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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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New post 10 Mar 2016, 05:23
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.


Hey all,
Why do we divide 250/2.4? Where does 2.4 come from?
Thanks!
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 10 Mar 2016, 21:23
HarveyKlaus 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.


Hey all,
Why do we divide 250/2.4? Where does 2.4 come from?
Thanks!


Cost to dealer = 250/1.2 (because 250 is the selling price which 20% markup)

Revenue from the 6 cameras is half the cost price. This is = (250/1.2)*(1/2) = 250/2.4

P.S. - Saw your post later Engr2012
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 22 May 2016, 18:57
vaidhaichaturvedi 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/



Hi Karishma,

Could you please explain me why there is negative sign,as in Weightage average problem has + sign in its formula. or whenever there is loss we have to take - sign in formula in numerator

I may sound silly but kindly explain me.

Thanks


Yes, loss and discount take negative signs.
Think about it - If you have two profits, one of 20% and the other of 5%, you will use
(.2*9 + .5*1)/10

But you actually have a profit of 20% and a loss of 5%. How will you depict the loss? With a negative sign because it reduces the cost price.
(.2*9 + (-.5)*1)/10
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 26 Oct 2016, 10:56
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


Cp of 60 cameras = 250*60/120*100 => 12500
CP of 6 cameras = 1250

6 were returned at 50% of CP, so amount received from return= 625

Net CP of 54 Cameras = Cp of 60 cameras - Amount received from return

Net CP of 54 Cameras = 12500 - 625 = 11875

SP of 54 cameras = 250*54 = 13500

So, Profit Percentage = \(\frac{( 13500 - 11875 )}{11875} *100\) ~ 13%

Hence correct answer must be (D)

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

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New post 28 Feb 2017, 03:05
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.

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

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New post 02 Mar 2017, 09:39
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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New post 02 Apr 2017, 04:45
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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New post 04 Apr 2017, 16:16
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%

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

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New post 12 Jun 2017, 21:13
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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New post 16 Jun 2017, 17:54
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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New post Updated on: 16 Oct 2017, 20:04
Hello diegocml,

I hope this is what you meant by scale method.

Attachment:
OG13_Q182.png
OG13_Q182.png [ 15.47 KiB | Viewed 1576 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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Originally posted by susheelh on 28 Jun 2017, 05:58.
Last edited by susheelh on 16 Oct 2017, 20:04, 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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Attached is a visual that should help.
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Screen Shot 2017-10-16 at 7.54.24 PM.png [ 192.37 KiB | Viewed 1355 times ]

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

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New post 04 Jan 2018, 05:03
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%

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

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New post 10 Feb 2018, 13:04
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
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 10 Feb 2018, 19:08
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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New post 10 Feb 2018, 20:23
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 \(CP = 60x\)
\(Profit = \frac{67.8x - 60x}{60x} = 13\)
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 05 May 2018, 07:36
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 :)
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 05 May 2018, 10:19
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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 :thumbup: :thumbup:

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

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New post 03 Jun 2018, 15:52
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%
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 15 Aug 2018, 02:25
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%

Answer: D.


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 :)
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Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 15 Aug 2018, 02:25

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