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

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06 May 2012, 10:53

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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

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

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24 Jun 2012, 22:20

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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

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

Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]

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16 Nov 2012, 23:48

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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

actually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations. Here is my 30 sec approach: Notice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss). Hence overall profit/loss = \((9*20 -1*50) /(9+1) = 13\) Ans D it is! _________________

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

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06 May 2012, 11:07

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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

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%

Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]

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16 Nov 2012, 23:15

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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

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. _________________

Did you find this post helpful?... Please let me know through the Kudos button.

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

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23 Jun 2012, 04:06

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kashishh wrote:

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.

Dear Bunuel, Ths is a doubt i face with %ages, in ths quest too. could you pls tell where i m going wrong? when 20% mark up for initial cost is given , how to calculate it? 250 - 20/100*250 or to take if 120 is 250 then how much is 100? how have you arrived at 1.2?

If it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} --> 1.2*{Cost}=$250 --> {Cost}=$250/1.2.

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

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27 May 2013, 12:19

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FTGNGU wrote:

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.

this must be kind of awkward question but i got no option other than asking you.

20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"

(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) = $250 (Cost per unit) = $250/1.2

Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.

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

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24 Jun 2012, 02:06

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MBAhereIcome wrote:

54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.

suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000

profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130

so, 130 profit from a cost of 1000. that is 13% profit.

Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this?

Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]

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16 Nov 2012, 23:27

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60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera = $208 => Cost of all 60 cameras = $60*208

6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 = $624 The rest of the cameras (60-6 = 54 in number) were sold => Revenue made from selling the cameras = $54 * 250

For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33)

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

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24 Aug 2014, 21:54

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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

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

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

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22 Apr 2015, 05:13

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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.

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:

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

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06 Sep 2015, 20:58

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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. _________________

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

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30 Dec 2015, 23:21

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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

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.

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

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22 Jun 2012, 14:14

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.

Dear Bunuel, Ths is a doubt i face with %ages, in ths quest too. could you pls tell where i m going wrong? when 20% mark up for initial cost is given , how to calculate it? 250 - 20/100*250 or to take if 120 is 250 then how much is 100? how have you arrived at 1.2?

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

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24 Jun 2012, 12:26

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gmatDeep wrote:

MBAhereIcome wrote:

54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.

suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000

profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130

so, 130 profit from a cost of 1000. that is 13% profit.

Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this?

Thanks.

yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20%

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

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24 Nov 2012, 11:16

Hi,

For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula: Income: 54*1.2C + 6*0.5C Cost: 60*C Then percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.

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

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27 May 2013, 12:12

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.

this must be kind of awkward question but i got no option other than asking you.

20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"

gmatclubot

Re: A photography dealer ordered 60 Model X cameras to be sold
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