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Re: A photography dealer ordered 60 Model X cameras to be sold
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02 Apr 2017, 23:17
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'. 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
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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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A photography dealer ordered 60 Model X cameras to be sold
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04 Apr 2017, 18:49
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 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% 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
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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 1509 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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28 Jun 2017, 19:45
susheelh wrote: Hello diegocml, I hope this is what you meant by scale method (See attachment). Based on the scale we get \(\frac{54}{6} = \frac{(x+50)}{(20x)}\) \(\frac{9}{1} = \frac{(x+50)}{(20x)}\) \(180  9x = x+50\) \(130 = 10x\) \(x = 13\) 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 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
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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 1290 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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04 Jan 2018, 10:43
dave13 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. Hi MacFauz, Why did you put 2.4 in the denominator instead of 1.2 when calculating revenue from 6 cameras ? thanks! Hi dave13It is given that on returning 6 unsold cameras 50% is refunded. Hence \(\frac{250}{1.2}*6*\frac{1}{2}=\frac{250}{2.4}*6\)



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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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Re: A photography dealer ordered 60 Model X cameras to be sold
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15 Nov 2018, 23:34
We are told that \($250\) is equal to \(120\%\) of original price. We can set \(X \ = \ the \ original \ price\): \((1.2)(X) = $250\) \(X =\frac{250}{1.2} = $208.33\) We need to find profit or loss. This can be calculated by using the formula: profit = (money earned from sales) – (money spent on purchasing) We can approximate the values: \(Spent \ = \ 60 \times $208.33 = approximately\ $12,500\) \(Earned \ = 54 \times $250 + 6 \times $208.33 \times 50\% = \ approximately \ $14,000\) \(Profit \ = $14,000  $12,500 = approximately \ $1500\) Profit as a percentage of initial cost \(= \frac{1500}{12500} = \frac{15}{125} = \frac{3}{25} = approximately \ 12\%\ profit\) The final answer is .



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Re: A photography dealer ordered 60 Model X cameras to be sold
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05 Oct 2019, 00:08
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 of weighted average gives (.2*9 .5*1)\10 simple and sweet logic or else u can use the longer method mentioned by bunnel that too is good



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Re: A photography dealer ordered 60 Model X cameras to be sold
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05 Oct 2019, 00:08
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 of weighted average gives (.2*9 .5*1)\10 simple and sweet logic or else u can use the longer method mentioned by bunnel that too is good




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