It is currently 18 Jan 2018, 15:24

### 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 car dealership sold two cars: the first car at a 10%

Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Feb 2010
Posts: 71

Kudos [?]: 142 [0], given: 4

A car dealership sold two cars: the first car at a 10% [#permalink]

### Show Tags

02 Sep 2010, 11:24
00:00

Difficulty:

55% (hard)

Question Stats:

61% (01:14) correct 39% (01:20) wrong based on 121 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 [Reveal] Spoiler: OA Kudos [?]: 142 [0], given: 4 GMAT Tutor Joined: 24 Jun 2008 Posts: 1347 Kudos [?]: 2082 [4], given: 6 Re: better way to solve than back solving [#permalink] ### Show Tags 02 Sep 2010, 17:00 4 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

Kudos [?]: 2082 [4], given: 6

Current Student
Status: Three Down.
Joined: 09 Jun 2010
Posts: 1914

Kudos [?]: 2263 [3], given: 210

Concentration: General Management, Nonprofit
Re: better way to solve than back solving [#permalink]

### Show Tags

02 Sep 2010, 15:32
3
KUDOS
1
This post was
BOOKMARKED
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. Kudos [?]: 2263 [3], given: 210 Senior Manager Status: Time to step up the tempo Joined: 24 Jun 2010 Posts: 403 Kudos [?]: 269 [1], given: 50 Location: Milky way Schools: ISB, Tepper - CMU, Chicago Booth, LSB Re: better way to solve than back solving [#permalink] ### Show Tags 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

Kudos [?]: 269 [1], given: 50

Math Expert
Joined: 02 Sep 2009
Posts: 43322

Kudos [?]: 139433 [1], given: 12790

Re: better way to solve than back solving [#permalink]

### Show Tags

11 Sep 2010, 21:20
1
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".
_________________

Kudos [?]: 139433 [1], given: 12790

Manager
Status: Keep fighting!
Joined: 31 Jul 2010
Posts: 220

Kudos [?]: 557 [0], given: 104

WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Re: better way to solve than back solving [#permalink]

### Show Tags

02 Sep 2010, 17:19
Very true Ian! Good catch.

Kudos [?]: 557 [0], given: 104

Manager
Joined: 26 Mar 2010
Posts: 116

Kudos [?]: 16 [0], given: 17

Re: better way to solve than back solving [#permalink]

### Show Tags

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: 1.10A+0.9B-(A+B)=1.05(A+B) please let me know where am I going wrong!!! Kudos [?]: 16 [0], given: 17 Senior Manager Joined: 20 Jul 2010 Posts: 253 Kudos [?]: 107 [0], given: 9 Re: better way to solve than back solving [#permalink] ### Show Tags 11 Sep 2010, 20:26 .05 of (a+b) is 1000. So A+B is 20000. Only two choices remain. I did calculation on ans choices to select D _________________ If you like my post, consider giving me some KUDOS !!!!! Like you I need them Kudos [?]: 107 [0], given: 9 Director Joined: 23 Apr 2010 Posts: 572 Kudos [?]: 104 [0], given: 7 Re: better way to solve than back solving [#permalink] ### Show Tags 15 Sep 2010, 02:26 Quote: I've never seen a GMAT question using the term "profit margin" That's exactly what I've thought when I read the question. Kudos [?]: 104 [0], given: 7 Director Joined: 01 Feb 2011 Posts: 703 Kudos [?]: 153 [0], given: 42 Re: better way to solve than back solving [#permalink] ### Show Tags 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 Overall profit = 1000 =>(5/100)(c1+c2) = 1000 => c1+c2 = 20000------equation 2 from 1 and 2 we can find out values of c1 and c2. c2=5000 c1 = 15000 Answer is D. Kudos [?]: 153 [0], given: 42 Senior Manager Joined: 19 Oct 2010 Posts: 251 Kudos [?]: 93 [0], given: 27 Location: India GMAT 1: 560 Q36 V31 GPA: 3 Re: better way to solve than back solving [#permalink] ### Show Tags 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?
_________________

petrifiedbutstanding

Kudos [?]: 93 [0], given: 27

Intern
Affiliations: IEEE, IAB
Joined: 09 Aug 2011
Posts: 15

Kudos [?]: 7 [0], given: 2

Mustafa: Golam
Re: better way to solve than back solving [#permalink]

### Show Tags

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.

_________________

*´¨)
¸.•´¸.•*´¨) ¸.•*¨)
(¸.•´ (¸.• *Mustafa Golam,'.',.

Kudos [?]: 7 [0], given: 2

Intern
Joined: 04 Aug 2010
Posts: 9

Kudos [?]: 6 [0], given: 12

Re: better way to solve than back solving [#permalink]

### Show Tags

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.

Kudos [?]: 6 [0], given: 12

Current Student
Joined: 21 Aug 2010
Posts: 203

Kudos [?]: 18 [0], given: 28

Re: better way to solve than back solving [#permalink]

### Show Tags

18 Sep 2011, 05:25
This question can very easily be done by back substitution also.

BR
Mandeep

Kudos [?]: 18 [0], given: 28

Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 261

Kudos [?]: 1278 [0], given: 48

Re: better way to solve than back solving [#permalink]

### Show Tags

20 Sep 2011, 04:14
$$\frac{x}{10} - \frac{y}{10} = 1000$$

$$\frac{105}{100}(x+y) = \frac{11}{10}x + \frac{9}{10}y$$

x=15k, y=3k

sale price of x = 1.1*15k = 16,500
that of y = 0.9*3k = 2,700

_________________

press +1 Kudos to appreciate posts

Kudos [?]: 1278 [0], given: 48

Manager
Joined: 15 Feb 2011
Posts: 240

Kudos [?]: 209 [0], given: 9

Re: better way to solve than back solving [#permalink]

### Show Tags

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

Kudos [?]: 209 [0], given: 9

Manager
Joined: 08 Sep 2011
Posts: 66

Kudos [?]: 5 [0], given: 5

Concentration: Finance, Strategy
Re: better way to solve than back solving [#permalink]

### Show Tags

06 Oct 2011, 07:10
This is how I looked at it and I maybe wrong.

1.10 (x) + .90 (y) = 1.05 (x + y)

.05x = .15y

x= 3y

So once I saw that the total was 20,000 and it was between c and d, only option d fit x=3y

Kudos [?]: 5 [0], given: 5

Re: better way to solve than back solving   [#permalink] 06 Oct 2011, 07:10
Display posts from previous: Sort by