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

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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28 Jun 2016, 05:50
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 We are given that a dealer ordered 60 cameras to be sold for$250 each. This is a 20% markup over the initial cost to the dealer. We can use this to determine the initial cost to the dealer.

We can let C = dealer’s cost per camera. So,

1.2C = 250

12C = 2,500

C ≈ 210

Thus, we have an approximate cost of $210 per camera. We are next given that 6 cameras were never sold and were returned to the manufacturer for a refund of 50 percent of the dealers cost. Since the approximate cost was$210 per camera, the dealer received a $105 refund per camera, or$630 total. We can now calculate the total cost of the cameras to the dealer as:

Cost = $210 x 60 –$630

Cost = $12,600 –$630

Cost = $11,970 Next, we need to determine the revenue. We know that the dealer sold 54 cameras for$250 each. So total revenue is:

Revenue = 54 x $250 =$13,500

Since profit = revenue – cost, profit = $13,500 –$11,970 = $1,530 Finally we need to determine the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras. (Profit/initial cost) x 100 1,530/12,600 x 100 153/1260 x 100 This is roughly equal to: 150/1200 x 100 15/120 x 100 1/8 x 100 = 12.5% profit The closest answer is answer choice D, 13% profit. _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Kudos [?]: 979 [0], given: 5 Intern Joined: 30 Nov 2012 Posts: 10 Kudos [?]: 3 [0], given: 1 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 26 Oct 2016, 07:51 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/

it is a great approach.
Pls explain why do we take o,2*9 which is 54 items. It is said " which represents a 20 percent markup over the dealer’s initial cost for each camera". I just confuswd 20% out of 54 or 60 items. Pls explain so i will not make he same mistake again.

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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26 Oct 2016, 11:56
pgmat wrote:

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. _________________ If you have any queries, you can always whatsapp on my number +91945412028 Ayush Kudos [?]: 9 [0], given: 56 Manager Joined: 17 May 2015 Posts: 209 Kudos [?]: 246 [0], given: 73 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 02 Mar 2017, 10:39 aayushamj wrote: Bunuel, I did not understand this solution at all. Can you please explain again. 60 Cameras were ordered Markup on CP was 20% Hence Profit would be 20%) 6 Were Returned at 50%, Hence There was a Loss of 50% on 6 Cameras, So the Profit was 20% on 54 Cameras and Loss of 50% on 6 Cameras How did You arrive at your answer from Here? 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*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. Hi aayushamj, Given Marked price = CP + 20%of CP = CP + 0.2CP = 1.2CP =$ 250 => CP = $250/1.2 Total cost price = (60*250)/1.2 = 50*250 =$12500

The selling price has two components: Cameras sold and Cameras returned.

Revenue from cameras sold = 54*250 = 13500

Quote:
6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost.

Please note that initial cost of each camera is same as CP, i.e. $250/1.2 Revenue from cameras returned =$$\frac{6\times 250 \times 0.5}{1.2}$$ =$625

Total income = 13500 + 625 = $14125 Profit = 14125 - 12500 =$1625

Profit percentage = 1625*100/12500 = 13 %.

Hope this helps.

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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13 Mar 2017, 01:43
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Cost of each camera would be $250/1.2 =$208.33
total cost of 60 cameras would be 60*$208.33$12500
Actual number of cameras sold is 6 less i.e 54
revenue generated 54*250
no of cameras returned 6 so refund cost is $6*208.3*50% So, total income 54*250+ 6*($208.33)*50%

The dealer's approximate profit is (54*250+ 6*($208.33)*0.5-50*250)/(50*250)*100=13% Kudos [?]: 35 [1], given: 11 Manager Status: On a 600-long battle Joined: 22 Apr 2016 Posts: 138 Kudos [?]: 25 [0], given: 392 Location: Hungary Concentration: Accounting, Leadership Schools: Erasmus '20 GMAT 1: 410 Q18 V27 GMAT 2: 490 Q35 V23 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 02 Apr 2017, 05:45 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'.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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02 Apr 2017, 07:04
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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 It's important to recognize that we really don't need to use the information about the cameras selling for$250 each. The question boils down to . . .

54 cameras were sold at a 20% markup, and 6 cameras were (essentially) sold at a 50% markdown. What was the approximate profit or loss as a percent of the dealer’s initial cost for all 60 cameras?

So, we can assign A NICE value of $100 to the initial cost per camera. This means the 60 cameras cost$6000 to buy.

54 cameras were sold at a 20% markup and 6 cameras were sold at a 50% markdown.
So, 54 cameras were sold for $120, and 6 cameras were sold for$50.
(54)($120) + (6)($50) = $6780 So, the cameras were sold for$6780

This represents a profit of $780 (eliminate A and B) If the initial cost was$6000, we must determine the percentage equivalent to $780/$6000

$780/$6000 = 78/600 = 13/100 = 13%

[Reveal] Spoiler:
D

Cheers,
Brent
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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
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Expert's post
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'. Yes, you have used the scale method successfully to find the average sale price. Assuming cost price of each camera is 1, 1.2 is the sale price of 54 cameras and .5 is the sale price of 6 cameras. Hence you get the average sale price as 1.13. Scale method is also depicted by the formula you used. You can use the same formula with profit/loss percent and get the same answer. w1/w2 = (A2 - Aavg)/(Aavg - A1) 1/9 = (20 - Aavg)/(Aavg - (-50)) Aavg + 50 = 9*(20 - Aavg) 10*Aavg = 130 Aavg = 13 Just that, when I have to find the average, I prefer to use the formula in the form Aavg = (A1*w1 + A2*w2)/(w1 + w2) It is just a realignment of the previous formula. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for199

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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04 Apr 2017, 17:16
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%

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%

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

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04 Apr 2017, 19:49
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Cez005 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%

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.

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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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) _________________ My Best is yet to come! Kudos [?]: 46 [0], given: 149 Retired Moderator Joined: 12 Aug 2015 Posts: 2209 Kudos [?]: 899 [0], given: 607 GRE 1: 323 Q169 V154 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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. _________________ Give me a hell yeah ...!!!!! Kudos [?]: 899 [0], given: 607 Manager Joined: 12 Jun 2016 Posts: 227 Kudos [?]: 46 [0], given: 149 Location: India Concentration: Technology, Leadership WE: Sales (Telecommunications) A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 28 Jun 2017, 06:58 Hello diegocml, I hope this is what you meant by scale method. Attachment: OG13_Q182.png [ 15.47 KiB | Viewed 340 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'.

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Last edited by susheelh on 16 Oct 2017, 21:04, edited 1 time in total.

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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28 Jun 2017, 20:45
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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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18103 [1], given: 236 Manager Joined: 11 Mar 2015 Posts: 51 Kudos [?]: 6 [0], given: 8 GMAT 1: 730 Q50 V38 GPA: 4 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 05 Aug 2017, 12:09 Wow! What a great explanation! First time I see someone's explanation better than Bunuel's explanation ) Bravo! 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 Sep 2017, 10:45
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%

Bunuel I have a conceptual doubt. If the dealer returned 6 cameras back to the manufacturer and received only 50% refund on his initial investment cost, doesn't that mean that the dealer actually got 50% of his money back i.e. he was at loss since he lost 50% of his money- let alone the fact that he would have made profits on these 6 cameras if he had sold them? If it was a loss on his initial investments on these 6 cameras, why are we adding the $he got refunded to his revenue. Isn't revenue specifically the$'s collected from whatever he sells? He didn't sell these 6 cameras; in fact, he suffered a loss on them.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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16 Oct 2017, 19:55
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Attached is a visual that should help.
Attachments

Screen Shot 2017-10-16 at 7.54.24 PM.png [ 192.37 KiB | Viewed 117 times ]

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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

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23 Oct 2017, 03:43
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/ Hi karishma you had the formula which I saw once a+b+ab/100 for calculating the percentage of change, can we solve this question whit that? Kudos [?]: 2 [0], given: 74 Intern Joined: 09 Sep 2014 Posts: 5 Kudos [?]: 1 [0], given: 12 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 23 Oct 2017, 14:17 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%

Hii.. It took me like 10 minutes to perform all these calculations without a calculator.. Isn't there an easier way to solve this question? Thanks..

Kudos [?]: 1 [0], given: 12

Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 23 Oct 2017, 14:17

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