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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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56% (03:53) correct 43% (02:45) wrong based on 158 sessions
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 Current Student Status: mba here i come! Joined: 07 Aug 2011 Posts: 270 Location: Pakistan Concentration: Strategy, Marketing Schools: Insead '13 (M) GMAT 1: 680 Q46 V37 GMAT 2: Q V Followers: 23 Kudos [?]: 623 [18] , given: 48 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 23 Jun 2012, 06:31 18 This post received KUDOS 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. _________________ press +1 Kudos to appreciate posts Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4054 Location: Pune, India Followers: 865 Kudos [?]: 3634 [12] , given: 144 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Jun 2012, 21:20 12 This post received KUDOS Expert's post 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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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 643 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Followers: 32 Kudos [?]: 325 [10] , given: 23 Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] 16 Nov 2012, 22:48 10 This post received KUDOS 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]  06 May 2012, 10:07
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KUDOS
Expert's post
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%

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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]  23 Jun 2012, 03:06
2
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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%

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. _________________ Moderator Joined: 02 Jul 2012 Posts: 1131 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Followers: 48 Kudos [?]: 508 [2] , given: 93 Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] 16 Nov 2012, 22:15 2 This post received KUDOS 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%

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]  27 May 2013, 11:19
2
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Expert's post
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%

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: 28 Feb 2012 Posts: 22 GMAT 1: 700 Q48 V39 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 7 [1] , given: 3 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Jun 2012, 01:06 1 This post received KUDOS 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. Manager Joined: 02 Jun 2011 Posts: 159 Followers: 1 Kudos [?]: 12 [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? Manager Joined: 02 Jun 2011 Posts: 159 Followers: 1 Kudos [?]: 12 [0], given: 11 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% Director Status: Joined: 24 Jul 2011 Posts: 509 GMAT 1: 780 Q51 V48 GRE 1: 1540 Q800 V740 Followers: 53 Kudos [?]: 224 [0], given: 9 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: 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.
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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%

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;" Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 27 May 2013, 11:12 Similar topics Replies Last post Similar Topics: The number of digital camera sold is twice the number of 1 08 Sep 2004, 05:54 According to a car dealer's report, 1/3 of the cars sold 2 17 Oct 2006, 13:38 1 The price of a model M camera is$209 and the price of a 3 16 Nov 2007, 11:15
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