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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

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

69% (03:19) correct
31% (02:46) wrong based on 51 sessions

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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

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

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02 Sep 2010, 16:32

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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

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.

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

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02 Sep 2010, 18:00

3

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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

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:

\((110-105A) = (105-90)B\) 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:

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

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11 Sep 2010, 20:31

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. _________________

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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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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". _________________

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

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? _________________

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.

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.

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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