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Manager  Joined: 18 Feb 2015
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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MacFauz wrote:
carcass wrote:
A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost 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 have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble. I would not show the OA but the rules are stringent. either way try on your own Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ =$12,500

Revenue for 54 cameras = 54*250 = $13,500 Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ =$625

Total Revenue = $14,125 Profit percent = $$\frac{14125-12500}{12500}$$ = 13% Answer is D. Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E. Kudos Please... If my post helped. Hey all, Why do we divide 250/2.4? Where does 2.4 come from? Thanks! Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 10226 Location: Pune, India A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags HarveyKlaus wrote: MacFauz wrote: carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost 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

have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.

I would not show the OA but the rules are stringent. either way try on your own Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ = $12,500 Revenue for 54 cameras = 54*250 =$13,500

Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ = $625 Total Revenue =$14,125

Profit percent = $$\frac{14125-12500}{12500}$$ = 13%

Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.

Kudos Please... If my post helped.

Hey all,
Why do we divide 250/2.4? Where does 2.4 come from?
Thanks!

Cost to dealer = 250/1.2 (because 250 is the selling price which 20% markup)

Revenue from the 6 cameras is half the cost price. This is = (250/1.2)*(1/2) = 250/2.4

P.S. - Saw your post later Engr2012
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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vaidhaichaturvedi 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/ Hi Karishma, Could you please explain me why there is negative sign,as in Weightage average problem has + sign in its formula. or whenever there is loss we have to take - sign in formula in numerator I may sound silly but kindly explain me. Thanks Yes, loss and discount take negative signs. Think about it - If you have two profits, one of 20% and the other of 5%, you will use (.2*9 + .5*1)/10 But you actually have a profit of 20% and a loss of 5%. How will you depict the loss? With a negative sign because it reduces the cost price. (.2*9 + (-.5)*1)/10 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Board of Directors D Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4880 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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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Bunuel,

I did not understand this solution at all.
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

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%

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

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aayushamj wrote:
Bunuel,

I did not understand this solution at all.
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

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%

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. Manager  S Status: On a 600-long battle Joined: 21 Apr 2016 Posts: 135 Location: Hungary Concentration: Accounting, Leadership Schools: Erasmus '19 GMAT 1: 410 Q18 V27 GMAT 2: 490 Q35 V23 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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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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%

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

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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) _________________ My Best is yet to come! Current Student D Joined: 12 Aug 2015 Posts: 2537 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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. _________________ Manager  G Joined: 12 Jun 2016 Posts: 200 Location: India Concentration: Technology, Leadership WE: Sales (Telecommunications) A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags Hello diegocml, I hope this is what you meant by scale method. Attachment: OG13_Q182.png [ 15.47 KiB | Viewed 1576 times ] 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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Originally posted by susheelh on 28 Jun 2017, 05:58.
Last edited by susheelh on 16 Oct 2017, 20:04, edited 1 time in total.
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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Top Contributor
Attached is a visual that should help.
Attachments Screen Shot 2017-10-16 at 7.54.24 PM.png [ 192.37 KiB | Viewed 1355 times ]

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

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MacFauz wrote:
carcass wrote:
A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost 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 have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble. I would not show the OA but the rules are stringent. either way try on your own Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ =$12,500

Revenue for 54 cameras = 54*250 = $13,500 Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ =$625

Total Revenue = $14,125 Profit percent = $$\frac{14125-12500}{12500}$$ = 13% Answer is D. Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E. Kudos Please... If my post helped. Hi MacFauz, Why did you put 2.4 in the denominator instead of 1.2 when calculating revenue from 6 cameras ? thanks! Intern  B Joined: 27 Apr 2016 Posts: 6 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags hi, can we substitute numbers? for example, I tried using the following: 10 is the cost price and therefore 12 is the sales price. Following the information given I get the following: 1- cost = 60*10 = 600 2- revenue = 54*12 = 648 3- refund = 6*5 = 30 therefore, I get the following: (648 + 30) / 600 = 1.13. Now I know that I should deduct 1 from this but not sure why. so two questions: 1. can we plug in a number for the cost? 2. why do we need to deduct 1 from the answer above (1.13)? thanks Retired Moderator D Joined: 25 Feb 2013 Posts: 1130 Location: India GPA: 3.82 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags hudhudaa wrote: hi, can we substitute numbers? for example, I tried using the following: 10 is the cost price and therefore 12 is the sales price. Following the information given I get the following: 1- cost = 60*10 = 600 2- revenue = 54*12 = 648 3- refund = 6*5 = 30 therefore, I get the following: (648 + 30) / 600 = 1.13. Now I know that I should deduct 1 from this but not sure why. so two questions: 1. can we plug in a number for the cost? 2. why do we need to deduct 1 from the answer above (1.13)? thanks Hi hudhudaa what you did in the highlighted portion is you calculated % of SP over CP i.e. SP is 113% of CP so Profit will be =113-100=13% i.e. 1.13-1 We are asked to calculate profit % so you got SP=648+30=678 And CP=600 Hence Profit = 678-600=78 Therefore Profit % = 78/600*100=13% Yes you can use substitution here and in fact SP of$250 given in the question might not be required because SP and CP are connected through a % and finally you have to calculate profit %
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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 Let x be the Original Price for Each Camera. Given, $$1.2x = 250.$$ Total $$SP = 54*1.2x + 6*(\frac{x}{2})$$ {This represents the the price which he got back) = 67.8x Total $$CP = 60x$$ $$Profit = \frac{67.8x - 60x}{60x} = 13$$ VP  D Joined: 09 Mar 2016 Posts: 1220 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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

pushpitkc, niks18, generis, yes its me again Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this "60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following Cost of one camera $$1.20x = 250$$ $$x$$ = $$208$$ Hence Mark up = $$42$$ Now 54 were sold ---> $$Profit = 42*54 = 2,268$$ 6 cameras unsold and returned at 0.50% of its cost = $$104*6 = 624$$ $$Total Profit$$ = $$2268-624 = 1,644$$ Dealers Initial Cost for the 60 cameras = $$208*60 = 12,480$$ So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is $$\frac{1,644}{12,480}$$ = $$0.13 %$$ Thank you Retired Moderator D Joined: 25 Feb 2013 Posts: 1130 Location: India GPA: 3.82 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 1 dave13 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

pushpitkc, niks18, generis, yes its me again Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this "60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following Cost of one camera $$1.20x = 250$$ $$x$$ = $$208$$ Hence Mark up = $$42$$ Now 54 were sold ---> $$Profit = 42*54 = 2,268$$ 6 cameras unsold and returned at 0.50% of its cost = $$104*6 = 624$$ $$Total Profit$$ = $$2268-624 = 1,644$$ Dealers Initial Cost for the 60 cameras = $$208*60 = 12,480$$ So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is $$\frac{1,644}{12,480}$$ = $$0.13 %$$ Thank you Hi dave13 Yes your method is absolutely correct  As a shortcut I would advice you to work with smart numbers. you actually don't need 250 here. Let the CP=100, so total cost = 100*60=6000 mark-up=20, hence Profit = 20*54=1080 loss on 6 cameras = 50*6=300 Hence net profit = 1080-300=780 therefore profit % = 780/6000=0.13 The calculation becomes very simple when you chose smart numbers such as 100 Intern  B Joined: 04 Mar 2018 Posts: 3 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags First, let's consider an easier version of the problem: If the dealer sold all 60 cameras at a profit of 20% on each camera, how much was the total profit? 20% - easy. Therefore: 20% profit per camera sold * ($$\frac{60 sold}{60 bought}$$) = 20% profit Now, the problem says that 54 cameras were sold. Using the equation above: PROFIT = 20% profit per camera sold * ($$\frac{54 sold}{60 bought}$$) = 18% profit. What about the cameras that didn't sell? Let's consider an easier problem: If the dealer returned all 60 cameras at a loss of 50% for each camera, how much was the total loss? 50% - easy. Therefore: 50% loss per camera returned * ($$\frac{60 returned}{60 bought}$$) = 50% loss. Now, the problem says that 6 cameras were returned, so let's use a similar method: LOSS = 50% loss per camera returned * ($$\frac{6 returned}{60 bought}$$) = 5% loss. Net = Profit - Loss = 18% - 5% = 13% Intern  B Joined: 06 Jun 2017 Posts: 10 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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%

Ca you share similar problems like this ? - I have searched the thread and found few but if you can link me to few more problems, it would help in firming up this concept  Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 15 Aug 2018, 02:25

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