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

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

67% (03:08) correct
33% (02:40) wrong based on 43 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

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

2

This post received KUDOS

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]
02 Sep 2010, 17:00

3

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

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.

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:

Re: better way to solve than back solving [#permalink]
11 Sep 2010, 19:31

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]
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". _________________

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

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? _________________

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.

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.

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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