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

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Question Stats: 73% (02:16) correct 27% (02:18) wrong based on 1060 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 V Joined: 02 Sep 2009 Posts: 60627 A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 14 15 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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Senior Manager  Joined: 27 May 2008
Posts: 381
Re: Easy percentage problem...but whats the easiest way to solve  [#permalink]

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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: 102 Location: Montreal Schools: Harvard, Yale, HEC I am angry at this question [#permalink] ### Show Tags 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: 102 Location: Montreal Schools: Harvard, Yale, HEC Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags oh yeah, it helps. Well done. Thanks +1 Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 10011 Location: Pune, India Re: Easy percentage problem...but whats the easiest way to solve [#permalink] ### Show Tags 16 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 Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Manager  Joined: 27 Oct 2011 Posts: 115 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 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. Intern  Joined: 02 Jan 2015 Posts: 31 GMAT Date: 02-08-2015 GPA: 3.7 WE: Management Consulting (Consulting) Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 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 Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 10011 Location: Pune, India Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 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. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Intern  Joined: 20 Dec 2014 Posts: 19 Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 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]

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 VP  D Joined: 14 Feb 2017 Posts: 1364 Location: Australia Concentration: Technology, Strategy GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31 GMAT 6: 600 Q38 V35 GMAT 7: 710 Q47 V41 GPA: 3 WE: Management Consulting (Consulting) Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags Profit = revenue - cost Need to determine cost, since we already have revenue Therefore cost = Profit - Revenue Cost1 * 5/4 = 20,000 5*Cost1 = 80,000 Cost1= 16,000 Revenue = 20,000 20,000-16,000 = 4000 profit Cost2 *4/5 = 20,000 (sold at a loss) 4*Cost2 = 100,000 Cost2 = 25,000 Revenue - cost2 = 20,000-25,000 = (5000) loss combined = (1000) loss _________________ Here's how I went from 430 to 710, and how you can do it yourself: https://www.youtube.com/watch?v=KGY5vxqMeYk&t= Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 10011 Location: Pune, India Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 1 VeritasKarishma wrote: 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 Quote: I can understand “1 part profit for 4 parts of cost price“” by applying formula profit%=\frac{proft}{cost} But I didn't fully understand how did you get "So selling price ($20,000) has total 5 parts (each being $4000) out of which 1 part is profit i.e.$4000"

If for every 4 parts cost price, we have 1 part profit, what is the selling price in terms of parts?

Selling price = Cost price + Profit
Selling price = 4 parts + 1 part = 5 parts.

These 5 parts are actually $20,000 so each part must be 20,000/5 =$4000

Since cost price is 4 parts, it must be 4000*4 = $16,000 and since profit is 1 part, it must be 4000*1 =$4000
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Karishma
Veritas Prep GMAT Instructor

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

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Car A Price =20000×4/5 =16000
Profit = 20000-16000 =4000 ... 25% of D.P.

Car B Price =20000x5/4 =25000
Loss = 20000-25000 =-5000 ... 20% of D.P.

Overall Loss/Gain = 4000+(-5000) = -1000

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

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 Revenue = Profit - Cost Profit = 40, we need to find cost of Car 1 + Car 2 Car 1 cost = x, x = 20(4/5) = 16 Car 2 cost = y, y = 20(5/4) = 25 Revenue = 40 - 41 = -1 Target Test Prep Representative V Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9125 Location: United States (CA) Re: A used car dealer sold one car at a profit of 25 percent of [#permalink] ### Show Tags 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

We are given that the dealer sold two cars at $20,000, and for one sale the dealer had a profit of 25 percent of the dealer’s purchase price while on the other sale the dealer had a loss of 20 percent of the dealer’s purchase price. If we can find the purchase price of each car, then we can find the profit or loss he made on the sale of both cars. Let’s first find the dealer’s purchase price for the sale in which he made a 25% profit. If we let x = the purchase price, p = profit, and 20,000 = sale price then we can create the following equation: p = 0.25x 20,000 – x = 0.25x 20,000 = 1.25x x = 16,000 Now let’s first find the dealer’s purchase price for the sale in which he sustained a 20% loss. If we let y = the purchase price, then we can create the following equation: p = -0.2y 20,000 – y = -0.2y 20,000 = 0.8y y = 25,000 The dealer purchased the cars for a total of$16,000 + $25,000 =$41,000 and sold them for a total of $20,000 +$20,000 = $40,000. Thus, the dealer sustained a$1,000 loss on the sale of the two cars.

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