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

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Intern
Joined: 16 Jul 2008
Posts: 12
A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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23 Jul 2008, 22:15
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35% (medium)

Question Stats:

71% (01:18) correct 29% (01:06) wrong based on 1187 sessions

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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 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 48041 A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 05 Jan 2011, 07:53 8 11 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 $$41,000-40,000=1,000$$.

Hope it's clear.
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Re: Easy percentage problem...but whats the easiest way to solve  [#permalink]

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23 Jul 2008, 23:23
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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 ##### General Discussion Manager Joined: 27 Oct 2009 Posts: 129 Location: Montreal Schools: Harvard, Yale, HEC I am angry at this question [#permalink] ### Show Tags 04 Jan 2011, 18:37 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. Manager Joined: 27 Oct 2009 Posts: 129 Location: Montreal Schools: Harvard, Yale, HEC Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 05 Jan 2011, 08:13 oh yeah, it helps. Well done. Thanks +1 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8196 Location: Pune, India Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 05 Jan 2011, 12:00 14 4 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 Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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

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18 Apr 2012, 19:59
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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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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24 Jan 2015, 06:50
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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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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26 Jan 2015, 04:12
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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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Intern Joined: 20 Dec 2014 Posts: 21 Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 16 Mar 2015, 14:10 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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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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25 Jun 2017, 18:09
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:
20000 = 1.25P

P = 16000

profit = 20000 - 16000 = 4000

Car 2:
20000 = 0.8P

200000 = 8P

100000 = 4P

25000 = P

profit = -5000

Combined 1+2:

4000 - 5000 = -1000
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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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27 Jul 2018, 18:44
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