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Quicker way

This topic has 4 expert replies and 0 member replies
eitijan Master | Next Rank: 500 Posts Default Avatar
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Quicker way

Post Mon Apr 25, 2016 4:55 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 took me a lot of time to solve this question involving all the calculations. Is there any quicker way to solve it?

Experts, please comment.
OA D

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GMAT/MBA Expert

Post Wed Aug 23, 2017 4:37 pm
eitijan 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
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

Cheers,
Brent

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Matt@VeritasPrep GMAT Instructor
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Post Wed Apr 27, 2016 1:49 pm
Since the answers are in terms of percentages, just stick with percentages and ignore the actual cost.

Suppose cost per camera = c. Then the dealer's cost = 60 * c.

The gross is 54*.2c - 6*.5c => 7.8c. (We made 20% of c on 54 cameras, and lost 50% of c on 6 cameras.)

That leaves us with a Profit/Cost ratio of 7.8c / 60c, or 7.8/60, or 13%.

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Post Mon Apr 25, 2016 5:08 am
eitijan 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
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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Post Mon Apr 25, 2016 5:09 am
Alternate 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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