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m03 Q 32

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m03 Q 32 [#permalink]

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New post 02 Apr 2011, 19:35
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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


x - Cost of car 1

y - Cost of car 2

1.1x + 0.9y - x - y = 1000

0.1x - 0.1y = 1000

x - y = 10000

1.1x + 0.9y = 1.05(x + y)

0.05x - 0.15y = 0

x - 3y = 0

=> 2y = 10000, y = 5000

x = 15000

I think the question should ask "Cost Price" instead of "sale price". Could someone please advise on this ?

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Subhash
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Kudos [?]: 600 [1], given: 40

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Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
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Re: m03 Q 32 [#permalink]

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New post 02 Apr 2011, 21:16
I think the question should ask "Cost Price" instead of "sale price". Could someone please advise on this ?

Yes you are correct!

I think elimination is the strategy here. Start from C and if the profit is less move down the options and if profit is more, move up the options. The overall profit is $1000
Option C
Profit from first car = 11k * 0.1=1100
loss from the second car 9k * 0.1= 900
overall profit =1100-900=$200

Option D
Profit from first car= 15k * 0.1= 1500
loss from the second car= 5k * 0.1=500
overall profit = 1500-500=$1000

I am done. Hence D

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Kudos [?]: 399 [0], given: 123

Re: m03 Q 32   [#permalink] 02 Apr 2011, 21:16
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