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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] New post 02 Sep 2010, 11:24
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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
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Re: better way to solve than back solving [#permalink] New post 02 Sep 2010, 15:32
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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


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] New post 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).
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Re: better way to solve than back solving [#permalink] New post 02 Sep 2010, 17:19
Very true Ian! Good catch.
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Re: better way to solve than back solving [#permalink] New post 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] New post 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.
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Re: better way to solve than back solving [#permalink] New post 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
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Re: better way to solve than back solving [#permalink] New post 11 Sep 2010, 21:20
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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".
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Re: better way to solve than back solving [#permalink] New post 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.
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Re: better way to solve than back solving [#permalink] New post 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.
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Re: better way to solve than back solving [#permalink] New post 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] New post 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.

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

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