A used car dealer sold one car at a profit of 25 percent of

Question

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. 1000 profit
B. 2000 profit
C. 1000 loss
D. 2000 loss
E. 3334 loss

Bunuel · Accepted Answer

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) 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.

KarishmaB · Answer

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

durgesh79 · Answer

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

ezinis · Answer

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.

My question is what determines which one is the reference point???

Thanks guys.

ezinis · Answer

oh yeah, it helps. Well done. Thanks +1

calreg11 · Answer

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.

ElCorazon · Answer

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

KarishmaB · Answer

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.

GMAT01 · Answer

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

mbaapp1234 · Answer

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

dcummins · Answer

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

KarishmaB · Answer

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

bhupenderpruthi · Answer

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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energetics · Answer

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

ScottTargetTestPrep · Answer

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

CrackverbalGMAT · Answer

Simple concept of Profit and Loss. All we need to remember that Profit and loss % are on the CP.

When the car is sold at a profit of 25%, then  CP * \frac{125}{100} = 20000

CP = \frac{20000 \space * \space 100}{125} = $16000

When the car is sold at a Loss of 20%, then  CP * \frac{80}{100} = 20000

CP = \frac{20000 \space * \space 100}{80} = $25000

Total CP = 16000 + 25000 = 41000

Total SP = 20000 + 20000 = 40000

Since CP is greater than SP; Loss = CP - SP = 41000 - 40000 = $1000

Option C

Arun Kumar

Abhishek009 · Answer

CP_1 = \frac{20000}{125}*100 = 16000

CP_2 =\frac{20000}{80}*100 = 25000

Total CP is 41000 , Total SP is 40000

Hence, there is a loss of 1000, Answer must be (C)

Basshead · Answer

20000/1.25 = 16,000

20,000 - 16,000 = 4,000 profit

20,000/0.8 = 25,000 - 20,000 = 5,000 loss

4,000 + (-5,000) = -$1,000

Answer is C.

himgkp1989 · Answer

KarishmaB  Is that possible I can solve this with weighted average concept. If yes then what would be the weights.

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

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