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

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A car dealership sold two cars: the first car at a 10% [#permalink]  02 Sep 2010, 11:24
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Difficulty:

55% (hard)

Question Stats:

67% (03:12) correct 33% (02:40) wrong based on 42 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 Ms. Big Fat Panda Status: Three Down. Joined: 09 Jun 2010 Posts: 1878 Concentration: Social Entrepreneurship, Organizational Behavior Followers: 347 Kudos [?]: 1414 [2] , given: 196 Re: better way to solve than back solving [#permalink] 02 Sep 2010, 15:32 2 This post received KUDOS 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.
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Re: better way to solve than back solving [#permalink]  02 Sep 2010, 17:00
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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). _________________ Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details. Private GMAT Tutor based in Toronto Manager Status: Keep fighting! Affiliations: IIT Madras Joined: 31 Jul 2010 Posts: 238 WE 1: 2+ years - Programming WE 2: 3+ years - Product developement, WE 3: 2+ years - Program management Followers: 4 Kudos [?]: 182 [0], given: 104 Re: better way to solve than back solving [#permalink] 02 Sep 2010, 17:19 Very true Ian! Good catch. Manager Joined: 26 Mar 2010 Posts: 125 Followers: 2 Kudos [?]: 6 [0], given: 17 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:

1.10A+0.9B-(A+B)=1.05(A+B)

please let me know where am I going wrong!!!
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Re: better way to solve than back solving [#permalink]  11 Sep 2010, 19:31
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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 Senior Manager Joined: 20 Jul 2010 Posts: 271 Followers: 2 Kudos [?]: 40 [0], given: 9 Re: better way to solve than back solving [#permalink] 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 Math Expert Joined: 02 Sep 2009 Posts: 23351 Followers: 3602 Kudos [?]: 28696 [1] , given: 2809 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". _________________ Director Joined: 23 Apr 2010 Posts: 583 Followers: 2 Kudos [?]: 26 [0], given: 7 Re: better way to solve than back solving [#permalink] 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. Director Joined: 01 Feb 2011 Posts: 770 Followers: 14 Kudos [?]: 58 [0], given: 42 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 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. Senior Manager Joined: 19 Oct 2010 Posts: 274 Location: India GMAT 1: 560 Q36 V31 GPA: 3 Followers: 6 Kudos [?]: 37 [0], given: 27 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?
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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.

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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.
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Re: better way to solve than back solving [#permalink]  18 Sep 2011, 05:25
This question can very easily be done by back substitution also.

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

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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
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Re: better way to solve than back solving [#permalink]  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
Re: better way to solve than back solving   [#permalink] 06 Oct 2011, 07:10
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