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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

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 ≈ 210

Thus, we have an approximate cost of $210 per camera.

We are next given that 6 cameras were never sold and were returned to the manufacturer for a refund of 50 percent of the dealers cost. Since the approximate cost was $210 per camera, the dealer received a $105 refund per camera, or $630 total. We can now calculate the total cost of the cameras to the dealer as:

Cost = $210 x 60 – $630

Cost = $12,600 – $630

Cost = $11,970

Next, we need to determine the revenue. We know that the dealer sold 54 cameras for $250 each. So total revenue is:

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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26 Oct 2016, 07:51

VeritasPrepKarishma 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

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

it is a great approach. Pls explain why do we take o,2*9 which is 54 items. It is said " which represents a 20 percent markup over the dealer’s initial cost for each camera". I just confuswd 20% out of 54 or 60 items. Pls explain so i will not make he same mistake again.

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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26 Oct 2016, 11:56

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

Cp of 60 cameras = 250*60/120*100 => 12500 CP of 6 cameras = 1250

6 were returned at 50% of CP, so amount received from return= 625

Net CP of 54 Cameras = Cp of 60 cameras - Amount received from return

Net CP of 54 Cameras = 12500 - 625 = 11875

SP of 54 cameras = 250*54 = 13500

So, Profit Percentage = \(\frac{( 13500 - 11875 )}{11875} *100\) ~ 13% Hence correct answer must be (D) _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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28 Feb 2017, 04:05

Bunuel,

I did not understand this solution at all. Can you please explain again. 60 Cameras were ordered Markup on CP was 20% Hence Profit would be 20%) 6 Were Returned at 50%, Hence There was a Loss of 50% on 6 Cameras, So the Profit was 20% on 54 Cameras and Loss of 50% on 6 Cameras How did You arrive at your answer from Here?

Bunuel 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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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02 Mar 2017, 10:39

aayushamj wrote:

Bunuel,

I did not understand this solution at all. Can you please explain again. 60 Cameras were ordered Markup on CP was 20% Hence Profit would be 20%) 6 Were Returned at 50%, Hence There was a Loss of 50% on 6 Cameras, So the Profit was 20% on 54 Cameras and Loss of 50% on 6 Cameras How did You arrive at your answer from Here?

Bunuel 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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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13 Mar 2017, 01:43

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Cost of each camera would be $250/1.2 =$208.33 total cost of 60 cameras would be 60*$208.33 $12500 Actual number of cameras sold is 6 less i.e 54 revenue generated 54*250 no of cameras returned 6 so refund cost is $6*208.3*50% So, total income 54*250+ 6*($208.33)*50%

The dealer's approximate profit is (54*250+ 6*($208.33)*0.5-50*250)/(50*250)*100=13%

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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02 Apr 2017, 05:45

VeritasPrepKarishma 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

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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02 Apr 2017, 07:04

Top Contributor

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

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

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

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

I'm a big fan of the scaled method but I can't figure out how to set it up as to find the 'combined average'.

Yes, you have used the scale method successfully to find the average sale price. Assuming cost price of each camera is 1, 1.2 is the sale price of 54 cameras and .5 is the sale price of 6 cameras. Hence you get the average sale price as 1.13.

Scale method is also depicted by the formula you used.

You can use the same formula with profit/loss percent and get the same answer.

w1/w2 = (A2 - Aavg)/(Aavg - A1)

1/9 = (20 - Aavg)/(Aavg - (-50))

Aavg + 50 = 9*(20 - Aavg)

10*Aavg = 130

Aavg = 13

Just that, when I have to find the average, I prefer to use the formula in the form

Aavg = (A1*w1 + A2*w2)/(w1 + w2)

It is just a realignment of the previous formula.
_________________

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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04 Apr 2017, 17:16

Bunuel 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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

Hi. Can you help point out why this method is incorrect?

Cost price is $208. So profit per camera is $42. Profit for the 54 cameras sold is 54*42=2,268. 2,268+625(refund)=2,893.

A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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04 Apr 2017, 19:49

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Cez005 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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

Hi. Can you help point out why this method is incorrect?

Cost price is $208. So profit per camera is $42. Profit for the 54 cameras sold is 54*42=2,268. 2,268+625(refund)=2,893.

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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12 Jun 2017, 22:13

Absolutely loved this approach! Probably the author will never read this, but I cant leave without saying thanks.

What I liked about this approach? - a) Identifying this is a Weighted average problem b) Using smaller manageable number (10) to ease calculations and c) Calculating profits/loss directly instead of calculating total SP and the subtracting CP.

Thanks for publishing this approach. +1 to you

MBAhereIcome wrote:

54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.

suppose total cameras = 10 and each costs $10 total profit = $2*9 -$5*1 = $13 or 13% (with $100 cost)

I hope this is what you meant by scale method (See attachment).

Based on the scale we get

\(\frac{54}{6} = \frac{(x+50)}{(20-x)}\)

\(\frac{9}{1} = \frac{(x+50)}{(20-x)}\)

\(180 - 9x = x+50\)

\(130 = 10x\)

\(x = 13\)

So, the dealer earns a profit of 13%

I learnt this approach after reading VeritasPrepKarishma's blog. Since the answer is correct, I am assuming the approach is correct

diegocml wrote:

VeritasPrepKarishma 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

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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16 Sep 2017, 10:45

Bunuel 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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

Bunuel I have a conceptual doubt. If the dealer returned 6 cameras back to the manufacturer and received only 50% refund on his initial investment cost, doesn't that mean that the dealer actually got 50% of his money back i.e. he was at loss since he lost 50% of his money- let alone the fact that he would have made profits on these 6 cameras if he had sold them? If it was a loss on his initial investments on these 6 cameras, why are we adding the $ he got refunded to his revenue. Isn't revenue specifically the $'s collected from whatever he sells? He didn't sell these 6 cameras; in fact, he suffered a loss on them.
_________________

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