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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Karishma
Veritas Prep GMAT Instructor
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