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A car dealership sold two cars: the first car at a 10% [#permalink]
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02 Sep 2010, 12:24
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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
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Re: better way to solve than back solving [#permalink]
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02 Sep 2010, 18:00
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]
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02 Sep 2010, 16:32
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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: \((110105A) = (10590)B\) 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]
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11 Sep 2010, 20:31
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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: \((110105A) = (10590)B\) 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]
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11 Sep 2010, 22:20



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Re: better way to solve than back solving [#permalink]
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02 Sep 2010, 18:19
Very true Ian! Good catch.



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Re: better way to solve than back solving [#permalink]
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11 Sep 2010, 18:56
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: \((110105A) = (10590)B\) 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]
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11 Sep 2010, 21:26
.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]
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15 Sep 2010, 03:26
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]
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22 Jun 2011, 14:56
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 = 20000equation 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]
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09 Sep 2011, 13:49
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]
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09 Sep 2011, 20:10
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]
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18 Sep 2011, 00:54
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]
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18 Sep 2011, 06:25
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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20 Sep 2011, 05:14
\(\frac{x}{10}  \frac{y}{10} = 1000\) \(\frac{105}{100}(x+y) = \frac{11}{10}x + \frac{9}{10}y\) 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]
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20 Sep 2011, 07:02
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]
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06 Oct 2011, 08:10
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
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