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Intern  Joined: 09 Feb 2010
Posts: 42
A car dealership sold two cars: the first car at a 10%  [#permalink]

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1 00:00

Difficulty:   55% (hard)

Question Stats: 68% (02:05) correct 32% (02:42) wrong based on 53 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 SVP  Status: Three Down. Joined: 09 Jun 2010 Posts: 1851 Concentration: General Management, Nonprofit Re: better way to solve than back solving [#permalink] ### Show Tags 3 1 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]

4
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 Manager  Status: Keep fighting! Affiliations: IIT Madras Joined: 31 Jul 2010 Posts: 192 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 Very true Ian! Good catch. Manager  Joined: 27 Mar 2010 Posts: 92 Re: better way to solve than back solving [#permalink] ### Show Tags 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!!!
Senior Manager  Status: Time to step up the tempo
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Re: better way to solve than back solving  [#permalink]

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1
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 Manager  Joined: 20 Jul 2010 Posts: 213 Re: better way to solve than back solving [#permalink] ### Show Tags .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 V Joined: 02 Sep 2009 Posts: 53716 Re: better way to solve than back solving [#permalink] ### Show Tags 1 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: 541 Re: better way to solve than back solving [#permalink] ### Show Tags 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: 653 Re: better way to solve than back solving [#permalink] ### Show Tags 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. Manager  Joined: 19 Oct 2010 Posts: 168 Location: India GMAT 1: 560 Q36 V31 GPA: 3 Re: better way to solve than back solving [#permalink] ### Show Tags 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]

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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.

So, Right Answer is D
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Re: better way to solve than back solving  [#permalink]

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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.
Current Student Joined: 21 Aug 2010
Posts: 186
Re: better way to solve than back solving  [#permalink]

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This question can very easily be done by back substitution also.

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

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$$\frac{x}{10} - \frac{y}{10} = 1000$$

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

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]

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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
Manager  Joined: 08 Sep 2011
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Re: better way to solve than back solving  [#permalink]

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

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