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An auto dealership sold the first car with a 10% profit and

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An auto dealership sold the first car with a 10% profit and [#permalink]

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10 Apr 2008, 20:52
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An auto dealership sold the first car with a 10% profit and the second car with a 10% loss, which gave it an overall profit of 5% from these sales. How much were the cars sold for if the dealership profit was $1000? A$11,000 and $9,000 B$15,000 and $5,000 C$9,450 and $10,550 D$11,550 and $8,450 E$10,000 and $10,000 Can you folks try this. Read question carefully. SVP Joined: 29 Aug 2007 Posts: 2492 Followers: 68 Kudos [?]: 735 [0], given: 19 Re: auto dealership [#permalink] Show Tags 10 Apr 2008, 21:22 kyatin wrote: An auto dealership sold the first car with a 10% profit and the second car with a 10% loss, which gave it an overall profit of 5% from these sales. How much were the cars sold for if the dealership profit was$1000?

A $11,000 and$9,000
B $15,000 and$5,000
C $9,450 and$10,550
D $11,550 and$8,450
E $10,000 and$10,000

Can you folks try this. Read question carefully.

0.10x - 0.10y = 1000 ..............i
0.05 (x + y) = 1000 ............. ii
0.20x = 3000
x = 3000/0.20 = 15000
y = 5000

B.
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10 Apr 2008, 23:24
GMAT TIGER wrote:
kyatin wrote:
An auto dealership sold the first car with a 10% profit and the second car with a 10% loss, which gave it an overall profit of 5% from these sales. How much were the cars sold for if the dealership profit was $1000? A$11,000 and $9,000 B$15,000 and $5,000 C$9,450 and $10,550 D$11,550 and $8,450 E$10,000 and $10,000 Can you folks try this. Read question carefully. 0.10x - 0.10y = 1000 ..............i 0.05 (x + y) = 1000 ............. ii 0.20x = 3000 x = 3000/0.20 = 15000 y = 5000 B. GMATTIGER, I was getting bit confused with terminologies here.....in my mind....% profit or loss is always based on Purchase price....but in this question it seems to be based on Selling price. ('Sold with 10% loss') So 15000 is selling price that includes 10% profit indicating purchase price was 13,500 and 1,500 is profit. Same token 5,000 is selling price of second car and 10% is loss off Selling price...so actual purchase price is 5,500 I guess we can easily twist this type of quesiton. Let me know if you have any other insight...or did I just read too much into the problem. GMAT Forum Moderator Status: Accepting donations for the mohater MBA debt repayment fund Joined: 05 Feb 2008 Posts: 1884 Location: United States Concentration: Operations, Finance Schools: Ross '14 (M) GMAT 1: 610 Q0 V0 GMAT 2: 710 Q48 V38 GPA: 3.54 WE: Accounting (Manufacturing) Followers: 58 Kudos [?]: 787 [0], given: 234 Re: auto dealership [#permalink] Show Tags 11 Apr 2008, 11:35 I've found with problems like this, instead of starting to solve or set up the equation, just figure out what 10% is, since that part is easy. Write them down, find the different of$1000.

Don't even need the equation.
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11 Apr 2008, 16:56
kyatin wrote:
GMAT TIGER wrote:
kyatin wrote:
An auto dealership sold the first car with a 10% profit and the second car with a 10% loss, which gave it an overall profit of 5% from these sales. How much were the cars sold for if the dealership profit was $1000? A$11,000 and $9,000 B$15,000 and $5,000 C$9,450 and $10,550 D$11,550 and $8,450 E$10,000 and \$10,000

Can you folks try this. Read question carefully.

0.10x - 0.10y = 1000 ..............i
0.05 (x + y) = 1000 ............. ii
0.20x = 3000
x = 3000/0.20 = 15000
y = 5000

B.

GMATTIGER,

I was getting bit confused with terminologies here.....in my mind....% profit or loss is always based on Purchase price....but in this question it seems to be based on Selling price. ('Sold with 10% loss')

So 15000 is selling price that includes 10% profit indicating purchase price was 13,500 and 1,500 is profit.

Same token 5,000 is selling price of second car and 10% is loss off Selling price...so actual purchase price is 5,500

I guess we can easily twist this type of quesiton.

Let me know if you have any other insight...or did I just read too much into the problem.

nope, there is no confusion.

x = cost = 15000
10% proft = 1500

y = cost = 5000
10% loss = 500

so net profit = 1000
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12 Apr 2008, 01:43
equation method is pretty good but finding out 10% of the two figures and then subtracting 2 from 1 is quicker. Saves time !
Re: auto dealership   [#permalink] 12 Apr 2008, 01:43
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