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

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Source: GMAT Prep

A used car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car. If the dealer sold each car for $20,000, what was the dealer's 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

The OA is C
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by Jay@ManhattanReview » Tue Jan 07, 2020 9:41 pm
BTGmoderatorLU wrote:Source: GMAT Prep

A used car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car. If the dealer sold each car for $20,000, what was the dealer's 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

The OA is C
Total sale price of the two cars = 2*20,000 = $40,000.

"¢ Purchase price of the first car = 20,000/(1 + 25%) = $16,000
"¢ Purchase price of the second car = 20,000/(1 - 20%) = $25,000

Total purchase price of the two cars =16,000 + 25,000 = $41,000

Total loss = 41,000 - 40,000 = $1,000

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Fri Jan 10, 2020 8:28 am
BTGmoderatorLU wrote:Source: GMAT Prep

A used car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car. If the dealer sold each car for $20,000, what was the dealer's 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

The OA is C
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 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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Selling price of each car - $20,000
Profit on first cat- 25% i.e cost price is increased by 25/100 = 1/4.
To get the cost price selling price should be reduced by 1/5 (n/n+d), which is the profit earned.
Profit earned = (1/5) * 20,000 = $4,000

Similarly, the other car is sold at loss of 20%
Which means it is sold by reducing (20/100) 1/5th of its cost value
To get the cost price selling price should be increased by (n/n-d) = 1/4 times, which is loss amount.
Loss amount = (1/4)*20,000 = $5000

Total profit or loss = 4000 + (-5000) = -$1000 or loss of $1000