Mitch is it possible to ballpark the % value by doing the following? -

If all the 60 cameras were sold, then Profit % = 20%

If 50 cameras were sold, then Profit % = 0%

If 55 cameras were sold, then Profit = 10% (Benchmark)

Since, more than 50 cameras were sold, a Loss is NOT possible. Eliminate A & B.

In this case, technically 57 cameras have been sold. (54 + Half Revenue for 6 cameras) So the profit % has to be > 10%. Eliminate C.

To get a 15% Profit > 57 cameras have to be sold. Eliminate E.

Only answer is 13%.

GMATGuruNY wrote:

GmatKiss 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

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.

Mitch is it possible to ballpark the % value by doing the following? -

If all the 60 cameras were sold, then Profit % = 20%

If 50 cameras were sold, then Profit % = 0%

If 55 cameras were sold, then Profit = 10% (Benchmark)

Since, more than 50 cameras were sold, a Loss is NOT possible. Eliminate A & B.

In this case, technically 57 cameras have been sold. (54 + Half Revenue for 6 cameras) So the profit % has to be > 10%. Eliminate C.

To get a 15% Profit > 57 cameras have to be sold. Eliminate E.

Only answer is 13%.

Jayanth2689 wrote:

GMATGuruNY wrote:

GmatKiss 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

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.

Mitch is it possible to ballpark the % value by doing the following? -

If all the 60 cameras were sold, then Profit % = 20%

If 50 cameras were sold, then Profit % = 0%

If 55 cameras were sold, then Profit = 10% (Benchmark)

Since, more than 50 cameras were sold, a Loss is NOT possible. Eliminate A & B.

In this case, technically 57 cameras have been sold. (54 + Half Revenue for 6 cameras) So the profit % has to be > 10%. Eliminate C.

To get a 15% Profit > 57 cameras have to be sold. Eliminate E.

Only answer is 13%.

Nice job.
One more approach:
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.

front a profit loss problem to weighted averages problem. Nice method

Honestly, how do you come up with this?

It's incredible! Thanks a lot! I've been going over the problem all night!

A big question I have is, why does following method give different answer though I believe there is no flaw in my method of attempting this problem -

Purchase price per camera = 250/1.2 = 1250/6.

Total purchase price = 60 * 1250/6 = 12500$

Got return money because of 6 cameras which were returned = 50% of 6 = [(6 * 250/1.2)/2] = 625$

Effective purchase price = 12500 - 625 = 11875$

Total sale = 54 * 250 = 13500$.

Hence % profit = [(Total sale - effective purchase price) / (effective purchase price)] x 100

= [(13500 - 11875) / 11875] x 100

= (1625/11875) x 100

= 13.684 Ans.

Dealer paid total cost of 60 cameras and then I deducted 50% cost of 6 cameras from that instead of adding it as a revenue.

So in this case, technically 54 cameras have been sold and 57 were bought.

Can anyone explain why can't I do that way?

TIA

GMATGuruNY wrote:

GmatKiss 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

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.

Dear GMATGuru,
In solving the problem, your logic here is that you treated the 6 cameras as if sold at loss. But I have another logic ( I will follow your numbers):

Net Total cost = 9 * 10 + 1 * 5 = 95..............Here, I added the half cost of 1 camera that I already paid as it is still a cost)
Total Revenue = 9 * 12 = 108
Profit =13

Profit% = 13/95 = 13. 68%

Why is there difference between both methods? where did I go wrong with my thinking?

Thanks

Mo2men wrote:Net Total cost = 9 * 10 + 1 * 5 = 95..............Here, I added the half cost of 1 camera that I already paid as it is still a cost)

where did I go wrong with my thinking?

The portion in red does not reflect the question at hand.
The question stem asks for the profit or loss as a percentage not of the NET cost but of the INITIAL cost.
Initially -- before the refund -- the dealer buys 10 cameras at a cost of $10 per camera.
Thus, the initial cost = 10*10 = $100.

GMATGuruNY wrote:

Mo2men wrote:Net Total cost = 9 * 10 + 1 * 5 = 95..............Here, I added the half cost of 1 camera that I already paid as it is still a cost)

where did I go wrong with my thinking?

The portion in red does not reflect the question at hand.
The question stem asks for the profit or loss as a percentage not of the NET cost but of the INITIAL cost.
Initially -- before the refund -- the dealer buys 10 cameras at a cost of $10 per camera.
Thus, the initial cost = 10*10 = $100.

Good hint.

Regardless of the question, why both methods give different results? shouldn't both reflect the same?
In my method, there is some cost incurred with the items returned to the manufacture with partial refund.

Mo2men wrote:Why both methods give different results? should both reflect the same?

In your solution, you calculate the following:
(profit)/(resulting cost after the refund)
In my solution, the following is calculated:
(profit)/(initial cost before the refund)
The resulting cost AFTER THE REFUND is not the same as the initial cost BEFORE THE REFUND.