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

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A photography dealer ordered 60 Model X cameras to be sold [#permalink]  06 May 2012, 09:53
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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 [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 18824 Followers: 3264 Kudos [?]: 22753 [7] , given: 2646 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 06 May 2012, 10:07 7 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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. _________________ Manager Joined: 02 Jun 2011 Posts: 160 Followers: 1 Kudos [?]: 17 [0], given: 11 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 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 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? Math Expert Joined: 02 Sep 2009 Posts: 18824 Followers: 3264 Kudos [?]: 22753 [2] , given: 2646 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 23 Jun 2012, 03:06 2 This post received KUDOS Expert's post 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]  23 Jun 2012, 06:31
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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, 01:06
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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 = 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, 11:23
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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 = 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, 11:26
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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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]  24 Jun 2012, 21:20
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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/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]  16 Nov 2012, 22:15
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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. _________________ Did you find this post helpful?... Please let me know through the Kudos button. Thanks To The Almighty - My GMAT Debrief GMAT Reading Comprehension: 7 Most Common Passage Types Director Status: Joined: 24 Jul 2011 Posts: 547 GMAT 1: 780 Q51 V48 GRE 1: 1540 Q800 V740 Followers: 60 Kudos [?]: 261 [0], given: 11 Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] 16 Nov 2012, 22:27 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]  16 Nov 2012, 22:48
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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) = 13 Ans D it is! _________________ Lets Kudos!!! Black Friday Debrief Intern Joined: 16 Nov 2009 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Nov 2012, 10:16 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. Manager Status: K... M. G... Joined: 22 Oct 2012 Posts: 51 Concentration: General Management, Leadership GMAT Date: 08-27-2013 GPA: 3.8 Followers: 0 Kudos [?]: 5 [0], given: 118 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 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 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. 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]  27 May 2013, 11:19
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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 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.

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*($250/1.2) = 50*250. Hope it's clear. _________________ Intern Joined: 02 Mar 2010 Posts: 15 Followers: 0 Kudos [?]: 8 [0], given: 16 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 07 Apr 2014, 18:23 1 This post was BOOKMARKED Because the answers are in percentages, I thought not to worry about the$ 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 (60-6)=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]  27 May 2014, 09:33
{(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]  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 ?

Posted from my mobile device
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]  30 Jun 2014, 01:50
Expert's post
kshitij89 wrote:
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 ?

Posted from my mobile device

The cost per unit is not 0.8*250 = 200, it's 250/1.2 = ~208. Markup is calculated on the cost value (check here: a-photography-dealer-ordered-60-model-x-cameras-to-be-sold-132077.html#p1229596).

Hope it helps.
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Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 30 Jun 2014, 01:50
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# A photography dealer ordered 60 Model X cameras to be sold

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