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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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New post 23 Jul 2008, 22:15
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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
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A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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New post 05 Jan 2011, 07:53
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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.
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Re: Easy percentage problem...but whats the easiest way to solve  [#permalink]

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New post 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
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I am angry at this question  [#permalink]

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New post 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.
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Re: Easy percentage problem...but whats the easiest way to solve  [#permalink]

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New post 05 Jan 2011, 08:13
oh yeah, it helps. Well done. Thanks +1
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Re: Easy percentage problem...but whats the easiest way to solve  [#permalink]

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New post 05 Jan 2011, 12:00
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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


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

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

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New post 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

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

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New post 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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New post 17 Jul 2019, 15:17
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
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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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New post 24 Sep 2019, 03:54
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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"

Can you please elaborate more?


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

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New post 24 Sep 2019, 04:20
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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A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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New post 30 Sep 2019, 11:45
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
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Re: A used car dealer sold one car at a profit of 25 percent of  [#permalink]

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New post 03 Oct 2019, 12:16
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.

Answer: C
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Re: A used car dealer sold one car at a profit of 25 percent of   [#permalink] 08 Oct 2019, 06:04
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