Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

06 May 2012, 10:53

7

This post received KUDOS

59

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

57% (03:00) correct
43% (03:18) wrong based on 1162 sessions

HideShow timer Statistics

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]

Show Tags

06 May 2012, 11:07

14

This post received KUDOS

Expert's post

10

This post was BOOKMARKED

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

Show Tags

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]

Show Tags

23 Jun 2012, 04:06

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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]

Show Tags

24 Jun 2012, 02:06

1

This post received KUDOS

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

Show Tags

24 Jun 2012, 12:26

1

This post was BOOKMARKED

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]

Show Tags

24 Jun 2012, 22:20

25

This post received KUDOS

Expert's post

15

This post was BOOKMARKED

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]

Show Tags

16 Nov 2012, 23:15

5

This post received KUDOS

1

This post was BOOKMARKED

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

Show Tags

16 Nov 2012, 23:27

1

This post received KUDOS

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

Show Tags

16 Nov 2012, 23:48

15

This post received KUDOS

5

This post was BOOKMARKED

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]

Show Tags

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]

Show Tags

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;"

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

Show Tags

27 May 2013, 12:19

2

This post received KUDOS

Expert's post

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]

Show Tags

27 May 2014, 10:33

{(54*20) + (6*(-50))}/ 60

This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding.

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

Show Tags

21 Aug 2014, 08:48

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

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

I’m a little delirious because I’m a little sleep deprived. But whatever. I have to write this post because... I’M IN! Funnily enough, I actually missed the acceptance phone...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...