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A photography dealer ordered 60 Model X cameras to be sold

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

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New post 20 Jan 2015, 02:26
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:
(14125-12500)/12500 = 13. ANS D
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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 22 Apr 2015, 05:13
1
Andrewbpaa wrote:
Suppose: what if I want to calculate the number of the total profit first. How can I get it? I tried to get it by "54^250/1.2. Is this is the correct way? Pls enlighten. Thanks in advance.


Hi Andrewbpaa,

For calculating the total net profit, you would need to know the cost price and the selling price of all the 60 cameras:

On a selling price of $250, the dealer makes a profit of 20%. So, his cost price for 1 camera would be \(= \frac{$250}{1.20}\)
His refund price for 6 cameras is 50% of his cost price i.e. \(0.5* \frac{$250}{1.20}\)

I. Profit from selling 54 cameras at $250 each\(= 54 * ( $ 250 - \frac{$ 250}{1.20})\)
II. Loss from refund of 6 cameras = \(6 *( \frac{$250}{1.20} - \frac{$ 250}{1.20} * 0.50)\) \(= 6*( \frac{$250}{1.20} * 0.50)\)

Adding I & II would give you the total net profit the dealer made on the 60 cameras. The net profit can then be divided by the total cost price of 60 cameras to arrive at the total profit% of the dealer.

However, I would suggest you use weighted average method to solve the question.

Points to Note
In such questions, it pays to be careful about the concept of markup% and discount%. In this question, there was no discount offered, hence the markup price and selling price were the same i.e. $250. Had there been a discount of lets say 10%, the selling price would have been \($250* 0.9 = $ 225\). The profit then would have been equal to \($225 - \frac{$ 250}{1.20}\)

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

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

if-the-original-price-of-an-item-in-a-retail-store-is-marked-163706.html?hilit=discount
a-pair-of-skis-originally-cost-160-after-discount-of-x-149918.html
a-jewelry-dealer-initially-offered-a-bracelet-for-sale-at-an-8215.html?hilit=discount%20&%20markup
henry-purchased-three-items-during-a-sale-he-received-a-20-discount-194280.html?hilit=discount

Hope its clear!

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

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New post 06 Sep 2015, 15:40
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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New post 06 Sep 2015, 20:58
1
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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New post 07 Sep 2015, 21:34
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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New post 30 Dec 2015, 23:21
1
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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New post 01 Mar 2016, 21:17
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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New post 01 Mar 2016, 23:48
1
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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New post 05 Mar 2016, 23:56
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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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New post 10 Mar 2016, 06:23
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, 22:23
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, 19:57
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 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.
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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, 11:56
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, 04:05
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, 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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New post 13 Mar 2017, 01:43
1
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%
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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, 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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New post 02 Apr 2017, 07:04
Top Contributor
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%

Answer:

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Re: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 02 Apr 2017, 07:04

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