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.

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:

gmatclub test question M03, Q32 : car dealership [#permalink]
19 Mar 2009, 09:53

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

60% (02:11) correct
40% (01:31) wrong based on 192 sessions

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car?

A. $5,000 and $1,000 B. $9,000 and $5,000 C. $11,000 and $9,000 D. $15,000 and $5,000 E. $20,000 and $10,000

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
08 Oct 2012, 04:37

6

This post received KUDOS

This question can be quickly solved using the alligation method. Second caR______________ First car -10%-----------------------5%-----10%

Ratio of distances 3 : 1

This means that the ratio of the Cost Price of First Car to that of the Second Car is 3:1.

Scanning the answer choice, we see that only "D" fulfills this 3:1 ratio constraint. Therefore, D is the answer.

P.S. In using the alligation method, we didn't even have to bother to use the $1000 total profit. GMAT is usually about finding the fastest (yet potent) method to solve questions. Of course, this question can just as well be solved algebraically. Take home message:Flexibility is key to beating GMAT Quant.

Cheers, Der alte Fritz.

_________________

+1 Kudos me - I'm half Irish, half Prussian.

Last edited by OldFritz on 08 Oct 2012, 05:11, edited 1 time in total.

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
08 Oct 2012, 01:35

4

This post received KUDOS

Expert's post

sanjay_gmat wrote:

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit margin of 5% from these two sales. If the dealership's total profit was $1000, what was the sale price of each car?

(A) $5,000 and $1,000 (B) $9,000 and $5,000 (C) $11,000 and $9,000 (D) $15,000 and $5,000 (E) $20,000 and $10,000

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car?

A. $5,000 and $1,000 B. $9,000 and $5,000 C. $11,000 and $9,000 D. $15,000 and $5,000 E. $20,000 and $10,000

Since 5% profit equals to $1,000 of profit, then total cost price of two cars was $20,000 (0.05*x=1,000 --> x=20,000).

Now, options C and D both fit. Let's check one of them:

C. $11,000 and $9,000 --> the profit from the first car would be 0.1*$11,000=$1,100 and the loss from the second car would be 0.1*$9,000=$900 --> overall profit: $1,100-$900=$200\neq{$1,000}. So, the answer must be D.

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
08 Oct 2012, 05:08

1

This post received KUDOS

OldFritz wrote:

This question can be quickly solved using the alligation method. Second caR______________ First car -10%-----------------------5%-----10%

Ratio of distances 3 : 1

This means that the ratio of the Selling Price of First Car to that of the Second Car is 3:1.

Scanning the answer choice, we see that only "D" fulfills this 3:1 ratio constraint. Therefore, D is the answer.

P.S. In using the alligation method, we didn't even have to bother to use the $1000 total profit. GMAT is usually about finding the fastest (yet potent) method to solve questions. Of course, this question can just as well be solved algebraically. Take home message:Flexibility is key to beating GMAT Quant.

Cheers, Der alte Fritz.

I was wondering what is alligation, although from the context it was pretty clear. Here is what I found on Wikipedia: http://en.wikipedia.org/wiki/Alligation

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
08 Oct 2012, 06:42

1

This post received KUDOS

EvaJager wrote:

Hi Fritz,

I am quite familiar with the alligation method, I just didn't know the term. Even now, my speller shouts that it should be "allegation" and the dictionary on my computer doesn't know the word either ) The subject was also stated incorrectly, as it can be seen in the above link.

And thanks for the article in your other post. Just sheer beauty of basic arithmetic!

Eva Jager

I guess this shows that the alligation (allegation) method is an oldie, since modern dictionaries can't define it....

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
11 Oct 2012, 14:30

1

This post received KUDOS

I guess alligation is the same as weighted average

Thats how I did it -10% and 10% are the two numbers whose average is 5% . -10% is 15 units away from our average of 5% whereas 10% is 5 units away. So the weights will be the inverse of the distance ratios in 5 and 15 ratios and only one answer meets the requirement.

Alternatively you can also use 90% (for 10% loss) and 110% (for 10% profit) with average as 105%.

The main thing to remember is to reverse the ratio of the "distances" in order to find the ratio of the actual "quantities" present. If you have any other questions, please feel free to post 'em.

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
08 Oct 2012, 05:29

EvaJager wrote:

I was wondering what is alligation, although from the context it was pretty clear. Here is what I found on Wikipedia: http://en.wikipedia.org/wiki/Alligation

Please see attached article. Enjoy reading it--I did.

The main thing to remember is to reverse the ratio of the "distances" in order to find the ratio of the actual "quantities" present. If you have any other questions, please feel free to post 'em.

Cheers, Der alte Fritz.

Hi Fritz,

I am quite familiar with the alligation method, I just didn't know the term. Even now, my speller shouts that it should be "allegation" and the dictionary on my computer doesn't know the word either ) The subject was also stated incorrectly, as it can be seen in the above link.

And thanks for the article in your other post. Just sheer beauty of basic arithmetic!

Eva Jager

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
03 Feb 2013, 01:57

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit margin of 5% from these two sales. If the dealership's total profit was $1000, what was the sale price of each car?

(A) $5,000 and $1,000 (B) $9,000 and $5,000 (C) $11,000 and $9,000 (D) $15,000 and $5,000 (E) $20,000 and $10,000

Solution: This is a question of weighted averages concept.

One thing to remember is the ratio of values of two points is reciprocal of the distance of two points from the average, will explain it with answer below:

-10%(car B) -------- 0 ---------5%(average profit) ------- 10%(car A)

so the distance is 15:5 or 3:1 from the weighted average

Therefore, values of cost prices of Car B and Car A are in ratio 1:3. Since 5% profit = 1000, total cost price 20000

So Car A cost price = 3/4* 20000 = 15000 Car B cost price = 1/4*20000 = 5000

Re: gmatclub test question M03, Q32 : car dealership [#permalink]
21 Oct 2013, 23:51

profit, 5% of (x+y)= 1000 n c.p.of two cars is (x+y)=20000 .......eqn (i) now, (1.10x-x) + (0.9y-y)=1000 (i.e. profit means s.p. - c.p.) solving this we get, x-y=10000 ....eqn (ii)

solving eqn i & ii, we get x=15000 & y= 5000 answer is D

gmatclubot

Re: gmatclub test question M03, Q32 : car dealership
[#permalink]
21 Oct 2013, 23:51