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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
 06 May 2012, 10:53
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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]
23 Jun 2012, 07:31
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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, 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 = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
24 Jun 2012, 22:20
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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]
16 Nov 2012, 23:48
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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) = 13Ans D it is!
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
06 May 2012, 11:07
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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 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D.
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
16 Nov 2012, 23:15
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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{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.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
22 Jun 2012, 14:14
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 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. 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]
23 Jun 2012, 04:06
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 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. 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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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
24 Jun 2012, 02:06
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 = 180-50 = 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]
24 Jun 2012, 12:23
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 = 180-50 = 130
so, 130 profit from a cost of 1000. that is 13% profit. hey ths is greatt ! i truly struggled for ths .. thnx for such easy way out. +1 for ths. thnx
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
24 Jun 2012, 12:26
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 = 180-50 = 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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Of the cameras ordered, 6 were never sold and were returned [#permalink]
16 Nov 2012, 21:35
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
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
16 Nov 2012, 23:27
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 (60-6 = 54 in number) were sold => Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/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]
17 Nov 2012, 05:23
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 Merging similar topics. Please search the forum before posting.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
17 Nov 2012, 15:39
true Bunuel sorry but sometimes the post are not properly tagged. Next time, I'll do
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
24 Nov 2012, 11:16
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 =(Income-cost)/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
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24 Nov 2012, 11:16
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