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An auto dealership sold one car making a 10% profit and one

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Director
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An auto dealership sold one car making a 10% profit and one [#permalink] New post 13 Feb 2007, 21:46
An auto dealership sold one car making a 10% profit and one car with a 10% loss. At the end of the month, the dealership figured that it has made a 5% profit from these two sales. How much was the cheapest car if the dealership earned a profit of $1000.00?

10K
9K
7K
5K
3K

I started like this: 1.1X + .90Y = 1000
Can anyone finish this off?
VP
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 [#permalink] New post 14 Feb 2007, 05:49
9K is the cheapest price of the car
Director
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 [#permalink] New post 14 Feb 2007, 06:55
x - the price of the more expensive car
y- the price of the cheaper car

We know that 5% of two sales equal 1000, thus
z*0.05=1000
z=20.000 this is the total revenue received after two sales

Construct two equations

0.1x-0.1y=1000
x+y=20000.....divide by 10

0.1x-0.1y=1000
0.1x+0.1y=2000

Add the two equations together

(2/10)x=3000
2x=30000
x=15.000

The more expensive car was sold for 15.000
The cheaper car was sold for 20.000-15.000=5.000

D
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5k [#permalink] New post 14 Feb 2007, 11:38
An auto dealership sold one car making a 10% profit and one car with a 10% loss. At the end of the month, the dealership figured that it has made a 5% profit from these two sales. How much was the cheapest car if the dealership earned a profit of $1000.00?

10K
9K
7K
5K
3K

Let Cost be X and Y

1.1X+0.9Y = 1.05 (X+Y) i.e. 5% of x+y

.05X=0.15Y==> x=3y

Profit 5% of (x+y) ==> 5(x+y)/100 = 1000
(3y+y)(5/100)=1000
20y/100=1000

y=5000
VP
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Re: 5k [#permalink] New post 14 Feb 2007, 15:36
Damager wrote:
An auto dealership sold one car making a 10% profit and one car with a 10% loss. At the end of the month, the dealership figured that it has made a 5% profit from these two sales. How much was the cheapest car if the dealership earned a profit of $1000.00?

10K
9K
7K
5K
3K

Let Cost be X and Y

1.1X+0.9Y = 1.05 (X+Y) i.e. 5% of x+y

.05X=0.15Y==> x=3y

Profit 5% of (x+y) ==> 5(x+y)/100 = 1000
(3y+y)(5/100)=1000
20y/100=1000

y=5000


This is good explanation
Director
Director
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 [#permalink] New post 17 Feb 2007, 23:45
Quote:
.05X=0.15Y==> x=3y
x=3y tells us that y is the cheaper car ... right?
Great solution by the way.
Director
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 [#permalink] New post 18 Feb 2007, 00:14
Also,
Would 1.1X+0.9Y = .05 (X+Y) or 1.1X+0.9Y = 1000 work? Why or why not? After all, 1.1X+0.9Y = a profit of 5%. Logically, shouldn't these other 2 alternatives work? I tried them. They don't seem to.
  [#permalink] 18 Feb 2007, 00:14
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