It is currently 17 Feb 2018, 17:04

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 10 Sep 2014
Posts: 4
Schools: LBS '18
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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 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%

I think the easiest way to solve this is to eliminate the costs of goods sold. Instead, look at "money-in" as a percent of "money-out". That number minus 100 will give you the percent profit:

Money-in= 54(250) + (6)(.5)(250/1.2) = 250(54+3/1.2)
Money-out= 60(250/1.2) = 250(50)

100(Money-in)/(Money-out) = [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.
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 432
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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.

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
Intern
Joined: 18 Apr 2015
Posts: 1
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 806
A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

22 Apr 2015, 04:13
1
KUDOS
Expert's post
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 _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Director Joined: 10 Mar 2013 Posts: 584 Location: Germany Concentration: Finance, Entrepreneurship GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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 60-6=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*250-50*250)/50*250)*100 = (6,5*250*100)/(50*250) = 6,5*2 = 13 (D)
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Intern
Joined: 16 Aug 2015
Posts: 23
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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 !
Intern
Joined: 16 Aug 2015
Posts: 26
Location: India
Concentration: General Management, Entrepreneurship
GMAT 1: 510 Q22 V19
GPA: 3.41
A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7934
Location: Pune, India
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

06 Sep 2015, 19:58
1
KUDOS
Expert's post
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 16 Aug 2015 Posts: 26 Location: India Concentration: General Management, Entrepreneurship GMAT 1: 510 Q22 V19 GPA: 3.41 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 07 Sep 2015, 03:26 VeritasPrepKarishma wrote: bablu1234 wrote: A big question I have is, why does following method give different answer though I believe there is no flaw in my method of attempting this problem - Purchase price per camera = 250/1.2 = 1250/6. Total purchase price = 60 * 1250/6 = 12500$

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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7934 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Current Student
Joined: 06 Mar 2014
Posts: 267
Location: India
GMAT Date: 04-30-2015
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7934
Location: Pune, India
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

30 Dec 2015, 22:21
1
KUDOS
Expert's post
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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 02 Feb 2016
Posts: 2
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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!
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7934
Location: Pune, India
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

01 Mar 2016, 22:48
1
KUDOS
Expert's post
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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 08 Feb 2015
Posts: 27
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

05 Mar 2016, 22:56
1
KUDOS
Two words: "weighted average"

[ 1.2(54) + 0.5(6) ] / 60 = 67.8 / 60 = 1.13.

Hence, 13% profit.
Manager
Joined: 18 Feb 2015
Posts: 87
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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! Current Student Joined: 20 Mar 2014 Posts: 2684 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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 = $$\frac{14125-12500}{12500}$$ = 13%

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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7934
Location: Pune, India
A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 13 Jun 2011
Posts: 24
Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]

### Show Tags

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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7934 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 22 May 2016, 18:57

Go to page   Previous    1   2   3   4    Next  [ 66 posts ]

Display posts from previous: Sort by