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A photography dealer ordered 60 Model X cameras to be sold

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

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New post 28 Jun 2016, 05:50
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


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:

Revenue = 54 x $250 = $13,500

Since profit = revenue – cost, profit = $13,500 – $11,970 = $1,530

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

(Profit/initial cost) x 100

1,530/12,600 x 100

153/1260 x 100

This is roughly equal to:

150/1200 x 100

15/120 x 100

1/8 x 100 = 12.5% profit

The closest answer is answer choice D, 13% profit.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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New post 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

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/


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.

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

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New post 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)

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

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New post 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%

Answer: D.

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

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New post 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%

Answer: D.


Hi aayushamj,

Given Marked price = CP + 20%of CP = CP + 0.2CP = 1.2CP = $ 250 => CP = $250/1.2

Total cost price = (60*250)/1.2 = 50*250 = $12500

The selling price has two components: Cameras sold and Cameras returned.

Revenue from cameras sold = 54*250 = 13500

Quote:
6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost.

Please note that initial cost of each camera is same as CP, i.e. $250/1.2
Revenue from cameras returned =\(\frac{6\times 250 \times 0.5}{1.2}\) = $625

Total income = 13500 + 625 = $14125

Profit = 14125 - 12500 = $1625

Profit percentage = 1625*100/12500 = 13 %.

Hope this helps.

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

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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%

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

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New post 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

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/


VeritasPrepKarishma, I think you method is pretty sweet - I just used another weighted average looking formula you shared in your blog:

\(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2-Aavg \right) }{ \left( Aavg-A1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2-Aavg \right) }{ \left( Aavg-0.5 \right) } \\ =Aavg-0.5=9\left( 1.2-Aavg \right) \\ =Aavg-0.5=10.8-9Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\)

13% profit

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

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New post 02 Apr 2017, 07:04
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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

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

Answer:
[Reveal] Spoiler:
D


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

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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

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/


VeritasPrepKarishma, I think you method is pretty sweet - I just used another weighted average looking formula you shared in your blog:

\(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2-Aavg \right) }{ \left( Aavg-A1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2-Aavg \right) }{ \left( Aavg-0.5 \right) } \\ =Aavg-0.5=9\left( 1.2-Aavg \right) \\ =Aavg-0.5=10.8-9Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\)

13% profit

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

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New post 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.

Total price to dealer is 12,480.

2893/12,480*100=23%

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

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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.

Total price to dealer is 12,480.

2893/12,480*100=23%


Hi Cez005,

Please note that when you return 6 cameras at 50% of CP, you are making a loss on these 6 pieces. So, in your method
Net profit = 2268 - 625 = 1643

Profit percentage = 1643/12480*100 = \(\approx\)13%.

Hope it helps.

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

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New post 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)

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

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New post 16 Jun 2017, 18:54
Here is wha I did for this question ->


Initial cost price => x (let)
1.2x=250
x=208(approx)
Total Selling price => 44*250 + 6* 104 => 11624
Total cost price => 50*208 => 10400

Profit => 1224

Profit percent => 1224/10400. * 100 => 11.76 percent


The closet value is 13 percent


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

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New post 28 Jun 2017, 06:58
Hello diegocml,

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

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/


VeritasPrepKarishma, I think you method is pretty sweet - I just used another weighted average looking formula you shared in your blog:

\(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2-Aavg \right) }{ \left( Aavg-A1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2-Aavg \right) }{ \left( Aavg-0.5 \right) } \\ =Aavg-0.5=9\left( 1.2-Aavg \right) \\ =Aavg-0.5=10.8-9Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\)

13% profit

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'.

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

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susheelh wrote:
Hello diegocml,

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 :)






This formula is an application of the scale method:

\(\frac{w1}{w2} = \frac{A2 - Aavg}{Aavg - A1}\)

It is faster and more direct than actually using the number line.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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New post 05 Aug 2017, 12:09
Wow! What a great explanation! First time I see someone's explanation better than Bunuel's explanation :)) Bravo!

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)

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

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New post 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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Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 16 Sep 2017, 10:45

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