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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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.

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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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

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

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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