A used car dealer sold one car at a profit of 25 percent of : GMAT Problem Solving (PS)
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# A used car dealer sold one car at a profit of 25 percent of

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A used car dealer sold one car at a profit of 25 percent of [#permalink]

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23 Jul 2008, 21:15
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A used car dealer sold one car at a profit of 25 percent of the dealers purchase price for that car and sold another car at a loss of 20 percent of the dealers purchase price for that car. If the dealer sold each car for $20,000 , what was the dealers total profit or loss, in dollars, for the two transactions combined? A. 1000 profit B. 2000 profit C. 1000 loss D. 2000 loss E. 3334 loss [Reveal] Spoiler: OA Director Joined: 27 May 2008 Posts: 549 Followers: 8 Kudos [?]: 300 [4] , given: 0 Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 23 Jul 2008, 22:23 4 This post received KUDOS 7 This post was BOOKMARKED djdela wrote: A used car dealer sold one car at a profit of 25 percent of the dealers purchase price for that car and sold another car at a loss of 20 percent of the dealers purchase price for that car. If the dealer sold each car for$20,000 , what was the dealers total profit or loss, in dollars, for the two transactions combined?

a) 1000 profit
b) 2000 profit
c) 1000 loss
d) 2000 loss
e) 3334 loss

car 1 :
P1*1.25 = 20000
P1 = 16000

car2
P2*0.8 = 20000
P2 = 25000

P1+P2 = 41000
Final sale = 40000

1000 loss .. option C
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04 Jan 2011, 17:37
For me, the key to solve this problem is to know which amount to subtract from. Its easy to calculate the two amount of 41000 and 40000 but if you subtract 40000 from 41000 then you will have a profit of 1000 and vice-versa.

My question is what determines which one is the reference point???

Thanks guys.
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05 Jan 2011, 06:53
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ezinis wrote:
For me, the key to solve this problem is to know which amount to subtract from. Its easy to calculate the two amount of 41000 and 40000 but if you subtract 40000 from 41000 then you will have a profit of 1000 and vice-versa.

My question is what determines which one is the reference point???

Thanks guys.

A used car dealer sold one car at a profit of 25 percent of the dealers purchase price for that car and sold another car at a loss of 20 percent of the dealers purchase price for that car. If the dealer sold each car for $20,000 , what was the dealers total profit or loss, in dollars, for the two transactions combined? a) 1000 profit b) 2000 profit c) 1000 loss d) 2000 loss e) 3334 loss A dealer sold one car at a profit of 25% of the dealers purchase price for that car, let's say $$p_1$$, for$20,000 --> $$p_1*1.25=20,000$$ --> $$p_1=16,000$$ --> $$profit=selling \ price-purchase \ price=20,000-16,000=4,000$$;

A dealer sold another car at a loss of 20% of the dealers purchase price for that car, let's say $$p_2$$, again for $20,000 --> $$p_2*0.8=20,000$$ --> $$p_2=25,000$$ --> $$loss=purchase \ price-selling \ price=25,000-20,000=5,000$$; Overall loss 5,000-4,000=1,000. Or the way you are doing: the cars were purchased for $$p_1+p_2=16,000+25,000=41,000$$, and sold for $$20,000+20,000=40,000$$ so the overall loss is $$41,000-40,000=1,000$$. Answer: C. Hope it's clear. _________________ Manager Joined: 27 Oct 2009 Posts: 149 Location: Montreal Schools: Harvard, Yale, HEC Followers: 1 Kudos [?]: 78 [0], given: 18 Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 05 Jan 2011, 07:13 oh yeah, it helps. Well done. Thanks +1 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7059 Location: Pune, India Followers: 2081 Kudos [?]: 13237 [11] , given: 221 Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 05 Jan 2011, 11:00 11 This post received KUDOS Expert's post 2 This post was BOOKMARKED djdela wrote: A used car dealer sold one car at a profit of 25 percent of the dealers purchase price for that car and sold another car at a loss of 20 percent of the dealers purchase price for that car. If the dealer sold each car for$20,000 , what was the dealers total profit or loss, in dollars, for the two transactions combined?

a) 1000 profit
b) 2000 profit
c) 1000 loss
d) 2000 loss
e) 3334 loss

Logical approach:

Car 1 - profit of 25% i.e. 1/4 i.e. 1 part profit for 4 parts of cost price. So selling price ($20,000) has total 5 parts (each being$4000) out of which 1 part is profit i.e. $4000. Car 2 - loss of 20% i.e. 1/5 i.e. 1 part of loss out of 5 parts of cost price was removed to get selling price ($20,000) which has 4 parts now. So each part is $5000 and the loss is 1 part i.e.$5000.

Overall, loss of $5000 -$4000 = $1000 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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18 Apr 2012, 18:59
Great answer guys, you have to be sure you set your equations right. I mistakenly did 20,000 - P2 = .2P2 which is wrong cause it should be -.2P2.
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14 Aug 2013, 01:44
Bumping for review and further discussion.
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01 Mar 2014, 13:05
Bunuel wrote:
Bumping for review and further discussion.

20.000 * (25%-20%) = 5% * 20.000 = 1.000
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08 Jul 2014, 21:08
mariofelix wrote:
Bunuel wrote:
Bumping for review and further discussion.

20.000 * (25%-20%) = 5% * 20.000 = 1.000

This method will not tell if its loss or profit; I believe we require to do the whole computation (as did by Bunuel, others)
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24 Jan 2015, 05:50
Would it be possible to use the answer choices in this question? It seems that backsolving would be cumbersome but it is possible? Thanks.

Incidentally, you can save a small amount of calculation time by solving directly for the profit / loss (see below).

20,000 = 1.25x
So 0.25x = the profit for this car
0.25 = 1.25/5
20,000/5 = 4,000 profit

20,000 = 0.8y
So 0.2y = the loss for this car
0.2y = 0.8y/4
20,000/4 = 5,000 loss

Overall profit / loss = 4,000 - 5,000 = 1,000
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26 Jan 2015, 03:12
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ElCorazon wrote:
Would it be possible to use the answer choices in this question? It seems that backsolving would be cumbersome but it is possible? Thanks.

Incidentally, you can save a small amount of calculation time by solving directly for the profit / loss (see below).

20,000 = 1.25x
So 0.25x = the profit for this car
0.25 = 1.25/5
20,000/5 = 4,000 profit

20,000 = 0.8y
So 0.2y = the loss for this car
0.2y = 0.8y/4
20,000/4 = 5,000 loss

Overall profit / loss = 4,000 - 5,000 = 1,000

There is no reason why you should use the options in this question since it is easy to see how to proceed from the given data in the question whereas it is very difficult to decide how to proceed from the options.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 20 Dec 2014 Posts: 20 Followers: 0 Kudos [?]: 2 [0], given: 25 Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 16 Mar 2015, 13:10 Let P = Purchase Price Car 1 Profit:$20,000= (25/100) P + P => $20,000 = (1/4)P+P =>$20,000 = 5/4P

P= $16,000$20,000-$16,000=$4,000 profit

Car 2 Profit: $20,000 = P - (20/100)P =>$20,000= P - (1/5)P => $20,000 = 4/5P P=$25,000

$20,000 -$25,000 = $5,000 loss$4,000 - $5,000 =$1,000 loss

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15 Apr 2016, 02:57
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Re: A used car dealer sold one car at a profit of 25 percent of   [#permalink] 15 Apr 2016, 02:57
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