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

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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03 Apr 2017, 00:17
1
diegocml wrote:
VeritasPrepKarishma wrote:
pgmat wrote:

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit

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. Hi. Can you help point out why this method is incorrect? Cost price is$208. So profit per camera is $42. Profit for the 54 cameras sold is 54*42=2,268. 2,268+625(refund)=2,893. Total price to dealer is 12,480. 2893/12,480*100=23% Hi Cez005, Please note that when you return 6 cameras at 50% of CP, you are making a loss on these 6 pieces. So, in your method Net profit = 2268 - 625 = 1643 Profit percentage = 1643/12480*100 = $$\approx$$13%. Hope it helps. Manager Joined: 12 Jun 2016 Posts: 218 Location: India Concentration: Technology, Leadership WE: Sales (Telecommunications) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 12 Jun 2017, 22:13 Absolutely loved this approach! Probably the author will never read this, but I cant leave without saying thanks. What I liked about this approach? - a) Identifying this is a Weighted average problem b) Using smaller manageable number (10) to ease calculations and c) Calculating profits/loss directly instead of calculating total SP and the subtracting CP. Thanks for publishing this approach. +1 to you MBAhereIcome wrote: 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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16 Jun 2017, 18:54
Here is wha I did for this question ->

Initial cost price => x (let)
1.2x=250
x=208(approx)
Total Selling price => 44*250 + 6* 104 => 11624
Total cost price => 50*208 => 10400

Profit => 1224

Profit percent => 1224/10400. * 100 => 11.76 percent

The closet value is 13 percent

Hence D.
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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Updated on: 16 Oct 2017, 21:04
Hello diegocml,

I hope this is what you meant by scale method.

Attachment:

OG13_Q182.png [ 15.47 KiB | Viewed 735 times ]

Based on the scale we get

$$\frac{54}{6} = \frac{(x+50)}{(20-x)}$$

$$\frac{9}{1} = \frac{(x+50)}{(20-x)}$$

$$180 - 9x = x+50$$

$$130 = 10x$$

$$x = 13$$

So, the dealer earns a profit of 13%

I learnt this approach after reading VeritasPrepKarishma's blog. Since the answer is correct, I am assuming the approach is correct

diegocml 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/ VeritasPrepKarishma, I think you method is pretty sweet - I just used another weighted average looking formula you shared in your blog: $$\cfrac { w1 }{ w2 } :=\cfrac { \left( A2-Aavg \right) }{ \left( Aavg-A1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2-Aavg \right) }{ \left( Aavg-0.5 \right) } \\ =Aavg-0.5=9\left( 1.2-Aavg \right) \\ =Aavg-0.5=10.8-9Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13$$ 13% profit I'm a big fan of the scaled method but I can't figure out how to set it up as to find the 'combined average'. _________________ My Best is yet to come! Originally posted by susheelh on 28 Jun 2017, 06:58. Last edited by susheelh on 16 Oct 2017, 21:04, edited 1 time in total. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8298 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 28 Jun 2017, 20:45 1 susheelh wrote: Hello diegocml, I hope this is what you meant by scale method (See attachment). Based on the scale we get $$\frac{54}{6} = \frac{(x+50)}{(20-x)}$$ $$\frac{9}{1} = \frac{(x+50)}{(20-x)}$$ $$180 - 9x = x+50$$ $$130 = 10x$$ $$x = 13$$ So, the dealer earns a profit of 13% I learnt this approach after reading VeritasPrepKarishma's blog. Since the answer is correct, I am assuming the approach is correct This formula is an application of the scale method: $$\frac{w1}{w2} = \frac{A2 - Aavg}{Aavg - A1}$$ It is faster and more direct than actually using the number line. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > GMAT self-study has never been more personalized or more fun. Try ORION Free! Director Status: Professional GMAT Tutor Affiliations: AB, cum laude, Harvard University (Class of '02) Joined: 10 Jul 2015 Posts: 669 Location: United States (CA) Age: 38 GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: Q168 V169 WE: Education (Education) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 16 Oct 2017, 19:55 Top Contributor Attached is a visual that should help. Attachments Screen Shot 2017-10-16 at 7.54.24 PM.png [ 192.37 KiB | Viewed 514 times ] _________________ Harvard grad and 99% GMAT scorer, offering expert, private GMAT tutoring and coaching, both in-person (San Diego, CA, USA) and online worldwide, since 2002. One of the only known humans to have taken the GMAT 5 times and scored in the 700s every time (700, 710, 730, 750, 770), including verified section scores of Q50 / V47, as well as personal bests of 8/8 IR (2 times), 6/6 AWA (4 times), 50/51Q and 48/51V (1 question wrong). You can download my official test-taker score report (all scores within the last 5 years) directly from the Pearson Vue website: https://tinyurl.com/y94hlarr Date of Birth: 09 December 1979. GMAT Action Plan and Free E-Book - McElroy Tutoring Contact: mcelroy@post.harvard.edu Director Joined: 09 Mar 2016 Posts: 884 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 04 Jan 2018, 06:03 MacFauz wrote: 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.

Hi MacFauz, Why did you put 2.4 in the denominator instead of 1.2 when calculating revenue from 6 cameras ? thanks!
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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04 Jan 2018, 11:43
1
dave13 wrote:
MacFauz wrote:
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. Hi MacFauz, Why did you put 2.4 in the denominator instead of 1.2 when calculating revenue from 6 cameras ? thanks! Hi dave13 It is given that on returning 6 unsold cameras 50% is refunded. Hence $$\frac{250}{1.2}*6*\frac{1}{2}=\frac{250}{2.4}*6$$ Intern Joined: 27 Apr 2016 Posts: 6 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 10 Feb 2018, 14:04 hi, can we substitute numbers? for example, I tried using the following: 10 is the cost price and therefore 12 is the sales price. Following the information given I get the following: 1- cost = 60*10 = 600 2- revenue = 54*12 = 648 3- refund = 6*5 = 30 therefore, I get the following: (648 + 30) / 600 = 1.13. Now I know that I should deduct 1 from this but not sure why. so two questions: 1. can we plug in a number for the cost? 2. why do we need to deduct 1 from the answer above (1.13)? thanks PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 Location: India GPA: 3.82 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 10 Feb 2018, 20:08 hudhudaa wrote: hi, can we substitute numbers? for example, I tried using the following: 10 is the cost price and therefore 12 is the sales price. Following the information given I get the following: 1- cost = 60*10 = 600 2- revenue = 54*12 = 648 3- refund = 6*5 = 30 therefore, I get the following: (648 + 30) / 600 = 1.13. Now I know that I should deduct 1 from this but not sure why. so two questions: 1. can we plug in a number for the cost? 2. why do we need to deduct 1 from the answer above (1.13)? thanks Hi hudhudaa what you did in the highlighted portion is you calculated % of SP over CP i.e. SP is 113% of CP so Profit will be =113-100=13% i.e. 1.13-1 We are asked to calculate profit % so you got SP=648+30=678 And CP=600 Hence Profit = 678-600=78 Therefore Profit % = 78/600*100=13% Yes you can use substitution here and in fact SP of$250 given in the question might not be required because SP and CP are connected through a % and finally you have to calculate profit %
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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10 Feb 2018, 21:23
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 Let x be the Original Price for Each Camera. Given, $$1.2x = 250.$$ Total $$SP = 54*1.2x + 6*(\frac{x}{2})$$ {This represents the the price which he got back) = 67.8x Total $$CP = 60x$$ $$Profit = \frac{67.8x - 60x}{60x} = 13$$ _________________ If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !! Director Joined: 09 Mar 2016 Posts: 884 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 05 May 2018, 08:36 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

pushpitkc, niks18, generis, yes its me again

Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this

"60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following Cost of one camera $$1.20x = 250$$ $$x$$ = $$208$$ Hence Mark up = $$42$$ Now 54 were sold ---> $$Profit = 42*54 = 2,268$$ 6 cameras unsold and returned at 0.50% of its cost = $$104*6 = 624$$ $$Total Profit$$ = $$2268-624 = 1,644$$ Dealers Initial Cost for the 60 cameras = $$208*60 = 12,480$$ So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is $$\frac{1,644}{12,480}$$ = $$0.13 %$$ Thank you _________________ In English I speak with a dictionary, and with people I am shy. PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 Location: India GPA: 3.82 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 05 May 2018, 11:19 1 dave13 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

pushpitkc, niks18, generis, yes its me again

Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this

"60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following Cost of one camera $$1.20x = 250$$ $$x$$ = $$208$$ Hence Mark up = $$42$$ Now 54 were sold ---> $$Profit = 42*54 = 2,268$$ 6 cameras unsold and returned at 0.50% of its cost = $$104*6 = 624$$ $$Total Profit$$ = $$2268-624 = 1,644$$ Dealers Initial Cost for the 60 cameras = $$208*60 = 12,480$$ So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is $$\frac{1,644}{12,480}$$ = $$0.13 %$$ Thank you Hi dave13 Yes your method is absolutely correct As a shortcut I would advice you to work with smart numbers. you actually don't need 250 here. Let the CP=100, so total cost = 100*60=6000 mark-up=20, hence Profit = 20*54=1080 loss on 6 cameras = 50*6=300 Hence net profit = 1080-300=780 therefore profit % = 780/6000=0.13 The calculation becomes very simple when you chose smart numbers such as 100 Intern Joined: 04 Mar 2018 Posts: 3 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 03 Jun 2018, 16:52 First, let's consider an easier version of the problem: If the dealer sold all 60 cameras at a profit of 20% on each camera, how much was the total profit? 20% - easy. Therefore: 20% profit per camera sold * ($$\frac{60 sold}{60 bought}$$) = 20% profit Now, the problem says that 54 cameras were sold. Using the equation above: PROFIT = 20% profit per camera sold * ($$\frac{54 sold}{60 bought}$$) = 18% profit. What about the cameras that didn't sell? Let's consider an easier problem: If the dealer returned all 60 cameras at a loss of 50% for each camera, how much was the total loss? 50% - easy. Therefore: 50% loss per camera returned * ($$\frac{60 returned}{60 bought}$$) = 50% loss. Now, the problem says that 6 cameras were returned, so let's use a similar method: LOSS = 50% loss per camera returned * ($$\frac{6 returned}{60 bought}$$) = 5% loss. Net = Profit - Loss = 18% - 5% = 13% Intern Joined: 06 Jun 2017 Posts: 6 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 15 Aug 2018, 03:25 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%

Ca you share similar problems like this ? - I have searched the thread and found few but if you can link me to few more problems, it would help in firming up this concept
Re: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 15 Aug 2018, 03:25

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