tue Sep 02, 2014 11:26 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
I think Mitch's solution using weighted averages is, by far, the best way to go here.
That said, 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 » Tue Sep 02, 2014 11:27 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
Plugging in approach:

Since the question asks for a PERCENTAGE, ignore the numbers given.
Plug in values that satisfy the given conditions and make the math easy.
The fraction of cameras not sold = 6/60 = 1/10.

Let the number of cameras ordered = 10.
Let the cost per camera = 10.
Total cost = 10*10 = 100.

The number not sold = (1/10)10 = 1.
For this one camera, the refund received = .5(10) = 5.

The number of cameras sold = 9.
With a markup of 20%, the selling price = 12.
Total revenue = 9*12 = 108.

Refund + revenue = 5+108 = 113, a profit of 13%.

The correct answer is D.
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by GMATinsight » Tue Sep 02, 2014 8:23 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
Cost + 20% of Cost = $250
i.e. 1.2 x Cost = $250
i.e. Cost of 1 camera = $250 / 1.2 = $208.3
Cost of 60 Cameras = 60 x 208.33 = $12500

Profit on 54 Camera = ($250 - $208.33) x 54 = 2250
Loss of 6 cameras = (50%) of (6 x 208.33) = 625

Total Profit = 2250 - 625 = $1625

% Profit = (1625/12500) x 100 = 13%

Answer: Option D
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by Jeff@TargetTestPrep » Fri Jul 10, 2015 4:29 am
verma.kumarrishikesh 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
(0 7% profit
(D) 13% profit
(E) 15% profit
The Solution in OG is lengthy
Possible Alternatives
Answer
Target Sales =250x60= 15000
Dealers Initial Cost= 15000X100/120=12500
Actual Sales =54x250= 13500
Actual Profit/LOSS =13500-12500+Refund =13500-12500+ (6x12500/60) =1000+1250=2250
Profit % on Initial cost =2250/12500= 13%
Alternatively
10% of Cameras were returned (6 put of 60) hence
90% sold at 20% profit (Planned markup)
10% sold at 50% loss (Recovered Half of invested amount)
What is profit or loss%?
Now assume Initial Cost is 10€. And he bought 10 Camera
Total Investment = 100 €
Total Sales = 9 (90%) at 12 €. (120% of 10r€) + 1(10%) at 5€ (loss of 50% on 10€) =108+5=113
Profit =113-100=13 on Rs.100=13%
Ans is D
Solution:

We are given that a dealer ordered 60 cameras to be sold for $250 each. This is a 20% markup over the initial cost to the dealer. We can use this to determine the initial cost to the dealer.

We can let C = dealer's cost per camera. So,

1.2C = 250

12C = 2,500

C = 2,500/12

C ≈ 210

Therefore the initial cost of the 60 cameras is 2,500/12 x 60 = 2,500 x 5 = $12,500.

Thus, we have an approximate cost of $210 per camera. (The exact cost is $208.33, but we use the more convenient number of $210 because the question asks just for an approximation.)

Of the 60 cameras, we are given that 6 were never sold. It follows that 54 were sold. Since each of these 54 cameras were sold at $250, the revenue from these 54 cameras is 250 x 54 = $13,500.

For the 6 cameras that were never sold, each was given a refund of 50 percent of the dealer's initial cost, that is, each is given a refund of ½(2,500/12) = 2,500/24. Therefore the total refund from these 6 cameras is 2,500/24 x 6 = 2,500/4 = $625.

Now, whether we think of the refund of $625 as part of the revenue or as part of the cost does not affect the profit. For example, if we think of the refund as part of the revenue, then it's an addition to the revenue. That is, our total revenue of the 60 cameras would be 13,500 + 625 = $14,125. Since our cost is $12,500, our profit would then be 14,125 - 12,500 = $1,625.

On the other hand, if we think of the refund as part of the cost, then it's a reduction to the cost. That is, our cost would be reduced by $625, so the cost would be 12,500 - 625 = $11,875. Since our revenue is $13,500, our profit would then be 13,500 - 11,875 = $1,625.

We see that the profit would be $1,625 either way.

Finally, we need to determine the dealer's approximate profit or loss as a percent of the dealer's initial cost of $12,500 for the 60 cameras.

(Profit/initial cost) x 100

(1,625/12,500) x 100 = 1,625/125 = 13

Thus, the dealer's profit was 13%.

Answer: D

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by Jim@StratusPrep » Fri Jul 10, 2015 4:38 am
It is best to think of this as a weighted average problem so that your calculations are to a minimum. The ratio of sold cameras to returned cameras is 9 : 1. This means that that weighted average will be 90% of the distance from the -50% for the returned cameras and the +20% of the distance for the sold cameras. The distance is 70 (20 - -50) and 90% of that is 63.

-50 + 63 = +13.


Know your Weighted Averages!
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