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

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Director
Joined: 12 Jun 2006
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An auto dealership sold one car making a 10% profit and one [#permalink]

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13 Feb 2007, 22:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

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VP
Joined: 28 Mar 2006
Posts: 1367

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14 Feb 2007, 06:49
9K is the cheapest price of the car

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Director
Joined: 06 Feb 2006
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14 Feb 2007, 07:55
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

(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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Senior Manager
Joined: 01 Sep 2006
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Location: Phoenix, AZ, USA

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14 Feb 2007, 12:38
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

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VP
Joined: 28 Mar 2006
Posts: 1367

Kudos [?]: 38 [0], given: 0

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14 Feb 2007, 16:36
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

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Director
Joined: 12 Jun 2006
Posts: 532

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18 Feb 2007, 00:45
Quote:
.05X=0.15Y==> x=3y
x=3y tells us that y is the cheaper car ... right?
Great solution by the way.

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Director
Joined: 12 Jun 2006
Posts: 532

Kudos [?]: 162 [0], given: 1

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18 Feb 2007, 01:14
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.

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18 Feb 2007, 01:14
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