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 car dealership sold two cars: the first car at a 10% [#permalink]
02 Sep 2010, 11:24
00:00
A
B
C
D
E
Difficulty:
55% (hard)
Question Stats:
69% (03:19) correct
31% (02:46) wrong based on 51 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 margin of 5% from these two sales. If the dealership's total profit was $1000, what was the sale price of each car?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
Re: better way to solve than back solving [#permalink]
02 Sep 2010, 15:32
2
This post received KUDOS
Expert's post
zest4mba 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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
Let the original price of the first car be A and that of the second car be B.
Sale Price of first car = \(\frac{110A}{100}\)
Sale Price of second car =\(\frac{90B}{100}\)
Overall purchase price of cars = A+B
Overall sale price = \(\frac{11A+9B}{10} = \frac{105}{100} (A+B)\)
Now if we solve for this, we get: \(10(11A+9B) = 105A + 105B\)
Solving the equation by taking like terms to one side:
\((110-105A) = (105-90)B\) which means \(5A = 15B\) which means \(A = 3B\).
Now the only answer choice that follows this relationship is answer choice D. And hence D is the right answer.
Re: better way to solve than back solving [#permalink]
02 Sep 2010, 17:00
3
This post received KUDOS
Expert's post
zest4mba 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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
There's something wrong with the question. While I imagine D is intended to be the right answer, $15,000 and $5,000 are not the *sale* prices of each car (the price at which the dealership sold the cars); they are the prices at which the dealership bought each car. The sale prices, which is what the question asks for, would be $16,500 (adding 10% profit to $15,000) and $4,500 (subtracting the 10% loss from $5000). _________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Re: better way to solve than back solving [#permalink]
11 Sep 2010, 17:56
whiplash2411 wrote:
zest4mba 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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
Let the original price of the first car be A and that of the second car be B.
Sale Price of first car = \(\frac{110A}{100}\)
Sale Price of second car =\(\frac{90B}{100}\)
Overall purchase price of cars = A+B
Overall sale price = \(\frac{11A+9B}{10} = \frac{105}{100} (A+B)\)
Now if we solve for this, we get: \(10(11A+9B) = 105A + 105B\)
Solving the equation by taking like terms to one side:
\((110-105A) = (105-90)B\) which means \(5A = 15B\) which means \(A = 3B\).
Now the only answer choice that follows this relationship is answer choice D. And hence D is the right answer.
Hi, It seems I am gettign stuck in a wrong equation: According to my understanding, equation should be:
Re: better way to solve than back solving [#permalink]
11 Sep 2010, 19:31
1
This post received KUDOS
utin wrote:
whiplash2411 wrote:
zest4mba 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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
Let the original price of the first car be A and that of the second car be B.
Sale Price of first car = \(\frac{110A}{100}\)
Sale Price of second car =\(\frac{90B}{100}\)
Overall purchase price of cars = A+B
Overall sale price = \(\frac{11A+9B}{10} = \frac{105}{100} (A+B)\)
Now if we solve for this, we get: \(10(11A+9B) = 105A + 105B\)
Solving the equation by taking like terms to one side:
\((110-105A) = (105-90)B\) which means \(5A = 15B\) which means \(A = 3B\).
Now the only answer choice that follows this relationship is answer choice D. And hence D is the right answer.
Hi, It seems I am gettign stuck in a wrong equation: According to my understanding, equation should be:
1.10A+0.9B-(A+B)=1.05(A+B)
please let me know where am I going wrong!!!
The R.H.S of the equation is incorrect. It should be 1.10A + 0.9B - (A+B) = 0.05 (A+B) since you are equating the profit on both sides.
1.10A + 0.9B - (A+B) represents the profit made on the sale of individual cars and hence the R.H.S should also be the profit expression. _________________
Support GMAT Club by putting a GMAT Club badge on your blog
Re: better way to solve than back solving [#permalink]
11 Sep 2010, 21:20
1
This post received KUDOS
Expert's post
This is not a good question because of the reasons noted above by Ian Stewart. Also I've never seen a GMAT question using the term "profit margin" in the quant questions, moreover as I remember "profit margin" is calculated on total revenue not on cost as "profit" or "loss" is.
Answer to be D the question should ask about cost prices of the cars and use the term "profit" instead of "profit margin". _________________
Re: better way to solve than back solving [#permalink]
22 Jun 2011, 13:56
lets denote cost of Car1 as c1 and cost car2 as c2 respectively.
overall sale resulted in a profit of 5% => profit from first sale - loss from second sale still yielded in overall 5% profit => (10/100)c1 - (10/100)c2 = (5/100)(c1+c2) => c1 = 3c2 ----equation 1
Re: better way to solve than back solving [#permalink]
09 Sep 2011, 12:49
IanStewart wrote:
zest4mba 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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
There's something wrong with the question. While I imagine D is intended to be the right answer, $15,000 and $5,000 are not the *sale* prices of each car (the price at which the dealership sold the cars); they are the prices at which the dealership bought each car. The sale prices, which is what the question asks for, would be $16,500 (adding 10% profit to $15,000) and $4,500 (subtracting the 10% loss from $5000).
This is what I did. So I guess going by what Bunuel says, this question is a douche bag? _________________
Re: better way to solve than back solving [#permalink]
09 Sep 2011, 19:10
Quote:
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?
$5,000 and $1,000 $9,000 and $5,000 $11,000 and $9,000 $15,000 and $5,000 $20,000 and $10,000
Relevant Information: 10% profit from 1st car = P(A) 10% loss from 2nd car = P(B Profit Margin is 5% i.e. 1/20th. i.e. sell should be $20000 for $1000 profit
Calculating Profit - Loss should be $1000
So, $5,000 and $1,000 --> 500 - 100 <>1000, Wrong $9,000 and $5,000 --> 900 - 500 <> 1000, Wrong $11,000 and $9,000 --> 1100 - 900 <> 1000, Wrong $15,000 and $5,000 --> 1500 - 500 = 1000, ding, Right Answer. $20,000 and $10,000 --> 2000 - 1000 = 1000, ding, Another Right Answer. However, overall profit margin is 5% i.e. $1000 profit for $20000 sell. Hence, Sum of P(A) and P(B) should be 20000.
Re: better way to solve than back solving [#permalink]
17 Sep 2011, 23:54
Revenue - Cost = Profit,
Given that Total Profit is 5%.
So 5% of Cost = 1000 => Cost= 20000.
So total cost of two car is 20,000.
Now see which answer choice gives 20000. We can see only answer choice D gives a total of 20000. BTW as Ian said the question is wrong. Seems like it asked for the individual cost of two cars. Sales price or Revenue would be Cost + Profit, so 20000+1000= 21000 or 16500 and 4500 individually.
Re: better way to solve than back solving [#permalink]
20 Sep 2011, 06:02
newmoon wrote:
Revenue - Cost = Profit,
Given that Total Profit is 5%.
So 5% of Cost = 1000 => Cost= 20000.
So total cost of two car is 20,000.
Now see which answer choice gives 20000. We can see only answer choice D gives a total of 20000. BTW as Ian said the question is wrong. Seems like it asked for the individual cost of two cars. Sales price or Revenue would be Cost + Profit, so 20000+1000= 21000 or 16500 and 4500 individually.
Ok well..both the options C and D give a total of 20000..help me understand why D
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...
Ninety-five percent of the Full-Time Class of 2015 received an offer by three months post-graduation, as reported today by Kellogg’s Career Management Center(CMC). Kellogg also saw an increase...