Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A used car dealer sold one car at a profit of 25 percent of [#permalink]
23 Jul 2008, 21:15

1

This post received KUDOS

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

70% (02:28) correct
30% (01:04) wrong based on 381 sessions

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

Re: Easy percentage problem...but whats the easiest way to solve [#permalink]
23 Jul 2008, 22:23

2

This post received KUDOS

4

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

I am angry at this question [#permalink]
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???

Re: I am angry at this question [#permalink]
05 Jan 2011, 06:53

5

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

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 \(41,000-40,000=1,000\).

Re: Easy percentage problem...but whats the easiest way to solve [#permalink]
05 Jan 2011, 11:00

9

This post received KUDOS

Expert's post

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 _________________

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

Re: A used car dealer sold one car at a profit of 25 percent of [#permalink]
26 Jan 2015, 03:12

Expert's post

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. _________________

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...