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
Percentage problems on the GMAT tend to involve very ROUND numbers.
Car 1 earns a 25% profit.
Here, the purchase price must be LESS than 20,000.
Make a list of options:
15000,
16000, 17000, 18000, 19000.
Adding a mark-up of 25% to the value in red yields a selling price of 20,000:
16000 + .25(16000) = 16000 + 4000 = 20,000.
Thus, the purchase price of Car 1 = 16,000.
Car 2 earns a 20% loss.
Here, the purchase price must be MORE than 20,000.
Make a list of options:
21000, 22000, 23000, 24000,
25000.
Subtracting 20% from the value in red yields a selling price of 20,000:
25000 - .2(25000) = 25000 - 5000 = 20,000.
Thus, the purchase price of Car 2 = 25,000.
Sum of the purchase prices = 16000 + 25000 = 41,000.
Sum of the selling prices = 20000 + 20000 = 40,000.
Resulting LOSS = 41000 - 40000 = 1000.
The correct answer is
C.
Algebraically:
A profit of 25% implies that the $20,000 selling price is equal to 125% of the purchase price:
20000 = (125/100)x
20000 = (5/4)x
80000 = 5x
x = 16,000.
A loss of 20% implies that the $20,000 selling price is equal to 80% of the purchase price:
20000 = (80/100)y
20000 = (4/5)y
100000 = 4y
y = 25,000.
Sum of the purchase prices = 16000 + 25000 = 41,000.
Sum of the selling prices = 20000 + 20000 = 40,000.
Resulting LOSS = 41000 - 40000 = 1000.
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