Agree this method is quite cumbersome as already suggested, but its an alternate method

Let the car dealer have ordered

c=10 cars each priced at

p=100Hence, the cost price that the cars cost the car dealer was

1000(10*100)It has been given that the dealer was able to sell

d=4 cars with a profit of 25 making

f=125 The selling price of the 4 cars is

4*(100+25) = 500The remaining cars(6 in number) were sold at the a loss of 25 dollars each(

m=75)

The selling price of these cars was

6*75 = 450, making the total selling price

450+500 = 950The selling of the cars leads to a loss of 50, profit is -50

Plugging the values into the equation, we get

A. df – cp – mp = 1250 - 1000 - 2500 (need not calculate, can't be -50)

B. df + f(d – m) – cp = 1250 + 25(-21) - 1000 (need not calculate, can't be -50)

C. d(f – m) + c(m – p) = 4(125-75) + 10(75-100) = 200 + 10(-25) = 200 - 250 = -50

Hence,

Option C {d(f – m) + c(m – p)} is the correct answer

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