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# m03 Q 32

Author Message
SVP
Joined: 16 Nov 2010
Posts: 1605

Kudos [?]: 581 [1], given: 36

Location: United States (IN)
Concentration: Strategy, Technology

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02 Apr 2011, 19:35
1
KUDOS
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

Regards,
Subhash
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Kudos [?]: 581 [1], given: 36

Director
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
Posts: 872

Kudos [?]: 389 [0], given: 123

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02 Apr 2011, 21:16

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

Posted from my mobile device

Kudos [?]: 389 [0], given: 123

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

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