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A used car dealer sold one car at a profit of 25 percent of
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23 Jul 2008, 22:15
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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
A used car dealer sold one car at a profit of 25 percent of
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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 \(41,000-40,000=1,000\).
Re: Easy percentage problem...but whats the easiest way to solve
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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
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: Easy percentage problem...but whats the easiest way to solve
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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
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26 Jan 2015, 04:12
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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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Re: A used car dealer sold one car at a profit of 25 percent of
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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
Re: A used car dealer sold one car at a profit of 25 percent of
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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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A used car dealer sold one car at a profit of 25 percent of
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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
Re: A used car dealer sold one car at a profit of 25 percent of
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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.
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08 Oct 2019, 06:04
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