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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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30 Sep 2014, 14:31
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 606=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)*0.550*250)/(50*250)*100=13% Answer: D. I think the easiest way to solve this is to eliminate the costs of goods sold. Instead, look at "moneyin" as a percent of "moneyout". That number minus 100 will give you the percent profit: Moneyin= 54(250) + (6)(.5)(250/1.2) = 250(54+3/1.2) Moneyout= 60(250/1.2) = 250(50) 100(Moneyin)/(Moneyout) = [100(250)(54+3/1.2)] / [250(50)] =2(54+3/1.2) =108 + 5 = 113% (13% profit) Cheers! and *Kudos* if you like the simplification.



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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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20 Jan 2015, 01:26
For those that get confused with the word "markup", I read it as increase.
So, in this sense it is calculated like this: 250*60=x + 0.20(x), # of cameras bought by cost for each is a 20% increase over the initial cost (x). 250*60=1.20x 15000=1.20x 12500=x. And this is how you get to the cost.
The revenue for the cameras is the # of cameras sold by the price for each: 54*250=13500.
6 of these cameras were returned and the amount the dealer received was 50% less than what he paid for them: 12500/60= 208 (about). This is the cost for each of the 60 cameras. 208*6=1248 (about), and this is the cost for the 6 cameras. Now, the dealer gave these 6 cameras back for 50% of the price he got them for: 1248  0.50(1248) = 624. So, he lost 624 and was left with 624.
List your findings: 12500 = cost 13500 = revenue 624 = revenue from the 6 cameras 14124 = total revenue
The question asks for the ratio of profit over cost. Profit means: revenue  cost. Then we have: (1412512500)/12500 = 13. ANS D



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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18 Apr 2015, 01:29
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.



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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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22 Apr 2015, 04:13
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 NoteIn 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: iftheoriginalpriceofaniteminaretailstoreismarked163706.html?hilit=discountapairofskisoriginallycost160afterdiscountofx149918.htmlajewelrydealerinitiallyofferedabraceletforsaleatan8215.html?hilit=discount%20&%20markuphenrypurchasedthreeitemsduringasalehereceiveda20discount194280.html?hilit=discountHope its clear! Regards Harsh
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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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22 May 2015, 04:09
Here is a way to simplify the calculation (based on Bunuel's solution) Total cost = 60*($250/1.2)=50*250 # of cameras sold is 606=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5 = 2,5*250 So, total income 54*250+ 2,5*250 ((54*250+2,5*250  50*250)/(50*250))*100 = ((56,5*25050*250)/50*250)*100 = (6,5*250*100)/(50*250) = 6,5*2 = 13 (D)
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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24 Aug 2015, 18:47
Hey,
I think that we can avoid such computations for this kind of problems. We have 20% profits for 54 cameras, and 50% loss on the remaining 6 cameras, we can simply use the weighted average (54x20% + 6x(50%))/ 60 = 13%. That's it !



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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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06 Sep 2015, 14:40
A big question I have is, why does following method give different answer though I believe there is no flaw in my method of attempting this problem 
Purchase price per camera = 250/1.2 = 1250/6.
Total purchase price = 60 * 1250/6 = 12500$
Got return money because of 6 cameras which were returned = 50% of 6 = [(6 * 250/1.2)/2] = 625$
Effective purchase price = 12500  625 = 11875$
Total sale = 54 * 250 = 13500$.
Hence % profit = [(Total sale  effective purchase price) / (effective purchase price)] x 100
= [(13500  11875) / 11875] x 100
= (1625/11875) x 100
= 13.684 Ans.
Dealer paid total cost of 60 cameras and then I deducted 50% cost of 6 cameras from that instead of adding it as a revenue. So in this case, technically 54 cameras have been sold and 57 were bought.
Can anyone explain why can't I do that way?
TIA



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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06 Sep 2015, 19:58
bablu1234 wrote: A big question I have is, why does following method give different answer though I believe there is no flaw in my method of attempting this problem 
Purchase price per camera = 250/1.2 = 1250/6.
Total purchase price = 60 * 1250/6 = 12500$
Got return money because of 6 cameras which were returned = 50% of 6 = [(6 * 250/1.2)/2] = 625$
Effective purchase price = 12500  625 = 11875$
Total sale = 54 * 250 = 13500$.
Hence % profit = [(Total sale  effective purchase price) / (effective purchase price)] x 100
= [(13500  11875) / 11875] x 100
= (1625/11875) x 100
= 13.684 Ans.
Dealer paid total cost of 60 cameras and then I deducted 50% cost of 6 cameras from that instead of adding it as a revenue. So in this case, technically 54 cameras have been sold and 57 were bought.
Can anyone explain why can't I do that way?
TIA You have assumed the situation to be a bit different from what it actually is. You have assumed that he bought 54 cameras at 250/1.2 each and 6 cameras at 125/1.2 each. Then he sold 54 cameras at 250 each and 6 cameras at 0. While actually this is the situation: He bought 60 cameras at 250/1.2 each. He sold 54 at 250 each and sold the rest of the 6 at 125/1.2 each. The difference (Revenue  Cost) in both cases will be the same but the cost price in the denominator will be different. So profit % obtained will be different. Please check Bunuel's solution on Page 1 to see the actual calculations of this method.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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07 Sep 2015, 03:26
VeritasPrepKarishma wrote: bablu1234 wrote: A big question I have is, why does following method give different answer though I believe there is no flaw in my method of attempting this problem 
Purchase price per camera = 250/1.2 = 1250/6.
Total purchase price = 60 * 1250/6 = 12500$
Got return money because of 6 cameras which were returned = 50% of 6 = [(6 * 250/1.2)/2] = 625$
Effective purchase price = 12500  625 = 11875$
Total sale = 54 * 250 = 13500$.
Hence % profit = [(Total sale  effective purchase price) / (effective purchase price)] x 100
= [(13500  11875) / 11875] x 100
= (1625/11875) x 100
= 13.684 Ans.
Dealer paid total cost of 60 cameras and then I deducted 50% cost of 6 cameras from that instead of adding it as a revenue. So in this case, technically 54 cameras have been sold and 57 were bought.
Can anyone explain why can't I do that way?
TIA You have assumed the situation to be a bit different from what it actually is. You have assumed that he bought 54 cameras at 250/1.2 each and 6 cameras at 125/1.2 each. Then he sold 54 cameras at 250 each and 6 cameras at 0. While actually this is the situation: He bought 60 cameras at 250/1.2 each. He sold 54 at 250 each and sold the rest of the 6 at 125/1.2 each. The difference (Revenue  Cost) in both cases will be the same but the cost price in the denominator will be different. So profit % obtained will be different. Please check Bunuel's solution on Page 1 to see the actual calculations of this method. Thanks, I understood what you mean. >>He bought 60 cameras at 250/1.2 each. >>He sold 54 at 250 each and sold the rest of the 6 at 125/1.2 each. From lines above, one can also assume that 57 Cameras were bought and 57 were sold. Total purchase price = 57 * 1250/6 = 11875 Total sale = 57 * 250 = 14250 That gives 20%. Don't take it otherwise about dragging but I am trying to understand, with similar questions, I can think or assume differently during the real test due to pressure of time.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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07 Sep 2015, 20:34
bablu1234 wrote: Thanks, I understood what you mean.
>>He bought 60 cameras at 250/1.2 each. >>He sold 54 at 250 each and sold the rest of the 6 at 125/1.2 each.
From lines above, one can also assume that 57 Cameras were bought and 57 were sold.
Total purchase price = 57 * 1250/6 = 11875 Total sale = 57 * 250 = 14250 That gives 20%.
Don't take it otherwise about dragging but I am trying to understand, with similar questions, I can think or assume differently during the real test due to pressure of time.
All I can say is that you have to carefully consider what the cost price is and what the selling price is.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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25 Sep 2015, 04:23
VeritasPrepKarishma wrote: sagnik242 wrote: 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.[/quote] VeritasPrepKarishma, As per my understanding, isn't the value of C2>C1 but here 0.5 is < 0.2 please let me know if there is anything wrong in my understanding. Thank you



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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30 Dec 2015, 22:21
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? The weights are 54 (20% profit on 54 cameras) and 6 (50% loss on 6 cameras). This is a ratio of 54:6 = 9:1 (in lowest terms). It is the same thing whether you use 54 and 6 as weights or 9 and 1 but 9 and 1 simplify the calculations. Avg = (A1 * w1 + A2 * w2)/(w1 + w2) = (A1 * 9 + A2 * 1)/(9 + 1) = (A1 * 9 + A2 * 1)/10
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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01 Mar 2016, 20:17
Does this question bother anyone else in formatting terms? Perhaps I'm misinterpreting, but let me explain:
"What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?"
Profit = Revenue  Costs
Thus, the question is looking for (Revenue  Cost) / Cost
In number terms this would look like, [54*(250) + 6*(250)*(4/5)*(1/2)  600*(250)*(4/5)]/[(60)*(250)*(4/5)]x100 ~~ 10.7%
I'd appreciate any help in clearing up this misunderstanding!



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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01 Mar 2016, 22:48
philipb wrote: Does this question bother anyone else in formatting terms? Perhaps I'm misinterpreting, but let me explain:
"What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?"
Profit = Revenue  Costs
Thus, the question is looking for (Revenue  Cost) / Cost
In number terms this would look like, [54*(250) + 6*(250)*(4/5)*(1/2)  600*(250)*(4/5)]/[(60)*(250)*(4/5)]x100 ~~ 10.7%
I'd appreciate any help in clearing up this misunderstanding! You are correct conceptually. The error lies here: "$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera." Marked price is 250 This is a 20% markup on cost. Cost * (1 + 20/100) = Marked price Cost * (6/5) = Marked Price Cost = (5/6) * 250 You have taken this to be (4/5)*250. Replace it by (5/6) and you should get the correct answer.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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05 Mar 2016, 22:56
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Two words: "weighted average"
[ 1.2(54) + 0.5(6) ] / 60 = 67.8 / 60 = 1.13.
Hence, 13% profit.



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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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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 = \(\frac{1412512500}{12500}\) = 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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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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10 Mar 2016, 05:37
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 = \(\frac{1412512500}{12500}\) = 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! It comes from the fact that the cost price for the dealer at which he purchased 60 cameras = 250/1.2 and the dealer had to return 6 unsold cameras at 50% of the original cost price > 0.5*(250/1.2) = 250/2.4 Hope this helps.



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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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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 = \(\frac{1412512500}{12500}\) = 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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22 May 2016, 04:56
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



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