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Re: A photography dealer ordered 60 Model X cameras to be sold
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10 Mar 2016, 05:23
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 = \(\frac{1412512500}{12500}\) = 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!



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A photography dealer ordered 60 Model X cameras to be sold
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10 Mar 2016, 21:23
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 = \(\frac{1412512500}{12500}\) = 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! 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
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22 May 2016, 18:57
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
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Re: A photography dealer ordered 60 Model X cameras to be sold
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26 Oct 2016, 10:56
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
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28 Feb 2017, 03:05
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 606=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)*0.550*250)/(50*250)*100=13% Answer: D.
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A photography dealer ordered 60 Model X cameras to be sold
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02 Mar 2017, 09:39
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 606=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)*0.550*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
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02 Apr 2017, 04:45
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: \(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2Aavg \right) }{ \left( AavgA1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2Aavg \right) }{ \left( Aavg0.5 \right) } \\ =Aavg0.5=9\left( 1.2Aavg \right) \\ =Aavg0.5=10.89Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\) 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
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04 Apr 2017, 16:16
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 606=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)*0.550*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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Re: A photography dealer ordered 60 Model X cameras to be sold
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12 Jun 2017, 21:13
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
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16 Jun 2017, 17:54
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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A photography dealer ordered 60 Model X cameras to be sold
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Updated on: 16 Oct 2017, 20:04
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 \(\frac{54}{6} = \frac{(x+50)}{(20x)}\) \(\frac{9}{1} = \frac{(x+50)}{(20x)}\) \(180  9x = x+50\) \(130 = 10x\) \(x = 13\) 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: \(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2Aavg \right) }{ \left( AavgA1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2Aavg \right) }{ \left( Aavg0.5 \right) } \\ =Aavg0.5=9\left( 1.2Aavg \right) \\ =Aavg0.5=10.89Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\) 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
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16 Oct 2017, 18:55
Attached is a visual that should help.
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Screen Shot 20171016 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
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04 Jan 2018, 05:03
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 = \(\frac{1412512500}{12500}\) = 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!



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Re: A photography dealer ordered 60 Model X cameras to be sold
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10 Feb 2018, 13:04
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



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A photography dealer ordered 60 Model X cameras to be sold
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10 Feb 2018, 19:08
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 hudhudaawhat you did in the highlighted portion is you calculated % of SP over CP i.e. SP is 113% of CP so Profit will be =113100=13% i.e. 1.131 We are asked to calculate profit % so you got SP=648+30=678 And CP=600 Hence Profit = 678600=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
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10 Feb 2018, 20:23
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) = 67.8xTotal \(CP = 60x\) \(Profit = \frac{67.8x  60x}{60x} = 13\)



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A photography dealer ordered 60 Model X cameras to be sold
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05 May 2018, 07:36
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 = \(104*6 = 624\) \(Total Profit\) = \(2268624 = 1,644\) 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



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Re: A photography dealer ordered 60 Model X cameras to be sold
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05 May 2018, 10:19
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 = \(104*6 = 624\) \(Total Profit\) = \(2268624 = 1,644\) 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 dave13Yes 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 markup=20, hence Profit = 20*54=1080 loss on 6 cameras = 50*6=300 Hence net profit = 1080300=780 therefore profit % = 780/6000=0.13 The calculation becomes very simple when you chose smart numbers such as 100



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Re: A photography dealer ordered 60 Model X cameras to be sold
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03 Jun 2018, 15:52
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%



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Re: A photography dealer ordered 60 Model X cameras to be sold
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15 Aug 2018, 02:25
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 606=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)*0.550*250)/(50*250)*100=13% Answer: D. 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




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