Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 May 2015, 23:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Intern
Joined: 26 Jul 2010
Posts: 24
Followers: 0

Kudos [?]: 44 [5] , given: 8

A photography dealer ordered 60 Model X cameras to be sold [#permalink]  06 May 2012, 09:53
5
KUDOS
29
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

56% (03:51) correct 44% (03:14) wrong based on 689 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 Senior Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 271 Location: Pakistan Concentration: Strategy, Marketing GMAT 1: 680 Q46 V37 GMAT 2: Q V Followers: 32 Kudos [?]: 790 [30] , given: 48 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 23 Jun 2012, 06:31 30 This post received KUDOS 3 This post was BOOKMARKED 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: 5548 Location: Pune, India Followers: 1372 Kudos [?]: 6972 [19] , given: 178 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Jun 2012, 21:20 19 This post received KUDOS Expert's post 8 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

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

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 648 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Followers: 38 Kudos [?]: 405 [14] , given: 23 Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] 16 Nov 2012, 22:48 14 This post received KUDOS 3 This post was BOOKMARKED 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

Math Expert
Joined: 02 Sep 2009
Posts: 27472
Followers: 4312

Kudos [?]: 42240 [9] , given: 5969

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]  06 May 2012, 10:07
9
KUDOS
Expert's post
3
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%

_________________
Moderator
Joined: 02 Jul 2012
Posts: 1228
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 80

Kudos [?]: 858 [4] , given: 116

Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]  16 Nov 2012, 22:15
4
KUDOS
1
This post was
BOOKMARKED
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 Manager Joined: 02 Jun 2011 Posts: 160 Followers: 1 Kudos [?]: 34 [3] , given: 11 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Jun 2012, 11:23 3 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. hey ths is greatt ! i truly struggled for ths .. thnx for such easy way out. +1 for ths. thnx Math Expert Joined: 02 Sep 2009 Posts: 27472 Followers: 4312 Kudos [?]: 42240 [2] , given: 5969 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.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 27472
Followers: 4312

Kudos [?]: 42240 [2] , given: 5969

Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]  27 May 2013, 11:19
2
KUDOS
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 [?]: 9 [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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5548 Location: Pune, India Followers: 1372 Kudos [?]: 6972 [1] , given: 178 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 24 Aug 2014, 20:54 1 This post received KUDOS Expert's post 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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews e-GMAT Representative Joined: 04 Jan 2015 Posts: 253 Followers: 34 Kudos [?]: 241 [1] , given: 74 A photography dealer ordered 60 Model X cameras to be sold [#permalink] 22 Apr 2015, 04:13 1 This post received KUDOS Expert's post Andrewbpaa wrote: Suppose: what if I want to calculate the number of the total profit first. How can I get it? I tried to get it by "54^250/1.2. Is this is the correct way? Pls enlighten. Thanks in advance. Hi Andrewbpaa, For calculating the total net profit, you would need to know the cost price and the selling price of all the 60 cameras: On a selling price of$250, the dealer makes a profit of 20%. So, his cost price for 1 camera would be $$= \frac{250}{1.20}$$
His refund price for 6 cameras is 50% of his cost price i.e. $$0.5* \frac{250}{1.20}$$

I. Profit from selling 54 cameras at $250 each$$= 54 * ( 250 - \frac{ 250}{1.20})$$ II. Loss from refund of 6 cameras = $$6 *( \frac{250}{1.20} - \frac{ 250}{1.20} * 0.50)$$ $$= 6*( \frac{250}{1.20} * 0.50)$$ Adding I & II would give you the total net profit the dealer made on the 60 cameras. The net profit can then be divided by the total cost price of 60 cameras to arrive at the total profit% of the dealer. However, I would suggest you use weighted average method to solve the question. Points to Note In such questions, it pays to be careful about the concept of markup% and discount%. In this question, there was no discount offered, hence the markup price and selling price were the same i.e.$250. Had there been a discount of lets say 10%, the selling price would have been $$250* 0.9 =  225$$. The profit then would have been equal to $$225 - \frac{ 250}{1.20}$$

Please remember that the markup% is calculated on the cost price while the discount % is calculated on the markup price.

Given below are few questions on discount & markup for your practice:

if-the-original-price-of-an-item-in-a-retail-store-is-marked-163706.html?hilit=discount
a-pair-of-skis-originally-cost-160-after-discount-of-x-149918.html
a-jewelry-dealer-initially-offered-a-bracelet-for-sale-at-an-8215.html?hilit=discount%20&%20markup

Hope its clear!

Regards
Harsh
_________________

Manager
Joined: 02 Jun 2011
Posts: 160
Followers: 1

Kudos [?]: 34 [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%

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: 160
Followers: 1

Kudos [?]: 34 [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: 604
GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Followers: 77

Kudos [?]: 332 [0], given: 14

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) _________________ GyanOne | http://www.GyanOne.com | +91 9899831738 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 [?]: 7 [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;"
Intern
Joined: 02 Mar 2010
Posts: 19
Followers: 0

Kudos [?]: 10 [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. Intern Status: 1st attempt: Can I do it? Joined: 09 Jun 2013 Posts: 1 Location: India Concentration: Healthcare, Marketing GMAT Date: 05-30-2014 GPA: 2.99 WE: Pharmaceuticals (Pharmaceuticals and Biotech) Followers: 0 Kudos [?]: 5 [0], given: 0 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. Intern Joined: 09 Feb 2013 Posts: 12 Followers: 0 Kudos [?]: 0 [0], given: 2 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 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] 29 Jun 2014, 22:52 Go to page 1 2 Next [ 27 posts ] Similar topics Replies Last post Similar Topics: 9 How many of the 60 cars sold last month by a certain dealer 9 14 Dec 2012, 06:34 1 model A sold 3000 units and accounted for .6 of sales. model 3 23 Dec 2007, 03:22 1 The price of a model M camera is$209 and the price of a 3 16 Nov 2007, 11:15
A used-car dealer sold one car at a profit of 25% of the 5 12 May 2007, 14:12
According to a car dealer's report, 1/3 of the cars sold 2 17 Oct 2006, 13:38
Display posts from previous: Sort by