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A car dealership sold two cars

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A car dealership sold two cars [#permalink] New post 29 Aug 2010, 15:40
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

66% (01:30) correct 33% (01:25) wrong based on 6 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


I got lost in options C and D.
[Reveal] Spoiler: OA
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Re: A car dealership sold two cars [#permalink] New post 29 Aug 2010, 17:03
seekmba 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

I got lost in options C and D.


my method is a long one, please any1 feel free to post
start from middle and try out answer choices going up or down accordingly
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Re: A car dealership sold two cars [#permalink] New post 29 Aug 2010, 17:54
5% profit is equal to 1000$, that means that the combined costs of the cars = 1000*100/5 = 20000 $
Now, lets assume that the first car cost x $, so the 2nd one costs (20000-x)$

So, x/10 - (20000-x)/10 = 1000
=> x-20000+x = 10000
=> x = 15000
so, 20000 -x = 5000

Hence, D.
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Re: A car dealership sold two cars [#permalink] New post 29 Aug 2010, 17:58
My approach:

Given profit in one sale is 10% and loss in another sale is 10%
Hence overall profit = Sum of the two profits or Net (Profit - loss)

Let the sale price of car1, car2 be x, y respectively.

Hence 0.1 x - 0.1y (since it is loss) = 0.05(x+y)

0.05 x - 0.15 y = 0

x/y = 0.15/0.05 => x/y = 3/1

The only choice is D. (15000, 5000)
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Re: A car dealership sold two cars [#permalink] New post 29 Aug 2010, 18:17
makes all the sense now...thanks guys.
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Re: A car dealership sold two cars [#permalink] New post 30 Aug 2010, 03:27
profit:
5% (X+Y)=1000 , x+y=20k

10%(X-Y)=1000 x-y=10k

hence D
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Re: A car dealership sold two cars [#permalink] New post 22 Jun 2011, 12:31
sorry to bump an old thread, but i have a question regarding this question's wording.

doesn't X denote the cost to the dealership, and not the sale price of the first car? i thought sale price of X would be 110/100 of X, since it is sold at a 10% profit.

let X denote cost to the dealership of 1st car, and Y for the 2nd car.

sale price minus cost of X + sale price minus cost of Y = profit
X*(110/100)-X + Y*(90/100)-Y = 1000
X*(10/100) - Y(10/100) = 1000

once we finally get X from the other equation, wouldn't the sale price of X be 15000*110/100 = 16500 and sale price of Y be 5000 * 90/100 = 4500?

so profit from X is 1500 and loss from Y is 500, total profit is 1000.

the question asked for sale price, not cost, that's why i was confused.
can someone clear up the confusion for me or am i just crazy? thanks
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Re: A car dealership sold two cars [#permalink] New post 22 Jun 2011, 13:06
anon123 wrote:
sorry to bump an old thread, but i have a question regarding this question's wording.

doesn't X denote the cost to the dealership, and not the sale price of the first car? i thought sale price of X would be 110/100 of X, since it is sold at a 10% profit.

let X denote cost to the dealership of 1st car, and Y for the 2nd car.

sale price minus cost of X + sale price minus cost of Y = profit
X*(110/100)-X + Y*(90/100)-Y = 1000
X*(10/100) - Y(10/100) = 1000

once we finally get X from the other equation, wouldn't the sale price of X be 15000*110/100 = 16500 and sale price of Y be 5000 * 90/100 = 4500?

so profit from X is 1500 and loss from Y is 500, total profit is 1000.

the question asked for sale price, not cost, that's why i was confused.
can someone clear up the confusion for me or am i just crazy? thanks


This is just one of the reasons why multiple threads discussing same question is discouraged in this forum.
better-way-to-solve-than-back-solving-100287.html
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Re: A car dealership sold two cars [#permalink] New post 22 Jun 2011, 13:50
Thanks fluke. I just did a quick google search and this was the thread that came up 1st, didn't know there was another thread on this already. but glad to know i was right, i never doubt the questions/answers of OA, but i spent way too much time thinking about this problem!
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Re: A car dealership sold two cars [#permalink] New post 25 Sep 2011, 02:10
USING AN MGMAT-style TABLE (see attached picture)

Now, take any extra relationships that don’t quite go in this table and write equations for them.
In this case, Cost = X + Y ... We’re told the profit margin is 5%. Therefore, profit is 5% of the cost price, which is 1,000. Therefore, 0.05(X + Y) = 1,000 so X + Y = 1000/0.05 = 20,000.

So, equation 1: X + Y = 20,000

Equation 2: 0.1X – 0.1Y = 1,000

Solve for x and y using elimination (or substitution, whatever you prefer):
X + Y = 20000
X – Y = 10000
0 + 2Y = 10,000

Y = 5000.
X = 20,000 – 5,000 = 15,000

This is the answer.
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Re: A car dealership sold two cars   [#permalink] 25 Sep 2011, 02:10
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