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# m03 Q 32

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Joined: 16 Nov 2010
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Location: United States (IN)
Concentration: Strategy, Technology
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Kudos [?]: 521 [1] , given: 36

m03 Q 32 [#permalink]

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02 Apr 2011, 18:35
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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 ? Regards, Subhash _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings 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: 920 Followers: 14 Kudos [?]: 346 [0], given: 123 Re: m03 Q 32 [#permalink] ### Show Tags 02 Apr 2011, 20:16 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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Re: m03 Q 32   [#permalink] 02 Apr 2011, 20:16
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# m03 Q 32

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