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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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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 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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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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22 Jun 2012, 13:14
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. Dear Bunuel, Ths is a doubt i face with %ages, in ths quest too. could you pls tell where i m going wrong? when 20% mark up for initial cost is given , how to calculate it? 250  20/100*250 or to take if 120 is 250 then how much is 100? how have you arrived at 1.2?



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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kashishh wrote: 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. Dear Bunuel, Ths is a doubt i face with %ages, in ths quest too. could you pls tell where i m going wrong? when 20% mark up for initial cost is given , how to calculate it? 250  20/100*250 or to take if 120 is 250 then how much is 100? how have you arrived at 1.2? If it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} > 1.2*{Cost}=$250 > {Cost}=$250/1.2. Hope it's clear.
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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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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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MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.
suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000
profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 18050 = 130
so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this  54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.Could you please explain, how you deduced this? Thanks.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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gmatDeep wrote: MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.
suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000
profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 18050 = 130
so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this  54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.Could you please explain, how you deduced this? Thanks. yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20%



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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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 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/
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
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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.
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
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60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera = $208 => Cost of all 60 cameras = $60*208 6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 = $624 The rest of the cameras (606 = 54 in number) were sold => Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250208) + (6241248)]/[(60*200)] =100* (1410012000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33) Option (D)
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
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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 actually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations. Here is my 30 sec approach: Notice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss). Hence overall profit/loss = \((9*20 1*50) /(9+1) = 13\) Ans D it is!
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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Hi,
For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula: Income: 54*1.2C + 6*0.5C Cost: 60*C Then percent =(Incomecost)/cost = (54*1.2C + 6*0.5C  60*C)/60*C. We can see that we can eliminate C from the fraction > percent = (54*1.2 + 6*0.5  60)/60 = (54*1.2 + 6*0.5)/60  1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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27 May 2013, 11:12
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. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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27 May 2013, 11:19
FTGNGU wrote: 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. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;" (Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) = $250 (Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250. Hope it's clear.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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Because the answers are in percentages, I thought not to worry about the $ amounts and just focus on the relationships:
Let C be the total cost of all 60 cameras. Originally, the dealer thought to sell all these 60 cameras at 20% profit or for (1.2)C.
However, he sold only (606)=54 or 90% of cameras at this price. So revenue from these cameras = (0.9)(1.2)C = (1.08)C
For the remaining 10%, he got a refund of 50% of cost or (0.5)C. So total refund = (0.1)(0.5)C = (0.05)C
Therefore, total revenue in terms of original cost = (1.08 + 0.05)C = 1.13C or 13% profit.
So D is the correct ans.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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{(54*20) + (6*(50))}/ 60
This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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29 Jun 2014, 22:52
My soln is :
Dealer's per unit initial cost =250 20/100*250 = 200 Total initial cost = 60*200=12000
Total profit = (54*250+6*100)  12000 = 2100
Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 %
Can you please tell me where I went wrong ?
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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21 Aug 2014, 07:48
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/why are we dividing by 10?



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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24 Aug 2014, 20:54
sagnik242 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/why are we dividing by 10? Weighted Average Formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) C1 and C2 represent the quantity which we want to average so they will be profit/loss here. C1 = 20% = .2 C2 = 50% (loss) = .5 The weights, given by w1 and w2, are the cost prices. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 So w1 =9 and w2 = 1 Avg Profit/Loss = (.2*9 + (.5)*1)/(9 + 1) = (.2*9 + (.5)*1)/10 Get more details on this concept from the link given in my post above.
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